LiAuto-DSR/GeoGramBench huggingface dataset:
In the diagram line segments  $AB$  and  $CD$  are of length 1 while angles  $ABC$  and  $CBD$  are  $90^\circ$  and  $30^\circ$  respectively.  Find  $AC$ .

[asy]
import geometry;
import graph;

unitsize(1.5 cm);

pair A, B, C, D;

B = (0,0);
D = (3,0);
A = 2*dir(120);
C = extension(B,dir(30),A,D);

draw(A--B--D--cycle);
draw(B--C);
draw(arc(B,0.5,0,30));

label(" $A$ ", A, NW);
label(" $B$ ", B, SW);
label(" $C$ ", C, NE);
label(" $D$ ", D, SE);
label(" $30^\circ$ ", (0.8,0.2));
label(" $90^\circ$ ", (0.1,0.5));

perpendicular(B,NE,C-B);
[/asy]

Answer: $\sqrt[3]{2}$
In the following diagram, a chord $AN$ intersects its diameter $MN$ at $B$. Given that  $MB : BN = 2 : 3$  and  $MN = 10$ . If  $AN = x$ , find  $x^2$ .

[asy]
size(120); defaultpen(linewidth(0.7)+fontsize(10)); 
pair D2(pair P) {
dot(P,linewidth(3)); return P;
}
real r = sqrt(80)/5;
pair M=(-1,0), N=(1,0), A=intersectionpoints(arc((M+N)/2, 1, 0, 180),circle(N,r))[0], C=intersectionpoints(circle(A,1),circle(N,1))[0], B=intersectionpoints(circle(C,1),M--N)[0];
draw(arc((M+N)/2, 1, 0, 180)--cycle); draw(A--N); draw(arc(C,1,180,180+2*aSin(r/2)));
label(" $A$ ",D2(A),NW);
label(" $B$ ",D2(B),SW);
label(" $M$ ",D2(M),S);
label(" $N$ ",D2(N),SE);
[/asy]

Answer: $80$
$DEB$  is a chord of a circle such that  $DE=6$  and  $EB=5$ . Let  $O$  be the centre of the circle. Join  $OE$  and extend  $OE$  to cut the circle at  $C$. Given  $EC=2$ , find the radius of the circle.

[asy]
size(6cm);

pair O = (0,0), B = dir(110), D = dir(30), E = 0.4 * B + 0.6 * D, C = intersectionpoint(O--2*E, unitcircle);

draw(unitcircle);
draw(O--C);
draw(B--D);

dot(O);
dot(B);
dot(C);
dot(D);
dot(E);

label(" $B$ ", B, B);
label(" $C$ ", C, C);
label(" $D$ ", D, D);
label(" $E$ ", E, dir(280));
label(" $O$ ", O, dir(270));
[/asy]

Answer: $14$
The figure below shows a  $9\times7$  arrangement of  $3\times3$  squares. As shown in the diagram, find the area of the shaded region in the grid.

[asy]
size(5cm);
defaultpen(linewidth(.6));
fill((0,1)--(1,1)--(1,0)--cycle^^(0,3)--(1,3)--(1,2)--cycle^^(1,2)--(2,2)--(2,1)--cycle^^(2,1)--(3,1)--(3,0)--cycle,rgb(.76,.76,.76));
fill((0,5)--(1,5)--(1,4)--cycle^^(1,4)--(2,4)--(2,3)--cycle^^(2,3)--(3,3)--(3,2)--cycle^^(3,2)--(4,2)--(4,1)--cycle^^(4,1)--(5,1)--(5,0)--cycle,rgb(.76,.76,.76));
fill((0,7)--(1,7)--(1,6)--cycle^^(1,6)--(2,6)--(2,5)--cycle^^(2,5)--(3,5)--(3,4)--cycle^^(3,4)--(4,4)--(4,3)--cycle^^(4,3)--(5,3)--(5,2)--cycle^^(5,2)--(6,2)--(6,1)--cycle^^(6,1)--(7,1)--(7,0)--cycle,rgb(.76,.76,.76));
fill((2,7)--(3,7)--(3,6)--cycle^^(3,6)--(4,6)--(4,5)--cycle^^(4,5)--(5,5)--(5,4)--cycle^^(5,4)--(6,4)--(6,3)--cycle^^(6,3)--(7,3)--(7,2)--cycle^^(7,2)--(8,2)--(8,1)--cycle^^(8,1)--(9,1)--(9,0)--cycle,rgb(.76,.76,.76));
fill((4,7)--(5,7)--(5,6)--cycle^^(5,6)--(6,6)--(6,5)--cycle^^(6,5)--(7,5)--(7,4)--cycle^^(7,4)--(8,4)--(8,3)--cycle^^(8,3)--(9,3)--(9,2)--cycle,rgb(.76,.76,.76));
fill((6,7)--(7,7)--(7,6)--cycle^^(7,6)--(8,6)--(8,5)--cycle^^(8,5)--(9,5)--(9,4)--cycle,rgb(.76,.76,.76));
fill((8,7)--(9,7)--(9,6)--cycle,rgb(.76,.76,.76));
draw((0,0)--(0,7)^^(1,0)--(1,7)^^(2,0)--(2,7)^^(3,0)--(3,7)^^(4,0)--(4,7)^^(5,0)--(5,7)^^(6,0)--(6,7)^^(7,0)--(7,7)^^(8,0)--(8,7)^^(9,0)--(9,7));
draw((0,0)--(9,0)^^(0,1)--(9,1)^^(0,2)--(9,2)^^(0,3)--(9,3)^^(0,4)--(9,4)^^(0,5)--(9,5)^^(0,6)--(9,6)^^(0,7)--(9,7));
draw((0,1)--(1,0)^^(0,3)--(3,0)^^(0,5)--(5,0)^^(0,7)--(7,0)^^(2,7)--(9,0)^^(4,7)--(9,2)^^(6,7)--(9,4)^^(8,7)--(9,6));
[/asy]

Answer: $144$
A semicircle with diameter length  $16$  contains a circle radius  $3$  tangent both to the inside of the semicircle and its diameter as shown. A second larger circle is tangent to the inside of the semicircle, the outside of the circle, and the diameter of the semicircle. The diameter of the second circle can be written as  $\frac{n + k\sqrt{2}}{m}$  where  $m$ ,  $n$ , and  $k$  are positive integers and  $m$  and  $n$  have no factors in common. Find  $m + n - k$ .

[asy]
size(200);
pair O=(0,0);
real R=10, r=4.7;
draw(arc(O,R,0,180)--cycle);
pair P=(sqrt((R-r)^2-r^2),r),Q;
draw(circle(P,r));
real a=0,b=r,c;
for(int k=0;k<20;++k)
{
c=(a+b)/2;
Q=(-sqrt((R-c)^2-c^2),c);
if(abs(P-Q)>c+r) a=c; else b=c;
}
draw(circle(Q,c));[/asy]

Answer: $169$
Three lines are drawn parallel to each of the three sides of  $\triangle ABC$  so that the three lines intersect in the interior of  $ABC$ . The resulting three smaller triangles have areas  $1$ ,  $4$ , and  $16$ . Find the area of  $\triangle ABC$ .

[asy]
defaultpen(linewidth(0.7)); size(120);
pair relpt(pair P, pair Q, real a, real b) { return (a*Q+b*P)/(a+b); }
pair B = (0,0), C = (1,0), A = (0.3, 0.8), D = relpt(relpt(A,B,3,3),relpt(A,C,3,3),1,2);
draw(A--B--C--cycle);
label(" $A$ ",A,N); label(" $B$ ",B,S); label(" $C$ ",C,S);
filldraw(relpt(A,B,2,4)--relpt(A,B,3,3)--D--cycle, gray(0.7));
filldraw(relpt(A,C,1,5)--relpt(A,C,3,3)--D--cycle, gray(0.7));
filldraw(relpt(C,B,2,4)--relpt(B,C,1,5)--D--cycle, gray(0.7));[/asy]

Answer: $49$
In the figure on the right,  $O$  is the center of the circle,  $OK$  and  $OA$  are perpendicular to one another,  $M$  is the midpoint of  $OK$ ,  $BN$   is parallel to  $OK$ , and  $\angle AMN=\angle NMO$ . Determine the measure of  $\angle ABN$  in degrees.

[asy]
size(120);defaultpen(linewidth(0.7)+fontsize(10));
pair O=origin, A=dir(90), K=dir(180), M=1/2*dir(180), N=0.4*dir(90), B=dir(degrees((2/5, sqrt(21/25)))+90);
draw(K--O--A--M--N--B--A^^circle((0,0),1));
label(" $A$ ", A, dir(O--A));
label(" $K$ ", K, dir(O--K));
label(" $B$ ", B, dir(O--B));
label(" $N$ ", N, E);
label(" $M$ ", M, S);
label(" $O$ ", O, SE);[/asy]

Answer: $36$
$\triangle ABC$  has area  $360$ . Points  $X, Y, Z$  lie on sides  $AB$ ,  $BC$ , and  $CA$ , respectively. Given that  $\frac{AX}{BX} = 3$ ,  $\frac{BY}{CY} = 4$ , and  $\frac{CZ}{AZ} = 5$ , find the area of  $\triangle XYZ$ .

[asy]
size(175);
defaultpen(linewidth(0.8));
pair A=(0,15),B=(0,-5),C=(25,0.5),X=origin,Y=(4C+B)/5,Z=(5A+C)/6;
draw(A--B--C--cycle^^X--Y--Z--cycle);
label(" $A$ ",A,N);
label(" $B$ ",B,S);
label(" $C$ ",C,E);
label(" $X$ ",X,W);
label(" $Y$ ",Y,S);
label(" $Z$ ",Z,NE);[/asy]

Answer: $183$
In acute triangle  $ABC$ ,  $AB > AC$ . Let  $M$  be the midpoint of side  $BC$ . The exterior angle bisector of  $\widehat{BAC}$  meet ray  $BC$  at  $P$ . Point  $K$  and  $F$  lie on line  $PA$  such that  $MF \perp BC$  and  $MK \perp PA$ . What is $\frac{BC^2}{PF\cdot AK}$ ?

[asy] pair incenter(pair A=(0,0), pair B=(0,0), pair C=(0,0)){pair P,Q;P=rotate((angle(C-A)-angle(B-A))*90/pi,A)*B;Q=rotate((angle(A-B)-angle(C-B))*90/pi,B)*C;return extension(A,P,B,Q);}pair foot(pair P,pair A, pair B){real s;s=dot(P-A,unit(B-A));return (scale(s)*unit(B-A)+A);}defaultpen(fontsize(10)); size(7cm);pair A = (5.6,5), B = (0,0), C = (5,0), M = midpoint(B--C), I = incenter(A,B,C), P = extension(A, A+dir(I--A)*dir(-90), B,C), K = foot(M,A,P), F = extension(M, (M.x, M.x+1), A,P);draw(K--M--F--P--B--A--C);pair point = I;pair[] p={A,B,C,M,P,F,K};string s = "A,B,C,M,P,F,K";int size = p.length;real[] d; real[] mult; for(int i = 0; i<size; ++i) { d[i] = 0; mult[i] = 1;}string[] k= split(s,",");for(int i = 0;i<p.length;++i) {label("$"+k[i]+"$",p[i],mult[i]*dir(point--p[i])*dir(d[i]));}[/asy]

Answer: $4$
Let  $A_1A_2A_3A_4A_5$  be a regular pentagon with side length 1. The sides of the pentagon are extended to form the 10-sided polygon shown in bold at right. How many times of the area of the 10-sided polygon compared to the shaded area?

[asy]
size(8cm);
defaultpen(fontsize(10pt)); 
pair A_2=(-0.4382971011,5.15554989475), B_4=(-2.1182971011,-0.0149584477027), B_5=(-4.8365942022,8.3510997895), A_3=(0.6,8.3510997895), B_1=(2.28,13.521608132), A_4=(3.96,8.3510997895), B_2=(9.3965942022,8.3510997895), A_5=(4.9982971011,5.15554989475), B_3=(6.6782971011,-0.0149584477027), A_1=(2.28,3.18059144705); 
filldraw(A_2--A_5--B_2--B_5--cycle,rgb(.8,.8,.8));
draw(B_1--A_4^^A_4--B_2^^B_2--A_5^^A_5--B_3^^B_3--A_1^^A_1--B_4^^B_4--A_2^^A_2--B_5^^B_5--A_3^^A_3--B_1,linewidth(1.2)); draw(A_1--A_2--A_3--A_4--A_5--cycle);

pair O = (A_1+A_2+A_3+A_4+A_5)/5;
label(" $A_1$ ",A_1, 2dir(A_1-O));
label(" $A_2$ ",A_2, 2dir(A_2-O));
label(" $A_3$ ",A_3, 2dir(A_3-O));
label(" $A_4$ ",A_4, 2dir(A_4-O));
label(" $A_5$ ",A_5, 2dir(A_5-O));
label(" $B_1$ ",B_1, 2dir(B_1-O));
label(" $B_2$ ",B_2, 2dir(B_2-O));
label(" $B_3$ ",B_3, 2dir(B_3-O));
label(" $B_4$ ",B_4, 2dir(B_4-O));
label(" $B_5$ ",B_5, 2dir(B_5-O));
[/asy]

Answer: $2$
Point  $P$  lies outside a circle, and two rays are drawn from  $P$  that intersect the circle as shown. One ray intersects the circle at points  $A$  and  $B$  while the other ray intersects the circle at  $M$  and  $N$ .  $AN$  and  $MB$  intersect at  $X$ . Given that  $\angle AXB$  measures  $127^{\circ}$  and the minor arc  $AM$  measures  $14^{\circ}$ , compute the measure of the angle at  $P$ .

[asy]
size(200);
defaultpen(fontsize(10pt));
pair P=(60,10),C=(-20,10),K=(-20,-10);
path CC=circle((0,0),20), PC=P--C, PK=P--K;
pair A=intersectionpoints(CC,PC)[0],
B=intersectionpoints(CC,PC)[1],
M=intersectionpoints(CC,PK)[0],
N=intersectionpoints(CC,PK)[1],
X=intersectionpoint(A--N,B--M);
draw(CC);draw(PC);draw(PK);draw(A--N);draw(B--M);
label(" $A$ ",A,plain.NE);label(" $B$ ",B,plain.NW);label(" $M$ ",M,SE);
label(" $P$ ",P,E);label(" $N$ ",N,dir(250));label(" $X$ ",X,plain.N);[/asy]

Answer: $39$
Equilateral triangle  $ T$  is inscribed in circle  $ A$ , which has radius  $ 10$ . Circle  $ B$  with radius  $ 3$  is internally tangent to circle  $ A$  at one vertex of  $ T$ . Circles  $ C$  and  $ D$ , both with radius  $ 2$ , are internally tangent to circle  $ A$  at the other two vertices of  $ T$ . Circles  $ B$ ,  $ C$ , and  $ D$  are all externally tangent to circle  $ E$ , which has radius  $ \frac {m}{n}$ , where  $ m$  and  $ n$  are relatively prime positive integers. Find  $ m - n$ .

[asy]unitsize(2.2mm);
defaultpen(linewidth(.8pt)+fontsize(8pt));
dotfactor=4;

pair A=(0,0), D=8*dir(330), C=8*dir(210), B=7*dir(90);
pair Ep=(0,4-27/5);
pair[] dotted={A,B,C,D,Ep};

draw(Circle(A,10));
draw(Circle(B,3));
draw(Circle(C,2));
draw(Circle(D,2));
draw(Circle(Ep,27/5));

dot(dotted);
label(" $E$ ",Ep,E);
label(" $A$ ",A,W);
label(" $B$ ",B,W);
label(" $C$ ",C,W);
label(" $D$ ",D,E);[/asy]

Answer: $22$
Let  $ABCD$  be a rectangle with  $AB=10$  and  $BC=66$ . Let  $\omega_1$  be the circle with diameter  $\overline{AB}$  and  $\omega_2$  be the circle with diameter  $\overline{CD}$ . Suppose  $\ell$  is a common internal tangent to  $\omega_1$  and  $\omega_2$  and that  $\ell$  intersects  $AD$  and  $BC$  at  $E$  and  $F$  respectively. What is  $EF$ ?

[asy]
        size(10cm);
        draw((0,0)--(26,0)--(26,10)--(0,10)--cycle);
        draw((1,0)--(25,10));
        draw(circle((0,5),5));
        draw(circle((26,5),5));
        dot((1,0));
        dot((25,10));
        label(" $E$ ",(1,0),SE);
        label(" $F$ ",(25,10),NW);
        label(" $A$ ", (0,0), SW);
        label(" $B$ ", (0,10), NW);
        label(" $C$ ", (26,10), NE);
        label(" $D$ ", (26,0), SE);
        dot((0,0));
        dot((0,10));
        dot((26,0));
        dot((26,10));
[/asy]

Answer: $66$
Point  $P$  is in the interior of  $\triangle ABC$ . The side lengths of  $ABC$  are  $AB = 7$ ,  $BC = 8$ ,  $CA = 9$ . The three foots of perpendicular lines from  $P$  to sides  $BC$ ,  $CA$ ,  $AB$  are  $D$ ,  $E$ ,  $F$  respectively. Suppose the minimal value of  $\frac{BC}{PD} + \frac{CA}{PE} + \frac{AB}{PF}$  can be written as  $\frac{a}{b}\sqrt{c}$ , where  $\gcd(a,b) = 1$  and  $c$  is square free, calculate  $a+b+c$ .

[asy]
size(120); pathpen = linewidth(0.7); pointfontpen = fontsize(10); // pointpen = black;
pair B=(0,0), C=(8,0), A=IP(CR(B,7),CR(C,9)), P = (2,1.6), D=foot(P,B,C), E=foot(P,A,C), F=foot(P,A,B);
D(A--B--C--cycle); D(P--D); D(P--E); D(P--F);
D(MP("A",A,N)); D(MP("B",B)); D(MP("C",C)); D(MP("D",D)); D(MP("E",E,NE)); D(MP("F",F,NW)); D(MP("P",P,SE));
[/asy]

Answer: $34$
A rectangular piece of paper  $ABCD$  has sides of lengths  $AB = 3$ ,  $BC = 6$ . The rectangle is folded in half such that  $AD$  coincides with  $BC$  and  $EF$  is the folding line. Then fold the paper along a line  $BM$  such that the corner  $A$  falls on line  $EF$ . How large, in degrees, is  $\angle AMB$ ?

[asy]
size(180); pathpen = rgb(0,0,0.6)+linewidth(1); pointpen = black+linewidth(3); pointfontpen = fontsize(10); pen dd = rgb(0,0,0.6) + linewidth(0.7) + linetype("4 4"), dr = rgb(0.8,0,0), dg = rgb(0,0.6,0), db = rgb(0,0,0.6)+linewidth(1);
pair A=(0,1), B=(0,0), C=(2,0), D=(2,1), E=A/2, F=(2,.5), M=(1/3^.5,1), N=reflect(B,M)*A;
D(B--M--D("N",N,NE)--B--D("C",C,SE)--D("D",D,NE)--M); D(D("M",M,plain.N)--D("A",A,NW)--D("B",B,SW),dd); D(D("E",E,W)--D("F",F,plain.E),dd); 
[/asy]

Answer: $60$
Two concentric circles have radii  $1$  and  $4$ .  Six congruent circles form a ring where each of the six circles is tangent to the two circles adjacent to it as shown.  The three lightly shaded circles are internally tangent to the circle with radius  $4$  while the three darkly shaded circles are externally tangent to the circle with radius  $1$ .  The radius of the six congruent circles can be written  $\frac{k+\sqrt m}{n}$ , where  $k,m,$  and  $n$  are integers with  $k$  and  $n$  relatively prime.  Find  $k	imes m	imes n$ .

[asy]
size(150);
defaultpen(linewidth(0.8));
real r = (sqrt(133)-9)/2;
draw(circle(origin,1)^^circle(origin,4));
for(int i=0;i<=2;i=i+1)
{
filldraw(circle(dir(90 + i*120)*(4-r),r),gray);
}
for(int j=0;j<=2;j=j+1)
{
filldraw(circle(dir(30+j*120)*(1+r),r),darkgray);
}
[/asy]

Answer: $-2394$
Inside the  $7\times 10$  rectangle below, one point is chosen a distance  $\sqrt2$  from the left side and a distance  $\sqrt7$  from the bottom side. The line segments from that point to the four vertices of the rectangle are drawn. Find the area of the shaded region.

[asy]
import graph;
size(4cm);
pair A = (0,0);
pair B = (9,0);
pair C = (9,7);
pair D = (0,7);
pair P = (1.5,3);
draw(A--B--C--D--cycle,linewidth(1.5));
filldraw(A--B--P--cycle,rgb(.76,.76,.76),linewidth(1.5));
filldraw(C--D--P--cycle,rgb(.76,.76,.76),linewidth(1.5));
[/asy]

Answer: $35$
The pattern in the figure below continues inward infinitely. The base of the biggest triangle is 1. All triangles are equilateral. The area of the shaded region can be written as $\frac{\sqrt{m}{n}}$ where $m$ and $n$ are relatively prime integers. Find the value of $m+n$.

[asy]
defaultpen(linewidth(0.8));
pen blu = rgb(0,112,191);
real r=sqrt(3);
fill((8,0)--(0,8r)--(-8,0)--cycle, blu);
fill(origin--(4,4r)--(-4,4r)--cycle, white);
fill((2,2r)--(0,4r)--(-2,2r)--cycle, blu);
fill((0,2r)--(1,3r)--(-1,3r)--cycle, white);[/asy]

Answer: $8$
The target below is made up of concentric circles with diameters  $2$ ,  $4$ ,  $8$ ,  $12$ , and  $16$ . The area of the dark region is  $n\pi$ . Find  $n$ .

[asy]
size(150);
defaultpen(linewidth(0.8));
int i;
for(i=5;i>=1;i=i-1)
{
if (floor(i/2)==i/2)
{
filldraw(circle(origin,4*i),white);
}
else
{
filldraw(circle(origin,4*i),red);
}
}
[/asy]

Answer: $41$
Find 600 times the area of a regular dodecagon inscribed in a unit circle. Round your answer to the nearest integer if necessary.

[asy]
defaultpen(linewidth(0.7)); real theta = 17; pen dr = rgb(0.8,0,0), dg = rgb(0,0.6,0), db = rgb(0,0,0.6)+linewidth(1);
draw(unitcircle,dg);
for(int i = 0; i < 12; ++i) {
 draw(dir(30*i+theta)--dir(30*(i+1)+theta), db);
 dot(dir(30*i+theta),Fill(rgb(0.8,0,0)));
} dot(dir(theta),Fill(dr)); dot((0,0),Fill(dr));
[/asy]

Answer: $1800$
Circles  $\omega$  and  $\gamma$  are drawn such that  $\omega$  is internally tangent to  $\gamma$ , the distance between their centers are  $12$ , and the area inside of  $\gamma$  but outside of  $\omega$  is  $360\pi$ . What is the sum of the radii of the circles?

[asy]
size(3cm);
real lw=0.4, dr=0.3;
real r1=14, r2=9;
pair A=(0,0), B=(r1-r2,0);
draw(A--B,dashed);
draw(circle(A,r1),linewidth(lw)); draw(circle(B,r2),linewidth(lw));
filldraw(circle(A,dr)); filldraw(circle(B,dr));
label(" $5$ ",(A+B)/2,dir(-90));
label(" $\gamma$ ",A+r1*dir(135),dir(135)); label(" $\omega$ ",B+r2*dir(135),dir(135));
[/asy]

Answer: $23$
The cross below is made up of five congruent squares. The perimeter of the cross is  $120$ . Find its area.

[asy]
import graph;
size(3cm);
pair A = (0,0);
pair temp = (1,0);
pair B = rotate(45,A)*temp;
pair C = rotate(90,B)*A;
pair D = rotate(270,C)*B;
pair E = rotate(270,D)*C;
pair F = rotate(90,E)*D;
pair G = rotate(270,F)*E;
pair H = rotate(270,G)*F;
pair I = rotate(90,H)*G;
pair J = rotate(270,I)*H;
pair K = rotate(270,J)*I;
pair L = rotate(90,K)*J;
draw(A--B--C--D--E--F--G--H--I--J--K--L--cycle);
[/asy]

Answer: $500$
As shown in figure, circle  $\Omega$ has $\overline{AB}=65$ and $CD=16$. Moreover, let  $AD, BC, AC,$  and  $BD$  all have integer lengths. Two other circles, $\omega_1$  and  $\omega_2$ , have  $\overline{AC}$  and  $\overline{BD}$  as their diameters, respectively. Let circle  $\omega_1$  intersect  $AB$  at a point  $E \neq A$  and let circle  $\omega_2$  intersect  $AB$  at a point  $F \neq B$ . Then  $EF=\frac{m}{n}$ , for relatively prime integers  $m$  and  $n$ . Find  $m+n$ .

[asy]
size(7cm);
pair A=(0,0), B=(64,0), C=(117/5,156/5), D=(125/13,300/13), E=(23.4,0), F=(9.615,0);
draw(A--B--C--D--cycle);
draw(A--C);
draw(B--D);
dot(" $A$ ", A, SW);
dot(" $B$ ", B, SE);
dot(" $C$ ", C, NE);
dot(" $D$ ", D, NW);
dot(" $E$ ", E, S);
dot(" $F$ ", F, S);
draw(circle((A + C)/2, abs(A - C)/2));
draw(circle((B + D)/2, abs(B - D)/2));
draw(circle((A + B)/2, abs(A - B)/2));
label(" $\mathcal P$ ", (A + B)/2 + abs(A - B)/2 * dir(-45), dir(-45));
label(" $\mathcal Q$ ", (A + C)/2 + abs(A - C)/2 * dir(-210), dir(-210));
label(" $\mathcal R$ ", (B + D)/2 + abs(B - D)/2 * dir(70), dir(70));
[/asy]

Answer: $961$
In an isosceles right-angled triangle AOB, points P; Q and S are chosen on sides OB, OA, and AB respectively such that a square PQRS is formed as shown. If the lengths of OP and OQ are a and b respectively, and the area of PQRS is $\frac{2}{5}$ that of triangle AOB. How many time of a compared to b?

[asy]

size(120);
pair A = (0,3);
pair B = (0,0);
pair C = (3,0);
pair D = (0,1.5);
pair E = (0.35,0);
pair F = (1.1,1.9);
pair J = (0.17,0);
pair Y = (0.17,0.75);

pair Z = (1.6,0.4);
draw(A--B);
draw(B--C);
draw(C--A);
draw(D--F--Z--E--D);
draw(" $O$ ", B, dir(180));
draw(" $B$ ", A, dir(45));
draw(" $A$ ", C, dir(45));
draw(" $Q$ ", E, dir(45));
draw(" $P$ ", D, dir(45));
draw(" $R$ ", Z, dir(45));
draw(" $S$ ", F, dir(45));
draw(" $a$ ", Y, dir(210));
draw(" $b$ ", J, dir(100));
[/asy]

Answer: $2$
Squares $ABCD$ and $EFGH$ are congruent, $G$ is the center of square $ABCD$ and $AB=5$. The area of the region in the plane covered by these squares is

[asy] draw((0,0)--(10,0)--(10,10)--(0,10)--cycle); draw((5,5)--(12,-2)--(5,-9)--(-2,-2)--cycle); label("A", (0,0), W); label("B", (10,0), E); label("C", (10,10), NE); label("D", (0,10), NW); label("G", (5,5), N); label("F", (12,-2), E); label("E", (5,-9), S); label("H", (-2,-2), W); dot((-2,-2)); dot((5,-9)); dot((12,-2)); dot((0,0)); dot((10,0)); dot((10,10)); dot((0,10)); dot((5,5)); [/asy]

Answer: $\frac{175}{4}$
Triangle $ABC$ is inscribed in a circle, and $\angle B = \angle C = 4\angle A$. If $B$ and $C$ are adjacent vertices of a regular polygon of $3n$ sides inscribed in this circle, then $n=$

[asy] draw(Circle((0,0), 10)); draw((0,10)--(6,-8)--(-6,-8)--cycle); label("A", (0,10), N); label("B", (-6,-8), SW); label("C", (6,-8), SE); dot((0,10)); dot((6,-8)); dot((-6,-8)); [/asy]

Answer: $3$
A regular polygon of $m$ sides is exactly enclosed (no overlaps, no gaps) by $m$ regular polygons of $n$ sides each. (Shown here for $m=4, n=8$.) If $n=5$, what is the value of $m$?

[asy] size(200); defaultpen(linewidth(0.8)); draw(unitsquare); path p=(0,1)--(1,1)--(1+sqrt(2)/2,1+sqrt(2)/2)--(1+sqrt(2)/2,2+sqrt(2)/2)--(1,2+sqrt(2))--(0,2+sqrt(2))--(-sqrt(2)/2,2+sqrt(2)/2)--(-sqrt(2)/2,1+sqrt(2)/2)--cycle; draw(p); draw(shift((1+sqrt(2)/2,-sqrt(2)/2-1))*p); draw(shift((0,-2-sqrt(2)))*p); draw(shift((-1-sqrt(2)/2,-sqrt(2)/2-1))*p);[/asy]

Answer: $10$
Three congruent circles with centers $P$, $Q$, and $R$ are tangent to the sides of rectangle $ABCD$ as shown. The circle centered at $Q$ has diameter $2$ and passes through points $P$ and $R$. The area of the rectangle is

[asy] pair A,B,C,D,P,Q,R; A = (0,4); B = (8,4); C = (8,0); D = (0,0); P = (2,2); Q = (4,2); R = (6,2); dot(A); dot(B); dot(C); dot(D); dot(P); dot(Q); dot(R); draw(A--B--C--D--cycle); draw(circle(P,2)); draw(circle(Q,2)); draw(circle(R,2)); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$P$",P,W); label("$Q$",Q,W); label("$R$",R,W); [/asy]

Answer: $8$
The horizontal and vertical distances between adjacent points equal 3 unit. What is the area of triangle $ABC$?

[asy] for (int a = 0; a < 5; ++a) { for (int b = 0; b < 4; ++b) { dot((a,b)); } } draw((0,0)--(3,2)--(4,3)--cycle); label("$A$",(0,0),SW); label("$B$",(3,2),SE); label("$C$",(4,3),NE); [/asy]

Answer: $\frac{9}{2}$
The measure of angle $ABC$ is $60^\circ$, $\overline{AD}$ bisects angle $BAC$, and $\overline{DC}$ bisects angle $BCA$. The measure of angle $ADC$ is

[asy] pair A,B,C,D; A = (0,0); B = (9,10); C = (10,0); D = (6.66,3); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--cycle); draw(A--D--C); label("$A$",A,SW); label("$B$",B,N); label("$C$",C,SE); label("$D$",D,N); label("$60^\circ $",(9.4,8.8),SW); [/asy]

Answer: $120$
Assume the square have side length of 3. Each side of the large square in the figure is trisected (divided into three equal parts). The corners of an inscribed square are at these trisection points, as shown. The area of the inscribed square is

[asy] draw((0,0)--(3,0)--(3,3)--(0,3)--cycle); draw((1,0)--(1,0.2)); draw((2,0)--(2,0.2)); draw((3,1)--(2.8,1)); draw((3,2)--(2.8,2)); draw((1,3)--(1,2.8)); draw((2,3)--(2,2.8)); draw((0,1)--(0.2,1)); draw((0,2)--(0.2,2)); draw((2,0)--(3,2)--(1,3)--(0,1)--cycle); [/asy]

Answer: $5$
In the figure,  $E$ is the midpoint of $\overline{AD}$, $ABCD$ is a $1 \times 1$ square, and $F$ is on $\overline{BE}$. The area of quadrilateral $CDEF$ is

[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = (0,2); pair B = origin; pair C = (2,0); pair D = (2,2); pair E = midpoint(A--D); pair F = foot(C,B,E); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F); label("$A$",A,N);label("$B$",B,S);label("$C$",C,S);label("$D$",D,N);label("$E$",E,N);label("$F$",F,NW); draw(A--B--C--D--cycle); draw(B--E); draw(C--F); draw(rightanglemark(B,F,C,4));[/asy]

Answer: $\frac{11}{20}$
$BD$ and $CE$ are medians of triangle $ABC$, $CE=6$, and $BD=4$. The area of triangle $ABC$ is

[asy] defaultpen(linewidth(.8pt)); dotfactor=4; pair A = origin; pair B = (1.25,1); pair C = (2,0); pair D = midpoint(A--C); pair E = midpoint(A--B); pair G = intersectionpoint(E--C,B--D); dot(A);dot(B);dot(C);dot(D);dot(E);dot(G); label("$A$",A,S);label("$B$",B,N);label("$C$",C,S);label("$D$",D,S);label("$E$",E,NW);label("$G$",G,NE); draw(A--B--C--cycle); draw(B--D); draw(E--C); draw(rightanglemark(C,G,D,3));[/asy]

Answer: $16$
A circle with center $O$ is tangent to the coordinate axes and to the hypotenuse of the $30^\circ$-$60^\circ$-$90^\circ$ triangle $ABC$ as shown. To the nearest hundredth, what is the radius of the circle?

[asy] defaultpen(linewidth(.8pt)); dotfactor=3; pair A = origin; pair B = (1,0); pair C = (0,sqrt(3)); pair O = (2.33,2.33); dot(A);dot(B);dot(C);dot(O); label("$A$",A,SW);label("$B$",B,SE);label("$C$",C,W);label("$O$",O,NW); label("$1$",midpoint(A--B),S);label("$60^\circ$",B,2W + N); draw((3,0)--A--(0,3)); draw(B--C); draw(Arc(O,2.33,163,288.5));[/asy]

Answer: $\frac{3}{2}+\frac{\sqrt{3}}{2}$