index
int64 1
500
| problem
stringlengths 37
987
| geo_code
stringlengths 81
4.63k
| answer
stringlengths 3
32
| category
stringclasses 3
values | source
stringclasses 7
values | problem_type
stringclasses 6
values |
|---|---|---|---|---|---|---|
101
|
The convex pentagon $ABCDE$ has $\angle{A}=\angle{B}=120^\circ,EA=AB=BC=4$ and $CD=DE=8$. What is the area of ABCDE?
|
[asy] draw((-1,0)--(1,0)--(1+sqrt(2),sqrt(2))--(0,sqrt(2)+sqrt(13-2*sqrt(2)))--(-1-sqrt(2),sqrt(2))--cycle,black+linewidth(.75)); MP("A",(-1,0),SW);MP("B",(1,0),SE);MP("C",(1+sqrt(2),sqrt(2)),E);MP("D",(0,sqrt(2)+sqrt(13-2*sqrt(2))),N);MP("E",(-1-sqrt(2),sqrt(2)),W); dot((-1,0));dot((1,0));dot((1+sqrt(2),sqrt(2)));dot((-1-sqrt(2),sqrt(2)));dot((0,sqrt(2)+sqrt(13-2*sqrt(2)))); [/asy]
|
$28\sqrt{3}$
|
Local Relation Composition
|
HARP
|
area
|
102
|
Amy painted a dartboard over a square clock face using the "hour positions" as boundaries.[See figure.] If $t$ is the area of one of the eight triangular regions such as that between 12 o'clock and 1 o'clock, and $q$ is the area of one of the four corner quadrilaterals such as that between 1 o'clock and 2 o'clock, then $\frac{t}{q}=$
|
[asy] draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle, black+linewidth(.75)); draw((0,-1)--(0,1), black+linewidth(.75)); draw((-1,0)--(1,0), black+linewidth(.75)); draw((-1,-1/sqrt(3))--(1,1/sqrt(3)), black+linewidth(.75)); draw((-1,1/sqrt(3))--(1,-1/sqrt(3)), black+linewidth(.75)); draw((-1/sqrt(3),-1)--(1/sqrt(3),1), black+linewidth(.75)); draw((1/sqrt(3),-1)--(-1/sqrt(3),1), black+linewidth(.75)); [/asy]
|
$\frac{\sqrt{3}+1}{4}$
|
Local Relation Composition
|
HARP
|
ratio
|
103
|
A large rectangle is partitioned into four rectangles by two segments parallel to its sides. The areas of three of the resulting rectangles are shown. What is the total area of the four rectangle?
|
[asy] draw((0,0)--(10,0)--(10,7)--(0,7)--cycle); draw((0,5)--(10,5)); draw((3,0)--(3,7)); label("6", (1.5,6)); label("?", (1.5,2.5)); label("14", (6.5,6)); label("35", (6.5,2.5)); [/asy]
|
$70$
|
Local Relation Composition
|
HARP
|
area
|
104
|
In the $xy$-plane, consider the L-shaped region bounded by horizontal and vertical segments with vertices at $(0,0), (0,3), (3,3), (3,1), (5,1)$ and $(5,0)$. Suppose there is a straight line that divides the area of the region into two equal parts. What is the tangent of the angle formed between this line and the y-axis?
|
[asy] Label l; l.p=fontsize(6); xaxis("$x$",0,6,Ticks(l,1.0,0.5),EndArrow); yaxis("$y$",0,4,Ticks(l,1.0,0.5),EndArrow); draw((0,3)--(3,3)--(3,1)--(5,1)--(5,0)--(0,0)--cycle,black+linewidth(2));[/asy]
|
$\frac{9}{7}$
|
Local Relation Composition
|
HARP
|
angle
|
105
|
Points $A, B$ and $C$ on a circle of radius $r$ are situated so that $AB=AC, AB>r$, and the length of minor arc $BC$ is $r$. If angles are measured in radians, then $BC/AB=$
|
[asy] draw(Circle((0,0), 13)); draw((-13,0)--(12,5)--(12,-5)--cycle); dot((-13,0)); dot((12,5)); dot((12,-5)); label("A", (-13,0), W); label("B", (12,5), NE); label("C", (12,-5), SE); [/asy]
|
$\2\sin\left(\frac{1}{4}\right)$
|
Local Relation Composition
|
HARP
|
ratio
|
106
|
Triangle $ABC$ is isosceles, with $AB=AC$ and altitude $AM=11.$ Suppose that there is a point $D$ on $\overline{AM}$ with $AD=10$ and $\angle BDC=3\angle BAC.$ Then the perimeter of $\triangle ABC$ is $
|
[asy] import graph; size(5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-1.55,xmax=7.95,ymin=-4.41,ymax=5.3; draw((1,3)--(0,0)); draw((0,0)--(2,0)); draw((2,0)--(1,3)); draw((1,3)--(1,0)); draw((1,0.7)--(0,0)); draw((1,0.7)--(2,0)); label("$11$",(1,1.63),W); dot((1,3),ds); label("$A$",(1,3),N); dot((0,0),ds); label("$B$",(0,0),SW); dot((2,0),ds); label("$C$",(2,0),SE); dot((1,0),ds); label("$M$",(1,0),S); dot((1,0.7),ds); label("$D$",(1,0.7),NE); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
|
$11+\sqrt{605}$
|
Local Relation Composition
|
HARP
|
length
|
107
|
Figure $OPQR$ is a square. Point $O$ is the origin, and point $Q$ has coordinates (2,2). For what value of the x-coordinate of point $T$ will the area of triangle $PQT$ be exactly twice the area of square $OPQR$?
|
[asy] pair O,P,Q,R,T; O = (0,0); P = (2,0); Q = (2,2); R = (0,2); T = (-4,0); draw((-5,0)--(3,0)); draw((0,-1)--(0,3)); draw(P--Q--R); draw((-0.2,-0.8)--(0,-1)--(0.2,-0.8)); draw((-0.2,2.8)--(0,3)--(0.2,2.8)); draw((-4.8,-0.2)--(-5,0)--(-4.8,0.2)); draw((2.8,-0.2)--(3,0)--(2.8,0.2)); draw(Q--T); label("$O$",O,SW); label("$P$",P,S); label("$Q$",Q,NE); label("$R$",R,W); label("$T$",T,S); [/asy]
|
$-6$
|
Local Relation Composition
|
HARP
|
length
|
108
|
On a $4\times 4\times 3$ rectangular parallelepiped, vertices $A$, $B$, and $C$ are adjacent to vertex $D$. The perpendicular distance from $D$ to the plane containing $A$, $B$, and $C$ is
|
[asy] size(120); import three; currentprojection=orthographic(1, 4/5, 1/3); draw(box(O, (4,4,3))); triple A=(0,4,3), B=(0,0,0) , C=(4,4,0), D=(0,4,0); draw(A--B--C--cycle, linewidth(0.9)); label("$A$", A, NE); label("$B$", B, NW); label("$C$", C, S); label("$D$", D, E); label("$4$", (4,2,0), SW); label("$4$", (2,4,0), SE); label("$3$", (0, 4, 1.5), E); [/asy]
|
$\frac{12}{\sqrt{34}}$
|
Local Relation Composition
|
HARP
|
length
|
109
|
What fraction of this square region is unshaded? Stripes are equal in width, and the figure is drawn to scale.
|
[asy] unitsize(8); fill((0,0)--(6,0)--(6,6)--(0,6)--cycle,black); fill((0,0)--(5,0)--(5,5)--(0,5)--cycle,white); fill((0,0)--(4,0)--(4,4)--(0,4)--cycle,black); fill((0,0)--(3,0)--(3,3)--(0,3)--cycle,white); fill((0,0)--(2,0)--(2,2)--(0,2)--cycle,black); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle,white); draw((0,6)--(0,0)--(6,0)); [/asy]
|
$\frac{5}{12}$
|
Local Relation Composition
|
HARP
|
ratio
|
110
|
$\angle 1 + \angle 2 = 180^\circ$ $\angle 3 = \angle 4$ Find $\angle 4.$
|
[asy] pair H,I,J,K,L; H = (0,0); I = 10*dir(70); J = I + 10*dir(290); K = J + 5*dir(110); L = J + 5*dir(0); draw(H--I--J--cycle); draw(K--L--J); draw(arc((0,0),dir(70),(1,0),CW)); label("$70^\circ$",dir(30),NE); draw(arc(I,I+dir(250),I+dir(290),CCW)); label("$40^\circ$",I+1.25*dir(270),S); label("$1$",J+0.25*dir(162.5),NW); label("$2$",J+0.25*dir(17.5),NE); label("$3$",L+dir(162.5),WNW); label("$4$",K+dir(-52.5),SE); [/asy]
|
$35$
|
Local Relation Composition
|
HARP
|
angle
|
111
|
Each corner cube is removed from this $6\text{ cm}\times 6\text{ cm}\times 6\text{ cm}$ cube. The surface area of the remaining figure in sq.cm is
|
[asy] draw((2.7,3.99)--(0,3)--(0,0)); draw((3.7,3.99)--(1,3)--(1,0)); draw((4.7,3.99)--(2,3)--(2,0)); draw((5.7,3.99)--(3,3)--(3,0)); draw((0,0)--(3,0)--(5.7,0.99)); draw((0,1)--(3,1)--(5.7,1.99)); draw((0,2)--(3,2)--(5.7,2.99)); draw((0,3)--(3,3)--(5.7,3.99)); draw((0,3)--(3,3)--(3,0)); draw((0.9,3.33)--(3.9,3.33)--(3.9,0.33)); draw((1.8,3.66)--(4.8,3.66)--(4.8,0.66)); draw((2.7,3.99)--(5.7,3.99)--(5.7,0.99)); [/asy]
|
$216$
|
Local Relation Composition
|
HARP
|
area
|
112
|
Diameter $ACE$ is divided at $C$ in the ratio $1:2$. The two semicircles, $ABC$ and $CDE$, divide the circular region into an upper (shaded) region and a lower region. The area of the upper region divided by that of the lower region is
|
[asy] pair A,B,C,D,EE; A = (0,0); B = (2,2); C = (4,0); D = (7,-3); EE = (10,0); fill(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)--arc((5,0),EE,A,CCW)--cycle,gray); draw(arc((2,0),A,C,CW)--arc((7,0),C,EE,CCW)); draw(circle((5,0),5)); dot(A); dot(B); dot(C); dot(D); dot(EE); label("$A$",A,W); label("$B$",B,N); label("$C$",C,E); label("$D$",D,N); label("$E$",EE,W); [/asy]
|
$2$
|
Local Relation Composition
|
HARP
|
ratio
|
113
|
In the figure, polygons $A$, $E$, and $F$ are isosceles right triangles; $B$, $C$, and $D$ are squares with sides of length $2$; and $G$ is an equilateral triangle. The figure can be folded along its edges to form a polyhedron having the polygons as faces. The volume of this polyhedron is
|
[asy] size(6cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); draw((-1,1)--(2,1)); draw((-1,0)--(1,0)); draw((-1,1)--(-1,0)); draw((0,-1)--(0,3)); draw((1,2)--(1,0)); draw((-1,1)--(1,1)); draw((0,2)--(1,2)); draw((0,3)--(1,2)); draw((0,-1)--(2,1)); draw((0,-1)--((0,-1) + sqrt(2)*dir(-15))); draw(((0,-1) + sqrt(2)*dir(-15))--(1,0)); label("$\textbf{A}$",foot((0,2),(0,3),(1,2)),SW); label("$\textbf{B}$",midpoint((0,1)--(1,2))); label("$\textbf{C}$",midpoint((-1,0)--(0,1))); label("$\textbf{D}$",midpoint((0,0)--(1,1))); label("$\textbf{E}$",midpoint((1,0)--(2,1)),NW); label("$\textbf{F}$",midpoint((0,-1)--(1,0)),NW); label("$\textbf{G}$",midpoint((0,-1)--(1,0)),2SE);[/asy]
|
$\frac{20}{3}$
|
Global Abstract Integration
|
HARP
|
volume
|
114
|
Triangle $ABC$ and point $P$ in the same plane are given. Point $P$ is equidistant from $A$ and $B$, angle $APB$ is twice angle $ACB$, and $\overline{AC}$ intersects $\overline{BP}$ at point $D$. If $PB = 6$ and $PD= 4$, then $AD\cdot CD =$
|
[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (2,0); pair C = (3,1); pair P = (1,2.25); pair D = intersectionpoint(P--B,C--A); dot(A);dot(B);dot(C);dot(P);dot(D); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,N);label("$D$",D,NE + N);label("$P$",P,N); draw(A--B--P--cycle); draw(A--C--B--cycle);[/asy]
|
$20$
|
Local Relation Composition
|
HARP
|
length
|
115
|
According to the figure, the degree measure of angle $A$ is
|
[asy] unitsize(12); draw((0,0)--(20,0)--(1,-10)--(9,5)--(18,-8)--cycle); draw(arc((1,-10),(1+19/sqrt(461),-10+10/sqrt(461)),(25/17,-155/17),CCW)); draw(arc((19/3,0),(19/3-8/17,-15/17),(22/3,0),CCW)); draw(arc((900/83,-400/83),(900/83+19/sqrt(461),-400/83+10/sqrt(461)),(900/83 - 9/sqrt(97),-400/83 + 4/sqrt(97)),CCW)); label(rotate(30)*"$40^\circ$",(2,-8.9),ENE); label("$100^\circ$",(21/3,-2/3),SE); label("$110^\circ$",(900/83,-317/83),NNW); label("$A$",(0,0),NW); [/asy]
|
$30$
|
Local Relation Composition
|
HARP
|
angle
|
116
|
If circular arcs $AC$ and $BC$ have centers at $B$ and $A$, respectively, then there exists a circle tangent to both $\overarc {AC}$ and $\overarc{BC}$, and to $\overline{AB}$. If the length of $\overarc{BC}$ is $12$, then the circumference of the circle is
|
[asy] label("A", (0,0), W); label("B", (64,0), E); label("C", (32, 32*sqrt(3)), N); draw(arc((0,0),64,0,60)); draw(arc((64,0),64,120,180)); draw((0,0)--(64,0)); draw(circle((32, 24), 24)); [/asy]
|
$27$
|
Local Relation Composition
|
HARP
|
length
|
117
|
A corner of a tiled floor is shown. If the entire floor is tiled in this way and each of the four corners looks like this one, then what fraction of the tiled floor is not made of darker tiles?
|
[asy] fill((0,2)--(1,3)--(2,3)--(2,4)--(3,5)--(4,4)--(4,3)--(5,3)--(6,2)--(5,1)--(4,1)--(4,0)--(2,0)--(2,1)--(1,1)--cycle, mediumgrey); fill((7,1)--(6,2)--(7,3)--(8,3)--(8,4)--(9,5)--(10,4)--(7,0)--cycle, mediumgrey); fill((3,5)--(2,6)--(2,7)--(1,7)--(0,8)--(1,9)--(2,9)--(2,10)--(3,11)--(4,10)--(4,9)--(5,9)--(6,8)--(5,7)--(4,7)--(4,6)--cycle, mediumgrey); fill((6,8)--(7,9)--(8,9)--(8,10)--(9,11)--(10,10)--(10,9)--(11,9)--(11,7)--(10,7)--(10,6)--(9,5)--(8,6)--(8,7)--(7,7)--cycle, mediumgrey); draw((0,0)--(0,11)--(11,11)); for ( int x = 1; x < 11; ++x ) { draw((x,11)--(x,0), linetype("4 4")); } for ( int y = 1; y < 11; ++y ) { draw((0,y)--(11,y), linetype("4 4")); } clip((0,0)--(0,11)--(11,11)--(11,5)--(4,1)--cycle);[/asy]
|
$\frac{5}{9}$
|
Local Relation Composition
|
HARP
|
ratio
|
118
|
Triangle $ABC$ is a right triangle with $\angle ACB$ as its right angle, $m\angle ABC = 60^\circ$ , and $AB = 10$. Let $P$ be randomly chosen inside $ABC$ , and extend $\overline{BP}$ to meet $\overline{AC}$ at $D$. What is the probability that $BD < 5\sqrt2$?
|
[asy] import math; unitsize(4mm); defaultpen(fontsize(8pt)+linewidth(0.7)); dotfactor=4; pair A=(10,0); pair C=(0,0); pair B=(0,10.0/sqrt(3)); pair P=(2,2); pair D=extension(A,C,B,P); draw(A--C--B--cycle); draw(B--D); dot(P); label("A",A,S); label("D",D,S); label("C",C,S); label("P",P,NE); label("B",B,N);[/asy]
|
$\frac{\sqrt3}{3}$
|
Local Relation Composition
|
HARP
|
ratio
|
119
|
What is the area enclosed by the geoboard quadrilateral below?
|
[asy] unitsize(3mm); defaultpen(linewidth(.8pt)); dotfactor=2; for(int a=0; a<=10; ++a) for(int b=0; b<=10; ++b) { dot((a,b)); }; draw((4,0)--(0,5)--(3,4)--(10,10)--cycle); [/asy]
|
$22\frac{1}{2}$
|
Local Relation Composition
|
HARP
|
area
|
120
|
What is the area of trapezoid $ABCD$?
|
[asy]size(3inch, 1.5inch); pair a=(0,0), b=(18,24), c=(68,24), d=(75,0), f=(68,0), e=(18,0); draw(a--b--c--d--cycle); draw(b--e); draw(shift(0,2)*e--shift(2,2)*e--shift(2,0)*e); label("30", (9,12), W); label("50", (43,24), N); label("25", (71.5, 12), E); label("24", (18, 12), E); label("$A$", a, SW); label("$B$", b, N); label("$C$", c, N); label("$D$", d, SE); label("$E$", e, S);[/asy]
|
$1500$
|
Local Relation Composition
|
HARP
|
area
|
121
|
A number of linked rings, each $2$ cm thick, are hanging on a peg. The top ring has an outside diameter of $20$ cm. The outside diameter of each of the outer rings is $1$ cm less than that of the ring above it. The bottom ring has an outside diameter of $3$ cm. What is the distance, in cm, from the top of the top ring to the bottom of the bottom ring?
|
[asy] size(7cm); pathpen = linewidth(0.7); D(CR((0,0),10)); D(CR((0,0),9.5)); D(CR((0,-18.5),9.5)); D(CR((0,-18.5),9)); MP("$\vdots$",(0,-31),(0,0)); D(CR((0,-39),3)); D(CR((0,-39),2.5)); D(CR((0,-43.5),2.5)); D(CR((0,-43.5),2)); D(CR((0,-47),2)); D(CR((0,-47),1.5)); D(CR((0,-49.5),1.5)); D(CR((0,-49.5),1.0)); D((12,-10)--(12,10)); MP('20',(12,0),E); D((12,-51)--(12,-48)); MP('3',(12,-49.5),E);[/asy]
|
$346$
|
Global Abstract Integration
|
HARP
|
length
|
122
|
An angle is drawn on a set of equally spaced parallel lines as shown. The ratio of the area of shaded region $B$ to the area of shaded region $C$ is 5/11. Find the ratio of shaded region $D$ to the area of shaded region $A.$
|
[asy] defaultpen(linewidth(0.7)+fontsize(10)); for(int i=0; i<4; i=i+1) { fill((2*i,0)--(2*i+1,0)--(2*i+1,6)--(2*i,6)--cycle, mediumgray); } pair A=(1/3,4), B=A+7.5*dir(-17), C=A+7*dir(10); draw(B--A--C); fill((7.3,0)--(7.8,0)--(7.8,6)--(7.3,6)--cycle, white); clip(B--A--C--cycle); for(int i=0; i<9; i=i+1) { draw((i,1)--(i,6)); } label("$\mathcal{A}$", A+0.2*dir(-17), S); label("$\mathcal{B}$", A+2.3*dir(-17), S); label("$\mathcal{C}$", A+4.4*dir(-17), S); label("$\mathcal{D}$", A+6.5*dir(-17), S);[/asy]
|
$408$
|
Local Relation Composition
|
HARP
|
ratio
|
123
|
Hexagon $ABCDEF$ is divided into five rhombuses, $\mathcal{P, Q, R, S,}$ and $\mathcal{T,}$ as shown. Rhombuses $\mathcal{P, Q, R,}$ and $\mathcal{S}$ are congruent, and each has area $\sqrt{2006}.$ Let $K$ be the area of rhombus $\mathcal{T}$. Given that $K$ is a positive integer, find the number of possible values for $K$.
|
[asy] size(8cm); pair A=(0,0), B=(4.2,0), C=(5.85,-1.6), D=(4.2,-3.2), EE=(0,-3.2), F=(-1.65,-1.6), G=(0.45,-1.6), H=(3.75,-1.6), I=(2.1,0), J=(2.1,-3.2), K=(2.1,-1.6); draw(A--B--C--D--EE--F--cycle); draw(F--G--(2.1,0)); draw(C--H--(2.1,0)); draw(G--(2.1,-3.2)); draw(H--(2.1,-3.2)); label("$\mathcal{T}$",(2.1,-1.6)); label("$\mathcal{P}$",(0,-1),NE); label("$\mathcal{Q}$",(4.2,-1),NW); label("$\mathcal{R}$",(0,-2.2),SE); label("$\mathcal{S}$",(4.2,-2.2),SW); [/asy]
|
$89$
|
Local Relation Composition
|
HARP
|
count
|
124
|
What is the area of the shaded pinwheel shown in the $5 \times 5$ grid if each grid has side length $2$?
|
[asy] filldraw((2.5,2.5)--(0,1)--(1,1)--(1,0)--(2.5,2.5)--(4,0)--(4,1)--(5,1)--(2.5,2.5)--(5,4)--(4,4)--(4,5)--(2.5,2.5)--(1,5)--(1,4)--(0,4)--cycle, gray, black); int i; for(i=0; i<6; i=i+1) { draw((i,0)--(i,5)); draw((0,i)--(5,i)); } [/asy]
|
$24$
|
Primitive Recognition
|
HARP
|
area
|
125
|
Two circles that share the same center have radii $5$ meters and $10$ meters. An aardvark runs along the path shown, starting at $A$ and ending at $K$. How many meters does the aardvark run?
|
[asy] size((150)); draw((10,0)..(0,10)..(-10,0)..(0,-10)..cycle); draw((20,0)..(0,20)..(-20,0)..(0,-20)..cycle); draw((20,0)--(-20,0)); draw((0,20)--(0,-20)); draw((-2,21.5)..(-15.4, 15.4)..(-22,0), EndArrow); draw((-18,1)--(-12, 1), EndArrow); draw((-12,0)..(-8.3,-8.3)..(0,-12), EndArrow); draw((1,-9)--(1,9), EndArrow); draw((0,12)..(8.3, 8.3)..(12,0), EndArrow); draw((12,-1)--(18,-1), EndArrow); label("$A$", (0,20), N); label("$K$", (20,0), E); [/asy]
|
$10\pi+20$
|
Local Relation Composition
|
HARP
|
length
|
126
|
Jerry cuts a wedge from a $10$-cm cylinder of bologna as shown by the dashed curve. The volume of his wedge in cubic centimeters is ?
|
[asy] defaultpen(linewidth(0.65)); real d=90-63.43494882; draw(ellipse((origin), 2, 4)); fill((0,4)--(0,-4)--(-8,-4)--(-8,4)--cycle, white); draw(ellipse((-4,0), 2, 4)); draw((0,4)--(-4,4)); draw((0,-4)--(-4,-4)); draw(shift(-2,0)*rotate(-d-5)*ellipse(origin, 1.82, 4.56), linetype("10 10")); draw((-4,4)--(-8,4), dashed); draw((-4,-4)--(-8,-4), dashed); draw((-4,4.3)--(-4,5)); draw((0,4.3)--(0,5)); draw((-7,4)--(-7,-4), Arrows(5)); draw((-4,4.7)--(0,4.7), Arrows(5)); label("$8$ cm", (-7,0), W); label("$6$ cm", (-2,4.7), N);[/asy]
|
$80\pi$
|
Local Relation Composition
|
HARP
|
area
|
127
|
A round table has radius $4$. Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x$. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x$?
|
[asy]unitsize(4mm); defaultpen(linewidth(.8)+fontsize(8)); draw(Circle((0,0),4)); path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle; draw(mat); draw(rotate(60)*mat); draw(rotate(120)*mat); draw(rotate(180)*mat); draw(rotate(240)*mat); draw(rotate(300)*mat); label("\(x\)",(-1.55,2.1),E); label("\(1\)",(-0.5,3.8),S);[/asy]
|
$\frac{3\sqrt{7}-\sqrt{3}}{2}$
|
Global Abstract Integration
|
HARP
|
length
|
128
|
The triangular plot of ACD lies between Aspen Road, Brown Road and a railroad. Main Street runs east and west, and the railroad runs north and south. The numbers in the diagram indicate distances in miles. The width of the railroad track can be ignored. How many square miles are in the plot of land ABD?
|
[asy] size(250); defaultpen(linewidth(0.55)); pair A=(-6,0), B=origin, C=(0,6), D=(0,12); pair ac=C+2.828*dir(45), ca=A+2.828*dir(225), ad=D+2.828*dir(A--D), da=A+2.828*dir(D--A), ab=(2.828,0), ba=(-6-2.828, 0); fill(A--C--D--cycle, gray); draw(ba--ab); draw(ac--ca); draw(ad--da); draw((0,-1)--(0,15)); draw((1/3, -1)--(1/3, 15)); int i; for(i=1; i<15; i=i+1) { draw((-1/10, i)--(13/30, i)); } label("$A$", A, SE); label("$B$", B, SE); label("$C$", C, SE); label("$D$", D, SE); label("$3$", (1/3,3), E); label("$3$", (1/3,9), E); label("$3$", (-3,0), S); label("Main", (-3,0), N); label(rotate(45)*"Aspen", A--C, SE); label(rotate(63.43494882)*"Brown", A--D, NW); [/asy]
|
$9$
|
Global Abstract Integration
|
HARP
|
area
|
129
|
Construct a square on one side of an equilateral triangle. On one non-adjacent side of the square, construct a regular pentagon, as shown. On a non-adjacent side of the pentagon, construct a hexagon. Continue to construct regular polygons in the same way, until you construct an octagon. How many sides does the resulting polygon have?
|
[asy] defaultpen(linewidth(0.6)); pair O=origin, A=(0,1), B=A+1*dir(60), C=(1,1), D=(1,0), E=D+1*dir(-72), F=E+1*dir(-144), G=O+1*dir(-108); draw(O--A--B--C--D--E--F--G--cycle); draw(O--D, dashed); draw(A--C, dashed);[/asy]
|
$23$
|
Global Abstract Integration
|
HARP
|
count
|
130
|
A one-cubic-foot cube is cut into four pieces by three cuts parallel to the top face of the cube. The first cut is $\frac{1}{2}$ foot from the top face. The second cut is $\frac{1}{3}$ foot below the first cut, and the third cut is $\frac{1}{17}$ foot below the second cut. From the top to the bottom the pieces are labeled A, B, C, and D. The pieces are then glued together end to end as shown in the second diagram. What is the total surface area of this solid in square feet?The first diagram: ; The second diagram:
|
[asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(2,8/15,7/15); int t=0; void f(real x) { path3 r=(t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--cycle; path3 f=(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)--cycle; path3 u=(t,1,x)--(t+1,1,x)--(t+1,0,x)--(t,0,x)--cycle; draw(surface(r), white, nolight); draw(surface(f), white, nolight); draw(surface(u), white, nolight); draw((t,1,x)--(t+1,1,x)--(t+1,1,0)--(t,1,0)--(t,1,x)--(t,0,x)--(t+1,0,x)--(t+1,1,x)--(t+1,1,0)--(t+1,0,0)--(t+1,0,x)); t=t+1; } f(d); f(1/2); f(1/3); f(1/17); label("D", (1/2, 1, 0), SE); label("A", (1+1/2, 1, 0), SE); label("B", (2+1/2, 1, 0), SE); label("C", (3+1/2, 1, 0), SE);[/asy][asy] import three; real d=11/102; defaultpen(fontsize(8)); defaultpen(linewidth(0.8)); currentprojection=orthographic(1,8/15,7/15); draw(unitcube, white, thick(), nolight); void f(real x) { draw((0,1,x)--(1,1,x)--(1,0,x)); } f(d); f(1/6); f(1/2); label("A", (1,0,3/4), W); label("B", (1,0,1/3), W); label("C", (1,0,1/6-d/4), W); label("D", (1,0,d/2), W); label("1/2", (1,1,3/4), E); label("1/3", (1,1,1/3), E); label("1/17", (0,1,1/6-d/4), E);[/asy]
|
$11$
|
Global Abstract Integration
|
HARP
|
area
|
131
|
A cubical cake with edge length $2$ inches is iced on the sides and the top. It is cut vertically into three pieces as shown in this top view, where $M$ is the midpoint of a top edge. The piece whose top is triangle $B$ contains $c$ cubic inches of cake and $s$ square inches of icing. What is $c+s$?
|
[asy] unitsize(1cm); defaultpen(linewidth(.8pt)+fontsize(8pt)); draw((-1,-1)--(1,-1)--(1,1)--(-1,1)--cycle); draw((1,1)--(-1,0)); pair P=foot((1,-1),(1,1),(-1,0)); draw((1,-1)--P); draw(rightanglemark((-1,0),P,(1,-1),4)); label("$M$",(-1,0),W); label("$C$",(-0.1,-0.3)); label("$A$",(-0.4,0.7)); label("$B$",(0.7,0.4)); [/asy]
|
$\frac{32}{5}$
|
Local Relation Composition
|
HARP
|
count
|
132
|
In $\triangle ABC$ shown in the figure, $AB=7$, $BC=8$, $CA=9$, and $\overline{AH}$ is an altitude. Points $D$ and $E$ lie on sides $\overline{AC}$ and $\overline{AB}$, respectively, so that $\overline{BD}$ and $\overline{CE}$ are angle bisectors, intersecting $\overline{AH}$ at $Q$ and $P$, respectively. What is the value of the ratio: $PQ/AP$?
|
[asy] import graph; size(9cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -4.381056062031275, xmax = 15.020004395092375, ymin = -4.051697595316909, ymax = 10.663513514111651; /* image dimensions */ draw((0.,0.)--(4.714285714285714,7.666518779999279)--(7.,0.)--cycle); /* draw figures */ draw((0.,0.)--(4.714285714285714,7.666518779999279)); draw((4.714285714285714,7.666518779999279)--(7.,0.)); draw((7.,0.)--(0.,0.)); label("7",(3.2916797119724284,-0.07831656949355523),SE*labelscalefactor); label("9",(2.0037562070503783,4.196493361737088),SE*labelscalefactor); label("8",(6.114150371695219,3.785453945272603),SE*labelscalefactor); draw((0.,0.)--(6.428571428571427,1.9166296949998194)); draw((7.,0.)--(2.2,3.5777087639996634)); draw((4.714285714285714,7.666518779999279)--(3.7058823529411766,0.)); /* dots and labels */ dot((0.,0.),dotstyle); label("$A$", (-0.2432592696221352,-0.5715638692509372), NE * labelscalefactor); dot((7.,0.),dotstyle); label("$B$", (7.0458397156813835,-0.48935598595804014), NE * labelscalefactor); dot((3.7058823529411766,0.),dotstyle); label("$E$", (3.8123296394941084,0.16830708038513573), NE * labelscalefactor); dot((4.714285714285714,7.666518779999279),dotstyle); label("$C$", (4.579603216894479,7.895848109917452), NE * labelscalefactor); dot((2.2,3.5777087639996634),linewidth(3.pt) + dotstyle); label("$D$", (2.1407693458718726,3.127790878929427), NE * labelscalefactor); dot((6.428571428571427,1.9166296949998194),linewidth(3.pt) + dotstyle); label("$H$", (6.004539860638023,1.9494778850645704), NE * labelscalefactor); dot((5.,1.49071198499986),linewidth(3.pt) + dotstyle); label("$Q$", (4.935837377830365,1.7302568629501784), NE * labelscalefactor); dot((3.857142857142857,1.1499778169998918),linewidth(3.pt) + dotstyle); label("$P$", (3.538303361851119,1.2370095631927964), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]
|
$\frac{8}{27}$
|
Primitive Recognition
|
HARP
|
ratio
|
133
|
In $\triangle ABC$, $AB = 6$, $BC = 7$, and $CA = 8$. Point $D$ lies on $\overline{BC}$, and $\overline{AD}$ bisects $\angle BAC$. Point $E$ lies on $\overline{AC}$, and $\overline{BE}$ bisects $\angle ABC$. The bisectors intersect at $F$. What is the value of the ratio $AF$ : $AD$?
|
[asy] pair A = (0,0), B=(6,0), C=intersectionpoints(Circle(A,8),Circle(B,7))[0], F=incenter(A,B,C), D=extension(A,F,B,C),E=extension(B,F,A,C); draw(A--B--C--A--D^^B--E); label("$A$",A,SW); label("$B$",B,SE); label("$C$",C,N); label("$D$",D,NE); label("$E$",E,NW); label("$F$",F,1.5*N); [/asy]
|
$\frac{2}{3}$
|
Local Relation Composition
|
HARP
|
ratio
|
134
|
In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $FGHJ$ is 539. Find the area of $KLMN$.
|
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]
|
$99$
|
Local Relation Composition
|
HARP
|
area
|
135
|
A collection of circles in the upper half-plane, all tangent to the $x$-axis, is constructed in layers as follows. Layer $L_0$ consists of two circles of radii $70^2$ and $73^2$ that are externally tangent. For $k \ge 1$, the circles in $\bigcup_{j=0}^{k-1}L_j$ are ordered according to their points of tangency with the $x$-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer $L_k$ consists of the $2^{k-1}$ circles constructed in this way. Let $S=\bigcup_{j=0}^{6}L_j$, and for every circle $C$ denote by $r(C)$ its radius. What is \[\sum_{C\in S} \frac{1}{\sqrt{r(C)}}?\]
|
[asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64 - skip; k = k + 2*skip) { pair cent1 = coord[k-skip], cent2 = coord[k+skip]; real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2); real shiftx = cent1.x + sqrt(4*r1*rn); coord[k] = (shiftx,rn); } // Draw the remaining 63 circles } for(int i=1;i<=63;i=i+1) { filldraw(circle(coord[i],coord[i].y),gray(0.75)); }[/asy]
|
$\frac{143}{14}$
|
Global Abstract Integration
|
HARP
|
count
|
136
|
The shaded region below is called a shark's fin falcata, a figure studied by Leonardo da Vinci. It is bounded by the portion of the circle of radius $3$ and center $(0,0)$ that lies in the first quadrant, the portion of the circle with radius $\tfrac{3}{2}$ and center $(0,\tfrac{3}{2})$ that lies in the first quadrant, and the line segment from $(0,0)$ to $(3,0)$. What is the area of the shark's fin falcata?
|
[asy] import cse5;pathpen=black;pointpen=black; size(1.5inch); D(MP("x",(3.5,0),S)--(0,0)--MP("\frac{3}{2}",(0,3/2),W)--MP("y",(0,3.5),W)); path P=(0,0)--MP("3",(3,0),S)..(3*dir(45))..MP("3",(0,3),W)--(0,3)..(3/2,3/2)..cycle; draw(P,linewidth(2)); fill(P,gray); [/asy]
|
$\frac{9\pi}{8}$
|
Local Relation Composition
|
HARP
|
area
|
137
|
The diagram below shows the circular face of a clock with radius $20$ cm and a circular disk with radius $10$ cm externally tangent to the clock face at $12$ o' clock. The disk has an arrow painted on it, initially pointing in the upward vertical direction. Let the disk roll clockwise around the clock face. At what numerical point on the clock face will the disk be tangent when the arrow is next pointing in the upward vertical direction?
|
[asy]size(170);defaultpen(linewidth(0.9)+fontsize(13pt));draw(unitcircle^^circle((0,1.5),0.5)); path arrow = origin--(-0.13,-0.35)--(-0.06,-0.35)--(-0.06,-0.7)--(0.06,-0.7)--(0.06,-0.35)--(0.13,-0.35)--cycle; for(int i=1;i<=12;i=i+1){draw(0.9*dir(90-30*i)--dir(90-30*i));label("$"+(string) i+"$",0.78*dir(90-30*i));} dot(origin);draw(shift((0,1.87))*arrow);draw(arc(origin,1.5,68,30),EndArrow(size=12));[/asy]
|
$4$
|
Global Abstract Integration
|
HARP
|
count
|
138
|
Point $O$ is the center of the regular octagon $ABCDEFGH$, and $X$ is the midpoint of the side $\overline{AB}.$ What fraction of the area of the octagon is shaded?
|
[asy] pair A,B,C,D,E,F,G,H,O,X; A=dir(45); B=dir(90); C=dir(135); D=dir(180); E=dir(-135); F=dir(-90); G=dir(-45); H=dir(0); O=(0,0); X=midpoint(A--B); fill(X--B--C--D--E--O--cycle,rgb(0.75,0.75,0.75)); draw(A--B--C--D--E--F--G--H--cycle); dot("$A$",A,dir(45)); dot("$B$",B,dir(90)); dot("$C$",C,dir(135)); dot("$D$",D,dir(180)); dot("$E$",E,dir(-135)); dot("$F$",F,dir(-90)); dot("$G$",G,dir(-45)); dot("$H$",H,dir(0)); dot("$X$",X,dir(135/2)); dot("$O$",O,dir(0)); draw(E--O--X); [/asy]
|
$\frac{7}{16}$
|
Primitive Recognition
|
HARP
|
ratio
|
139
|
A rectangle has sides of length $a$ and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find $a$.
|
[asy] pair A,B,C,D,E,F,R,S,T,X,Y,Z; dotfactor = 2; unitsize(.1cm); A = (0,0); B = (0,18); C = (0,36); // don't look here D = (12*2.236, 36); E = (12*2.236, 18); F = (12*2.236, 0); draw(A--B--C--D--E--F--cycle); dot(" ",A,NW); dot(" ",B,NW); dot(" ",C,NW); dot(" ",D,NW); dot(" ",E,NW); dot(" ",F,NW); //don't look here R = (12*2.236 +22,0); S = (12*2.236 + 22 - 13.4164,12); T = (12*2.236 + 22,24); X = (12*4.472+ 22,24); Y = (12*4.472+ 22 + 13.4164,12); Z = (12*4.472+ 22,0); draw(R--S--T--X--Y--Z--cycle); dot(" ",R,NW); dot(" ",S,NW); dot(" ",T,NW); dot(" ",X,NW); dot(" ",Y,NW); dot(" ",Z,NW); // sqrt180 = 13.4164 // sqrt5 = 2.236[/asy]
|
$12\sqrt{5}$
|
Local Relation Composition
|
HARP
|
length
|
140
|
In the figure, $ABCD$ is a square of side length $2$. The rectangles $JKHG$ and $EBCF$ are congruent. What is $AE$?
|
[asy] pair A=(1,0), B=(0,0), C=(0,1), D=(1,1), E=(2-sqrt(3),0), F=(2-sqrt(3),1), G=(1,sqrt(3)/2), H=(2.5-sqrt(3),1), J=(.5,0), K=(2-sqrt(3),1-sqrt(3)/2); draw(A--B--C--D--cycle); draw(K--H--G--J--cycle); draw(F--E); label("$A$",A,SE); label("$B$",B,SW); label("$C$",C,NW); label("$D$",D,NE); label("$E$",E,S); label("$F$",F,N); label("$G$",G,E); label("$H$",H,N); label("$J$",J,S); label("$K$",K,W); [/asy]
|
$2\sqrt{3}-2$
|
Local Relation Composition
|
HARP
|
length
|
141
|
Six regular hexagons surround a regular hexagon of side length $2$ as shown. What is the area of $\triangle{ABC}$?
|
[asy] draw((0,0)--(-5,8.66025404)--(0, 17.3205081)--(10, 17.3205081)--(15,8.66025404)--(10, 0)--(0, 0)); draw((30,0)--(25,8.66025404)--(30, 17.3205081)--(40, 17.3205081)--(45, 8.66025404)--(40, 0)--(30, 0)); draw((30,0)--(25,-8.66025404)--(30, -17.3205081)--(40, -17.3205081)--(45, -8.66025404)--(40, 0)--(30, 0)); draw((0,0)--(-5, -8.66025404)--(0, -17.3205081)--(10, -17.3205081)--(15, -8.66025404)--(10, 0)--(0, 0)); draw((15,8.66025404)--(10, 17.3205081)--(15, 25.9807621)--(25, 25.9807621)--(30, 17.3205081)--(25, 8.66025404)--(15, 8.66025404)); draw((15,-8.66025404)--(10, -17.3205081)--(15, -25.9807621)--(25, -25.9807621)--(30, -17.3205081)--(25, -8.66025404)--(15, -8.66025404)); label("A", (0,0), W); label("B", (30, 17.3205081), NE); label("C", (30, -17.3205081), SE); draw((0,0)--(30, 17.3205081)--(30, -17.3205081)--(0, 0)); //(Diagram Creds-DivideBy0) [/asy]
|
$12\sqrt{3}$
|
Local Relation Composition
|
HARP
|
area
|
142
|
In triangle $ABC$, determine the measure of angle A. Note that all line segments marked with short ticks in the diagram are of equal length.
|
[asy]
import graph; size(9.115122858763474cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7.9216637359954705, xmax = 10.308581981531479, ymin = -6.062398124651168, ymax = 9.377503860601273; /* image dimensions */
/* draw figures */
draw((1.1862495478417192,2.0592342833377844)--(3.0842,-3.6348), linewidth(1.6));
draw((2.412546402365528,-0.6953452852662508)--(1.8579031454761883,-0.8802204313959623), linewidth(1.6));
draw((-3.696094000229639,5.5502174997511595)--(1.1862495478417192,2.0592342833377844), linewidth(1.6));
draw((-1.0848977580797177,4.042515022414228)--(-1.4249466943082025,3.5669367606747184), linewidth(1.6));
draw((1.1862495478417192,2.0592342833377844)--(9.086047353374928,-3.589295214974483), linewidth(1.6));
draw((9.086047353374928,-3.589295214974483)--(3.0842,-3.6348), linewidth(1.6));
draw((6.087339936804166,-3.904360930324946)--(6.082907416570757,-3.319734284649538), linewidth(1.6));
draw((3.0842,-3.6348)--(-2.9176473533749285,-3.6803047850255166), linewidth(1.6));
draw((0.08549258342923806,-3.9498657153504615)--(0.08106006319583221,-3.3652390696750536), linewidth(1.6));
draw((-2.9176473533749285,-3.6803047850255166)--(-6.62699301304923,-3.7084282888220432), linewidth(1.6));
draw((-6.62699301304923,-3.7084282888220432)--(-4.815597805533209,2.0137294983122676), linewidth(1.6));
draw((-5.999986761815922,-0.759127399441624)--(-5.442604056766517,-0.9355713910681529), linewidth(1.6));
draw((-4.815597805533209,2.0137294983122676)--(-3.696094000229639,5.5502174997511595), linewidth(1.6));
draw((-4.815597805533209,2.0137294983122676)--(1.1862495478417192,2.0592342833377844), linewidth(1.6));
draw((-1.8168903889624484,2.3287952136627297)--(-1.8124578687290425,1.744168567987322), linewidth(1.6));
draw((-4.815597805533209,2.0137294983122676)--(-2.9176473533749285,-3.6803047850255166), linewidth(1.6));
draw((-3.5893009510093994,-0.7408500702917692)--(-4.1439442078987385,-0.9257252164214806), linewidth(1.6));
label("$A$",(-4.440377746205339,7.118654172569505),SE*labelscalefactor,fontsize(14));
label("$B$",(-7.868514331571194,-3.218904987952353),SE*labelscalefactor,fontsize(14));
label("$C$",(9.165869786409527,-3.0594567746795223),SE*labelscalefactor,fontsize(14));
/* dots and labels */
dot((3.0842,-3.6348),linewidth(3.pt) + dotstyle);
dot((9.086047353374928,-3.589295214974483),linewidth(3.pt) + dotstyle);
dot((1.1862495478417192,2.0592342833377844),linewidth(3.pt) + dotstyle);
dot((-2.9176473533749285,-3.6803047850255166),linewidth(3.pt) + dotstyle);
dot((-4.815597805533209,2.0137294983122676),linewidth(3.pt) + dotstyle);
dot((-6.62699301304923,-3.7084282888220432),linewidth(3.pt) + dotstyle);
dot((-3.696094000229639,5.5502174997511595),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]
|
$72$
|
Global Abstract Integration
|
olympiads
|
angle
|
143
|
In the diagram below, angle $ABC$ is a right angle. Point $D$ is on $\overline{BC}$ , and $\overline{AD}$ bisects angle $CAB$ . Points $E$ and $F$ are on $\overline{AB}$ and $\overline{AC}$ , respectively, so that $AE=3$ and $AF=10$ . Given that $EB=9$ and $FC=27$ , find the area of quadrilateral $DCFG$. The final answer should discards the decimal part and retains only the integer.
|
[asy] size(250); pair A=(0,12), E=(0,8), B=origin, C=(24*sqrt(2),0), D=(6*sqrt(2),0), F=A+10*dir(A--C), G=intersectionpoint(E--F, A--D); draw(A--B--C--A--D^^E--F); pair point=G+1*dir(250); label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); label("$E$", E, dir(point--E)); label("$F$", F, dir(point--F)); label("$G$", G, dir(point--G)); markscalefactor=0.1; draw(rightanglemark(A,B,C)); label("10", A--F, dir(90)*dir(A--F)); label("27", F--C, dir(90)*dir(F--C)); label("3", (0,10), W); label("9", (0,4), W); [/asy]
|
$147$
|
Local Relation Composition
|
AIME-83-24
|
area
|
144
|
In $\triangle{ABC}$, $AX=XY=YB=BC$, and $m\angle{ABC}=120^{\circ}$. What is the measure of $m\angle{BCA}$?
|
[asy]
pair A, B, C, X, Y;
A = origin;
X = dir(30);
Y = X + dir(0);
B = Y + dir(60);
C = B + dir(330);
draw(A--B--C--cycle);
draw(X--Y--B);
label("$A$",A,W);
label("$B$",B,N);
label("$C$",C,E);
label("$X$",X,NW);
label("$Y$",Y,SE);
[/asy]
|
$45$
|
Local Relation Composition
|
olympiads
|
angle
|
145
|
In triangle $ABC$, determine the measure of angle C. Note that all line segments marked with short ticks in the diagram are of equal length.
|
[asy]
import graph; size(9.115122858763474cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7.9216637359954705, xmax = 10.308581981531479, ymin = -6.062398124651168, ymax = 9.377503860601273; /* image dimensions */
/* draw figures */
draw((1.1862495478417192,2.0592342833377844)--(3.0842,-3.6348), linewidth(1.6));
draw((2.412546402365528,-0.6953452852662508)--(1.8579031454761883,-0.8802204313959623), linewidth(1.6));
draw((-3.696094000229639,5.5502174997511595)--(1.1862495478417192,2.0592342833377844), linewidth(1.6));
draw((-1.0848977580797177,4.042515022414228)--(-1.4249466943082025,3.5669367606747184), linewidth(1.6));
draw((1.1862495478417192,2.0592342833377844)--(9.086047353374928,-3.589295214974483), linewidth(1.6));
draw((9.086047353374928,-3.589295214974483)--(3.0842,-3.6348), linewidth(1.6));
draw((6.087339936804166,-3.904360930324946)--(6.082907416570757,-3.319734284649538), linewidth(1.6));
draw((3.0842,-3.6348)--(-2.9176473533749285,-3.6803047850255166), linewidth(1.6));
draw((0.08549258342923806,-3.9498657153504615)--(0.08106006319583221,-3.3652390696750536), linewidth(1.6));
draw((-2.9176473533749285,-3.6803047850255166)--(-6.62699301304923,-3.7084282888220432), linewidth(1.6));
draw((-6.62699301304923,-3.7084282888220432)--(-4.815597805533209,2.0137294983122676), linewidth(1.6));
draw((-5.999986761815922,-0.759127399441624)--(-5.442604056766517,-0.9355713910681529), linewidth(1.6));
draw((-4.815597805533209,2.0137294983122676)--(-3.696094000229639,5.5502174997511595), linewidth(1.6));
draw((-4.815597805533209,2.0137294983122676)--(1.1862495478417192,2.0592342833377844), linewidth(1.6));
draw((-1.8168903889624484,2.3287952136627297)--(-1.8124578687290425,1.744168567987322), linewidth(1.6));
draw((-4.815597805533209,2.0137294983122676)--(-2.9176473533749285,-3.6803047850255166), linewidth(1.6));
draw((-3.5893009510093994,-0.7408500702917692)--(-4.1439442078987385,-0.9257252164214806), linewidth(1.6));
label("$A$",(-4.440377746205339,7.118654172569505),SE*labelscalefactor,fontsize(14));
label("$B$",(-7.868514331571194,-3.218904987952353),SE*labelscalefactor,fontsize(14));
label("$C$",(9.165869786409527,-3.0594567746795223),SE*labelscalefactor,fontsize(14));
/* dots and labels */
dot((3.0842,-3.6348),linewidth(3.pt) + dotstyle);
dot((9.086047353374928,-3.589295214974483),linewidth(3.pt) + dotstyle);
dot((1.1862495478417192,2.0592342833377844),linewidth(3.pt) + dotstyle);
dot((-2.9176473533749285,-3.6803047850255166),linewidth(3.pt) + dotstyle);
dot((-4.815597805533209,2.0137294983122676),linewidth(3.pt) + dotstyle);
dot((-6.62699301304923,-3.7084282888220432),linewidth(3.pt) + dotstyle);
dot((-3.696094000229639,5.5502174997511595),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
/* end of picture */
[/asy]
|
$36$
|
Local Relation Composition
|
olympiads
|
angle
|
146
|
The diagram shows an octagon consisting of $10$ unit squares. The portion below $\overline{PQ}$ is a unit square and a triangle with base $5$. If $\overline{PQ}$ bisects the area of the octagon, what is the ratio $\dfrac{XQ}{XY}$?
|
[asy] import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((0,0)--(6,0),linewidth(1.2pt)); draw((0,0)--(0,1),linewidth(1.2pt)); draw((0,1)--(1,1),linewidth(1.2pt)); draw((1,1)--(1,2),linewidth(1.2pt)); draw((1,2)--(5,2),linewidth(1.2pt)); draw((5,2)--(5,1),linewidth(1.2pt)); draw((5,1)--(6,1),linewidth(1.2pt)); draw((6,1)--(6,0),linewidth(1.2pt)); draw((1,1)--(5,1),linewidth(1.2pt)); draw((1,1)--(1,0),linewidth(1.2pt)); draw((2,2)--(2,0),linewidth(1.2pt)); draw((3,2)--(3,0),linewidth(1.2pt)); draw((4,2)--(4,0),linewidth(1.2pt)); draw((5,1)--(5,0),linewidth(1.2pt)); draw((0,0)--(5,1.5),linewidth(1.2pt)); dot((0,0),ds); label("$P$", (-0.23,-0.26),NE*lsf); dot((0,1),ds); dot((1,1),ds); dot((1,2),ds); dot((5,2),ds); label("$X$", (5.14,2.02),NE*lsf); dot((5,1),ds); label("$Y$", (5.12,1.14),NE*lsf); dot((6,1),ds); dot((6,0),ds); dot((1,0),ds); dot((2,0),ds); dot((3,0),ds); dot((4,0),ds); dot((5,0),ds); dot((2,2),ds); dot((3,2),ds); dot((4,2),ds); dot((5,1.5),ds); label("$Q$", (5.14,1.51),NE*lsf); clip((-4.19,-5.52)--(-4.19,6.5)--(10.08,6.5)--(10.08,-5.52)--cycle); [/asy]
|
$\frac{2}{5}$
|
Local Relation Composition
|
HARP
|
ratio
|
147
|
In the accompanying figure, the outer square $S$ has side length $20$. A second square $S'$ of side length $7.5$ is constructed inside $S$ with the same center as $S$ and with sides parallel to those of $S$. From each midpoint of a side of $S$, segments are drawn to the two closest vertices of $S'$. The result is a four-pointed starlike figure inscribed in $S$. The star figure is cut out and then folded to form a pyramid with base $S'$. Find the volume of this pyramid.
|
[asy] pair S1 = (20, 20), S2 = (-20, 20), S3 = (-20, -20), S4 = (20, -20); pair M1 = (S1+S2)/2, M2 = (S2+S3)/2, M3=(S3+S4)/2, M4=(S4+S1)/2; pair Sp1 = (7.5, 7.5), Sp2=(-7.5, 7.5), Sp3 = (-7.5, -7.5), Sp4 = (7.5, -7.5); draw(S1--S2--S3--S4--cycle); draw(Sp1--Sp2--Sp3--Sp4--cycle); draw(Sp1--M1--Sp2--M2--Sp3--M3--Sp4--M4--cycle); [/asy]
|
$\frac{375}{4}$
|
Global Abstract Integration
|
HARP
|
volume
|
148
|
Semicircles $POQ$ and $ROS$ pass through the center $O$. What is the area of the two semicircles?
|
[asy] import graph; size(7.5cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.27,xmax=10.01,ymin=-5.65,ymax=10.98; draw(circle((0,0),2)); draw((-3,0)--(3,0),EndArrow(6)); draw((0,-3)--(0,3),EndArrow(6)); draw(shift((0.01,1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,179.76,359.76)); draw(shift((-0.01,-1.42))*xscale(1.41)*yscale(1.41)*arc((0,0),1,-0.38,179.62)); draw((-1.4,1.43)--(1.41,1.41)); draw((-1.42,-1.41)--(1.4,-1.42)); label("$ P(-1,1) $",(-2.57,2.17),SE*lsf); label("$ Q(1,1) $",(1.55,2.21),SE*lsf); label("$ R(-1,-1) $",(-2.72,-1.45),SE*lsf); label("$S(1,-1)$",(1.59,-1.49),SE*lsf); dot((0,0),ds); label("$O$",(-0.24,-0.35),NE*lsf); dot((1.41,1.41),ds); dot((-1.4,1.43),ds); dot((1.4,-1.42),ds); dot((-1.42,-1.41),ds); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]
|
$\pi$
|
Primitive Recognition
|
HARP
|
area
|
149
|
A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths $20$ and $30$ meters. What fraction of the yard is occupied by the flower beds?
$\mathrm{(A)}\frac {1}{8}\qquad \mathrm{(B)}\frac {1}{6}\qquad \mathrm{(C)}\frac {1}{5}\qquad \mathrm{(D)}\frac {1}{4}\qquad \mathrm{(E)}\frac {1}{3}$
|
[asy] unitsize(2mm); defaultpen(linewidth(.8pt)); fill((0,0)--(0,5)--(5,5)--cycle,gray); fill((25,0)--(25,5)--(20,5)--cycle,gray); draw((0,0)--(0,5)--(25,5)--(25,0)--cycle); draw((0,0)--(5,5)); draw((20,5)--(25,0)); [/asy]
|
$\frac{1}{6}$
|
Local Relation Composition
|
olympiads
|
ratio
|
150
|
Circle $C_1$ and $C_2$ each have radius $1$, and the distance between their centers is $\frac{1}{2}$. Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2$. Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and externally tangent to $C_3$. How many times the diameter of $C_2$ is the diameter of $C_4$?
|
[asy] import olympiad; size(10cm); draw(circle((0,0),0.75)); draw(circle((-0.25,0),1)); draw(circle((0.25,0),1)); draw(circle((0,6/7),3/28)); pair A = (0,0), B = (-0.25,0), C = (0.25,0), D = (0,6/7), E = (-0.95710678118, 0.70710678118), F = (0.95710678118, -0.70710678118); dot(B^^C); draw(B--E, dashed); draw(C--F, dashed); draw(B--C); label("$C_4$", D); label("$C_1$", (-1.375, 0)); label("$C_2$", (1.375,0)); label("$\frac{1}{2}$", (0, -.125)); label("$C_3$", (-0.4, -0.4)); label("$1$", (-.85, 0.70)); label("$1$", (.85, -.7)); import olympiad; markscalefactor=0.005; [/asy]
|
$\frac{28}{3}$
|
Local Relation Composition
|
HARP
|
ratio
|
151
|
Usain is walking for exercise by zigzagging across a $100$-meter by $30$-meter rectangular field, beginning at point $A$ and ending on the segment $\overline{BC}$. He wants to increase the distance walked by zigzagging as shown in the figure below $(APQRS)$. What is the measure of angle $\theta = \angle PAB=\angle QPC=\angle RQB=\cdots$ will produce in a length that is $200$ meters? (This figure is not drawn to scale. Do not assume that he zigzag path has exactly four segments as shown; there could be more or fewer.)
|
[asy] import olympiad; draw((-50,15)--(50,15)); draw((50,15)--(50,-15)); draw((50,-15)--(-50,-15)); draw((-50,-15)--(-50,15)); draw((-50,-15)--(-22.5,15)); draw((-22.5,15)--(5,-15)); draw((5,-15)--(32.5,15)); draw((32.5,15)--(50,-4.090909090909)); label("$\theta$", (-41.5,-10.5)); label("$\theta$", (-13,10.5)); label("$\theta$", (15.5,-10.5)); label("$\theta$", (43,10.5)); dot((-50,15)); dot((-50,-15)); dot((50,15)); dot((50,-15)); dot((50,-4.09090909090909)); label("$D$",(-58,15)); label("$A$",(-58,-15)); label("$C$",(58,15)); label("$B$",(58,-15)); label("$S$",(58,-4.0909090909)); dot((-22.5,15)); dot((5,-15)); dot((32.5,15)); label("$P$",(-22.5,23)); label("$Q$",(5,-23)); label("$R$",(32.5,23)); [/asy]
|
$60$
|
Global Abstract Integration
|
HARP
|
angle
|
152
|
Six regular hexagonal blocks of side length 2 unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied by the blocks?
|
[asy] unitsize(1cm); draw(scale(3)*polygon(6)); filldraw(shift(dir(0)*2+dir(120)*0.4)*polygon(6), lightgray); filldraw(shift(dir(60)*2+dir(180)*0.4)*polygon(6), lightgray); filldraw(shift(dir(120)*2+dir(240)*0.4)*polygon(6), lightgray); filldraw(shift(dir(180)*2+dir(300)*0.4)*polygon(6), lightgray); filldraw(shift(dir(240)*2+dir(360)*0.4)*polygon(6), lightgray); filldraw(shift(dir(300)*2+dir(420)*0.4)*polygon(6), lightgray); [/asy]
|
$18\sqrt{3}}$
|
Local Relation Composition
|
HARP
|
area
|
153
|
Isosceles $\triangle ABC$ has equal side lengths $AB$ and $BC$. In the figure below, segments are drawn parallel to $\overline{AC}$ so that the shaded portions of $\triangle ABC$ have the same area. The heights of the two unshaded portions are 11 and 5 units, respectively. What is the height of $h$ of $\triangle ABC$?
|
[asy] size(12cm); real h = 2.5; // height real g=4; //c2c space real s = 0.65; //Xcord of Hline real adj = 0.08; //adjust line diffs pair A,B,C; B=(0,h); C=(1,0); A=-conj(C); pair PONE=(s,h*(1-s)); //Endpoint of Hline ONE pair PTWO=(s+adj,h*(1-s-adj)); //Endpoint of Hline ONE path LONE=PONE--(-conj(PONE)); //Hline ONE path LTWO=PTWO--(-conj(PTWO)); path T=A--B--C--cycle; //Triangle fill (shift(g,0)*(LTWO--B--cycle),mediumgrey); fill (LONE--A--C--cycle,mediumgrey); draw(LONE); draw(T); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); draw(shift(g,0)*LTWO); draw(shift(g,0)*T); label("$A$",shift(g,0)*A,SW); label("$B$",shift(g,0)*B,N); label("$C$",shift(g,0)*C,SE); draw(B--shift(g,0)*B,dashed); draw(C--shift(g,0)*A,dashed); draw((g/2,0)--(g/2,h),dashed); draw((0,h*(1-s))--B,dashed); draw((g,h*(1-s-adj))--(g,0),dashed); label("$5$", midpoint((g,h*(1-s-adj))--(g,0)),UnFill); label("$h$", midpoint((g/2,0)--(g/2,h)),UnFill); label("$11$", midpoint((0,h*(1-s))--B),UnFill); [/asy]
|
$14.6$
|
Local Relation Composition
|
HARP
|
length
|
154
|
A <i>regular octahedron</i> has eight equilateral triangle faces with four faces meeting at each vertex. Jun will make the regular octahedron shown on the right by folding the piece of paper shown on the left. Which numbered face will end up to the left of $Q$?
|
[asy] import graph; size(15cm); pair Fr, Lf, Rt, Tp, Bt, Bk; Lf=(0,0); Rt=(12,1); Fr=(7,-1); Bk=(5,2); Tp=(6,6.7); Bt=(6,-5.2); draw(Lf--Fr--Rt); draw(Lf--Tp--Rt); draw(Lf--Bt--Rt); draw(Tp--Fr--Bt); draw(Lf--Bk--Rt,dashed); draw(Tp--Bk--Bt,dashed); label(rotate(-8.13010235)*slant(0.1)*"$Q$", (4.2,1.6)); label(rotate(21.8014095)*slant(-0.2)*"$?$", (8.5,2.05)); pair g = (-8,0); real a = 8; draw(g+(-a/2,1)--g+(a/2,1), Arrow()); pair DA,DB,DC,CD,O; DA = (4*sqrt(3),0); DB = (2*sqrt(3),6); DC = (DA+DB)/3; CD = conj(DC); O=(0,0); transform trf=shift(3g+(0,3)); path NET = O--(-2*DA)--(-2DB)--(-DB)--(2DA-DB)--DB--O--DA--(DA-DB)--O--(-DB)--(-DA)--(-DA-DB)--(-DB); draw(trf*NET); label("$7$",trf*DC); label("$Q$",trf*DC+DA-DB); label("$5$",trf*DC-DB); label("$3$",trf*DC-DA-DB); label("$6$",trf*CD); label("$4$",trf*CD-DA); label("$2$",trf*CD-DA-DB); label("$1$",trf*CD-2DA); [/asy]
|
$6$
|
Global Abstract Integration
|
HARP
|
count
|
155
|
The figure below shows a large white circle with a number of smaller white and shaded circles in its interior. What fraction of the interior of the large white circle is not shaded?
|
[asy] size(6cm); draw(circle((3,3),3)); filldraw(circle((2,3),2),lightgrey); filldraw(circle((3,3),1),white); filldraw(circle((1,3),1),white); filldraw(circle((5.5,3),0.5),lightgrey); filldraw(circle((4.5,4.5),0.5),lightgrey); filldraw(circle((4.5,1.5),0.5),lightgrey); int i, j; for(i=0; i<7; i=i+1) { draw((0,i)--(6,i), dashed+grey); draw((i,0)--(i,6), dashed+grey); } [/asy]
|
$\frac{25}{36}$
|
Local Relation Composition
|
HARP
|
ratio
|
156
|
A bowl is formed by attaching four regular hexagons of side $3$ to a square of side $3$. The edges of the adjacent hexagons coincide, as shown in the figure. What is the area of the octagon obtained by joining the top eight vertices of the four hexagons, situated on the rim of the bowl?
|
[asy] import three; size(225); currentprojection= orthographic(camera=(-5.52541796301147,-2.61548797564715,1.6545450372312), up=(0.00247902062334861,0.000877141782387748,0.00966536329192992), target=(0,0,0), zoom=0.570588560870951); currentpen = black+1.5bp; triple A = O; triple M = (X+Y)/2; triple B = (-1/2,-1/2,1/sqrt(2)); triple C = (-1,0,sqrt(2)); triple D = (0,-1,sqrt(2)); transform3 rho = rotate(90,M,M+Z); //arrays of vertices for the lower level (the square), the middle level, //and the interleaves vertices of the upper level (the octagon) triple[] lVs = {A}; triple[] mVs = {B}; triple[] uVsl = {C}; triple[] uVsr = {D}; for(int i = 0; i < 3; ++i){ lVs.push(rho*lVs[i]); mVs.push(rho*mVs[i]); uVsl.push(rho*uVsl[i]); uVsr.push(rho*uVsr[i]); } lVs.cyclic = true; uVsl.cyclic = true; for(int i : new int[] {0,1,2,3}){ draw(uVsl[i]--uVsr[i]); draw(uVsr[i]--uVsl[i+1]); } draw(lVs[0]--lVs[1]^^lVs[0]--lVs[3]); for(int i : new int[] {0,1,3}){ draw(lVs[0]--lVs[i]); draw(lVs[i]--mVs[i]); draw(mVs[i]--uVsl[i]); } for(int i : new int[] {0,3}){ draw(mVs[i]--uVsr[i]); } for(int i : new int[] {1,3}) draw(lVs[2]--lVs[i],dashed); draw(lVs[2]--mVs[2],dashed); draw(mVs[2]--uVsl[2]^^mVs[2]--uVsr[2],dashed); draw(mVs[1]--uVsr[1],dashed); //Comment two lines below to remove red edges //draw(lVs[1]--lVs[3],red+2bp); //draw(uVsl[0]--uVsr[0],red+2bp); [/asy]
|
$63$
|
Global Abstract Integration
|
HARP
|
area
|
157
|
A ball with diameter 8 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are $R_1 = 120$ inches, $R_2 = 80$ inches, and $R_3 = 100$ inches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?
|
[asy] pair A,B; size(8cm); A=(0,0); B=(480,0); draw((0,0)--(480,0),linetype("3 4")); filldraw(circle((8,0),8),black); draw((0,0)..(100,-100)..(200,0)); draw((200,0)..(260,60)..(320,0)); draw((320,0)..(400,-80)..(480,0)); draw((100,0)--(150,-50sqrt(3)),Arrow(size=4)); draw((260,0)--(290,30sqrt(3)),Arrow(size=4)); draw((400,0)--(440,-40sqrt(3)),Arrow(size=4)); label("$A$", A, SW); label("$B$", B, SE); label("$R_1$", (100,-40), W); label("$R_2$", (260,40), SW); label("$R_3$", (400,-40), W);[/asy]
|
$296\pi$
|
Local Relation Composition
|
HARP
|
length
|
158
|
Four identical squares and one rectangle are placed together to form one large square as shown. The width of the rectangle is how many times as large as its length?
|
[asy] unitsize(8mm); defaultpen(linewidth(.8pt)); draw((0,0)--(4,0)--(4,4)--(0,4)--cycle); draw((0,3)--(0,4)--(1,4)--(1,3)--cycle); draw((1,3)--(1,4)--(2,4)--(2,3)--cycle); draw((2,3)--(2,4)--(3,4)--(3,3)--cycle); draw((3,3)--(3,4)--(4,4)--(4,3)--cycle); [/asy]
|
$\frac{3}{4}$
|
Local Relation Composition
|
HARP
|
ratio
|
159
|
In triangle $ABC$, determine the measure of angle B. Note that all line segments marked with short ticks in the diagram are of equal length.
|
[asy]
import graph; size(9.115122858763474cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* point style */
real xmin = -7.9216637359954705, xmax = 10.308581981531479, ymin = -6.062398124651168, ymax = 9.377503860601273; /* image dimensions */
/* draw figures */
draw((1.1862495478417192,2.0592342833377844)--(3.0842,-3.6348), linewidth(1.6));
draw((2.412546402365528,-0.6953452852662508)--(1.8579031454761883,-0.8802204313959623), linewidth(1.6));
draw((-3.696094000229639,5.5502174997511595)--(1.1862495478417192,2.0592342833377844), linewidth(1.6));
draw((-1.0848977580797177,4.042515022414228)--(-1.4249466943082025,3.5669367606747184), linewidth(1.6));
draw((1.1862495478417192,2.0592342833377844)--(9.086047353374928,-3.589295214974483), linewidth(1.6));
draw((9.086047353374928,-3.589295214974483)--(3.0842,-3.6348), linewidth(1.6));
draw((6.087339936804166,-3.904360930324946)--(6.082907416570757,-3.319734284649538), linewidth(1.6));
draw((3.0842,-3.6348)--(-2.9176473533749285,-3.6803047850255166), linewidth(1.6));
draw((0.08549258342923806,-3.9498657153504615)--(0.08106006319583221,-3.3652390696750536), linewidth(1.6));
draw((-2.9176473533749285,-3.6803047850255166)--(-6.62699301304923,-3.7084282888220432), linewidth(1.6));
draw((-6.62699301304923,-3.7084282888220432)--(-4.815597805533209,2.0137294983122676), linewidth(1.6));
draw((-5.999986761815922,-0.759127399441624)--(-5.442604056766517,-0.9355713910681529), linewidth(1.6));
draw((-4.815597805533209,2.0137294983122676)--(-3.696094000229639,5.5502174997511595), linewidth(1.6));
draw((-4.815597805533209,2.0137294983122676)--(1.1862495478417192,2.0592342833377844), linewidth(1.6));
draw((-1.8168903889624484,2.3287952136627297)--(-1.8124578687290425,1.744168567987322), linewidth(1.6));
draw((-4.815597805533209,2.0137294983122676)--(-2.9176473533749285,-3.6803047850255166), linewidth(1.6));
draw((-3.5893009510093994,-0.7408500702917692)--(-4.1439442078987385,-0.9257252164214806), linewidth(1.6));
label("$A$",(-4.440377746205339,7.118654172569505),SE*labelscalefactor,fontsize(14));
label("$B$",(-7.868514331571194,-3.218904987952353),SE*labelscalefactor,fontsize(14));
label("$C$",(9.165869786409527,-3.0594567746795223),SE*labelscalefactor,fontsize(14));
/* dots and labels */
dot((3.0842,-3.6348),linewidth(3.pt) + dotstyle);
dot((9.086047353374928,-3.589295214974483),linewidth(3.pt) + dotstyle);
dot((1.1862495478417192,2.0592342833377844),linewidth(3.pt) + dotstyle);
dot((-2.9176473533749285,-3.6803047850255166),linewidth(3.pt) + dotstyle);
dot((-4.815597805533209,2.0137294983122676),linewidth(3.pt) + dotstyle);
dot((-6.62699301304923,-3.7084282888220432),linewidth(3.pt) + dotstyle);
dot((-3.696094000229639,5.5502174997511595),linewidth(3.pt) + dotstyle);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy]
|
$72$
|
Primitive Recognition
|
olympiads
|
angle
|
160
|
Rectangle $ABCD$ is half the area of right triangle. They are joined to form a trapezoid, as shown. What is $DE$?
|
[asy] size(250); defaultpen(linewidth(0.8)); pair A=(0,5),B=origin,C=(6,0),D=(6,5),E=(18,0); draw(A--B--E--D--cycle^^C--D); draw(rightanglemark(D,C,E,30)); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,S); label("$D$",D,N); label("$E$",E,S); label("$5$",A/2,W); label("$6$",(A+D)/2,N); [/asy]
|
$\sqrt{601}$
|
Local Relation Composition
|
HARP
|
length
|
161
|
A circle with radius $2$ is inscribed in a square and circumscribed about another square as shown. What is the ratio of the area between the two squares to circle's shaded area ?
|
[asy] filldraw((-1,-1)--(-1,1)--(1,1)--(1,-1)--cycle,gray,black); filldraw(Circle((0,0),1), mediumgray,black); filldraw((-1,0)--(0,1)--(1,0)--(0,-1)--cycle,white,black);[/asy]
|
$\frac{2}{\pi-1}$
|
Local Relation Composition
|
HARP
|
ratio
|
162
|
The figure below shows a polygon $ABCDEFGH$, consisting of rectangles and right triangles. When cut out and folded on the dotted lines, the polygon forms a triangular prism. Suppose that $AH = EF = 8$ and $GH = 14$. What is the volume of the prism?
|
[asy] usepackage("mathptmx"); size(275); defaultpen(linewidth(0.8)); real r = 2, s = 2.5, theta = 14; pair G = (0,0), F = (r,0), C = (r,s), B = (0,s), M = (C+F)/2, I = M + s/2 * dir(-theta); pair N = (B+G)/2, J = N + s/2 * dir(180+theta); pair E = F + r * dir(- 45 - theta/2), D = I+E-F; pair H = J + r * dir(135 + theta/2), A = B+H-J; draw(A--B--C--I--D--E--F--G--J--H--cycle^^rightanglemark(F,I,C)^^rightanglemark(G,J,B)); draw(J--B--G^^C--F--I,linetype ("4 4")); dot("$A$",A,N); dot("$B$",B,1.2*N); dot("$C$",C,N); dot("$D$",D,dir(0)); dot("$E$",E,S); dot("$F$",F,1.5*dir(-100)); dot("$G$",G,S); dot("$H$",H,W); dot("$I$",I,NE); dot("$J$",J,1.5*S); [/asy]
|
$192$
|
Local Relation Composition
|
HARP
|
area
|
163
|
The Math Team designed a logo shaped like a multiplication symbol, shown below on a grid of 1-inch squares. What is the side length of the logo in square inches?
|
[asy] defaultpen(linewidth(0.5)); size(5cm); defaultpen(fontsize(14pt)); label("$\textbf{Math}$", (2.1,3.7)--(3.9,3.7)); label("$\textbf{Team}$", (2.1,3)--(3.9,3)); filldraw((1,2)--(2,1)--(3,2)--(4,1)--(5,2)--(4,3)--(5,4)--(4,5)--(3,4)--(2,5)--(1,4)--(2,3)--(1,2)--cycle, mediumgray*0.5 + lightgray*0.5); draw((0,0)--(6,0), gray); draw((0,1)--(6,1), gray); draw((0,2)--(6,2), gray); draw((0,3)--(6,3), gray); draw((0,4)--(6,4), gray); draw((0,5)--(6,5), gray); draw((0,6)--(6,6), gray); draw((0,0)--(0,6), gray); draw((1,0)--(1,6), gray); draw((2,0)--(2,6), gray); draw((3,0)--(3,6), gray); draw((4,0)--(4,6), gray); draw((5,0)--(5,6), gray); draw((6,0)--(6,6), gray); [/asy]
|
$12\sqrt{2}$
|
Primitive Recognition
|
HARP
|
length
|
164
|
What is the area of the shaded figure shown below?
|
[asy] size(200); defaultpen(linewidth(0.4)+fontsize(12)); pen s = linewidth(0.8)+fontsize(8); pair O,X,Y; O = origin; X = (6,0); Y = (0,5); fill((1,0)--(3,5)--(5,0)--(3,2)--cycle, palegray+opacity(0.2)); for (int i=1; i<7; ++i) { draw((i,0)--(i,5), gray+dashed); label("${"+string(i)+"}$", (i,0), 2*S); if (i<6) { draw((0,i)--(6,i), gray+dashed); label("${"+string(i)+"}$", (0,i), 2*W); } } label("$0$", O, 2*SW); draw(O--X+(0.35,0), black+1.5, EndArrow(10)); draw(O--Y+(0,0.35), black+1.5, EndArrow(10)); draw((1,0)--(3,5)--(5,0)--(3,2)--(1,0), black+1.5); [/asy]
|
$6$
|
Local Relation Composition
|
HARP
|
area
|
165
|
Let $ABCD$ be an isosceles trapezoid with $\overline{BC}\parallel \overline{AD}$ and $AB=CD$. Points $X$ and $Y$ lie on diagonal $\overline{AC}$ with $X$ between $A$ and $Y$, as shown in the figure. Suppose $\angle AXD = \angle BYC = 90^\circ$, $AX = 3$, $XY = 1$, and $YC = 2$. What is the area of $ABC?$
|
[asy] size(10cm); usepackage("mathptmx"); import geometry; void perp(picture pic=currentpicture, pair O, pair M, pair B, real size=5, pen p=currentpen, filltype filltype = NoFill){ perpendicularmark(pic, M,unit(unit(O-M)+unit(B-M)),size,p,filltype); } pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),Y=(2,0),X=(3,0),A=(6,0),B=(2,sqrt(5.6)),D=(3,-sqrt(12.6)); draw(A--B--C--D--cycle,p); draw(A--C,p); draw(B--Y,p); draw(D--X,p); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(X,q); dot(Y,q); label("2",C--Y,S); label("1",Y--X,S); label("3",X--A,S); label("$A$",A,2*E); label("$B$",B,2*N); label("$C$",C,2*W); label("$D$",D,2*S); label("$Y$",Y,2*sqrt(2)*NE); label("$X$",X,2*N); perp(B,Y,C,8,p); perp(A,X,D,8,p); [/asy]
|
$\frac{6\sqrt{35}}{5}$
|
Local Relation Composition
|
HARP
|
area
|
166
|
In the figure, equilateral hexagon $ABCDEF$ has three nonadjacent acute interior angles that each measure $30^\circ$. The enclosed area of the hexagon is $24\sqrt{3}$. What is the perimeter of the hexagon?
|
[asy] size(10cm); pen p=black+linewidth(1),q=black+linewidth(5); pair C=(0,0),D=(cos(pi/12),sin(pi/12)),E=rotate(150,D)*C,F=rotate(-30,E)*D,A=rotate(150,F)*E,B=rotate(-30,A)*F; draw(C--D--E--F--A--B--cycle,p); dot(A,q); dot(B,q); dot(C,q); dot(D,q); dot(E,q); dot(F,q); label("$C$",C,2*S); label("$D$",D,2*S); label("$E$",E,2*S); label("$F$",F,2*dir(0)); label("$A$",A,2*N); label("$B$",B,2*W); [/asy]
|
$24\sqrt{3}$
|
Local Relation Composition
|
HARP
|
length
|
167
|
As shown in the figure below, point $E$ lies on the opposite half-plane determined by line $CD$ from point $A$ so that $\angle CDE = 110^\circ$. Point $F$ lies on $\overline{AD}$ so that $DE=DF$, and $ABCD$ is a square. What is the degree measure of $\angle EFD$?
|
[asy] size(6cm); pair A = (0,10); label("$A$", A, N); pair B = (0,0); label("$B$", B, S); pair C = (10,0); label("$C$", C, S); pair D = (10,10); label("$D$", D, SW); pair EE = (15,11.8); label("$E$", EE, N); pair F = (3,10); label("$F$", F, N); filldraw(D--arc(D,2.5,270,380)--cycle,lightgray); dot(A^^B^^C^^D^^EE^^F); draw(A--B--C--D--cycle); draw(D--EE--F--cycle); label("$110^\circ$", (15,9), SW); [/asy]
|
$10$
|
Local Relation Composition
|
HARP
|
angle
|
168
|
In square $ABCD$, points $P$ and $Q$ lie on $\overline{AD}$ and $\overline{AB}$, respectively. Segments $\overline{BP}$ and $\overline{CQ}$ intersect at right angles at $R$, with $BR = 6$ and $PR = 7$. What is the area of the square?
|
[asy] size(170); defaultpen(linewidth(0.6)+fontsize(10)); real r = 3.5; pair A = origin, B = (5,0), C = (5,5), D = (0,5), P = (0,r), Q = (5-r,0), R = intersectionpoint(B--P,C--Q); draw(A--B--C--D--A^^B--P^^C--Q^^rightanglemark(P,R,C,7)); dot("$A$",A,S); dot("$B$",B,S); dot("$C$",C,N); dot("$D$",D,N); dot("$Q$",Q,S); dot("$P$",P,W); dot("$R$",R,1.3*S); label("$7$",(P+R)/2,NE); label("$6$",(R+B)/2,NE); [/asy]
|
$117$
|
Local Relation Composition
|
HARP
|
area
|
169
|
Equilateral triangle $ABC$ has side length $840$. Point $D$ lies on the same side of line $BC$ as $A$ such that $\overline{BD} \perp \overline{BC}$. The line $\ell$ through $D$ parallel to line $BC$ intersects sides $\overline{AB}$ and $\overline{AC}$ at points $E$ and $F$, respectively. Point $G$ lies on $\ell$ such that $F$ is between $E$ and $G$, $\triangle AFG$ is isosceles, and the ratio of the area of $\triangle AFG$ to the area of $\triangle BED$ is $8:9$. Find $FC$.
|
[asy] pair A,B,C,D,E,F,G; B=origin; A=5*dir(60); C=(5,0); E=0.6*A+0.4*B; F=0.6*A+0.4*C; G=rotate(240,F)*A; D=extension(E,F,B,dir(90)); draw(D--G--A,grey); draw(B--0.5*A+rotate(60,B)*A*0.5,grey); draw(A--B--C--cycle,linewidth(1.5)); dot(A^^B^^C^^D^^E^^F^^G); label("$A$",A,dir(90)); label("$B$",B,dir(225)); label("$C$",C,dir(-45)); label("$D$",D,dir(180)); label("$E$",E,dir(-45)); label("$F$",F,dir(225)); label("$G$",G,dir(0)); label("$\ell$",midpoint(E--F),dir(90)); [/asy]
|
$504$
|
Local Relation Composition
|
HARP
|
length
|
170
|
Let $ABCD$ be an isoceles trapezoid having parallel bases $\overline{AB}$ and $\overline{CD}$ with $AB>CD.$ Line segments from a point inside $ABCD$ to the vertices divide the trapezoid into four triangles whose areas are $2, 3, 4,$ and $5$ starting with the triangle with base $\overline{CD}$ and moving clockwise as shown in the diagram below. What is the ratio $\frac{CD}{AB}?$
|
[asy] unitsize(100); pair A=(-1, 0), B=(1, 0), C=(0.3, 0.9), D=(-0.3, 0.9), P=(0.2, 0.5), E=(0.1, 0.75), F=(0.4, 0.5), G=(0.15, 0.2), H=(-0.3, 0.5); draw(A--B--C--D--cycle, black); draw(A--P, black); draw(B--P, black); draw(C--P, black); draw(D--P, black); label("$A$",A,(-1,0)); label("$B$",B,(1,0)); label("$C$",C,(1,-0)); label("$D$",D,(-1,0)); label("$2$",E,(0,0)); label("$3$",F,(0,0)); label("$4$",G,(0,0)); label("$5$",H,(0,0)); dot(A^^B^^C^^D^^P); [/asy]
|
$2-\sqrt{2}$
|
Local Relation Composition
|
HARP
|
ratio
|
171
|
The figure is constructed from $11$ line segments, each of which has length $2$. Find the area of pentagon $ABCDE$.
|
[asy] pair A=(-2.4638,4.10658); pair B=(-4,2.6567453480756127); pair C=(-3.47132,0.6335248637894945); pair D=(-1.464483379039766,0.6335248637894945); pair E=(-0.956630463955801,2.6567453480756127); pair F=(-2,2); pair G=(-3,2); draw(A--B--C--D--E--A); draw(A--F--A--G); draw(B--F--C); draw(E--G--D); label("A",A,N); label("B",B,W); label("C",C,S); label("D",D,S); label("E",E,dir(0)); dot(A^^B^^C^^D^^E^^F^^G); [/asy]
|
$\sqrt{12}+\sqrt{11}$
|
Local Relation Composition
|
HARP
|
area
|
172
|
Equilateral $\triangle ABC$ has side length $3$, and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$?
|
[asy] import graph; size(6cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,W); label("$F$",F,E); label("$G$",G,E); label("$H$",H,SE); label("$I$",I,SW); [/asy]
|
$27+9\sqrt{3}$
|
Local Relation Composition
|
HARP
|
area
|
173
|
Three unit squares and two line segments connecting two pairs of vertices are shown. Each square has side length 5. What is the area of $\triangle ABC$?
|
[asy] unitsize(2cm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=4; pair A=(0,0), B=(1,0); pair C=(0.8,-0.4); draw(A--(2,0)); draw((0,-1)--(2,-1)); draw((0,-2)--(1,-2)); draw(A--(0,-2)); draw(B--(1,-2)); draw((2,0)--(2,-1)); draw(A--(2,-1)); draw(B--(0,-2)); pair[] ps={A,B,C}; dot(ps); label("$A$",A,N); label("$B$",B,N); label("$C$",C,W); [/asy]
|
$5$
|
Local Relation Composition
|
HARP
|
area
|
174
|
Square $AXYZ$ is inscribed in equiangular hexagon $ABCDEF$ with $X$ on $\overline{BC}$, $Y$ on $\overline{DE}$, and $Z$ on $\overline{EF}$. Suppose that $AB=40$, and $EF=41(\sqrt{3}-1)$. What is the perimeter of the square?
|
[asy] size(200); defaultpen(linewidth(1)); pair A=origin,B=(2.5,0),C=B+2.5*dir(60), D=C+1.75*dir(120),E=D-(3.19,0),F=E-1.8*dir(60); pair X=waypoint(B--C,0.345),Z=rotate(90,A)*X,Y=rotate(90,Z)*A; draw(A--B--C--D--E--F--cycle); draw(A--X--Y--Z--cycle,linewidth(0.9)+linetype("2 2")); dot("$A$",A,W,linewidth(4)); dot("$B$",B,dir(0),linewidth(4)); dot("$C$",C,dir(0),linewidth(4)); dot("$D$",D,dir(20),linewidth(4)); dot("$E$",E,dir(100),linewidth(4)); dot("$F$",F,W,linewidth(4)); dot("$X$",X,dir(0),linewidth(4)); dot("$Y$",Y,N,linewidth(4)); dot("$Z$",Z,W,linewidth(4)); [/asy]
|
$116\sqrt{3}$
|
Local Relation Composition
|
HARP
|
length
|
175
|
Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are $3$ cm and $6$ cm. Into each cone is dropped a spherical marble of radius $1$ cm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the value of the ratio of the rise of the liquid level in the wide cone to the rise of the liquid level in the narrow cone?
|
[asy] size(350); defaultpen(linewidth(0.8)); real h1 = 10, r = 3.1, s=0.75; pair P = (r,h1), Q = (-r,h1), Pp = s * P, Qp = s * Q; path e = ellipse((0,h1),r,0.9), ep = ellipse((0,h1*s),r*s,0.9); draw(ellipse(origin,r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill(origin--Pp--Qp--cycle,gray(0.8)); draw((-r,h1)--(0,0)--(r,h1)^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(0,Qp.y),Arrows(size=8)); draw(origin--(0,12),linetype("4 4")); draw(origin--(r*(s-0.1),0)); label("$3$",(-0.9,h1*s),N,fontsize(10)); real h2 = 7.5, r = 6, s=0.6, d = 14; pair P = (d+r-0.05,h2-0.15), Q = (d-r+0.05,h2-0.15), Pp = s * P + (1-s)*(d,0), Qp = s * Q + (1-s)*(d,0); path e = ellipse((d,h2),r,1), ep = ellipse((d,h2*s+0.09),r*s,1); draw(ellipse((d,0),r*(s-0.1),0.8)); fill(ep,gray(0.8)); fill((d,0)--Pp--Qp--cycle,gray(0.8)); draw(P--(d,0)--Q^^e); draw(subpath(ep,0,reltime(ep,0.5)),linetype("4 4")); draw(subpath(ep,reltime(ep,0.5),reltime(ep,1))); draw(Qp--(d,Qp.y),Arrows(size=8)); draw((d,0)--(d,10),linetype("4 4")); draw((d,0)--(d+r*(s-0.1),0)); label("$6$",(d-r/4,h2*s-0.06),N,fontsize(10)); [/asy]
|
$\frac{1}{4}$
|
Global Abstract Integration
|
HARP
|
ratio
|
176
|
A square piece of paper has side length $6$ and vertices $A,B,C,$ and $D$ in that order. As shown in the figure, the paper is folded so that vertex $C$ meets edge $\overline{AD}$ at point $C'$, and edge $\overline{BC}$ intersects edge $\overline{AB}$ at point $E$. Suppose that $C'D = \frac{1}{3}$. What is the perimeter of triangle $\bigtriangleup AEC' ?$
|
[asy] pair A=(0,1); pair CC=(0.666666666666,1); pair D=(1,1); pair F=(1,0.440062); pair C=(1,0); pair B=(0,0); pair G=(0,0.22005); pair H=(-0.13,0.41); pair E=(0,0.5); dot(A^^CC^^D^^C^^B^^E); draw(E--A--D--F); draw(G--B--C--F, dashed); fill(E--CC--F--G--H--E--CC--cycle, gray); draw(E--CC--F--G--H--E--CC); label("A",A,NW); label("B",B,SW); label("C",C,SE); label("D",D,NE); label("E",E,NW); label("C'",CC,N); label("F",F,NE); [/asy]
|
$12$
|
Global Abstract Integration
|
HARP
|
length
|
177
|
In the plane figure shown below, $3$ of the unit squares have been shaded. What is the least number of additional unit squares that must be shaded so that the resulting figure has two lines of symmetry?
|
[asy] import olympiad; unitsize(25); filldraw((1,3)--(1,4)--(2,4)--(2,3)--cycle, gray(0.7)); filldraw((2,1)--(2,2)--(3,2)--(3,1)--cycle, gray(0.7)); filldraw((4,0)--(5,0)--(5,1)--(4,1)--cycle, gray(0.7)); for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { pair A = (j,i); } } for (int i = 0; i < 5; ++i) { for (int j = 0; j < 6; ++j) { if (j != 5) { draw((j,i)--(j+1,i)); } if (i != 4) { draw((j,i)--(j,i+1)); } } } [/asy]
|
$7$
|
Local Relation Composition
|
HARP
|
count
|
178
|
Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the perimeter of $S_2$ in units?
|
[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("$R_1$",(3/2,1/2)); label("$S_3$",(4,1)); label("$S_2$",(5/2,3/2)); label("$S_1$",(1,2)); label("$R_2$",(7/2,5/2)); [/asy]
|
$2604$
|
Local Relation Composition
|
HARP
|
length
|
179
|
Octagon $ABCDEFGH$ with side lengths $AB = CD = EF = GH = 10$ and $BC = DE = FG = HA = 11$ is formed by removing 6-8-10 triangles from the corners of a $23$ $\times$ $27$ rectangle with side $\overline{AH}$ on a short side of the rectangle, as shown. Let $J$ be the midpoint of $\overline{AH}$, and partition the octagon into 7 triangles by drawing segments $\overline{JB}$, $\overline{JC}$, $\overline{JD}$, $\overline{JE}$, $\overline{JF}$, and $\overline{JG}$. Find the area of the convex polygon whose vertices are the centroids of these 7 triangles.
|
[asy] unitsize(6); pair P = (0, 0), Q = (0, 23), R = (27, 23), SS = (27, 0); pair A = (0, 6), B = (8, 0), C = (19, 0), D = (27, 6), EE = (27, 17), F = (19, 23), G = (8, 23), J = (0, 23/2), H = (0, 17); draw(P--Q--R--SS--cycle); draw(J--B); draw(J--C); draw(J--D); draw(J--EE); draw(J--F); draw(J--G); draw(A--B); draw(H--G); real dark = 0.6; filldraw(A--B--P--cycle, gray(dark)); filldraw(H--G--Q--cycle, gray(dark)); filldraw(F--EE--R--cycle, gray(dark)); filldraw(D--C--SS--cycle, gray(dark)); dot(A); dot(B); dot(C); dot(D); dot(EE); dot(F); dot(G); dot(H); dot(J); dot(H); defaultpen(fontsize(10pt)); real r = 1.3; label("$A$", A, W*r); label("$B$", B, S*r); label("$C$", C, S*r); label("$D$", D, E*r); label("$E$", EE, E*r); label("$F$", F, N*r); label("$G$", G, N*r); label("$H$", H, W*r); label("$J$", J, W*r); [/asy]
|
$184$
|
Local Relation Composition
|
HARP
|
area
|
180
|
A closed box with a square base is to be wrapped with a square sheet of wrapping paper. The box is centered on the wrapping paper with the vertices of the base lying on the midlines of the square sheet of paper, as shown in the figure on the left. The four corners of the wrapping paper are to be folded up over the sides and brought together to meet at the center of the top of the box, point $A$ in the figure on the right. The box has base length $w$ and height $h$. What is the area of the sheet of wrapping paper? Assume $w$=2 and $h$=3.
|
[asy] size(270pt); defaultpen(fontsize(10pt)); filldraw(((3,3)--(-3,3)--(-3,-3)--(3,-3)--cycle),lightgrey); dot((-3,3)); label("$A$",(-3,3),NW); draw((1,3)--(-3,-1),dashed+linewidth(.5)); draw((-1,3)--(3,-1),dashed+linewidth(.5)); draw((-1,-3)--(3,1),dashed+linewidth(.5)); draw((1,-3)--(-3,1),dashed+linewidth(.5)); draw((0,2)--(2,0)--(0,-2)--(-2,0)--cycle,linewidth(.5)); draw((0,3)--(0,-3),linetype("2.5 2.5")+linewidth(.5)); draw((3,0)--(-3,0),linetype("2.5 2.5")+linewidth(.5)); label('$w$',(-1,-1),SW); label('$w$',(1,-1),SE); draw((4.5,0)--(6.5,2)--(8.5,0)--(6.5,-2)--cycle); draw((4.5,0)--(8.5,0)); draw((6.5,2)--(6.5,-2)); label("$A$",(6.5,0),NW); dot((6.5,0)); [/asy]
|
$50$
|
Global Abstract Integration
|
HARP
|
area
|
181
|
The twelve-sided figure shown has been drawn on $3 \text{ cm}\times 3 \text{ cm}$ graph paper. What is the area of the figure in $\text{cm}^2$?
|
[asy] unitsize(8mm); for (int i=0; i<7; ++i) { draw((i,0)--(i,7),gray); draw((0,i+1)--(7,i+1),gray); } draw((1,3)--(2,4)--(2,5)--(3,6)--(4,5)--(5,5)--(6,4)--(5,3)--(5,2)--(4,1)--(3,2)--(2,2)--cycle,black+2bp); [/asy]
|
$117$
|
Primitive Recognition
|
HARP
|
area
|
182
|
In the figure shown, $\overline{US}$ and $\overline{UT}$ are line segments each of length 1, and $m\angle TUS = 60^\circ$. Arcs $\overarc{TR}$ and $\overarc{SR}$ are each one-sixth of a circle with radius 1. What is the area of the region shown?
|
[asy]draw((1,1.732)--(2,3.464)--(3,1.732)); draw(arc((0,0),(2,0),(1,1.732))); draw(arc((4,0),(3,1.732),(2,0))); label("$U$", (2,3.464), N); label("$S$", (1,1.732), W); label("$T$", (3,1.732), E); label("$R$", (2,0), S);[/asy]
|
$\sqrt{3}-\frac{\pi}{3}$
|
Local Relation Composition
|
HARP
|
area
|
183
|
In the non-convex quadrilateral $ABCD$ shown below, $\angle BCD$ is a right angle, $AB=24$, $BC=8$, $CD=6$, and $AD=26$. What is the area of quadrilateral $ABCD$?
|
[asy]draw((0,0)--(2.4,3.6)--(0,5)--(12,0)--(0,0)); label("$B$", (0, 0), SW); label("$A$", (12, 0), ESE); label("$C$", (2.4, 3.6), SE); label("$D$", (0, 5), N);[/asy]
|
$96$
|
Local Relation Composition
|
HARP
|
area
|
184
|
In the figure below, choose point $D$ on $\overline{BC}$ so that $\triangle ACD$ and $\triangle ABD$ have equal perimeters. What is the area of $\triangle ACD$?
|
[asy]draw((0,0)--(4,0)--(0,3)--(0,0)); label("$A$", (0,0), SW); label("$B$", (4,0), ESE); label("$C$", (0, 3), N); label("$3$", (0, 1.5), W); label("$4$", (2, 0), S); label("$5$", (2, 1.5), NE);[/asy]
|
$\frac{18}{5}$
|
Local Relation Composition
|
HARP
|
area
|
185
|
In the given figure, $BD = DC$ and $\angle BCD = 70^\circ$. By how many times does the measure of $\angle ADB$ exceed that of $\angle BCD$?
|
[asy] size(300); defaultpen(linewidth(0.8)); pair A=(-1,0),C=(1,0),B=dir(40),D=origin; draw(A--B--C--A); draw(D--B); dot("$A$", A, SW); dot("$B$", B, NE); dot("$C$", C, SE); dot("$D$", D, S); label("$70^\circ$",C,2*dir(180-35));[/asy]
|
$2$
|
Primitive Recognition
|
HARP
|
ratio
|
186
|
A straight one-mile stretch of highway, which has a width of 40 feet, has been shut down. Robert rides his bike on a path composed of semicircles as shown. If he rides at 2.5 miles per hour, how many hours will it take to cover the one-mile stretch? Note: 1 mile = 5280 feet
|
[asy]size(10cm); pathpen=black; pointpen=black; D(arc((-2,0),1,300,360)); D(arc((0,0),1,0,180)); D(arc((2,0),1,180,360)); D(arc((4,0),1,0,180)); D(arc((6,0),1,180,240)); D((-1.5,-1)--(5.5,-1));[/asy]
|
$\frac{\pi}{5}$
|
Local Relation Composition
|
HARP
|
count
|
187
|
Equilateral $\triangle ABC$ has side length \$1$, and squares $ABDE$, $BCHI$, and $CAFG$ are constructed externally on its sides. What is the value of the area of hexagon $DEFGHI$ minus the area of the three squares?
|
[asy] import graph; size(6cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,W); label("$F$",F,E); label("$G$",G,E); label("$H$",H,SE); label("$I$",I,SW); [/asy]
|
$\sqrt{3}$
|
Local Relation Composition
|
HARP
|
area
|
188
|
In a stack of four cubes arranged as shown, what is twice the length of the portion of $\overline{XY}$ that is contained in the cube with edge length \$3$?
|
[asy] dotfactor = 3; size(10cm); dot((0, 10)); label("$X$", (0,10),W,fontsize(8pt)); dot((6,2)); label("$Y$", (6,2),E,fontsize(8pt)); draw((0, 0)--(0, 10)--(1, 10)--(1, 9)--(2, 9)--(2, 7)--(3, 7)--(3,4)--(4, 4)--(4, 0)--cycle); draw((0,9)--(1, 9)--(1.5, 9.5)--(1.5, 10.5)--(0.5, 10.5)--(0, 10)); draw((1, 10)--(1.5,10.5)); draw((1.5, 10)--(3,10)--(3,8)--(2,7)--(0,7)); draw((2,9)--(3,10)); draw((3,8.5)--(4.5,8.5)--(4.5,5.5)--(3,4)--(0,4)); draw((3,7)--(4.5,8.5)); draw((4.5,6)--(6,6)--(6,2)--(4,0)); draw((4,4)--(6,6)); label("$1$", (1,9.5), W,fontsize(8pt)); label("$2$", (2,8), W,fontsize(8pt)); label("$3$", (3,5.5), W,fontsize(8pt)); label("$4$", (4,2), W,fontsize(8pt)); [/asy]
|
$\frac{6\sqrt{33}}{5}$
|
Global Abstract Integration
|
HARP
|
length
|
189
|
Six regular hexagons are arranged around a central regular hexagon with side length \$1$, as illustrated. What is twice the area of $\triangle ABC$?
|
[asy] draw((0,0)--(-5,8.66025404)--(0, 17.3205081)--(10, 17.3205081)--(15,8.66025404)--(10, 0)--(0, 0)); draw((30,0)--(25,8.66025404)--(30, 17.3205081)--(40, 17.3205081)--(45, 8.66025404)--(40, 0)--(30, 0)); draw((30,0)--(25,-8.66025404)--(30, -17.3205081)--(40, -17.3205081)--(45, -8.66025404)--(40, 0)--(30, 0)); draw((0,0)--(-5, -8.66025404)--(0, -17.3205081)--(10, -17.3205081)--(15, -8.66025404)--(10, 0)--(0, 0)); draw((15,8.66025404)--(10, 17.3205081)--(15, 25.9807621)--(25, 25.9807621)--(30, 17.3205081)--(25, 8.66025404)--(15, 8.66025404)); draw((15,-8.66025404)--(10, -17.3205081)--(15, -25.9807621)--(25, -25.9807621)--(30, -17.3205081)--(25, -8.66025404)--(15, -8.66025404)); label("A", (0,0), W); label("B", (30, 17.3205081), NE); label("C", (30, -17.3205081), SE); draw((0,0)--(30, 17.3205081)--(30, -17.3205081)--(0, 0)); //(Diagram Creds-DivideBy0) [/asy]
|
$6\sqrt{3}$
|
Local Relation Composition
|
HARP
|
area
|
190
|
Doug constructs a square window using $8$ equal-size panes of glass, as shown. The ratio of the height to width for each pane is $5 : 2$, and the borders around and between the panes are $2$ inches wide. In inches, what is the area of the square window?
|
[asy] fill((0,0)--(2,0)--(2,26)--(0,26)--cycle,gray); fill((6,0)--(8,0)--(8,26)--(6,26)--cycle,gray); fill((12,0)--(14,0)--(14,26)--(12,26)--cycle,gray); fill((18,0)--(20,0)--(20,26)--(18,26)--cycle,gray); fill((24,0)--(26,0)--(26,26)--(24,26)--cycle,gray); fill((0,0)--(26,0)--(26,2)--(0,2)--cycle,gray); fill((0,12)--(26,12)--(26,14)--(0,14)--cycle,gray); fill((0,24)--(26,24)--(26,26)--(0,26)--cycle,gray); [/asy]
|
$676$
|
Local Relation Composition
|
HARP
|
area
|
191
|
The radii of the circles in the following figure are all 1cm, and they are tangent to each other. Calculate the area of the bounding rectangle.
|
[asy] real r = 3;
pair A = (r, -r);
pair B = (3*r, -r);
pair C = (6, -8.2);
real width = 4*r;
real top = 0;
real bottom = C.y - r;
draw(box((0,top),(width,bottom)), linewidth(0.8bp));
draw(circle(A, r), blue);
draw(circle(B, r), blue);
draw(circle(C, r), blue);
draw(A--(A+(0,r)), red+linewidth(1));
label("$1,\mathrm{cm}$", A+(0,r/2), W, red);
draw(B--(B+(0,r)), red+linewidth(1));
label("$1,\mathrm{cm}$", B+(0,r/2), E, red);
draw(C--(C+(0,-r)), red+linewidth(1));
label("$1,\mathrm{cm}$", C+(0,-r/2), E, red);
dot(midpoint(A--B));
dot(midpoint(A--C));
dot(midpoint(B--C));
dot(A); label("$A$", A, W);
dot(B); label("$B$", B, E);
dot(C); label("$C$", C, S);[/asy]
|
$8+4\sqrt{3}$
|
Global Abstract Integration
|
HARP
|
area
|
192
|
How many pairs of parallel edges does a cube have, excluding the pairs $\overline{AB}$ and $\overline{GH}$, and $\overline{EH}$ and $\overline{FG}$?
|
[asy] import three; currentprojection=orthographic(1/2,-1,1/2); /* three - currentprojection, orthographic */ draw((0,0,0)--(1,0,0)--(1,1,0)--(0,1,0)--cycle); draw((0,0,0)--(0,0,1)); draw((0,1,0)--(0,1,1)); draw((1,1,0)--(1,1,1)); draw((1,0,0)--(1,0,1)); draw((0,0,1)--(1,0,1)--(1,1,1)--(0,1,1)--cycle); label("$D$",(0,0,0),S); label("$A$",(0,0,1),N); label("$H$",(0,1,0),S); label("$E$",(0,1,1),N); label("$C$",(1,0,0),S); label("$B$",(1,0,1),N); label("$G$",(1,1,0),S); label("$F$",(1,1,1),N); [/asy]
|
$16$
|
Primitive Recognition
|
HARP
|
count
|
193
|
A triangle with vertices at $A=(1,3)$, $B=(5,1)$, and $C=(4,4)$ is plotted on a \$6\times5$ grid. How many times as large is the area of the grid compared to the area of the triangle?
|
[asy] draw((1,0)--(1,5),linewidth(.5)); draw((2,0)--(2,5),linewidth(.5)); draw((3,0)--(3,5),linewidth(.5)); draw((4,0)--(4,5),linewidth(.5)); draw((5,0)--(5,5),linewidth(.5)); draw((6,0)--(6,5),linewidth(.5)); draw((0,1)--(6,1),linewidth(.5)); draw((0,2)--(6,2),linewidth(.5)); draw((0,3)--(6,3),linewidth(.5)); draw((0,4)--(6,4),linewidth(.5)); draw((0,5)--(6,5),linewidth(.5)); draw((0,0)--(0,6),EndArrow); draw((0,0)--(7,0),EndArrow); draw((1,3)--(4,4)--(5,1)--cycle); label("$y$",(0,6),W); label("$x$",(7,0),S); label("$A$",(1,3),dir(210)); label("$B$",(5,1),SE); label("$C$",(4,4),dir(100)); [/asy]
|
$6$
|
Primitive Recognition
|
HARP
|
ratio
|
194
|
In the given figure, hexagon $ABCDEF$ is equiangular, $ABJI$ and $FEHG$ are squares with areas \$18$ and \$32$ respectively, $\triangle JBK$ is equilateral, and $FE = BC$. How many times as large is the area of $ABJI$ compared to the area of $\triangle KBC$?
|
[asy] draw((-4,6*sqrt(2))--(4,6*sqrt(2))); draw((-4,-6*sqrt(2))--(4,-6*sqrt(2))); draw((-8,0)--(-4,6*sqrt(2))); draw((-8,0)--(-4,-6*sqrt(2))); draw((4,6*sqrt(2))--(8,0)); draw((8,0)--(4,-6*sqrt(2))); draw((-4,6*sqrt(2))--(4,6*sqrt(2))--(4,8+6*sqrt(2))--(-4,8+6*sqrt(2))--cycle); draw((-8,0)--(-4,-6*sqrt(2))--(-4-6*sqrt(2),-4-6*sqrt(2))--(-8-6*sqrt(2),-4)--cycle); label("$I$",(-4,8+6*sqrt(2)),dir(100)); label("$J$",(4,8+6*sqrt(2)),dir(80)); label("$A$",(-4,6*sqrt(2)),dir(280)); label("$B$",(4,6*sqrt(2)),dir(250)); label("$C$",(8,0),W); label("$D$",(4,-6*sqrt(2)),NW); label("$E$",(-4,-6*sqrt(2)),NE); label("$F$",(-8,0),E); draw((4,8+6*sqrt(2))--(4,6*sqrt(2))--(4+4*sqrt(3),4+6*sqrt(2))--cycle); label("$K$",(4+4*sqrt(3),4+6*sqrt(2)),E); draw((4+4*sqrt(3),4+6*sqrt(2))--(8,0),dashed); label("$H$",(-4-6*sqrt(2),-4-6*sqrt(2)),S); label("$G$",(-8-6*sqrt(2),-4),W); label("$32$",(-10,-8),N); label("$18$",(0,6*sqrt(2)+2),N); [/asy]
|
$\frac{3}{2}$
|
Local Relation Composition
|
HARP
|
ratio
|
195
|
The shaded region below is called a shark's fin falcata, a figure studied by Leonardo da Vinci. It is bounded by the portion of the circle of radius $3$ and center $(0,0)$ that lies in the first quadrant, the portion of the circle with radius $\tfrac{3}{2}$ and center $(0,\tfrac{3}{2})$ that lies in the first quadrant, and the line segment from $(0,0)$ to $(3,0)$. What is the twice the area of the shark's fin falcata?
|
[asy] import cse5;pathpen=black;pointpen=black; size(1.5inch); D(MP("x",(3.5,0),S)--(0,0)--MP("\frac{3}{2}",(0,3/2),W)--MP("y",(0,3.5),W)); path P=(0,0)--MP("3",(3,0),S)..(3*dir(45))..MP("3",(0,3),W)--(0,3)..(3/2,3/2)..cycle; draw(P,linewidth(2)); fill(P,gray); [/asy]
|
$\frac{9\pi}{4}$
|
Local Relation Composition
|
HARP
|
area
|
196
|
In the upper half-plane, consider a configuration of circles, each tangent to the $x$-axis, built in successive layers as follows. The initial layer $L_0$ consists of two mutually externally tangent circles with radii \$70^2$ and \$73^2$. For each layer $k \ge 1$, take all circles from the union $\bigcup_{j=0}^{k-1}L_j$ and order them by the $x$-coordinates of their points of tangency with the $x$-axis. For every pair of adjacent circles in this sequence, construct a new circle that is externally tangent to both circles in the pair and also tangent to the $x$-axis. Define layer $L_k$ as the set of \$2^{k-1}$ new circles added in this way. Let $S = \bigcup_{j=0}^6 L_j$, and let $r(C)$ denote the radius of the circle $C$. Compute
\[
\sum_{C \in S} \frac{1}{\sqrt{r(C)}}.
\]
|
[asy] import olympiad; size(350); defaultpen(linewidth(0.7)); // define a bunch of arrays and starting points pair[] coord = new pair[65]; int[] trav = {32,16,8,4,2,1}; coord[0] = (0,73^2); coord[64] = (2*73*70,70^2); // draw the big circles and the bottom line path arc1 = arc(coord[0],coord[0].y,260,360); path arc2 = arc(coord[64],coord[64].y,175,280); fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.75)); fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.75)); draw(arc1^^arc2); draw((-930,0)--(70^2+73^2+850,0)); // We now apply the findCenter function 63 times to get // the location of the centers of all 63 constructed circles. // The complicated array setup ensures that all the circles // will be taken in the right order for(int i = 0;i<=5;i=i+1) { int skip = trav[i]; for(int k=skip;k<=64 - skip; k = k + 2*skip) { pair cent1 = coord[k-skip], cent2 = coord[k+skip]; real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2); real shiftx = cent1.x + sqrt(4*r1*rn); coord[k] = (shiftx,rn); } // Draw the remaining 63 circles } for(int i=1;i<=63;i=i+1) { filldraw(circle(coord[i],coord[i].y),gray(0.75)); }[/asy]
|
$\frac{143}{14}$
|
Global Abstract Integration
|
HARP
|
count
|
197
|
In the diagram below, $ABCD$ is a square. Point $E$ is the midpoint of $\overline{AD}$. Points $F$ and $G$ lie on $\overline{CE}$, and $H$ and $J$ lie on $\overline{AB}$ and $\overline{BC}$, respectively, so that $FGHJ$ is a square. Points $K$ and $L$ lie on $\overline{GH}$, and $M$ and $N$ lie on $\overline{AD}$ and $\overline{AB}$, respectively, so that $KLMN$ is a square. The area of $KLMN$ is 99. How many times as large is the area of $FGHJ$ as that of $KLMN$?.
|
[asy] pair A,B,C,D,E,F,G,H,J,K,L,M,N; B=(0,0); real m=7*sqrt(55)/5; J=(m,0); C=(7*m/2,0); A=(0,7*m/2); D=(7*m/2,7*m/2); E=(A+D)/2; H=(0,2m); N=(0,2m+3*sqrt(55)/2); G=foot(H,E,C); F=foot(J,E,C); draw(A--B--C--D--cycle); draw(C--E); draw(G--H--J--F); pair X=foot(N,E,C); M=extension(N,X,A,D); K=foot(N,H,G); L=foot(M,H,G); draw(K--N--M--L); label("$A$",A,NW); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,NE); label("$E$",E,dir(90)); label("$F$",F,NE); label("$G$",G,NE); label("$H$",H,W); label("$J$",J,S); label("$K$",K,SE); label("$L$",L,SE); label("$M$",M,dir(90)); label("$N$",N,dir(180)); [/asy]
|
$\frac{49}{9}$
|
Local Relation Composition
|
HARP
|
ratio
|
198
|
In rectangle $ABCD$, point $M$ is the midpoint of $\overline{AD}$. The area of $\triangle AMC$ is \$12$, and $\frac{AD}{AB} = \frac{3}{2}$. Find the length of side $AD$.
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[asy]size(4cm);draw((0,2)--(0,0)--(3,0)--(3,4)--(0,4)--(0,2)--(3,4)--(0,0));label("$A$", (0,0), SW); label("$B$", (3, 0), SE); label("$C$", (3,4), NE); label("$D$", (0, 4), NW); label("$M$", (0, 2), W); [/asy]
|
$8$
|
Local Relation Composition
|
HARP
|
length
|
199
|
Rectangle $DEFA$ below is a $3 \times 4$ rectangle with $DC=CB=BA=1$. What is the area of rectangle $DEFA$ minus the area of the shaded "bat wings" region?
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[asy] draw((0,0)--(3,0)--(3,4)--(0,4)--(0,0)--(2,4)--(3,0)); draw((3,0)--(1,4)--(0,0)); fill((0,0)--(1,4)--(1.5,3)--cycle, black); fill((3,0)--(2,4)--(1.5,3)--cycle, black); label("$A$",(3.05,4.2)); label("$B$",(2,4.2)); label("$C$",(1,4.2)); label("$D$",(0,4.2)); label("$E$", (0,-0.2)); label("$F$", (3,-0.2)); label("$1$", (0.5, 4), N); label("$1$", (1.5, 4), N); label("$1$", (2.5, 4), N); label("$4$", (3.2, 2), E); [/asy]
|
$9$
|
Primitive Recognition
|
HARP
|
area
|
200
|
A rug is made with three different colors as shown. The areas of the three differently colored regions form an arithmetic progression. The inner rectangle is one foot wide, and each of the two shaded regions is $1$ foot wide on all four sides. What is the perimeter, in feet, of the inner rectangle
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[asy] size(6cm); defaultpen(fontsize(9pt)); path rectangle(pair X, pair Y){ return X--(X.x,Y.y)--Y--(Y.x,X.y)--cycle; } filldraw(rectangle((0,0),(7,5)),gray(0.5)); filldraw(rectangle((1,1),(6,4)),gray(0.75)); filldraw(rectangle((2,2),(5,3)),white); label("$1$",(0.5,2.5)); draw((0.3,2.5)--(0,2.5),EndArrow(TeXHead)); draw((0.7,2.5)--(1,2.5),EndArrow(TeXHead)); label("$1$",(1.5,2.5)); draw((1.3,2.5)--(1,2.5),EndArrow(TeXHead)); draw((1.7,2.5)--(2,2.5),EndArrow(TeXHead)); label("$1$",(4.5,2.5)); draw((4.5,2.7)--(4.5,3),EndArrow(TeXHead)); draw((4.5,2.3)--(4.5,2),EndArrow(TeXHead)); label("$1$",(4.1,1.5)); draw((4.1,1.7)--(4.1,2),EndArrow(TeXHead)); draw((4.1,1.3)--(4.1,1),EndArrow(TeXHead)); label("$1$",(3.7,0.5)); draw((3.7,0.7)--(3.7,1),EndArrow(TeXHead)); draw((3.7,0.3)--(3.7,0),EndArrow(TeXHead)); [/asy]
|
$6$
|
Local Relation Composition
|
HARP
|
length
|
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