LiAuto-DSR/GeoGramBench huggingface dataset:
In the figure below, \(\angle ABC\) is a right angle. Point \(D\) lies on segment \(\overline{BC}\), and segment \(\overline{AD}\) is the angle bisector of \(\angle CAB\). Points \(E\) and \(F\) are located on segments \(\overline{AB}\) and \(\overline{AC}\), respectively, such that \(AE = 3\), \(AF = 10\), \(EB = 9\), and \(FC = 27\). Let \(G\) denote the point where \(\overline{DF}\) intersects \(\overline{CE}\). Determine the integer nearest to the area of quadrilateral \(DCFG\).

[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: $148$
Consider a fixed regular  $n$ -gon of unit side. When a second regular  $n$ -gon of unit size rolls around the first one, one of its vertices successively pinpoints the vertices of a closed broken line  $\kappa$  as in the figure.



Let  $10$  be the area of a regular  $n$ -gon of unit side, and let  $5$  be the area of a regular  $n$ -gon of unit circumradius. Calculate the area enclosed by  $\kappa$.

[asy]
int n=9;
draw(polygon(n));
for (int i = 0; i<n;++i) {
 draw(reflect(dir(360*i/n + 90), dir(360*(i+1)/n + 90))*polygon(n), dashed+linewidth(0.4));
 draw(reflect(dir(360*i/n + 90),dir(360*(i+1)/n + 90))*(0,1)--reflect(dir(360*(i-1)/n + 90),dir(360*i/n + 90))*(0,1), linewidth(1.2));
}
[/asy]

Answer: $50$
Let  $n\ge 2$  be an integer. Thibaud the Tiger lays  $n$   $2\times 2$  overlapping squares out on a table, such that the centers of the squares are equally spaced along the line  $y=x$  from  $(0,0)$  to  $(1,1)$  (including the two endpoints). For example, for  $n=4$  the resulting figure is shown below, and it covers a total area of  $\frac{23}{3}$ .

Find the minimum  $n$  such that the figure covers an area of at least  $\sqrt{63}$ .

[asy]
fill((0,0)--(2,0)--(2,.333333333333)--(0.333333333333,0.333333333333)--(0.333333333333,2)--(0,2)--cycle, lightgrey);
fill((0.333333333333,0.333333333333)--(2.333333333333,0.333333333333)--(2.333333333333,.6666666666666)--(0.666666666666,0.666666666666666)--(0.66666666666,2.33333333333)--(.333333333333,2.3333333333333)--cycle, lightgrey);
fill((0.6666666666666,.6666666666666)--(2.6666666666666,.6666666666)--(2.6666666666666,.6666666666666)--(2.6666666666666,1)--(1,1)--(1,2.6666666666666)--(0.6666666666666,2.6666666666666)--cycle, lightgrey);
fill((1,1)--(3,1)--(3,3)--(1,3)--cycle, lightgrey);
draw((0.33333333333333,2)--(2,2)--(2,0.333333333333), dashed+grey+linewidth(0.4));
draw((0.66666666666666,2.3333333333333)--(2.3333333333333,2.3333333333333)--(2.3333333333333,0.66666666666), dashed+grey+linewidth(0.4));
draw((1,2.666666666666)--(2.666666666666,2.666666666666)--(2.666666666666,1), dashed+grey+linewidth(0.4));
draw((0,0)--(2,0)--(2,.333333333333)--(0.333333333333,0.333333333333)--(0.333333333333,2)--(0,2)--(0,0),linewidth(0.4));
draw((0.333333333333,0.333333333333)--(2.333333333333,0.333333333333)--(2.333333333333,.6666666666666)--(0.666666666666,0.666666666666666)--(0.66666666666,2.33333333333)--(.333333333333,2.3333333333333)--(0.333333333333,.333333333333),linewidth(0.4));
draw((0.6666666666666,.6666666666666)--(2.6666666666666,.6666666666)--(2.6666666666666,.6666666666666)--(2.6666666666666,1)--(1,1)--(1,2.6666666666666)--(0.6666666666666,2.6666666666666)--(0.6666666666666,0.6666666666666),linewidth(0.4));
draw((1,1)--(3,1)--(3,3)--(1,3)--cycle,linewidth(0.4));
[/asy]

Answer: $17$
In the figure, AQPB and ASRC are squares, and AQS is an equilateral triangle. If QS  = 12 and  BC  = x, what is the value of x?

[asy]
unitsize(16);
pair A,B,C,P,Q,R,T;
A=(3.4641016151377544, 2);
B=(0, 0);
C=(6.928203230275509, 0);
P=(-1.9999999999999991, 3.464101615137755);
Q=(1.4641016151377544, 5.464101615137754);
R=(8.928203230275509, 3.4641016151377544);
T=(5.464101615137754, 5.464101615137754);
dot(A);dot(B);dot(C);dot(P);
dot(Q);dot(R);dot(T);
label(" $A$ ", (3.4641016151377544, 2),E);
label(" $B$ ", (0, 0),S);
label(" $C$ ", (6.928203230275509, 0),S);
label(" $P$ ", (-1.9999999999999991, 3.464101615137755), W);
label(" $Q$ ", (1.4641016151377544, 5.464101615137754),N);
label(" $R$ ", (8.928203230275509, 3.4641016151377544),E);
label(" $S$ ", (5.464101615137754, 5.464101615137754),N);
draw(B--C--A--B);
draw(B--P--Q--A--B);
draw(A--C--R--T--A);
draw(Q--T--A--Q);
label(" $x$ ", (3.4641016151377544, 0), S);
label(" $4$ ", (Q+T)/2, N);[/asy]

Answer: $12\sqrt{3}$
In the figure below, each side of the rhombus has length 8 centimeters.

The circle lies entirely within the rhombus.  The area of the circle is  $n$  square centimeters, where  $n$  is a positive integer.  Compute the number of possible values of  $n$ .

[asy]
  import graph;
  unitsize(2.0cm);
  real w = sqrt(3);
  draw((w, 0) -- (0, 1) -- (-w, 0) -- (0, -1) -- cycle);
  filldraw(Circle((-0.5, 0), 0.8 / sqrt(pi)), gray);
  label(" $60^\circ$ ", (w - 0.1, 0), W);
[/asy]

Answer: $37$
The grid below contains six rows with six points in each row. Points that are adjacent either horizontally or vertically with equal distance 13. Find the area of the irregularly shaped ten sided figure shown.

[asy]
import graph; size(5cm); 
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
pen dotstyle = black;
draw((-2,5)--(-3,4), linewidth(1.6)); 
draw((-3,4)--(-2,1), linewidth(1.6)); 
draw((-2,1)--(1,0), linewidth(1.6)); 
draw((1,0)--(2,1), linewidth(1.6)); 
draw((2,1)--(1,3), linewidth(1.6)); 
draw((1,3)--(1,4), linewidth(1.6)); 
draw((1,4)--(2,5), linewidth(1.6)); 
draw((2,5)--(0,5), linewidth(1.6)); 
draw((-2,5)--(-1,4), linewidth(1.6)); 
draw((-1,4)--(0,5), linewidth(1.6)); 
dot((-3,5),linewidth(6pt) + dotstyle); 
dot((-2,5),linewidth(6pt) + dotstyle); 
dot((-1,5),linewidth(6pt) + dotstyle); 
dot((0,5),linewidth(6pt) + dotstyle); 
dot((1,5),linewidth(6pt) + dotstyle); 
dot((2,5),linewidth(6pt) + dotstyle); 
dot((2,4),linewidth(6pt) + dotstyle); 
dot((2,3),linewidth(6pt) + dotstyle); 
dot((2,2),linewidth(6pt) + dotstyle); 
dot((2,1),linewidth(6pt) + dotstyle); 
dot((2,0),linewidth(6pt) + dotstyle); 
dot((-3,4),linewidth(6pt) + dotstyle); 
dot((-3,3),linewidth(6pt) + dotstyle); 
dot((-3,2),linewidth(6pt) + dotstyle); 
dot((-3,1),linewidth(6pt) + dotstyle); 
dot((-3,0),linewidth(6pt) + dotstyle); 
dot((-2,0),linewidth(6pt) + dotstyle); 
dot((-2,1),linewidth(6pt) + dotstyle); 
dot((-2,2),linewidth(6pt) + dotstyle); 
dot((-2,3),linewidth(6pt) + dotstyle); 
dot((-2,4),linewidth(6pt) + dotstyle); 
dot((-1,4),linewidth(6pt) + dotstyle); 
dot((0,4),linewidth(6pt) + dotstyle); 
dot((1,4),linewidth(6pt) + dotstyle); 
dot((1,3),linewidth(6pt) + dotstyle); 
dot((0,3),linewidth(6pt) + dotstyle); 
dot((-1,3),linewidth(6pt) + dotstyle); 
dot((-1,2),linewidth(6pt) + dotstyle); 
dot((-1,1),linewidth(6pt) + dotstyle); 
dot((-1,0),linewidth(6pt) + dotstyle); 
dot((0,0),linewidth(6pt) + dotstyle); 
dot((1,0),linewidth(6pt) + dotstyle); 
dot((1,1),linewidth(6pt) + dotstyle); 
dot((1,2),linewidth(6pt) + dotstyle); 
dot((0,2),linewidth(6pt) + dotstyle); 
dot((0,1),linewidth(6pt) + dotstyle); [/asy]

Answer: $214.5$
In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If  $AG=2$ ,  $GF=13$ ,  $FC=1$ , and  $HJ=7$, the length of $DE$ is ?

[asy]
size(200);
defaultpen(fontsize(10));
real r=sqrt(22);
pair B=origin, A=16*dir(60), C=(16,0), D=(10-r,0), E=(10+r,0), F=C+1*dir(120), G=C+14*dir(120), H=13*dir(60), J=6*dir(60), O=circumcenter(G,H,J);
dot(A^^B^^C^^D^^E^^F^^G^^H^^J);
draw(Circle(O, abs(O-D))^^A--B--C--cycle, linewidth(0.7));
label(" $A$ ", A, N);
label(" $B$ ", B, dir(210));
label(" $C$ ", C, dir(330));
label(" $D$ ", D, SW);
label(" $E$ ", E, SE);
label(" $F$ ", F, dir(170));
label(" $G$ ", G, dir(250));
label(" $H$ ", H, SE);
label(" $J$ ", J, dir(0));
label("2", A--G, dir(30));
label("13", F--G, dir(180+30));
label("1", F--C, dir(30));
label("7", H--J, dir(-30));[/asy]

Answer: $2\sqrt{22}$
An equilateral triangle has sides of length  $6$ cm. At each vertex, a circle with radius  $3$ cm is drawn, as shown in the figure below. The total area of the unshaded regions of the three circles is  $a\cdot \pi \text{cm}^2$ . Determine  $a$ .

[center][/center]

[asy]
size(2.5cm);
draw(circle((0,2sqrt(3)/3),1));
draw(circle((1,-sqrt(3)/3),1));
draw(circle((-1,-sqrt(3)/3),1));
draw((0,2sqrt(3)/3) -- arc((0,2sqrt(3)/3), 1, 240, 300) -- cycle);
fill(((0,2sqrt(3)/3) -- arc((0,2sqrt(3)/3), 1, 240, 300) -- cycle),mediumgray);
draw((1,-sqrt(3)/3) -- arc((1,-sqrt(3)/3), 1, 180, 120) -- cycle);
fill(((1,-sqrt(3)/3) -- arc((1,-sqrt(3)/3), 1, 180, 120) -- cycle),mediumgray);
draw((-1,-sqrt(3)/3) -- arc((-1,-sqrt(3)/3), 1, 0, 60) -- cycle);
fill(((-1,-sqrt(3)/3) -- arc((-1,-sqrt(3)/3), 1, 0, 60) -- cycle),mediumgray);
[/asy]

Answer: $22.5$
Suppose two circles  $\Omega_1$  and  $\Omega_2$  with centers  $O_1$  and  $O_2$  have radii  $6$  and  $8$ , respectively. Suppose that points  $A$  and  $B$  lie on circles  $\Omega_1$  and  $\Omega_2$ , respectively, such that segments  $AB$  and  $O_1O_2$  intersect and that  $AB$  is tangent to  $\Omega_1$  and  $\Omega_2$ . If  $O_1O_2=50$ , find the lenght of line $AB$

[asy]
import graph; size(12cm); 
real labelscalefactor = 0.5; 
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); 
pen dotstyle = black; 
real xmin = -12.81977592804657, xmax = 32.13023014338037, ymin = -14.185056097058798, ymax = 12.56855801985179;  
draw(circle((-3.4277328104418046,-1.4524996726688195), 3), linewidth(1.2)); 
draw(circle((21.572267189558197,-1.4524996726688195), 4), linewidth(1.2)); 
draw((-2.5877328104418034,1.4275003273311748)--(20.452267189558192,-5.2924996726687885), linewidth(1.2)); 

dot((-3.4277328104418046,-1.4524996726688195),linewidth(3pt) + dotstyle); 
label(" $O_1$ ", (-4.252707018231291,-1.545940604327141), N * labelscalefactor); 
dot((21.572267189558197,-1.4524996726688195),linewidth(3pt) + dotstyle); 
label(" $O_2$ ", (21.704189347819636,-1.250863978037686), NE * labelscalefactor); 
dot((-2.5877328104418034,1.4275003273311748),linewidth(3pt) + dotstyle); 
label(" $A$ ", (-2.3937351324858342,1.6999022848568643), NE * labelscalefactor); 
dot((20.452267189558192,-5.2924996726687885),linewidth(3pt) + dotstyle); 
label(" $B$ ", (20.671421155806545,-4.9885012443707835), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]

Answer: $48$
Suppose circles  $\mathit{W}_1$  and  $\mathit{W}2$ , with centres  $\mathit{O}_1$  and  $\mathit{O}_2$  respectively, intersect at points  $\mathit{M}$  and  $\mathit{N}$ . Let the tangent on  $\mathit{W}_2$  at point  $\mathit{N}$  intersect  $\mathit{W}_1$  for the second time at  $\mathit{B}_1$ . Similarly, let the tangent on  $\mathit{W}_1$  at point  $\mathit{N}$  intersect  $\mathit{W}_2$  for the second time at  $\mathit{B}_2$ . Let  $\mathit{A}_1$  be a point on  $\mathit{W}_1$  which is on arc  $\mathit{B}_1\mathit{N}$  not containing  $\mathit{M}$  and suppose line  $\mathit{A}_1\mathit{N}$  intersects  $\mathit{W}_2$  at point  $\mathit{A}_2$ . Denote the incentres of triangles  $\mathit{B}_1\mathit{A}_1\mathit{N}$  and  $\mathit{B}_2\mathit{A}_2\mathit{N}$  by  $\mathit{I}_1$  and  $\mathit{I}_2$ , respectively.If the degree of $\angle\mathit{I}_1\mathit{MI}_2$ is 37, find the degree of angle $\angle\mathit{O}_1\mathit{MO}_2 + \angle\mathit{I}_1\mathit{MI}_2$.

[asy]import graph; size(10.1cm); 
real labelscalefactor = 0.5; 
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); 
pen dotstyle = black; 
real xmin = -0.9748626324969808, xmax = 13.38440254515721, ymin = 0.5680051903627492, ymax = 10.99430986899034; 
pair O_2 = (7.682929606970993,6.084708172218866), O_1 = (2.180000000000002,6.760000000000007), M = (4.560858774883258,8.585242858926296), B_2 = (10.07334553576748,9.291873850408265), A_2 = (11.49301008867042,4.866805580476367), B_1 = (2.113311869970955,9.759258690628950), A_1 = (0.2203184186713625,4.488514120712773); 
draw(circle(O_2, 4.000000000000000)); 
draw(circle(O_1, 3.000000000000000)); 
draw((4.048892687647541,4.413249028538064)--B_2); 
draw(B_2--A_2); 
draw(A_2--(4.048892687647541,4.413249028538064)); 
draw((4.048892687647541,4.413249028538064)--B_1); 
draw(B_1--A_1); 
draw(A_1--(4.048892687647541,4.413249028538064)); 
dot(O_2,dotstyle); 
label(" $O_2$ ", (7.788512439159622,6.243082420501817), NE * labelscalefactor); 
dot(O_1,dotstyle); 
label(" $O_1$ ", (2.298205165350667,6.929370829727937), NE * labelscalefactor); 
dot(M,dotstyle); 
label(" $M$ ", (4.383466101076183,8.935444641311980), NE * labelscalefactor); 
dot((4.048892687647541,4.413249028538064),dotstyle); 
label(" $N$ ", (3.855551940133015,3.761885864068922), NE * labelscalefactor); 
dot(B_2,dotstyle); 
label(" $B_2$ ", (10.19052187145104,9.463358802255147), NE * labelscalefactor); 
dot(A_2,dotstyle); 
label(" $A_2$ ", (11.80066006232771,4.659339937672310), NE * labelscalefactor); 
dot(B_1,dotstyle); 
label(" $B_1$ ", (1.981456668784765,10.09685579538695), NE * labelscalefactor); 
dot(A_1,dotstyle); 
label(" $A_1$ ", (0.08096568938935705,3.973051528446190), NE * labelscalefactor); 
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);[/asy]

Answer: $74$
The diagram below shows nine points on a circle where  $AB=BC=CD=DE=EF=FG=GH$ .  Given that  $\angle GHJ=117^\circ$  and  $\overline{BH}$  is perpendicular to  $\overline{EJ}$ , there are relatively prime positive integers  $m$  and  $n$  so that the degree measure of  $\angle AJB$  is  $\frac{m}{n}$ .  Find  $m	imes n$ .

[asy]
size(175);
defaultpen(linewidth(0.6));
draw(unitcircle,linewidth(0.9));
string labels[] = {"A","B","C","D","E","F","G"};
int start=110,increment=20;
pair J=dir(210),x[],H=dir(start-7*increment);
for(int i=0;i<=6;i=i+1)
{
x[i]=dir(start-increment*i);
draw(J--x[i]--H);
dot(x[i]);
label(" $"+labels[i]+"$ ",x[i],dir(origin--x[i]));
}
draw(J--H);
dot(H^^J);
label(" $H$ ",H,dir(origin--H));
label(" $J$ ",J,dir(origin--J));
[/asy]

Answer: $54$
In the following diagram,  $O$  is the center of the circle. If three angles  $\alpha, \beta$  and  $\gamma$  be equal, find the degree of angle $\gamma.$

[asy]
unitsize(40);
import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen ttttff = rgb(0.2,0.2,1); pen ffttww = rgb(1,0.2,0.4); pen qqwuqq = rgb(0,0.39,0);
draw(circle((0,0),2.33),ttttff+linewidth(2.8pt)); draw((-1.95,-1.27)--(0.64,2.24),ffttww+linewidth(2pt)); draw((0.64,2.24)--(1.67,-1.63),ffttww+linewidth(2pt)); draw((-1.95,-1.27)--(1.06,0.67),ffttww+linewidth(2pt)); draw((1.67,-1.63)--(-0.6,0.56),ffttww+linewidth(2pt)); draw((-0.6,0.56)--(1.06,0.67),ffttww+linewidth(2pt)); pair parametricplot0_cus(real t){
return (0.6*cos(t)+0.64,0.6*sin(t)+2.24);
}
draw(graph(parametricplot0_cus,-2.2073069497794027,-1.3111498158746024)--(0.64,2.24)--cycle,qqwuqq); pair parametricplot1_cus(real t){
return (0.6*cos(t)+-0.6,0.6*sin(t)+0.56);
}
draw(graph(parametricplot1_cus,0.06654165390165974,0.9342857038103908)--(-0.6,0.56)--cycle,qqwuqq); pair parametricplot2_cus(real t){
return (0.6*cos(t)+-0.6,0.6*sin(t)+0.56);
}
draw(graph(parametricplot2_cus,-0.766242589858673,0.06654165390165967)--(-0.6,0.56)--cycle,qqwuqq);
dot((0,0),ds); label(" $O$ ", (-0.2,-0.38), NE*lsf); dot((0.64,2.24),ds); label(" $A$ ", (0.72,2.36), NE*lsf); dot((-1.95,-1.27),ds); label(" $B$ ", (-2.2,-1.58), NE*lsf); dot((1.67,-1.63),ds); label(" $C$ ", (1.78,-1.96), NE*lsf); dot((1.06,0.67),ds); label(" $E$ ", (1.14,0.78), NE*lsf); dot((-0.6,0.56),ds); label(" $D$ ", (-0.92,0.7), NE*lsf); label(" $\alpha$ ", (0.48,1.38),NE*lsf); label(" $\beta$ ", (-0.02,0.94),NE*lsf); label(" $\gamma$ ", (0.04,0.22),NE*lsf); clip((-8.84,-9.24)--(-8.84,8)--(11.64,8)--(11.64,-9.24)--cycle);
[/asy]

Answer: $50$
Square  $ABCD$  is divided into four rectangles by  $EF$  and  $GH$ .  $EF$  is parallel to  $AB$  and  $GH$  parallel to  $BC$ .   $\angle BAF = 18^\circ$ .  $EF$  and  $GH$  meet at point  $P$ . The area of rectangle  $PFCH$  is twice that of rectangle  $AGPE$ . Given that the value of  $\angle FAH$  in degrees is  $3y$ , find the nearest integer to  $y$ .

[asy]
size(100); defaultpen(linewidth(0.7)+fontsize(10)); 
pair D2(pair P) {
dot(P,linewidth(3)); return P;
}
pair A=(0,1), B=(0,0), C=(1,0), D=(1,1), F=intersectionpoints(A--A+2*dir(-76),B--C)[0], H=intersectionpoints(A--A+2*dir(-76+55),D--C)[0], E=F+(0,1), G=H-(1,0), P=intersectionpoints(E--F,G--H)[0];
draw(A--B--C--D--cycle);
draw(F--A--H); draw(E--F); draw(G--H); 
label(" $A$ ",D2(A),NW);
label(" $B$ ",D2(B),SW);
label(" $C$ ",D2(C),SE);
label(" $D$ ",D2(D),NE);
label(" $E$ ",D2(E),plain.N);
label(" $F$ ",D2(F),S);
label(" $G$ ",D2(G),W);
label(" $H$ ",D2(H),plain.E);
label(" $P$ ",D2(P),SE);
[/asy]

Answer: $15$
In the following diagram (not to scale),  $A$ ,  $B$ ,  $C$ ,  $D$  are four consecutive vertices of an 18-sided regular polygon with center  $O$ . Let  $P$  be the midpoint of  $AC$  and  $Q$  be the midpoint of  $DO$ . Find the degree of $\angle OPQ$.

[asy]
pathpen = rgb(0,0,0.6)+linewidth(0.7); pointpen = black+linewidth(3); pointfontpen = fontsize(10); pen dd = rgb(0,0,0.6)+ linewidth(0.7) + linetype("4 4"); real n = 10, start = 360/n*6-15;
pair O=(0,0), A=dir(start), B=dir(start+360/n), C=dir(start+2*360/n), D=dir(start+3*360/n), P=(A+C)/2, Q=(O+D)/2; D(D("O",O,NE)--D("A",A,W)--D("B",B,SW)--D("C",C,S)--D("D",D,SE)--O--D("P",P,1.6*dir(95))--D("Q",Q,NE)); D(A--C); D(A--(A+dir(start-360/n))/2, dd); D(D--(D+dir(start+4*360/n))/2, dd);
[/asy]

Answer: $30$
In the diagram below  $ \angle CAB, \angle CBD$ , and  $\angle CDE$  are all right angles with side lengths  $AC = 3$ ,  $BC = 5$ ,  $BD = 12$ , and  $DE = 84$ . The distance from point  $E$  to the line  $AB$  can be expressed as the ratio of two relatively prime positive integers,  $m$  and  $n$ . Find  $m	imes n$ .

[asy]
size(300);
defaultpen(linewidth(0.8));
draw(origin--(3,0)--(0,4)--cycle^^(0,4)--(6,8)--(3,0)--(30,-4)--(6,8));
label(" $A$ ",origin,SW);
label(" $B$ ",(0,4),dir(160));
label(" $C$ ",(3,0),S);
label(" $D$ ",(6,8),dir(80));
label(" $E$ ",(30,-4),E);[/asy]

Answer: $346320$
As in the following diagram, square  $ABCD$  and square  $CEFG$  are placed side by side. If  $CE = 18$ ,  $AB > 19$ , compute the minimal area of  $\triangle AEG$ .

[asy]
size(120); defaultpen(linewidth(0.7)+fontsize(10)); 
pair D2(real x, real y) {
 pair P = (x,y);
 dot(P,linewidth(3)); return P;
}
int big = 30, small = 14;
filldraw((0,big)--(big+small,0)--(big,small)--cycle, rgb(0.9,0.5,0.5));
draw(scale(big)*unitsquare); draw(shift(big,0)*scale(small)*unitsquare);
label(" $A$ ",D2(0,big),NW);
label(" $B$ ",D2(0,0),SW);
label(" $C$ ",D2(big,0),SW);
label(" $D$ ",D2(big,big),N);
label(" $E$ ",D2(big+small,0),SE);
label(" $F$ ",D2(big+small,small),NE);
label(" $G$ ",D2(big,small),NE);
[/asy]

Answer: $162$
In the diagram, the circle has radius  $\sqrt{\frac{31}{3}}$  and and centre  $O.$  Points  $A, B$  and  $C$  are on the circle. If  $\angle BOC=120^\circ$  and  $AC = AB + 1,$  determine the length of  $AB.$

[asy]
import graph; size(120); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttff = rgb(0,0.2,1); pen xdxdff = rgb(0.49,0.49,1); pen fftttt = rgb(1,0.2,0.2);
draw(circle((2.34,2.4),2.01),qqttff); draw((2.34,2.4)--(1.09,0.82),fftttt); draw((2.34,2.4)--(4.1,1.41),fftttt); draw((1.09,0.82)--(1.4,4.18),fftttt); draw((4.1,1.41)--(1.4,4.18),fftttt);
dot((2.34,2.4),ds); label(" $O$ ", (2.1,2.66),NE*lsf); dot((1.09,0.82),ds); label(" $B$ ", (0.86,0.46),NE*lsf); dot((4.1,1.41),ds); label(" $C$ ", (4.2,1.08),NE*lsf); dot((1.4,4.18),ds); label(" $A$ ", (1.22,4.48),NE*lsf); clip((-4.34,-10.94)--(-4.34,6.3)--(16.14,6.3)--(16.14,-10.94)--cycle);
[/asy]

Answer: $5$
The diagram below shows the regular hexagon  $BCEGHJ$  surrounded by the rectangle  $ADFI$ . Let  $\theta$ be the measure of the acute angle between the side  $\overline{EG}$  of the hexagon and the diagonal of the rectangle  $\overline{AF}$ . There are relatively prime positive integers  $m$  and  $n$  so that  $\sin^2\theta = \frac{m}{n}$ . Find  $n - m$ .

[asy]
import graph; size(3.2cm); 
real labelscalefactor = 0.5; 
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps);
draw((-1,3)--(-1,2)--(-0.13,1.5)--(0.73,2)--(0.73,3)--(-0.13,3.5)--cycle); 
draw((-1,3)--(-1,2)); 
draw((-1,2)--(-0.13,1.5)); 
draw((-0.13,1.5)--(0.73,2)); 
draw((0.73,2)--(0.73,3)); 
draw((0.73,3)--(-0.13,3.5)); 
draw((-0.13,3.5)--(-1,3)); 
draw((-1,3.5)--(0.73,3.5)); 
draw((0.73,3.5)--(0.73,1.5)); 
draw((-1,1.5)--(0.73,1.5)); 
draw((-1,3.5)--(-1,1.5)); 
label(" $ A $ ",(-1.4,3.9),SE*labelscalefactor); 
label(" $ B $ ",(-1.4,3.28),SE*labelscalefactor); 
label(" $ C $ ",(-1.4,2.29),SE*labelscalefactor); 
label(" $ D $ ",(-1.4,1.45),SE*labelscalefactor); 
label(" $ E $ ",(-0.3,1.4),SE*labelscalefactor); 
label(" $ F $ ",(0.8,1.45),SE*labelscalefactor); 
label(" $ G $ ",(0.8,2.24),SE*labelscalefactor); 
label(" $ H $ ",(0.8,3.26),SE*labelscalefactor); 
label(" $ I $ ",(0.8,3.9),SE*labelscalefactor); 
label(" $ J $ ",(-0.25,3.9),SE*labelscalefactor); [/asy]

Answer: $1$
The diagram below shows an isosceles triangle with base  $42$  and height  $56$ .  Inscribed in the triangle is a square.  Find the area of the shaded regione.

[asy]
size(170);
defaultpen(linewidth(0.8));
draw((0,0)--(1,1));
pair A=(5,0),B=(-5,0),C=(0,14), invis[]={(1,2),(-1,2)};
pair intsquare[]={extension(origin,invis[0],A,C),extension(origin,invis[1],B,C)};
path triangle=A--B--C--cycle,square=(intsquare[0]--intsquare[1]--(intsquare[1].x,0)--(intsquare[0].x,0)--cycle);
fill(triangle,gray);
unfill(square);
draw(triangle^^square);
[/asy]

Answer: $600$
A square is inscribed in a circle of radius  $8$ . A quarter circle is inscribed in the square, as shown in the diagram below. Given the area of the region inside the circle but outside the quarter circle is  $n\pi$  for some positive integer  $n$ , what is  $n$ ?

[asy]
        size(5 cm);
        draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
        draw(circle((1,1),1.41));
        draw(arc((0,0),2,0,90));[/asy]

Answer: $32$
In the following diagram, let  $ABCD$  be a square and let  $M,N,P$  and  $Q$  be the midpoints of its sides. How many times of the area of square $ABCD$ compared to the area of quadrilateral $A'B'C'D'$

[asy]
import graph; size(200); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen qqttzz = rgb(0,0.2,0.6); pen qqzzff = rgb(0,0.6,1);
draw((0,4)--(4,4),qqttzz+linewidth(1.6pt)); draw((4,4)--(4,0),qqttzz+linewidth(1.6pt)); draw((4,0)--(0,0),qqttzz+linewidth(1.6pt)); draw((0,0)--(0,4),qqttzz+linewidth(1.6pt)); draw((0,4)--(2,0),qqzzff+linewidth(1.2pt)); draw((2,4)--(4,0),qqzzff+linewidth(1.2pt)); draw((0,2)--(4,4),qqzzff+linewidth(1.2pt)); draw((0,0)--(4,2),qqzzff+linewidth(1.2pt));
dot((0,4),ds); label(" $A$ ", (0.07,4.12), NE*lsf); dot((0,0),ds); label(" $D$ ", (-0.27,-0.37), NE*lsf); dot((4,0),ds); label(" $C$ ", (4.14,-0.39), NE*lsf); dot((4,4),ds); label(" $B$ ", (4.08,4.12), NE*lsf); dot((2,4),ds); label(" $M$ ", (2.08,4.12), NE*lsf); dot((4,2),ds); label(" $N$ ", (4.2,1.98), NE*lsf); dot((2,0),ds); label(" $P$ ", (1.99,-0.49), NE*lsf); dot((0,2),ds); label(" $Q$ ", (-0.48,1.9), NE*lsf); dot((0.8,2.4),ds); label(" $A'$ ", (0.81,2.61), NE*lsf); dot((2.4,3.2),ds); label(" $B'$ ", (2.46,3.47), NE*lsf); dot((3.2,1.6),ds); label(" $C'$ ", (3.22,1.9), NE*lsf); dot((1.6,0.8),ds); label(" $D'$ ", (1.14,0.79), NE*lsf); clip((-4.44,-11.2)--(-4.44,6.41)--(16.48,6.41)--(16.48,-11.2)--cycle);
[/asy]

Answer: $5$