LiAuto-DSR/GeoGramBench huggingface dataset:
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]

Answer: $\frac{7}{16}$
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]

Answer: $12\sqrt{5}$
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]

Answer: $2\sqrt{3}-2$
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]

Answer: $12\sqrt{3}$
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]

Answer: $72$
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]

Answer: $147$
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]

Answer: $45$
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]

Answer: $36$
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]

Answer: $\frac{2}{5}$
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]

Answer: $\frac{375}{4}$
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]

Answer: $\pi$
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]

Answer: $\frac{1}{6}$
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]

Answer: $\frac{28}{3}$
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]

Answer: $60$
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]

Answer: $18\sqrt{3}}$
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]

Answer: $14.6$
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]

Answer: $6$
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]

Answer: $\frac{25}{36}$
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]

Answer: $63$
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]

Answer: $296\pi$
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]

Answer: $\frac{3}{4}$