problem_id int64 1 978 | question stringlengths 86 2.11k | source stringlengths 19 76 | solution stringlengths 94 14.7k | asymptote_code stringlengths 44 17.8k | solution_image_url stringclasses 16
values |
|---|---|---|---|---|---|
512 | Regular octagon $A_1A_2A_3A_4A_5A_6A_7A_8$ is inscribed in a circle of area $1.$ Point $P$ lies inside the circle so that the region bounded by $\overline{PA_1},\overline{PA_2},$ and the minor arc $\widehat{A_1A_2}$ of the circle has area $\tfrac{1}{7},$ while the region bounded by $\overline{PA_3},\overline{PA_4},$ an... | 2019 AIME II Problems/Problem 13 | Instead of considering the actual values of the areas, consider only the changes in the areas that result from moving point $P$ from the center of the circle. We will proceed by coordinates. Set the origin at the center of the circle and refer to the following diagram, where the octagon is oriented so as $A_1A_2$ is ho... | // Block 1
size(7cm);
draw(Circle((0,0),1));
pair P = (0.1,-0.15);
filldraw(P--dir(112.5)--dir(112.5-45)--cycle,yellow,red);
filldraw(P--dir(112.5-90)--dir(112.5-135)--cycle,yellow,red);
filldraw(P--dir(112.5-225)--dir(112.5-270)--cycle,green,red);
dot(P);
for(int i=0; i<8; ++i)
{
draw(dir(22.5+45i)--dir(67.5+45i)... | [] |
513 | Triangle $ABC$ has side lengths $AB=120,BC=220$, and $AC=180$. Lines $\ell_A,\ell_B$, and $\ell_C$ are drawn parallel to $\overline{BC},\overline{AC}$, and $\overline{AB}$, respectively, such that the intersections of $\ell_A,\ell_B$, and $\ell_C$ with the interior of $\triangle ABC$ are segments of lengths $55,45$, an... | 2019 AIME II Problems/Problem 7 | Let's squish a triangle with side lengths 15, 22.5, and 27.5 into a equilateral triangle with side length 1. Then, the original triangle gets turned into a equilateral triangle with side length 8. Since 15 is one eighth of 120, it has a length of one. Since 45 and 55 are one fourth of 180 and 220 respectively, they are... | // Block 1
pair A,B,C; B = (0,0); C = (1,0); A = intersectionpoints(circle(B,3/2),circle(C,11/6))[0]; draw(A--B--C--cycle); draw((3/2,3/4)--(5/2,3/4)); draw((3/2,1/4)--(5/2,1/4)); draw((9/4,1)--(11/4,1/2)--(9/4,0)); draw(shift(dir(0)*13/4)*shift(dir(30))*polygon(3));
// Block 2
for (int i=0; i<8; ++i) { for (int j=0; ... | [] |
514 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | 2020 AIME I Problems/Problem 13 | Let $M_A$, $M_B$, $M_C$ be the midpoints of arcs $BC$, $CA$, $AB$. By Fact 5, we know that $M_AB = M_AC = M_AI$, and so by Ptolmey's theorem, we deduce that \[AB\cdot M_AC + AC\cdot M_AB = BC\cdot M_AA \implies M_AA = 2M_AI.\]
In particular, we have $AI = IM_A$.
Now the key claim is that:
Claim: $\triangle DEF$ and... | // Block 1
defaultpen(fontsize(10pt));
size(200);
pair A, B, C, D, E, F, I, P, MA, MB, MC;
B = (0,0);
C = (5,0);
A = IP(Circle(B, 4), Circle(C, 6), 0);
I = incenter(A, B, C);
D = extension(A, I, B, C);
P = midpoint(A--D);
E = extension(P, rotate(90, P)*A, B, I);
F = extension(P, rotate(90, P)*A, C, I);
MA = IP(Line(A, ... | [] |
514 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | 2020 AIME I Problems/Problem 13 | Let $B=(0,0)$ and $BC$ be the line $y=0$.
We compute that $\cos{\angle{ABC}}=\frac{1}{8}$, so $\tan{\angle{ABC}}=3\sqrt{7}$.
Thus, $A$ lies on the line $y=3x\sqrt{7}$. The length of $AB$ at a point $x$ is $8x$, so $x=\frac{1}{2}$.
We now have the coordinates $A=\left(\frac{1}{2},\frac{3\sqrt{7}}{2}\right)$, $B=(0,0)$ ... | // Block 1
size(8cm); defaultpen(fontsize(10pt));
pair A,B,C,I,D,M,T,Y,Z,EE,F;
A=(0,3sqrt(7));
B=(-1,0);
C=(9,0);
I=incenter(A,B,C);
D=extension(A,I,B,C);
M=(A+D)/2;
draw(B--EE,gray+dashed);
draw(C--F,gray+dashed);
draw(A--B--C--A);
... | [] |
514 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | 2020 AIME I Problems/Problem 13 | As usual, we will use homogenized barycentric coordinates.
We have that $AD$ will have form $3z=2y$. Similarly, $CF$ has form $5y=6x$ and $BE$ has form $5z=4x$.
Since $A=(1,0,0)$ and $D=\left(0,\frac{3}{5},\frac{2}{5}\right)$, we also have $M=\left(\frac{1}{2},\frac{3}{10},\frac{1}{5}\right)$.
It remains to determine ... | // Block 1
size(8cm); defaultpen(fontsize(10pt));
pair A,B,C,I,D,M,T,Y,Z,EE,F;
A=(0,3sqrt(7));
B=(-1,0);
C=(9,0);
I=incenter(A,B,C);
D=extension(A,I,B,C);
M=(A+D)/2;
draw(B--EE,gray+dashed);
draw(C--F,gray+dashed);
draw(A--B--C--A);
... | [] |
514 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | 2020 AIME I Problems/Problem 13 | To get the area of $\triangle AEF$, we try to find $AM$ and $\angle EAF$.
Since $AD$ is the angle bisector, we can get that $BD=2$ and $CD=3$. By applying Stewart's Theorem, we can get that $AD=3\sqrt{2}$. Therefore $AM=\frac{3\sqrt{2}}{2}$.
Since $EF$ is the perpendicular bisector of $AD$, we know that $AE = DE$. Si... | // Block 1
size(8cm); defaultpen(fontsize(10pt));
pair A,B,C,I,D,M,T,Y,Z,EE,F;
A=(0,3sqrt(7));
B=(-1,0);
C=(9,0);
I=incenter(A,B,C);
D=extension(A,I,B,C);
M=(A+D)/2;
draw(B--EE,gray+dashed);
draw(C--F,gray+dashed);
draw(A--B--C--A);
... | [] |
514 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | 2020 AIME I Problems/Problem 13 | Let $x = \angle BAD = \angle CAD$, $y = \angle CBE = \angle ABE$, and $z = \angle BCF = \angle ACF$. Notice that $x+y+z = 90^\circ$.
In $\triangle ABD$, segment $\overline{BE}$ is the bisector of $\angle ABD$, and $E$ lies on the perpendicular bisector of side $\overline{AD}$. Therefore $E$ is the midpoint of arc $\st... | // Block 1
unitsize(0.8 cm);
pair A, B, C, D, E, F, I;
real angleC = aCos(3/4);
C = (0,0);
A = 6*dir(270 - angleC/2);
B = 5*dir(270 + angleC/2);
I = incenter(A,B,C);
D = extension(A, I, B, C);
E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I);
F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C, I... | [] |
514 | Point $D$ lies on side $\overline{BC}$ of $\triangle ABC$ so that $\overline{AD}$ bisects $\angle BAC.$ The perpendicular bisector of $\overline{AD}$ intersects the bisectors of $\angle ABC$ and $\angle ACB$ in points $E$ and $F,$ respectively. Given that $AB=4,BC=5,$ and $CA=6,$ the area of $\triangle AEF$ can be writ... | 2020 AIME I Problems/Problem 13 | Let the point $M$ be the midpoint of $\overline{AD}$, let $I$ be the incenter of $\triangle ABC$ which is the common point of lines $AD$, $BE$, and $CF$, and let $r$ be the inradius of $\triangle ABC$. The semiperimeter of $\triangle ABC$ is\[s = \frac{AB + BC + CA}2 = \frac{15}2,\]and Heron's Formula gives the area of... | // Block 1
unitsize(0.8 cm);
pair A, B, C, D, E, F, I, M;
real angleC = aCos(3/4);
C = (0,0);
A = 6*dir(270 - angleC/2);
B = 5*dir(270 + angleC/2);
I = incenter(A,B,C);
D = extension(A, I, B, C);
E = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), B, I);
F = extension((A + D)/2, (A + D)/2 + rotate(90)*(A - D), C... | [] |
515 | Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written i... | 2020 AIME I Problems/Problem 15 | The following is a power of a point solution to this menace of a problem:
Let points be what they appear as in the diagram below. Note that $3HX = HY$ is not insignificant; from here, we set $XH = HE = \frac{1}{2} EY = HL = 2$ by PoP and trivial construction. Now, $D$ is the reflection of $A$ over $H$. Note $AO \perp... | // Block 1
defaultpen(fontsize(12)+0.6); size(250);
pen p=fontsize(10)+gray+0.4;
var phi=75.5, theta=130, r=4.8;
pair A=r*dir(270-phi-57), C=r*dir(270+phi-57), B=r*dir(theta-57), H=extension(A,foot(A,B,C),B,foot(B,C,A));
path omega=circumcircle(A,B,C), c=circumcircle(H,B,C);
pair O=circumcenter(H,B,C), Op=2*H-O, Z=bis... | [] |
515 | Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written i... | 2020 AIME I Problems/Problem 15 | Diagram not to scale.
We first observe that $H'$, the image of the reflection of $H$ over line $BC$, lies on circle $O$. This is because $\angle HBC = 90 - \angle C = \angle H'AC = \angle H'BC$. This is a well known lemma. The result of this observation is that circle $O'$, the circumcircle of $\triangle BHC$ is ... | // Block 1
size(10cm);
pair A, B, C, D, H, K, O, P, L, M, X, Y;
A = (-15, 27);
B = (-24, 0);
C = (24, 0);
D = (-8.28, 18.04);
O = (0, 7);
P = (0, -7);
H = (-15, 13);
K = (-15, -13);
M = (0, 0);
L = (-15, 0);
X = (-24.9569, 5.53234);
Y = (8.39688, 30.5477);
draw(circle(O, 25));
draw(circle(P, 25));
draw(A--B--C--cycle);... | [] |
515 | Let $\triangle ABC$ be an acute triangle with circumcircle $\omega,$ and let $H$ be the intersection of the altitudes of $\triangle ABC.$ Suppose the tangent to the circumcircle of $\triangle HBC$ at $H$ intersects $\omega$ at points $X$ and $Y$ with $HA=3,HX=2,$ and $HY=6.$ The area of $\triangle ABC$ can be written i... | 2020 AIME I Problems/Problem 15 | Extend $\overline{AH}$ to intersect $\omega$ again at $P$. The Power of a Point Theorem yields $HP = \tfrac{HX \cdot HY}{HA} = 4$. Because $\angle CAP=\angle CBP$, and $\angle CAP$ and $\angle CBH$ are both complements to $\angle C$, it follows that $\angle CBP = \angle CBH$, implying that $\overline{BC}$ bisects $\ove... | // Block 1
unitsize(0.6 cm);
pair A, B, C, D, H, M, O, Op, P, T, X, Y, Z;
real R = 21/(2*sqrt(5));
A = (7/sqrt(5),7/2);
O = (0,0);
B = intersectionpoint((0,-3/2)--(R,-3/2),Circle(O,R));
C = intersectionpoint((0,-3/2)--(-R,-3/2),Circle(O,R));
H = A + B + C;
P = reflect(B,C)*(H);
D = (H + P)/2;
Op = reflect(B,C)*(O);
X... | [] |
516 | Convex pentagon $ABCDE$ has side lengths $AB=5$, $BC=CD=DE=6$, and $EA=7$. Moreover, the pentagon has an inscribed circle (a circle tangent to each side of the pentagon). Find the area of $ABCDE$. | 2020 AIME II Problems/Problem 13 | Let $\omega$ be the inscribed circle, $I$ be its center, and $r$ be its radius. The area of $ABCDE$ is equal to its semiperimeter, $15,$ times $r$, so the problem is reduced to finding $r$. Let $a$ be the length of the tangent segment from $A$ to $\omega$, and analogously define $b$, $c$, $d$, and $e$. Then $a+b=5$, $b... | // Block 1
defaultpen(fontsize(8pt));
unitsize(0.025cm);
pair[] vertices = {(0,0), (5,0), (8.6,4.8), (3.8,8.4), (-1.96, 6.72)};
string[] labels = {"$A$", "$B$", "$C$", "$D$", "$E$"};
pair[] dirs = {SW, SE,E, N, NW};
string[] smallLabels = {"$a$", "$b$", "$c$", "$d$", "$e$"};
pair I = (3,4);
real rad = 4;
pair Q =... | [] |
517 | Let $\triangle ABC$ be an acute scalene triangle with circumcircle $\omega$. The tangents to $\omega$ at $B$ and $C$ intersect at $T$. Let $X$ and $Y$ be the projections of $T$ onto lines $AB$ and $AC$, respectively. Suppose $BT = CT = 16$, $BC = 22$, and $TX^2 + TY^2 + XY^2 = 1143$. Find $XY^2$. | 2020 AIME II Problems/Problem 15 | Let $M$ denote the midpoint of $\overline{BC}$. The critical claim is that $M$ is the orthocenter of $\triangle AXY$, which has the circle with diameter $\overline{AT}$ as its circumcircle. To see this, note that because $\angle BXT = \angle BMT = 90^\circ$, the quadrilateral $MBXT$ is cyclic, it follows that
\[\angle ... | // Block 1
defaultpen(fontsize(8pt));
unitsize(0.8cm);
pair A = (0,0);
pair B = (-1.26,-4.43);
pair C = (-1.26+3.89, -4.43);
pair M = (B+C)/2;
pair O = circumcenter(A,B,C);
pair T = (0.68, -6.49);
pair X = foot(T,A,B);
pair Y = foot(T,A,C);
path omega = circumcircle(A,B,C);
real rad = circumradius(A,B,C);
filld... | [] |
518 | Let $P$ be a point chosen uniformly at random in the interior of the unit square with vertices at $(0,0), (1,0), (1,1)$, and $(0,1)$. The probability that the slope of the line determined by $P$ and the point $\left(\frac58, \frac38 \right)$ is greater than or equal to $\frac12$ can be written as $\frac{m}{n}$, where $... | 2020 AIME II Problems/Problem 2 | The line through the fixed point $\left(\frac58,\frac38\right)$ with slope $\frac12$ has equation $y=\frac12 x + \frac1{16}$. The slope between $P$ and the fixed point exceeds $\frac12$ if $P$ falls within the shaded region in the diagram below consisting of two trapezoids with area
\[\frac{\frac1{16}+\frac38}2\cdot\fr... | // Block 1
defaultpen(fontsize(8pt));
unitsize(4cm);
pair A = (0,0);
pair B = (1,0);
pair C = (1,1);
pair D = (0,1);
pair F = (0, 1/16);
pair G = (1, 9/16);
pair H = (5/8, 0);
pair J = (5/8, 1);
pair K = IP(H--J, F--G);
pair P = (13/16, 12/16);
pair Q = extension(P,K,A,B);
pair R = extension(K,P,C,D);
draw... | [] |
519 | Triangles $\triangle ABC$ and $\triangle A'B'C'$ lie in the coordinate plane with vertices $A(0,0)$, $B(0,12)$, $C(16,0)$, $A'(24,18)$, $B'(36,18)$, $C'(24,2)$. A rotation of $m$ degrees clockwise around the point $(x,y)$ where $0<m<180$, will transform $\triangle ABC$ to $\triangle A'B'C'$. Find $m+x+y$. | 2020 AIME II Problems/Problem 4 | We first draw a diagram with the correct Cartesian coordinates and a center of rotation $P$. Note that $PC=PC'$ because $P$ lies on the perpendicular bisector of $CC'$ (it must be equidistant from $C$ and $C'$ by properties of a rotation).
Since $AB$ is vertical while $A'B'$ is horizontal, we have that the angle of r... | // Block 1
/* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* poin... | [] |
520 | Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$, w... | 2020 AIME II Problems/Problem 7 | Consider the cross section of the cones and sphere by a plane that contains the two axes of symmetry of the cones as shown below. The sphere with maximum radius will be tangent to the sides of each of the cones. The center of that sphere must be on the axis of symmetry of each of the cones and thus must be at the inter... | // Block 1
unitsize(0.6cm);
pair A = (0,0);
pair TriangleOneLeft = (-6,0);
pair TriangleOneDown = (-3,-8);
pair TriangleOneMid = (-3,0);
pair D = (0,-3);
pair TriangleTwoDown = (0,-6);
pair B = (-8,-3);
pair C = IP(TriangleOneMid -- TriangleOneDown, B--D);
pair EE = foot(C, A, B);
real radius = arclength(C--EE)... | [] |
520 | Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$, w... | 2020 AIME II Problems/Problem 7 | Let $A$ be the origin in the above diagram. Then $B$ is $(6,0)$, $C$ is $(0,6)$, $D$ is $(3,8)$, and $E$ is $(8,3)$. Also, it is easy to see that the inscribed sphere is simply the inscribed circle of $AEF$. Then we want to find the intersection of $AD$ and $CE$ to determine the coordinates of point $F$. Note that $AD... | // Block 1
unitsize(0.6cm);
// Coordinates
pair A = (0,0);
pair B = (6,0);
pair C = (0,6);
// Calculate point C
pair D = (3,8);
pair E = (8,3);
pair F = (144/73,384/73);
// Draw triangles (cones)
draw(A--B--D--cycle);
draw(A--C--E--cycle);
draw(incircle(A,E,F));
pair EE = foot(C, A, B);
real radius = arclength(C--E... | [] |
520 | Two congruent right circular cones each with base radius $3$ and height $8$ have the axes of symmetry that intersect at right angles at a point in the interior of the cones a distance $3$ from the base of each cone. A sphere with radius $r$ lies within both cones. The maximum possible value of $r^2$ is $\frac{m}{n}$, w... | 2020 AIME II Problems/Problem 7 | We graph this on graph paper, with the scale of $\sqrt{2}:1$. So, we can find $OT$ then divide by $\sqrt{2}$ to convert to our desired units, then square the result. With 5 minutes' worth of coordbashing, we finally arrive at $298$.
~samrocksnature | // Block 1
/* Geogebra to Asymptote conversion by samrocksnature, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black; /* poin... | [] |
521 | In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. ... | 2020 AMC 10B Problems/Problem 21 | Easily, we can find that: quadrilateral $BFIE$ and $DHJG$ are congruent with each other, so we can move $DHJG$ to the shaded area ($F$ and $G$, $B$ and $D$ overlapping) to form a square $FIKJ'$ ($DG$ = $FB$, $CG$ = $FC$, ${\angle} CGF$ = ${\angle}CFG$ = $45^{\circ}$ so ${\angle} IFJ'= 90^{\circ}$). Then we can solve $A... | // Block 1
real x=2sqrt(2);
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);
real z=2sqrt(8-4sqrt(2));
real k= 8-2sqrt(2);
real l= 2sqrt(2)-4;
pair A, B, C, D, E, F, G, H, I, J, L, M, K;
A = (0,0);
B = (4,0);
C = (4,4);
D = (0,4);
E = (x,0);
F = (4,y);
G = (y,4);
H = (0,x);
I = F + z * dir(225);
L = (k,0);
M = F + z * dir(315);
K... | [] |
521 | In square $ABCD$, points $E$ and $H$ lie on $\overline{AB}$ and $\overline{DA}$, respectively, so that $AE=AH.$ Points $F$ and $G$ lie on $\overline{BC}$ and $\overline{CD}$, respectively, and points $I$ and $J$ lie on $\overline{EH}$ so that $\overline{FI} \perp \overline{EH}$ and $\overline{GJ} \perp \overline{EH}$. ... | 2020 AMC 10B Problems/Problem 21 | $[ABCD] = 4$, $AB = 2$, $[AHE] = 1$, $AH = AE = \sqrt{2}$, $DH = 2 - \sqrt{2}$, $JL = HK = \sqrt{2} \cdot DH = 2 \sqrt{2} - 2$
Because $ABCD$ is a square and $AH = AE$, $AC$ is the line of symmetry of pentagon $CDHEB$. Because $[DHJG] = [BFIE]$, $DHJG$ is the reflection of $BFIE$ about line $AC$
Let $FI = GJ = x$, $K... | // Block 1
real x=2sqrt(2);
real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);
real z=2sqrt(8-4sqrt(2));
pair A, B, C, D, E, F, G, H, I, J, K, L;
A = (0,0);
B = (4,0);
C = (4,4);
D = (0,4);
E = (x,0);
F = (4,y);
G = (y,4);
H = (0,x);
I = F + z * dir(225);
J = G + z * dir(225);
K = (4-x,4);
L = J + 1.68 * dir(45);
draw(A--B--C--D--... | [] |
522 | Let $ABCD$ be a cyclic quadrilateral with $AB=6, AC=8, BD=5, CD=2$. Let $P$ be the point on $\overline{AD}$ such that $\angle APB = \angle CPD$. Then $\frac{BP}{CP}$ can be expressed in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$. | 2020 CIME I Problems/Problem 9 | Let $C'$ be the reflection of $C$ over line $AD$. Since $\angle APB = \angle CPD = \angle C'PD$, $B, P, C$ are collinear. Suppose $X$ and $Y$ are the projections of $B$ and $C$ onto line $AD$, respectively. We want to find $\frac{BP}{CP}$ which by similar triangles is also equal to $\frac{BX}{C'Y}$ from $\triangle BPX ... | // Block 1
size(4cm);
/* draw figures */
draw((0,0)--(1,0));
draw(circle((0.5,-0.12046156424028276), 0.5143063177321622));
draw((0.29472641365670604,0.35110364144073225)--(0,0));
draw((0.29472641365670604,0.35110364144073225)--(0.7999672347109121,0.297307087718076));
draw((0.7999672347109121,0.297307087718076)--(1... | [] |
523 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120... | 2021 AIME I Problems/Problem 13 | Let $O_i$ and $r_i$ be the center and radius of $\omega_i$, and let $O$ and $r$ be the center and radius of $\omega$.
Since $\overline{AB}$ extends to an arc with arc $120^\circ$, the distance from $O$ to $\overline{AB}$ is $r/2$. Let $X=\overline{AB}\cap \overline{O_1O_2}$. Consider $\triangle OO_1O_2$. The line $\ov... | // Block 1
unitsize(3cm);
pointpen=black; pointfontpen=fontsize(9);
pair A=dir(110), B=dir(230), C=dir(310);
DPA(A--B--C--A);
pair H = foot(A, B, C);
draw(A--H);
pair X = 0.3*B + 0.7*C;
pair Y = A+X-H;
draw(X--1.3*Y-0.3*X);
draw(A--Y, dotted);
pair R1 = 1.3*X-0.3*Y;
pair R2 = 0.7*X+0.3*Y;
draw(R1--X);
D("O",A... | [] |
523 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120... | 2021 AIME I Problems/Problem 13 | Denote by $O_1$, $O_2$, and $O$ the centers of $\omega_1$, $\omega_2$, and $\omega$, respectively. Let $R_1 = 961$ and $R_2 = 625$ denote the radii of $\omega_1$ and $\omega_2$ respectively, $r$ be the radius of $\omega$, and $\ell$ the distance from $O$ to the line $AB$. We claim that\[\dfrac{\ell}{r} = \dfrac{R_2-R_1... | // Block 1
import olympiad;
size(230pt);
defaultpen(linewidth(0.8)+fontsize(10pt));
real r1 = 17, r2 = 27, d = 35, r = 18;
pair O1 = origin, O2 = (d,0);
path w1 = circle(origin,r1), w2 = circle((d,0),r2), w1p = circle(origin,r1+r), w2p = circle((d,0), r2 + r);
pair[] X = intersectionpoints(w1,w2), Y = intersectionpoint... | [] |
523 | Circles $\omega_1$ and $\omega_2$ with radii $961$ and $625$, respectively, intersect at distinct points $A$ and $B$. A third circle $\omega$ is externally tangent to both $\omega_1$ and $\omega_2$. Suppose line $AB$ intersects $\omega$ at two points $P$ and $Q$ such that the measure of minor arc $\widehat{PQ}$ is $120... | 2021 AIME I Problems/Problem 13 | Suppose we label the points as shown below.
By radical axis, the tangents to $\omega$ at $D$ and $E$ intersect on $AB$. Thus $PDQE$ is harmonic, so the tangents to $\omega$ at $P$ and $Q$ intersect at $X \in DE$. Moreover, $OX \parallel O_1O_2$ because both $OX$ and $O_1O_2$ are perpendicular to $AB$, and $OX = 2OP$ ... | // Block 1
defaultpen(fontsize(12)+0.6); size(300);
pen p=fontsize(10)+royalblue+0.4;
var r=1200;
pair O1=origin, O2=(672,0), O=OP(CR(O1,961+r),CR(O2,625+r));
path c1=CR(O1,961), c2=CR(O2,625), c=CR(O,r);
pair
A=IP(CR(O1,961),CR(O2,625)), B=OP(CR(O1,961),CR(O2,625)),
P=IP(L(A,B,0,0.2),c),
Q=IP(L(A,B,0,200),c),
F=IP... | [] |
524 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | 2021 AIME I Problems/Problem 9 | Let $\overline{AE}, \overline{AF},$ and $\overline{AG}$ be the perpendiculars from $A$ to $\overleftrightarrow{BC}, \overleftrightarrow{CD},$ and $\overleftrightarrow{BD},$ respectively. Next, let $H$ be the intersection of $\overline{AF}$ and $\overline{BD}.$
We set $AB=x$ and $AH=y,$ as shown below.
From here, we o... | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair A, B, C, D, E, F, G, H;
A = (-45sqrt(2)/8,18);
B = (45sqrt(2)/8,18);
C = (81sqrt(2)/8,0);
D = (-81sqrt(2)/8,0);
E = foot(A,C,B);
F = foot(A,C,D);
G = foot(A,B,D);
H = intersectionpoint(A--F,B--D);
markscalefactor=0.1;
draw(rightanglemark(A,E,B),red);
draw(rightangle... | [] |
524 | Let $ABCD$ be an isosceles trapezoid with $AD=BC$ and $AB<CD.$ Suppose that the distances from $A$ to the lines $BC,CD,$ and $BD$ are $15,18,$ and $10,$ respectively. Let $K$ be the area of $ABCD.$ Find $\sqrt2 \cdot K.$ | 2021 AIME I Problems/Problem 9 | Let the points formed by dropping altitudes from $A$ to the lines $BC$, $CD$, and $BD$ be $E$, $F$, and $G$, respectively.
We have
\[\triangle ABE \sim \triangle ADF \implies \frac{AD}{18} = \frac{AB}{15} \implies AD = \frac{6}{5}AB\]
and
\[BD\cdot10 = 2[ABD] = AB\cdot18 \implies BD = \frac{9}{5}AB.\]
For convenience,... | // Block 1
size(250);
pair A, B, C, D, E, F, G, H;
A = (-45sqrt(2)/8,18);
B = (45sqrt(2)/8,18);
C = (81sqrt(2)/8,0);
D = (-81sqrt(2)/8,0);
E = foot(A,C,B);
F = foot(A,C,D);
G = foot(A,B,D);
H = intersectionpoint(A--F,B--D);
markscalefactor=0.1;
draw(rightanglemark(A,E,B),red);
draw(rightanglemark(A,F,C),red);
draw(righ... | [] |
525 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime... | 2021 AMC 12A Problems/Problem 24 | Let $O=\Gamma$ be the center of the semicircle and $X=\Omega$ be the center of the circle.
Applying the Extended Law of Sines to $\triangle PQR,$ we find the radius of $\odot X:$ \[XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3.\]
Alternatively, by the Inscribed Angle Theorem, $\triangle... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair O, X, A, B, P, Q, R, M, C, D;
O = (0,0);
X = (4,3);
A = (-7,0);
B = (7,0);
P = (4,0);
Q = intersectionpoints(Circle(O,7),Circle(X,3))[0];
R = intersectionpoints(Circle(O,7),Circle(X,3))[1];
M = midpoint(Q--R);
C = foot(P,Q,R);
D = foot(X,P,C);
fill(P--Q--R--cycle,ye... | [] |
525 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime... | 2021 AMC 12A Problems/Problem 24 | Suppose we label the points as shown in the diagram above, where $C$ is the center of the semicircle and $O$ is the center of the circle tangent to $\overline{AB}$. Since $\angle QPR = 60^{\circ}$, we have $\angle QOR = 2\cdot 60^{\circ}=120^{\circ}$ and $\triangle QOR$ is a $30-30-120$ triangle, which can be split int... | // Block 1
size(150);
draw(circle((7,0),7));
pair A = (0, 0);
pair B = (14, 0);
draw(A--B);
draw(circle((11,3),3));
label("$C$", (7, 0), S);
label("$O$", (11, 3), E);
label("$P$", (11, 0), S);
pair C = (7, 0);
pair O = (11, 3);
pair P = (11, 0);
pair Q = intersectionpoints(circle(C, 7), circle(O, 3))[1];
pair R = inter... | [] |
525 | Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime... | 2021 AMC 12A Problems/Problem 24 | Define points as shown above, where $N=\overleftrightarrow{PA}\cap\overleftrightarrow{QR}$. The area of $\triangle PQR$ is simply\[\dfrac{1}{2}PX\cdot QR=\dfrac{3\sqrt{3}}{2}PX;\]
it remains to compute the value of $PX$. Note that $PX$ is simply a weighted average of $BT$ and $CS;$ it is $\dfrac{CP}{BP}$ times closer t... | size(300); pair C = (7, 0); draw(arc(C, 7, 0, 180)); pair A = (0, 0), B = (14, 0); draw(A--B); draw(circle((11,3),3)); label("$A$", A, SSE); label("$B$", B, SSW); label("$C$", (A+B)/2, S); label("$O$", (11, 3), E); label("$P$", (11, 0), S); pair O = (11, 3), P = (11, 0), Q = intersectionpoints(circle(C, 7), circle(O, ... | [] |
526 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positi... | 2021 AMC 12B Problems/Problem 25 | Based on area ratios between a square of side length $30$ and a triangle with base $30$, we estimate that the slope $m$ of the line we want is approximately $\frac{2}{3}$. Following this estimate, we see if there are approximately $30 \cdot 30 - 300 = 600$ lattice points above the line $y=\frac{2}{3}x$.
Counting the n... | /* Created by Brendanb4321 */ import graph; size(16cm); defaultpen(fontsize(9pt)); xaxis(0,30,Ticks(1.0)); yaxis(0,25,Ticks(1.0)); draw((0,0)--(30,20)); draw((0,0)--(30,30/28*19), dotted); for (int i = 1; i<=30; ++i) { for (int j = 1; j<=2/3*i+1; ++j) { dot((i,j)); } } dot((28,19), red); label("$m=2/3$", (32,20)); lab... | [] |
526 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positi... | 2021 AMC 12B Problems/Problem 25 | It's easier to calculate the number of lattice points inside a rectangle with vertices $(0,0)$, $(p,0)$, $(p,q)$, $(0,q)$. Those lattice points are divided by the diagonal $y = \frac{p}{q} \cdot x$ into $2$ halves. In this problem the number of lattice points on or below the diagonal and $x \ge 1$ is
$\frac{1}{2} [(p+... | // Block 1
/* Created by Brendanb4321, modified by isabelchen */
import graph;
size(18cm);
defaultpen(fontsize(9pt));
xaxis(0,31,Ticks(1.0));
yaxis(0,22,Ticks(1.0));
draw((0,0)--(30,20));
draw((0,0)--(30,30*19/28), dotted);
draw((0,0)--(31,31*21/31), dotted);
for (int i = 1; i<=31; ++i) {
for (int j = 1; j<=2/3*i+1; ... | [] |
526 | Let $S$ be the set of lattice points in the coordinate plane, both of whose coordinates are integers between $1$ and $30,$ inclusive. Exactly $300$ points in $S$ lie on or below a line with equation $y=mx.$ The possible values of $m$ lie in an interval of length $\frac ab,$ where $a$ and $b$ are relatively prime positi... | 2021 AMC 12B Problems/Problem 25 | The lower bound of $m$ is $\frac23 = \frac{20}{30}$. Inside the rectangle with vertices $(0,0)$, $(30,0)$, $(30,20)$, $(0, 20)$ and diagonal $y = \frac23 x$, there are $(30-1)(20-1) = 551$ lattice points inside, not including the edges. There are $9$ lattice points on diagonal $y = \frac23 x$ inside the rectangle, $551... | // Block 1
/* Created by isabelchen */
import graph;
size(8cm);
defaultpen(fontsize(9pt));
xaxis(0,8);
yaxis(0,6);
draw((0,0)--(7,5));
draw((7,0)--(7,5));
draw((0,5)--(7,5));
dot((0,0)); dot((1,0)); dot((2,0)); dot((3,0)); dot((4,0)); dot((5,0)); dot((6,0)); dot((7,0)); dot((8,0)); dot((0,1)); dot((1,1)); dot((2,1));... | [] |
527 | Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects lines $BH$ and $CH$ at $X$ and $Y$, and $BY\parallel CX$. Let line $AO$ intersect $\overline{BC}$ at $D$. Suppose that $AO=25, BC=49$, and $AD=a-b\sqrt{c}$ for positive integers... | 2021 CIME I Problems/Problem 14 | Let $H$ be the orthocenter of $\triangle ABC$, and let $E$, $F$ be the feet of the altitudes from $B, C$. Also let $A'$ be the antipode of $A$ on the circumcircle and let $S=\overline{AH}\cap\overline{EF}$, as shown below:
Disregarding the condition $\overline{BY}\parallel\overline{CX}$, we contend:
$\textbf{Claim}:... | // Block 1
size(6cm); defaultpen(fontsize(10pt)); pen pri=blue; pen sec=lightblue; pen tri=heavycyan; pen qua=paleblue; pen fil=invisible; pen sfil=invisible; pen tfil=invisible;
pair O,A,B,C,H,X,Y,EE,F,Ap,SS,D; O=(0,0); A=dir(147.55); B=dir(195); C=dir(345); H=A+B+C; X=extension(B,H,A,A+rotate(90)*A); Y=extension(C,H... | [] |
527 | Let $ABC$ be an acute triangle with orthocenter $H$ and circumcenter $O$. The tangent to the circumcircle of $\triangle ABC$ at $A$ intersects lines $BH$ and $CH$ at $X$ and $Y$, and $BY\parallel CX$. Let line $AO$ intersect $\overline{BC}$ at $D$. Suppose that $AO=25, BC=49$, and $AD=a-b\sqrt{c}$ for positive integers... | 2021 CIME I Problems/Problem 14 | Let $E$ and $F$ be the feet of the altitudes from $B$ to $\overline{AC}$ and from $C$ to $\overline{AB}$, respectively. It is clear that $\overline{EF} \parallel \overline{XY}$, as $\angle XAE = \angle AEF = \angle B$ and $\angle YAF = \angle AFE = \angle C$. We conjecture that quadrilateral $BCXY$ is cyclic because it... | // Block 1
import olympiad;
size(7cm); defaultpen(linewidth(0.7));
usepackage("mathptmx");
pair A = (-15, 20), B = (-24, -7), C = (24, -7), O = (0, 0), H = A + B + C, Ap = -A, X = extension(B, H, A, A + rotate(90, O) * A), Y = extension(C, H, A, A + rotate(90, O) * A), D = extension(A, O, B, C), E = extension(A, C, B, ... | [] |
528 | Right triangle $ABC$ has side lengths $BC=6$, $AC=8$, and $AB=10$. A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A$. A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B$. What is $OP$?
$\textbf{(A)}\ \frac{23}{8} \qquad\textbf{(B)}\ \frac{29}{10} \qquad\textbf{(... | 2021 Fall AMC 12B Problems/Problem 22 | In a Cartesian plane, let $C, B,$ and $A$ be $(0,0),(0,6),(8,0)$ respectively.
~(Diagram By MaPhyCom)
By analyzing the behaviors of the two circles, we set $O$ to be $(a,6)$ and $P$ be $(b,8)$.
Hence derive the two equations:
$(x-a)^2+(y-6)^2=a^2$
$(x-8)^2+(y-b)^2=b^2$
Considering the coordinates of $A$ and $B... | // Block 1
size(8cm);
pair A = (0,8);
pair B = (6,0);
pair C = (0,0);
pair O = (6,6.25);
pair P = (8.3333,8);
draw(A--B--C--cycle);
draw(Circle(O, 6.2498));
draw(Circle(P, 8.3331));
draw(O--P);
draw(O--B, red+dotted+1.2bp);
draw(A--P, blue+dotted+1.2bp);
filldraw(O--(0,6.25)--A--cycle, red+opacity(0.3), invisibl... | [] |
529 | Three spheres with radii $11$, $13$, and $19$ are mutually externally tangent. A plane intersects the spheres in three congruent circles centered at $A$, $B$, and $C$, respectively, and the centers of the spheres all lie on the same side of this plane. Suppose that $AB^2 = 560$. Find $AC^2$. | 2022 AIME I Problems/Problem 10 | This solution refers to the Diagram section.
We let $\ell$ be the plane that passes through the spheres and $O_A$ and $O_B$ be the centers of the spheres with radii $11$ and $13$. We take a cross-section that contains $A$ and $B$, which contains these two spheres but not the third, as shown below:
Because the plane c... | // Block 1
size(400);
pair A, B, OA, OB;
B = (0,0);
A = (-23.6643191,0);
OB = (0,8);
OA = (-23.6643191,4);
draw(circle(OB,13));
draw(circle(OA,11));
draw((-48,0)--(24,0));
label("$\ell$",(-42,0),S);
label("$A$",A,S);
label("$B$",B,S);
label("$O_A$",OA,N);
label("$O_B$",OB,N);
draw(A--OA);
draw(B--OB);
draw(OA--OB)... | [] |
530 | Let $ABCD$ be a parallelogram with $\angle BAD < 90^\circ.$ A circle tangent to sides $\overline{DA},$ $\overline{AB},$ and $\overline{BC}$ intersects diagonal $\overline{AC}$ at points $P$ and $Q$ with $AP < AQ,$ as shown. Suppose that $AP=3,$ $PQ=9,$ and $QC=16.$ Then the area of $ABCD$ can be expressed in the form $... | 2022 AIME I Problems/Problem 11 | Let's redraw the diagram, but extend some helpful lines.
We obviously see that we must use power of a point since they've given us lengths in a circle and there are intersection points. Let $T_1, T_2, T_3$ be our tangents from the circle to the parallelogram. By the secant power of a point, the power of $A = 3 \cdot... | // Block 1
size(10cm);
pair A,B,C,D,EE,F,P,Q,O;
A=(0,0);
EE = (24,15);
F = (30,0);
O = (10.5,7.5);
label("$A$", A, SW);
B=(6,15);
label("$B$", B, NW);
C=(30,15);
label("$C$", C, NE);
D=(24,0);
label("$D$", D, SE);
P=(5.2,2.6);
label("$P$", (5.8,2.6), N);
Q=(18.3,9.1);
label("$Q$", (18.1,9.7), W);
draw(A--B--C--D--cycle... | [] |
531 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problems/Problem 3 | We have the following diagram:
Let $X$ and $W$ be the points where $AP$ and $BQ$ extend to meet $CD$, and $YZ$ be the height of $\triangle AZB$. As proven in Solution 2, triangles $APD$ and $DPW$ are congruent right triangles. Therefore, $AD = DW = 333$. We can apply this logic to triangles $BCQ$ and $XCQ$ as well, g... | // Block 1
/* Made by MRENTHUSIASM , modified by Cytronical */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint... | [] |
531 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problems/Problem 3 | Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this:
Because the trapezoid is isosceles, by symmetry $PQ$ is parallel to $AB$ and $CD$. Therefore, $\angle PAB \cong \angle APP'$ by interior angles and $\angle PAB \cong \angle PAD$ by the problem statement. Thus, $\triangle P'AP$ is isos... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersectionpoin... | [] |
531 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problems/Problem 3 | We have the following diagram:
Extend lines $AP$ and $BQ$ to meet line $DC$ at points $W$ and $X$, respectively, and extend lines $DP$ and $CQ$ to meet $AB$ at points $Z$ and $Y$, respectively.
Claim: quadrilaterals $AZWD$ and $BYXC$ are rhombuses.
Proof: Since $\angle DAB + \angle ADC = 180^{\circ}$, $\angle ADP + ... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersection... | [] |
531 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problems/Problem 3 | Let $X$ and $Y$ be the feet of the altitudes from $P$ and $Q$, respectively, to $AB$, and let $Z$ and $W$ be the feet of the altitudes from $P$ and $Q$, respectively, to $CD$. Side $AB$ is parallel to side $CD$, so $XYWZ$ is a rectangle with width $PQ$. Furthermore, because $CD - AB = 650-500 = 150$ and trapezoid $ABCD... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersection... | [] |
531 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problems/Problem 3 | Extend line $PQ$ to meet $AD$ at $P'$ and $BC$ at $Q'$. The diagram looks like this:
Since $\angle A + \angle D=\angle B + \angle C = 180^{\circ}$, it follows that $\angle P'AP+\angle P'DP = \angle Q'BQ + \angle Q'CQ = 90^{\circ}$. Thus, $\angle APD = \angle BQC = 90^{\circ}$, implying that $\triangle APD$ and $\tria... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, P1, Q1;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersectionpoin... | [] |
531 | In isosceles trapezoid $ABCD$, parallel bases $\overline{AB}$ and $\overline{CD}$ have lengths $500$ and $650$, respectively, and $AD=BC=333$. The angle bisectors of $\angle{A}$ and $\angle{D}$ meet at $P$, and the angle bisectors of $\angle{B}$ and $\angle{C}$ meet at $Q$. Find $PQ$. | 2022 AIME I Problems/Problem 3 | As in solution 4, $\angle APD = 90^{\circ}$. Set $k = AX$ and $x = DP$.
We know that $DZ = AX + \frac{DC-AB}{2}$, so $DZ = k + \frac{650-500}{2} = k + 75$.
$\triangle DPZ \sim \triangle APD$ by AA, so we have $\frac{PD}{AD} = \frac{ZD}{PD}$, resulting in
\[\frac{x}{333} = \frac{k+75}{x} \text{ (1)}\]
$\triangle APX... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, Q, X, Y, Z, W;
A = (-250,6*sqrt(731));
B = (250,6*sqrt(731));
C = (325,-6*sqrt(731));
D = (-325,-6*sqrt(731));
A1 = bisectorpoint(B,A,D);
B1 = bisectorpoint(A,B,C);
C1 = bisectorpoint(B,C,D);
D1 = bisectorpoint(A,D,C);
P = intersection... | [] |
532 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six po... | 2022 AIME I Problems/Problem 8 | We can extend $AB$ and $AC$ to $B'$ and $C'$ respectively such that circle $\omega_A$ is the incircle of $\triangle AB'C'$.
Since the diameter of the circle is the height of this triangle, the height of this triangle is $36$. We can use inradius or equilateral triangle properties to get the inradius of this triangle i... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, B1, C1, W, WA, WB, WC, X, Y, Z;
A = 18*dir(90);
B = 18*dir(210);
C = 18*dir(330);
B1 = A+24*sqrt(3)*dir(B-A);
C1 = A+24*sqrt(3)*dir(C-A);
W = (0,0);
WA = 6*dir(270);
WB = 6*dir(30);
WC = 6*dir(150);
X = (sqrt(117)-3)*dir(270);
Y = (sqrt(117)-3)*dir(30);
Z =... | [] |
532 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18.$ Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega.$ Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A,$ $\omega_B,$ and $\omega_C$ meet in six po... | 2022 AIME I Problems/Problem 8 | For equilateral triangle with side length $l$, height $h$, and circumradius $r$, there are relationships: $h = \frac{\sqrt{3}}{2} l$, $r = \frac{2}{3} h = \frac{\sqrt{3}}{3} l$, and $l = \sqrt{3}r$.
There is a lot of symmetry in the figure. The radius of the big circle $\odot \omega$ is $R = 18$, let the radius of the... | // Block 1
/* Made by MRENTHUSIASM */
/* Modified by isabelchen */
size(250);
pair A, B, C, W, WA, WB, WC, X, Y, Z, D, E;
A = 18*dir(90);
B = 18*dir(210);
C = 18*dir(330);
W = (0,0);
WA = 6*dir(270);
WB = 6*dir(30);
WC = 6*dir(150);
X = (sqrt(117)-3)*dir(270);
Y = (sqrt(117)-3)*dir(30);
Z = (sqrt(117)-3)*dir(150);
D = ... | [] |
533 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = 2$, $O_1O_2 = 15$, $CD = 16$, and $ABO_1CDO_2$ is a convex hexagon... | 2022 AIME II Problems/Problem 15 | First observe that $AO_2 = O_2D$ and $BO_1 = O_1C$. Let points $A'$ and $B'$ be the reflections of $A$ and $B$, respectively, about the perpendicular bisector of $\overline{O_1O_2}$. Then quadrilaterals $ABO_1O_2$ and $B'A'O_2O_1$ are congruent, so hexagons $ABO_1CDO_2$ and $A'B'O_1CDO_2$ have the same area. Furthermor... | // Block 1
import olympiad;
size(180);
defaultpen(linewidth(0.7));
pair Ap = dir(105), Bp = dir(75), O1 = dir(25), C = dir(320), D = dir(220), O2 = dir(175);
draw(unitcircle^^Ap--Bp--O1--C--D--O2--cycle);
label("$A'$",Ap,dir(origin--Ap));
label("$B'$",Bp,dir(origin--Bp));
label("$O_1$",O1,dir(origin--O1));
label("$C$",... | ["https://artofproblemsolving.com/wiki/images/thumb/9/98/Actual_Graph_for_Solution_1.PNG/500px-Actual_Graph_for_Solution_1.PNG"] |
533 | Two externally tangent circles $\omega_1$ and $\omega_2$ have centers $O_1$ and $O_2$, respectively. A third circle $\Omega$ passing through $O_1$ and $O_2$ intersects $\omega_1$ at $B$ and $C$ and $\omega_2$ at $A$ and $D$, as shown. Suppose that $AB = 2$, $O_1O_2 = 15$, $CD = 16$, and $ABO_1CDO_2$ is a convex hexagon... | 2022 AIME II Problems/Problem 15 | Start by noting that the radius of $\omega_1$ and $\omega_2$ sum up to 15.
Let the radius of $\omega_1$ be $x$, then the radius of $\omega_2$ will be $15-x$. We can find the area of the hexagon as the sum of the area of the quadrilateral $[O_1BAO_2]$ and $[O_1O_2DC]$, which is given by the Brahmagupta's Formula as: $\... | // Block 1
import geometry;
size(10cm);
point O1=(0,0),O2=(15,0),B=9*dir(30);
circle w1=circle(O1,9),w2=circle(O2,6),o=circle(O1,O2,B);
point A=intersectionpoints(o,w2)[1],D=intersectionpoints(o,w2)[0],C=intersectionpoints(o,w1)[0];
filldraw(A--B--O1--C--D--O2--cycle,0.2*fuchsia+white,black);
draw(w1);
draw(w2);
draw(O... | [] |
534 | Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$
$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }... | 2022 AMC 10A Problems/Problem 23 | Consider the reflection $P^{\prime}$ of $P$ over the perpendicular bisector of $\overline{BC}$, creating two new isosceles trapezoids $DAPP^{\prime}$ and $CBPP^{\prime}$. Under this reflection, $P^{\prime}A=PD=4$, $P^{\prime}D=PA=1$, $P^{\prime}C=PB=2$, and $P^{\prime}B=PC=3$.
Since $DAPP'$ and $CBPP'$ are isosceles ... | // Block 1
size(300);
pair A = (0,0);
pair B = (1, 2);
pair C = (2,2);
pair D = (3,0);
label("$A$", A, SW);
label("$B$", B, NW);
label("$C$", C, NE);
label("$D$", D, SE);
draw(A--B--C--D--cycle, blue);
pair P = (0.8, 0.6);
dot("$P$", P, NW);
draw(P--A, magenta);
draw(P--B, magenta);
draw(P--C);
draw(P--D);
label("$1$",... | [] |
534 | Isosceles trapezoid $ABCD$ has parallel sides $\overline{AD}$ and $\overline{BC},$ with $BC < AD$ and $AB = CD.$ There is a point $P$ in the plane such that $PA=1, PB=2, PC=3,$ and $PD=4.$ What is $\tfrac{BC}{AD}?$
$\textbf{(A) }\frac{1}{4}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{1}{2}\qquad\textbf{(D) }... | 2022 AMC 10A Problems/Problem 23 | Extend $AB$ and $CD$ to a point $Q$ as shown, and let $PQ = s$. Then let $BQ=CQ = a$ and $AB=DC = b$. Notice that $\frac{BC}{AD} = \frac{QC}{QD} = \frac{a}{a+b}$ by similar triangles.
By Stewart's theorem on $APQ$ and $DPQ$, we have \begin{align*} ab(a+b) + 9(a+b) &= 16a + s^2b \\ ab(a+b) + 4(a+b) &= a + s^2b \\ \end{... | // Block 1
size(7.5cm);
draw((0,0)--(4.2,0));
draw((0,0)--(1.4,2)--(2.8,2)--(4.2,0));
draw((1.4,2)--(2.1,3)--(2.8,2));
draw((0,0)--(1,0.5)--(1.4,2)--(1,0.5)--(2.8,2)--(1,0.5)--(4.2,0));
label("$A$",(0,0),SW);
label("$B$",(1.4,2),NW);
label("$C$",(2.8,2),NE);
label("$D$",(4.2,0),SE);
label("$P$",(1,0.5),NW);
label("$Q$"... | [] |
535 | Let $c$ be a real number, and let $z_1$ and $z_2$ be the two complex numbers satisfying the equation
$z^2 - cz + 10 = 0$. Points $z_1$, $z_2$, $\frac{1}{z_1}$, and $\frac{1}{z_2}$ are the vertices of (convex) quadrilateral $\mathcal{Q}$ in the complex plane. When the area of $\mathcal{Q}$ obtains its maximum possible v... | 2022 AMC 12A Problems/Problem 22 | Note that $z=\dfrac c2\pm\dfrac{\sqrt{c^2-40}}2$, so let $z_1=\dfrac c2+\dfrac{\sqrt{c^2-40}}2$ and $z_2=\dfrac c2-\dfrac{\sqrt{c^2-40}}2$. Taking a look at the answer choices, they range between $c=4.5$ to $c=6.5$, and in that range, $c^2$ is always less than $40$. Thus, $c^2-40<0$ for our possible answer choices; we ... | // Block 1
import graph;
unitsize(0.5cm);
/*xaxis(Ticks, xmin=-1,xmax=8);
yaxis(Ticks, ymin=-11,ymax=11);*/
draw((-1,0)--(8,0));
draw((0,-11)--(0,11));
dot((7,10));dot((7,-10));dot((0.7,1));dot((0.7,-1));
draw((7,10)--(7,-10)--(0.7,-1)--(0.7,1)--cycle);
label("$z_1$", (7,10), E);
label("$z_2$", (7,-10), E);
label("... | [] |
536 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\a... | 2023 AIME I Problems/Problem 12 | By Miquel's theorem, $P=(AEF)\cap(BFD)\cap(CDE)$ (intersection of circles)$\text{*}$. The law of cosines can be used to compute $DE=42$, $EF=35$, and $FD=13$. Toss the points on the coordinate plane; let $B=(-7, 0)$, $D=(0, 0)$, and $C=(48, 0)$, where we will find $\tan^{2}\left(\measuredangle CDP\right)$ with $P=(BFD)... | // Block 1
/* Made by MRENTHUSIASM */
size(400);
pair A, B, C, D, E, F, P;
A = 55*sqrt(3)/3 * dir(90);
B = 55*sqrt(3)/3 * dir(210);
C = 55*sqrt(3)/3 * dir(330);
D = B + 7*dir(0);
E = A + 25*dir(C-A);
F = A + 40*dir(B-A);
P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0];
filldraw(D--E--F--c... | [] |
536 | Let $\triangle ABC$ be an equilateral triangle with side length $55.$ Points $D,$ $E,$ and $F$ lie on $\overline{BC},$ $\overline{CA},$ and $\overline{AB},$ respectively, with $BD = 7,$ $CE=30,$ and $AF=40.$ Point $P$ inside $\triangle ABC$ has the property that \[\angle AEP = \angle BFP = \angle CDP.\] Find $\tan^2(\a... | 2023 AIME I Problems/Problem 12 | By the law of cosines,
\[FE=\sqrt{AF^2+AE^2-2AF\cdot AE\cos\angle FAE}=35.\]
Similarly we get $FD=13$ and $DE=42$. $\angle AEP=\angle BFP=\angle CDP\overset\triangle=\theta$ implies that $AFPE$, $BDPF$, and $CDPE$ are three cyclic quadrilaterals, as shown below:
Using the law of sines in each,
\[\frac{AP}{35}=\frac{AP... | // Block 1
/* Made by MRENTHUSIASM */
size(400);
pair A, B, C, D, E, F, P;
A = 55*sqrt(3)/3 * dir(90);
B = 55*sqrt(3)/3 * dir(210);
C = 55*sqrt(3)/3 * dir(330);
D = B + 7*dir(0);
E = A + 25*dir(C-A);
F = A + 40*dir(B-A);
P = intersectionpoints(Circle(D,54*sqrt(19)/19),Circle(F,5*sqrt(19)/19))[0];
draw(anglemark(A,E,... | [] |
537 | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$.
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$, where $m$ and $n$
are relatively prime positive integers. Find $m + n$. A parallelepiped is a sol... | 2023 AIME I Problems/Problem 13 | Since the two parallelepipeds have the same base, all we need to do is to find their respective heights.
As illustrated in the above diagram, drop a perpendicular from $D$ onto the base at $P$. Denote the center of the base by $O$. By symmetry, $P$ must be on $AO$. Now we need to find $DP$.
Apply Pythagorean theore... | // Block 1
unitsize(2cm);
pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2;
label("A",a,S);
label("B",b,S);
label("C",c,S);
label("D",d,N);
label("P",p,S);
label("O",o,E);
draw(a--b--(b+c));
draw(a--c--(b+c), dotted);
draw(shift(d)*(a--b--(b+c)--c--cycle));
draw(a--d);
draw(b--... | [] |
537 | Each face of two noncongruent parallelepipeds is a rhombus whose diagonals have lengths $\sqrt{21}$ and $\sqrt{31}$.
The ratio of the volume of the larger of the two polyhedra to the volume of the smaller is $\frac{m}{n}$, where $m$ and $n$
are relatively prime positive integers. Find $m + n$. A parallelepiped is a sol... | 2023 AIME I Problems/Problem 13 | Let $\angle CAB$ be acute and let $X$ be the foot of the altitude from $C$ to $AB$. Notice that this also implies that $X$ is the foot of the altitude from $D$ to $AB$. Now $\sqrt{13} \cdot CX = AB \cdot CX = \frac{1}{2} \sqrt{21} \cdot \sqrt{31}$ so $CX = \frac{\sqrt{21} \cdot \sqrt{31}}{2 \sqrt{13}}$ and $DX$ is the ... | // Block 1
unitsize(2cm);
pair a = (0, 0), b = (1, 0), c = 0.8*dir(40), d = dir(70), p = 0.33*dir(20), o = (b+c)/2, x = (0.2, 0);
label("A",a,S);
label("B",b,S);
label("C",c,S);
label("D",d,N);
label("X",x,S);
draw(a--b--(b+c));
draw(a--c--(b+c), dotted);
draw(shift(d)*(a--b--(b+c)--c--cycle));
draw(a--d);
draw(b--(b... | [] |
538 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | 2023 AIME I Problems/Problem 5 | By the Inscribed Angle Theorem, we conclude that $\triangle PAC$ and $\triangle PBD$ are right triangles.
Let the brackets denote areas. We are given that
\begin{alignat*}{8} 2[PAC] &= PA \cdot PC &&= 56, \\ 2[PBD] &= PB \cdot PD &&= 90. \end{alignat*}
Let $O$ be the center of the circle, $X$ be the foot of the perpen... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair A, B, C, D, O, P, X, Y;
A = (-sqrt(106)/2,sqrt(106)/2);
B = (-sqrt(106)/2,-sqrt(106)/2);
C = (sqrt(106)/2,-sqrt(106)/2);
D = (sqrt(106)/2,sqrt(106)/2);
O = origin;
path p;
p = Circle(O,sqrt(212)/2);
draw(p);
P = intersectionpoints(Circle(A,4),p)[1];
X = foot(P,A,... | [] |
538 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | 2023 AIME I Problems/Problem 5 | Let the center of the circle be $O$, and the radius of the circle be $r$. Since $ABCD$ is a rhombus with diagonals $2r$ and $2r$, its area is $\dfrac{1}{2}(2r)(2r) = 2r^2$. Since $AC$ and $BD$ are diameters of the circle, $\triangle APC$ and $\triangle BPD$ are right triangles. Let $X$ and $Y$ be the foot of the altitu... | // Block 1
/* Made by MRENTHUSIASM */
size(200);
pair A, B, C, D, O, P, X, Y;
A = (-sqrt(106)/2,sqrt(106)/2);
B = (-sqrt(106)/2,-sqrt(106)/2);
C = (sqrt(106)/2,-sqrt(106)/2);
D = (sqrt(106)/2,sqrt(106)/2);
O = origin;
path p;
p = Circle(O,sqrt(212)/2);
draw(p);
P = intersectionpoints(Circle(A,4),p)[1];
X = foot(P,A,... | [] |
538 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | 2023 AIME I Problems/Problem 5 | First draw a diagram.
Let's say that the radius is $r$. Then the area of the $ABCD$ is $(\sqrt2r)^2 = 2r^2$
Using the formula for the length of a chord subtended by an angle, we get
\[PA = 2r\sin\left(\dfrac{\theta}2\right)\]
\[PC = 2r\sin\left(\dfrac{180-\theta}2\right) = 2r\sin\left(90 - \dfrac{\theta}2\right) = 2r\... | // Block 1
pair A, B, C, D, O, P;
A = (0,sqrt(106));
B = (0,0);
C = (sqrt(106),0);
D = (sqrt(106),sqrt(106));
O = (sqrt(106)/2, sqrt(106)/2);
P = intersectionpoint(circle(A, sqrt(212)*sin(atan(28/45)/2)), circle(O, sqrt(212)/2));
draw(A--B--C--D--cycle);
draw(circle(O, sqrt(212)/2));
label("$A$", A, NW);
label("$B$", B... | [] |
538 | Let $P$ be a point on the circle circumscribing square $ABCD$ that satisfies $PA \cdot PC = 56$ and $PB \cdot PD = 90.$ Find the area of $ABCD.$ | 2023 AIME I Problems/Problem 5 | Let $P=(a,b)$ on the upper quarter of the circle, and let $k$ be the side length of the square. Hence, we want to find $k^2$. Let the center of the circle be $(0,0)$.
The two equations would thus become:
\[\left(\left(a+\dfrac{k}2\right)^2+\left(b-\dfrac{k}2\right)^2\right)\left(\left(a-\dfrac{k}2\right)^2+\left(b+\dfr... | // Block 1
pair A,B,C,D,P;
A=(-3,3);
B=(3,3);
C=(3,-3);
D=(-3,-3);
draw(A--B--C--D--cycle);
label(A,"$A$",NW);
label(B,"$B$",NE);
label(C,"$C$",SE);
label(D,"$D$",SW);
draw(circle((0,0),4.24264068712));
P=(-1,4.12310562562);
label(P,"$P$", NW);
pen k=red+dashed;
draw(P--A,k);
draw(P--B,k);
draw(P--C,k);
draw(P--D,k);
d... | [] |
539 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | 2023 AIME I Problems/Problem 8 | This solution refers to the Diagram section.
Let $O$ be the incenter of $ABCD$ for which $\odot O$ is tangent to $\overline{DA},\overline{AB},$ and $\overline{BC}$ at $X,Y,$ and $Z,$ respectively. Moreover, suppose that $R,S,$ and $T$ are the feet of the perpendiculars from $P$ to $\overleftrightarrow{DA},\overleftrig... | // Block 1
/* Made by MRENTHUSIASM; inspired by Math Jams. */
size(300);
pair A, B, C, D, O, P, R, S, T, X, Y, Z, Q;
A = origin;
B = (125/4,0);
C = B + 125/4 * dir((3,4));
D = A + 125/4 * dir((3,4));
O = (25,25/2);
P = (15,5);
R = foot(P,A,D);
S = foot(P,A,B);
T = foot(P,B,C);
X = (15,20);
Y = (25,0);
Z = (35,5);
Q = ... | [] |
539 | Rhombus $ABCD$ has $\angle BAD < 90^\circ.$ There is a point $P$ on the incircle of the rhombus such that the distances from $P$ to the lines $DA,AB,$ and $BC$ are $9,$ $5,$ and $16,$ respectively. Find the perimeter of $ABCD.$ | 2023 AIME I Problems/Problem 8 | This solution refers to the Diagram section.
Define points $O,R,S,$ and $T$ as Solution 1 does. Moreover, let
$H$ be the foot of the perpendicular from $P$ to $\overleftrightarrow{CD},$
$M$ be the foot of the perpendicular from $O$ to $\overleftrightarrow{HS},$ and
$N$ be the foot of the perpendicular from $O$ to $\o... | // Block 1
/* Made by MRENTHUSIASM; inspired by Math Jams. */
size(300);
pair A, B, C, D, O, P, R, S, T, H, M, N;
A = origin;
B = (125/4,0);
C = B + 125/4 * dir((3,4));
D = A + 125/4 * dir((3,4));
O = (25,25/2);
P = (15,5);
R = foot(P,A,D);
S = foot(P,A,B);
T = foot(P,B,C);
H = foot(S,C,D);
M = foot(O,S,H);
N = foot(O... | [] |
540 | A cube-shaped container has vertices $A,$ $B,$ $C,$ and $D,$ where $\overline{AB}$ and $\overline{CD}$ are parallel edges of the cube, and $\overline{AC}$ and $\overline{BD}$ are diagonals of faces of the cube, as shown. Vertex $A$ of the cube is set on a horizontal plane $\mathcal{P}$ so that the plane of the rectangl... | 2023 AIME II Problems/Problem 14 | Let's first view the cube from a direction perpendicular to $ABDC$, as illustrated above. Let $x$ be the cube's side length. Since $\triangle CHA \sim \triangle AGB$, we have
\[\frac{CA}{CH} = \frac{AB}{AG}.\]
We know $AB = x$, $AG = \sqrt{x^2-2^2}$, $AC = \sqrt{2}x$, $CH = 8$. Plug them into the above equation, we get... | // Block 1
defaultpen(linewidth(0.6));
pair A = (0, 0), B = (5.656,2), C = (-2.828, 8), D = B+C, P = 0.875*C, Q = B+0.625*C, H = (-2.828, 0), G = (5.656, 0);
pair P1 = (-5, 0), P2 = (10, 0);
draw(A--B--D--C--A);
filldraw(A--B--Q--P--cycle,rgb(0.35,0.7,0.9));
draw(C--H, dotted);
draw(B--G, dotted);
draw(P1--P2);
dot("$A... | [] |
541 | Let $\triangle ABC$ be an isosceles triangle with $\angle A = 90^\circ.$ There exists a point $P$ inside $\triangle ABC$ such that $\angle PAB = \angle PBC = \angle PCA$ and $AP = 10.$ Find the area of $\triangle ABC.$ | 2023 AIME II Problems/Problem 3 | This solution refers to the Diagram section.
Let $\angle PAB = \angle PBC = \angle PCA = \theta,$ from which $\angle PAC = 90^\circ-\theta,$ and $\angle APC = 90^\circ.$
Moreover, we have $\angle PBA = \angle PCB = 45^\circ-\theta,$ as shown below:
Note that $\triangle PAB \sim \triangle PBC$ by AA Similarity. The r... | // Block 1
/* Made by MRENTHUSIASM */
size(250);
pair A, B, C, P;
A = origin;
B = (0,10*sqrt(5));
C = (10*sqrt(5),0);
P = intersectionpoints(Circle(A,10),Circle(C,20))[0];
dot("$A$",A,1.5*SW,linewidth(4));
dot("$B$",B,1.5*NW,linewidth(4));
dot("$C$",C,1.5*SE,linewidth(4));
dot("$P$",P,1.5*NE,linewidth(4));
markscal... | [] |
542 | A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
$\textbf{(A)}~4-\sqrt{5}\qquad\textbf{(B)}~\sqrt{5}-1\qquad\textbf{(C)}~8-3\sqrt{5}\qquad\textbf{(D... | 2023 AMC 12B Problems/Problem 25 | We can find the area of the red pentagon by taking the area of the total pentagon and subtracting the area outside the red pentagon.
The area outside the red pentagon is the sum of the larger isosceles triangles, but this double counts the overlapping regions of the small isosceles triangles, so we have to subtract t... | // Block 1
unitsize(5cm);
// Define the vertices of the pentagons
pair A, B, C, D, E;
pair F, G, H, I, J;
// Calculate the vertices of the larger pentagon
A = dir(90);
B = dir(90 - 72);
C = dir(90 - 2*72);
D = dir(90 - 3*72);
E = dir(90 - 4*72);
// Draw the larger pentagon
draw(A--B--C--D--E--cycle);
pair O = (A+B... | [] |
542 | A regular pentagon with area $\sqrt{5}+1$ is printed on paper and cut out. The five vertices of the pentagon are folded into the center of the pentagon, creating a smaller pentagon. What is the area of the new pentagon?
$\textbf{(A)}~4-\sqrt{5}\qquad\textbf{(B)}~\sqrt{5}-1\qquad\textbf{(C)}~8-3\sqrt{5}\qquad\textbf{(D... | 2023 AMC 12B Problems/Problem 25 | Let $r_b$ and $r_s$ be the circumradius of the big and small pentagon, respectively. Let $a_s$ be the apothem of the smaller pentagon and $A_s$ and $A_b$ be the areas of the smaller and larger pentagon, respectively.
From the diagram:
\begin{align*} \cos{36^\circ} &= \dfrac{a_s}{r_s} = \dfrac{\phi}{2} = \dfrac{\s... | // Block 1
unitsize(5cm);
// Define the vertices of the pentagons
pair A, B, C, D, E;
pair F, G, H, I, J;
// Calculate the vertices of the larger pentagon
A = dir(90);
B = dir(90 - 72);
C = dir(90 - 2*72);
D = dir(90 - 3*72);
E = dir(90 - 4*72);
// Draw the larger pentagon
draw(A--B--C--D--E--cycle);
pair O = (A+B+... | [] |
543 | Let $ABCD$ be a tetrahedron such that $AB=CD= \sqrt{41}$, $AC=BD= \sqrt{80}$, and $BC=AD= \sqrt{89}$. There exists a point $I$ inside the tetrahedron such that the distances from $I$ to each of the faces of the tetrahedron are all equal. This distance can be written in the form $\frac{m \sqrt n}{p}$, where $m$, $n$, an... | 2024 AIME I Problems/Problem 14 | Inscribe tetrahedron $ABCD$ in an rectangular prism as shown above.
By the Pythagorean theorem, we note
\[OA^2 + OB^2 = AB^2 = 41,\]
\[OA^2 + OC^2 = AC^2 = 80, \text{and}\]
\[OB^2 + OC^2 = BC^2 = 89.\]
Solving yields $OA = 4, OB = 5,$ and $OC = 8.$
Since each face of the tetrahedron is congruent, we know the point ... | // Block 1
import three;
currentprojection = orthographic(1,1,1);
triple O = (0,0,0);
triple A = (0,2,0);
triple B = (0,0,1);
triple C = (3,0,0);
triple D = (3,2,1);
triple E = (3,2,0);
triple F = (0,2,1);
triple G = (3,0,1);
draw(A--B--C--cycle, red);
draw(A--B--D--cycle, red);
draw(A--C--D--cycle, red);
draw(B--C-... | [] |
544 | Let $\triangle$$ABC$ have incenter $I$, circumcenter $O$, inradius $6$, and circumradius $13$. Suppose that $\overline{IA} \perp \overline{OI}$. Find $AB \cdot AC$. | 2024 AIME II Problems/Problem 10 | Start off by (of course) drawing a diagram! Let $I$ and $O$ be the incenter and circumcenters of triangle $ABC$, respectively. Furthermore, extend $AI$ to meet $BC$ at $L$ and the circumcircle of triangle $ABC$ at $D$.
We'll tackle the initial steps of the problem in two different manners, both leading us to the s... | // Block 1
size(300);
import olympiad;
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26);
pair A = (c/3,8.65*c/10);
draw(circumcircle(A,B,C));
pair I=incenter(A,B,C);
pair O=circumcenter(A,B,C);
pair L=extension(A,I,C,B);
dot(I^^O^^A^^B^^C^^D^^L);
draw(A--L);
draw(A--... | [] |
544 | Let $\triangle$$ABC$ have incenter $I$, circumcenter $O$, inradius $6$, and circumradius $13$. Suppose that $\overline{IA} \perp \overline{OI}$. Find $AB \cdot AC$. | 2024 AIME II Problems/Problem 10 | Now we have some sort of expression for $AB \cdot AC$ in terms of $ID$ and $AL$. Let's try to find $AL$ first.
Drop an altitude from $D$ to $BC$, $I$ to $AC$, and $I$ to $BC$:
Since $\angle DBE \cong \angle IAF$ and $\angle BED \cong \angle IFA$, $\triangle BDE \sim \triangle AIF$.
Furthermore, we know $BD = ID$ a... | // Block 1
size(300);
import olympiad;
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26), E = (c/2-0.01,0);
pair A = (c/3,8.65*c/10);
pair F = (2*c/3-0.14, 4-0.29);
pair G = (c/2-0.68,0);
draw(circumcircle(A,B,C));
pair I=incenter(A,B,C);
pair O=circumcenter(A,B,C);
p... | [] |
545 | Let \(O=(0,0)\), \(A=\left(\tfrac{1}{2},0\right)\), and \(B=\left(0,\tfrac{\sqrt{3}}{2}\right)\) be points in the coordinate plane. Let \(\mathcal{F}\) be the family of segments \(\overline{PQ}\) of unit length lying in the first quadrant with \(P\) on the \(x\)-axis and \(Q\) on the \(y\)-axis. There is a unique point... | 2024 AIME II Problems/Problem 12 | Let's move a little bit from $A$ to $A_1$, then $B$ must move to $B_1$ to keep $A_1B_1 = 1$. $AB$ intersects with $A_1B_1$ at $C$. Pick points $A_2$ and $B_2$ on $CA_1$ and $CB$ such that $CA_2 = CA$, $CB_2 = CB_1$, we have $A_1A_2 = BB_2$. Since $AA_1$ is very small, $\angle CA_1A \approx 60^\circ$, $\angle CBB_1 \app... | // Block 1
pair O=(0,0);
pair X=(1,0);
pair Y=(0,1);
pair A=(0.5,0); pair B=(0,sin(pi/3));
pair A1=(0.6,0); pair B1=(0,0.8);
pair A2=(0.575,0.04); pair B2=(0.03,0.816);
dot(O);
dot(X);
dot(Y); dot(A); dot(B);
dot(A1); dot(B1);
dot(A2); dot(B2);
draw(X--O--Y);
draw(A--B);
draw(A1--B1);
draw(A--A2);
draw(B1--B2);
label("... | [] |
546 | Torus $T$ is the surface produced by revolving a circle with radius $3$ around an axis in the plane of the circle that is a distance $6$ from the center of the circle (so like a donut). Let $S$ be a sphere with a radius $11$. When $T$ rests on the inside of $S$, it is internally tangent to $S$ along a circle with radiu... | 2024 AIME II Problems/Problem 8 | First, let's consider a section $\mathcal{P}$ of the solids, along the axis.
By some 3D-Geomerty thinking, we can simply know that the axis crosses the sphere center. So, that is saying, the $\mathcal{P}$ we took crosses one of the equator of the sphere.
Here I drew two graphs, the first one is the case when $T$ is in... | // Block 1
unitsize(0.35cm);
pair O = (0, 0);
real r1 = 11;
real r2 = 3;
draw(circle(O, r1));
pair A = O + (0, -r1);
pair B = O + (0, r1);
draw(A--B);
pair C = O + (0, -1.25*r1);
pair D = O + (0, 1.25*r1);
draw(C--D, dashed);
dot(O);
pair E = (2 * r2, -sqrt((r1 - r2) * (r1 - r2) - 4 * r2 * r2));
pair F = (0, -sqrt((r1 ... | [] |
547 | Let $ABCDE$ be a convex pentagon with $AB=14,$ $BC=7,$ $CD=24,$ $DE=13,$ $EA=26,$ and $\angle B=\angle E=60^{\circ}.$ For each point $X$ in the plane, define $f(X)=AX+BX+CX+DX+EX.$ The least possible value of $f(X)$ can be expressed as $m+n\sqrt{p},$ where $m$ and $n$ are positive integers and $p$ is not divisible by t... | 2025 AIME I Problems/Problem 14 | Firstly, note that $\triangle ABC$ and $\triangle ADE$ are just 30-60-90 triangles. Let $X$ be the Fermat point of $\triangle ACD$, with motivation stemming from considering the pentagon as $\triangle ACD$ with the two 30-60-90 extensions. Note that $AX+CX+DX$ is minimized at this point when $\angle AXC=\angle CXD=\ang... | // Block 1
size(10cm);
import math; import geometry; import olympiad;
point A,B,C,D,F,P,X; A=(0,-7sqrt(3)); B=(-7,0); C=(0,0); D=(156/7,-36sqrt(3)/7); F=(169/7,-88sqrt(3)/7); P=(132/7,60sqrt(3)/7); X=(8580/2527,-10604sqrt(3)/2527);
draw(A--B--C--P--D--F--A--C--D--A--P); draw(B--F); draw(circumcircle(A,B,C)); draw(circ... | [] |
548 | $\triangle DEF$ is inscribed inside $\triangle ABC$ such that $D,E,F$ lie on $BC, AC, AB$, respectively. The circumcircles of $\triangle DEC, \triangle BFD, \triangle AFE$ have centers $O_1,O_2,O_3$, respectively. Also, $AB = 23, BC = 25, AC=24$, and $\stackrel{\frown}{BF} = \stackrel{\frown}{EC},\ \stackrel{\frown}{AF... | Mock AIME 1 2007-2008 Problems/Problem 11 | From adjacent sides, the following relationships can be derived:
\begin{align*} DC &= EC + 1\\ AE &= AF + 1\\ BD &= BF + 2 \end{align*}
Since $BF = EC$, and $DC = BF + 1$, $BD = DC + 1$. Thus, $BC = BD + DC = BD + (BD - 1)$. $26 = 2BD$. Thus, $BD = 13/1$. Thus, the answer is $\boxed{014}$. | // Block 1
size(150); defaultpen(linewidth(0.8)); import markers;
pair B = (0,0), C = (25,0), A = (578/50,19.8838);
draw(A--B--C--cycle);
label("$B$",B,SW); label("$C$",C,SE); label("$A$",A,N);
pair D = (13,0), E = (11*A + 13*C)/24, F = (12*B + 11*A)/23;
draw(D--E--F--cycle);
label("$D$",D,dir(-90));
label("$E$",E,dir(... | [] |
549 | Points $A$ and $B$ lie on $\odot O$, with radius $r$, so that $\angle OAB$ is acute. Extend $AB$ to point $C$ so that $AB = BC$. Let $D$ be the intersection of $\odot O$ and $OC$ such that $CD = \frac {1}{18}$ and $\cos(2\angle OAC) = \frac 38$. If $r$ can be written as $\frac{a+b\sqrt{c}}{d}$, where $a,b,$ and $d$ are... | Mock AIME 1 2007-2008 Problems/Problem 14 | EDIT: The following solution is wrong. See the discussion page.
By the cosine double-angle formula,
\[\cos(2\angle OAC) = 2\cos^2(\angle OAC) - 1 = \frac 38\ \Longrightarrow\ \cos \angle OAC = \sqrt{\frac{11}{16}}.\]
The Law of Cosines on $\triangle BAO$ with respect to $\angle BAO$ yields
\begin{align*}r^2 &= r... | // Block 1
size(220); defaultpen(fontsize(10));
real r = (2+26^0.5)/198;
pair O = (0,0), A=(r,0), B= A*dir(112.02432), C=2*B-A;
pair[] D = intersectionpoints(C--O,Circle(O,r));
dot(O); dot(A); dot(B); dot(C); dot(D[0]);
draw(O--A--C--cycle); draw(Circle(O,r));
draw(O--B,dashed);
draw(anglemark(B,A,O,0.15));
label("\(A\... | [] |
550 | Suppose $\triangle ABC$ is inscribed in circle $\Gamma$. $B_1$ and $C_1$ are the feet of the altitude from $B$ to $CA$ and $C$ to $AB$, respectively. Let $D$ be the intersection of lines $\overline{B_1 C_1}$ and $\overline{BC}$, let $E$ be the point of intersection of $\Gamma$ and line $\overline{DA}$ distinct from $A$... | Mock AIME 1 2010 Problems/Problem 13 | Let $\measuredangle BED=\theta$. Because the problem gives us $ED^2+EB^2=3050$, we think to use the Law of Cosines in $\triangle BED$, which yields $BD^2=ED^2+EB^2-2ED\cdot EB\cos\theta$. Subtituting the values given by the problem, we get $28^2=3050-2ED\cdot EB\cos\theta$, which gives $ED\cdot EB=\tfrac{3050-784}{2\co... | // Block 1
import geometry;
size(8cm);
point B = origin;
point A = (3,8);
point C = (12,0);
triangle t = triangle(A,B,C);
circle c = circumcircle(t);
point B1 = foot(t.VB);
point C1 = foot(t.VC);
point D = intersectionpoint(line(B1,C1), line(B,C));
pair[] e = intersectionpoints(line(A,D), c);
point E = e[0];
// ... | [] |
551 | Let $AB$ be a line segment of length $20 \sqrt{2}$, and let $S$ be the set of all points $P$ such that $m \angle APB \geq 45^{\circ}$. Find the last three digits of the largest integer less than the area of $S$. | Mock AIME 1 2010 Problems/Problem 3 | Let point $O$ be on the perpendicular bisector of $\overline{AB}$ such that $\overline{AO} \perp \overline{OB}$. We know that $AO=OB$ (because $O$ is on the perpendicular bisector of $\overline{AB}$), so we can draw circle $O_A$. Do the same on the other side of $AB$ with point $Q$ and draw circle $Q_A$. The central an... | // Block 1
import geometry;
point A = origin;
point B = A + (20*sqrt(2),0);
point O = (A+B)/2+(0,10*sqrt(2));
point Q = O-(0,20*sqrt(2));
// Segment AB and Circles
draw(A--B);
draw(circle(O,length(segment(O,A))));
draw(circle(Q,length(segment(Q,A))));
// Radii
draw(O--A);
draw(O--B);
draw(Q--A);
draw(Q--B);
// Lab... | [] |
552 | Let $\omega_1$ and $\omega_2$ be circles of radii 5 and 7, respectively, and suppose that the distance between their centers is 10. There exists a circle $\omega_3$ that is internally tangent to both $\omega_1$ and $\omega_2$, and tangent to the line joining the centers of $\omega_1$ and $\omega_2$. If the radius of $\... | Mock AIME 1 2010 Problems/Problem 9 | Let $\omega_1$ have center $A$, $\omega_2$ have center $B$, and $\omega_3$ have center $C$. Further, let $\overline{AB}$ intersect $\omega_1$ at $D$, $\omega_2$ at $E$, and $\omega_3$ at $T$, as in the diagram. Let $r$ be the radius of $\omega_3$ and let $TE=x$.
Because $\overline{AB}$ is tangent to $\omega_3$, $\over... | // Block 1
import geometry;
size(10cm);
point A = origin;
point B = (10,0);
point C = (5-(30-2*sqrt(210)),10*sqrt(210)-144);
point D,E,T;
circle a = circle(A,5);
circle b = circle(B,7);
circle c = circle(C,10*sqrt(210)-144);
// Defining D, E, and T
pair[] d = intersectionpoints(b,A--B);
D = d[0];
pair[] e = inters... | [] |
553 | In rectangle $ABCD$, $AD=1$, $P$ is on $\overline{AB}$, and $\overline{DB}$ and $\overline{DP}$ trisect $\angle ADC$. What is the perimeter of $\triangle BDP$?
$\textbf{(A)}\ 3+\frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ 2+\frac{4\sqrt{3}}{3} \qquad\textbf{(C)}\ 2+2\sqrt{2} \qquad\textbf{(D)}\ \frac{3+3\sqrt{5}}{2} \qqu... | 2000 AMC 10 Problem 7 | $AD=1$.
Since $\angle ADC$ is trisected, $\angle ADP= \angle PDB= \angle BDC=30^\circ$.
Thus, $PD=\frac{2\sqrt{3}}{3}$
$DB=2$
$BP=\sqrt{3}-\frac{\sqrt{3}}{3}=\frac{2\sqrt{3}}{3}$.
Adding, we get $\boxed{\textbf{(B) } 2+\frac{4\sqrt{3}}{3}}$. | // Block 1
draw((0,2)--(3.4,2)--(3.4,0)--(0,0)--cycle);
draw((0,0)--(1.3,2));
draw((0,0)--(3.4,2));
dot((0,0));
dot((0,2));
dot((3.4,2));
dot((3.4,0));
dot((1.3,2));
label("$A$",(0,2),NW);
label("$B$",(3.4,2),NE);
label("$C$",(3.4,0),SE);
label("$D$",(0,0),SW);
label("$P$",(1.3,2),N);
label("$1$",(0,1),W);
label("$2$",... | [] |
554 | Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\t... | 2000 AMC 10 Problem 12 | By taking figure $n$, putting the centers of each square on a grid, and rotating the figure 45 degrees, we get the following shape (figure 3 is shown for reference):
There are two lattices of points (shown in red): one of side length $n$ (blue) and one of $n+1$ (green). The rest follows like in solution 1.
-Integrala... | import olympiad; int fig = 3; for (int i = 0; i <= 2*fig; ++i) { for (int j = 0; j <= 2*fig; ++j) { pair p = (i, j); //do we need this point? if ((i+j)%2 == 1) { continue; } //draw squares draw((p+N)--(p+E)--(p+S)--(p+W)--cycle); //draw lattices if (i < 2*fig-1) { draw(p-... | [] |
555 | There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?
$\textbf{(A)}\ 0 \qquad\textbf{(B)}\ 1 \qquad\textbf{(C)}\ 5!\cdot 4... | 2000 AMC 10 Problem 13 | In each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.
In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for th... | // Block 1
unitsize(20);
dot((0,0));
dot((1,0));
dot((2,0));
dot((3,0));
label("Y",(4,-.35),N);
dot((0,1));
dot((1,1));
dot((2,1));
label("Y",(3,.6),N);
dot((0,2));
dot((1,2));
label("Y",(2,1.6),N);
dot((0,3));
label("Y",(1,2.6),N);
label("Y",(0,3.6),N);
// Block 2
unitsize(20);
label("O",(0,-.35),N);
label("B",(1,-.35... | [] |
556 | The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$.
$\textbf{(A)}\ \frac{4\sqrt{5}}{3} \qquad\textbf{(B)}\ \frac{5\sqrt{5}}{3} \qquad\textbf{(C)}\ \frac{12\sqrt{5}}{7} \qquad\textbf{(D)}\ 2\sqrt{5} \qquad\textbf{(E)... | 2000 AMC 10 Problem 16 | Draw the perpendiculars from $A$ and $B$ to $CD$, respectively. As it turns out, $BC \perp CD$. Let $F$ be the point on $CD$ for which $AF\perp CD$.
$m\angle AFE=m\angle BCE=90^\circ$, and $m\angle AEF=m\angle CEB$, so by AA similarity, \[\triangle AFE\sim \triangle BCE \Rightarrow \frac{AF}{AE}=\frac{BC}{BE}\]
By th... | // Block 1
path seg1, seg2;
seg1=(6,0)--(0,3);
seg2=(2,0)--(4,2);
dot((0,0));
dot((1,0));
fill(circle((2,0),0.1),black);
dot((3,0));
dot((4,0));
dot((5,0));
fill(circle((6,0),0.1),black);
dot((0,1));
dot((1,1));
dot((2,1));
dot((3,1));
dot((4,1));
dot((5,1));
dot((6,1));
dot((0,2));
dot((1,2));
dot((2,2));
dot((3,2));
... | [] |
556 | The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$.
$\textbf{(A)}\ \frac{4\sqrt{5}}{3} \qquad\textbf{(B)}\ \frac{5\sqrt{5}}{3} \qquad\textbf{(C)}\ \frac{12\sqrt{5}}{7} \qquad\textbf{(D)}\ 2\sqrt{5} \qquad\textbf{(E)... | 2000 AMC 10 Problem 16 | Drawing line $\overline{BD}$ and parallel line $\overline{CF}$, we see that $\triangle FCE \sim \triangle BDE$ by AA similarity. Thus $\frac{FE}{EB} = \frac{FC}{DB} = \frac{2}{4} = \frac{1}{2}$. Reciprocating, we know that $\frac{EB}{FE} = 2$ so $\frac{EB+FE}{FE} = 2+1 \Rightarrow \frac{FB}{FE} = 3$. Reciprocating agai... | // Block 1
path seg1, seg2;
seg1=(6,0)--(0,3);
seg2=(2,0)--(4,2);
dot((0,0));
dot((1,0));
fill(circle((2,0),0.1),black);
dot((3,0));
dot((4,0));
dot((5,0));
fill(circle((6,0),0.1),black);
dot((0,1));
dot((1,1));
dot((2,1));
dot((3,1));
dot((4,1));
dot((5,1));
dot((6,1));
dot((0,2));
dot((1,2));
dot((2,2));
dot((3,2));
... | [] |
556 | The diagram shows $28$ lattice points, each one unit from its nearest neighbors. Segment $AB$ meets segment $CD$ at $E$. Find the length of segment $AE$.
$\textbf{(A)}\ \frac{4\sqrt{5}}{3} \qquad\textbf{(B)}\ \frac{5\sqrt{5}}{3} \qquad\textbf{(C)}\ \frac{12\sqrt{5}}{7} \qquad\textbf{(D)}\ 2\sqrt{5} \qquad\textbf{(E)... | 2000 AMC 10 Problem 16 | Extend line $\overline{DC}$ as above. This creates two similar triangles whose side lengths have the ratio $5:4$. Therefore $AE=\frac{5}{9}AB$. Using Pythagorean theorem to find $AB$ gives us:
\[AE=\frac{5}{9}AB=\frac{5}{9}\sqrt{3^2+6^2}=\frac{5}{9}\sqrt{45}= \boxed{\textbf{(B) }\frac{5\sqrt{5}}{3}}\]
~ proloto | // Block 1
// Coordinates
pair A = (0,3), B = (6,0), C = (4,2), D = (2,0);
path seg1 = B--A;
path seg2 = D--C;
pair[] intersectionPoints = intersectionpoints(seg1, seg2);
pair E = intersectionPoints[0];
for (int i = 0; i <= 6; i = i + 1) {
for (int j = 0; j <= 3; j = j + 1) {
dot((i,j));
}
}
// Draw
d... | [] |
557 | Charlyn walks completely around the boundary of a square whose sides are each $5$ km long. From any point on her path she can see exactly $1$ km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the near... | 2000 AMC 10 Problem 18 | The area she sees looks at follows:
The part inside the walk has area $5\cdot 5 - 3\cdot 3 = 16$. The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area $5\cdot 1=5$. The four arcs together form a circle with radius $1$.
Therefore the total area she can see is $16 + 4... | // Block 1
unitsize(0.8cm);
path p1 = (0,0)--(5,0)--(5,5)--(0,5)--cycle;
path p2 = (1,1)--(4,1)--(4,4)--(1,4)--cycle;
path p3 = arc((0,0),1,180,270) -- arc((5,0),1,270,360) -- arc((5,5),1,0,90) -- arc((0,5),1,90,180) -- cycle;
fill(p3,lightgray);
unfill(p2);
draw(p1,linewidth(bp)); draw(p2); draw(p3);
draw( (0,0)--(-... | [] |
558 | Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right ... | 2000 AMC 10 Problem 19 | Solution 1
WLOG, let a side of the square be $1$. Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since $A = \frac{1}{2}bh = \frac{h}{2}$, the base of the triangle with area $m$ is $2m$. Therefore $\frac{2m}{1} = \frac{1}{x}$ where $x$ is ... | // Block 1
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W);
// Block 2
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$b$",(2.5,0),S); label... | [] |
558 | Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right ... | 2000 AMC 10 Problem 19 | Because we know that the right triangles are always similar, in a figure
, x/y = y/a. Using cross products, we can say that xa= y^2. We are trying to solve for x because m is in terms of a and y; m= a/2y (the ratio of the area of one triangle to the area of the square). So solving for x, x= y^2/a. Then the question i... | unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$y$",(1,2),S); label("$y$",(2,1),W); label("$a$",(4,0),S); label("$x$",(0,2.5),W); | [] |
559 | Figures $0$, $1$, $2$, and $3$ consist of $1$, $5$, $13$, and $25$ nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?
$\textbf{(A)}\ 10401 \qquad\textbf{(B)}\ 19801 \qquad\textbf{(C)}\ 20201 \qquad\textbf{(D)}\ 39801 \qquad\t... | 2000 AMC 12 Problem 8 | By taking figure $n$, putting the centers of each square on a grid, and rotating the figure 45 degrees, we get the following shape (figure 3 is shown for reference):
There are two lattices of points (shown in red): one of side length $n$ (blue) and one of $n+1$ (green). The rest follows like in solution 1.
-Integrala... | // Block 1
import olympiad;
int fig = 3;
for (int i = 0; i <= 2*fig; ++i) {
for (int j = 0; j <= 2*fig; ++j) {
pair p = (i, j);
//do we need this point?
if ((i+j)%2 == 1) {
continue;
}
//draw squares
draw((p+N)--(p+E)--(p+S)--(p+W)--cycle);
//draw lattices
if (i < 2*fig-1) {
... | [] |
560 | Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right ... | 2000 AMC 12 Problem 21 | Solution 1
WLOG, let a side of the square be $1$. Simple angle chasing shows that the two right triangles are similar. Thus the ratio of the sides of the triangles are the same. Since $A = \frac{1}{2}bh = \frac{h}{2}$, the base of the triangle with area $m$ is $2m$. Therefore $\frac{2m}{1} = \frac{1}{x}$ where $x$ is ... | // Block 1
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$1$",(1,2),S); label("$1$",(2,1),W); label("$2m$",(4,0),S); label("$x$",(0,2.5),W);
// Block 2
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$b$",(2.5,0),S); label... | [] |
560 | Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is $m$ times the area of the square. The ratio of the area of the other small right ... | 2000 AMC 12 Problem 21 | Because we know that the right triangles are always similar, in a figure
, x/y = y/a. Using cross products, we can say that xa= y^2. We are trying to solve for x because m is in terms of a and y; m= a/2y (the ratio of the area of one triangle to the area of the square). So solving for x, x= y^2/a. Then the question i... | // Block 1
unitsize(36);
draw((0,0)--(6,0)--(0,3)--cycle);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
label("$y$",(1,2),S);
label("$y$",(2,1),W);
label("$a$",(4,0),S);
label("$x$",(0,2.5),W);
// Block 2
unitsize(36); draw((0,0)--(6,0)--(0,3)--cycle); draw((0,0)--(2,0)--(2,2)--(0,2)--cycle); label("$y$",(1,2),S); label("$... | [] |
561 | Eight congruent equilateral triangles, each of a different color, are used to construct a regular octahedron. How many distinguishable ways are there to construct the octahedron? (Two colored octahedrons are distinguishable if neither can be rotated to look just like the other.)
$\textbf {(A)}\ 210 \qquad \textbf {(... | 2000 AMC 12 Problem 25 | Since the octahedron is indistinguishable by rotations, without loss of generality fix a face to be red.
There are $7!$ ways to arrange the remaining seven colors, but there still are three possible rotations about the fixed face, so the answer is $7!/3 = 1680$. | // Block 1
size(8cm);
defaultpen(0.5);
import three;
import math;
currentprojection=orthographic(2,0.2,1);
triple A=(0,0,1);
triple B=(sqrt(2)/2,sqrt(2)/2,0);
triple C=(sqrt(2)/2,-sqrt(2)/2,0);
triple D=(-sqrt(2)/2,-sqrt(2)/2,0);
triple E=(-sqrt(2)/2,sqrt(2)/2,0);
triple F=(0,0,-1);
draw(A--B--E--cycle);
draw(A--C--D--... | [] |
562 | A street has parallel curbs $40$ feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is $15$ feet and each stripe is $50$ feet long. Find the distance, in feet, between the stripes.
$\textbf{(A)}\ 9 \qquad \textbf{(B)}\ 10 \qquad \textbf{(C... | 2001 AMC 10 Problem 15 | Drawing the problem out, we see we get a parallelogram with a height of $40$ and a base of $15$, giving an area of $600$.
If we look at it the other way, we see the distance between the stripes is the height and the base is $50$.
The area is still the same, so the distance between the stripes is $600/50 = \box... | // Block 1
draw((0,0)--(5,0));
draw((2.5,5)--(7.5,5));
draw((0,0)--(2.5,5));
draw((5,0)--(7.5,5),linewidth(2));
draw((2,4)--(6,2),dashed);
// Block 2
draw((0,0)--(5,0)); draw((2.5,5)--(7.5,5)); draw((0,0)--(2.5,5)); draw((5,0)--(7.5,5),linewidth(2)); draw((2,4)--(6,2),dashed); | [] |
563 | Which of the cones listed below can be formed from a $252^\circ$ sector of a circle of radius $10$ by aligning the two straight sides?
$\textbf{(A)} \text{ A cone with slant height of } 10 \text{ and radius } 6$
$\textbf{(B)} \text{ A cone with height of } 10 \text{ and radius } 6$
$\textbf{(C)} \text{ A cone wit... | 2001 AMC 10 Problem 17 | The blue lines will be joined together to form a single blue line on the surface of the cone, so $10$ will be the slant height of the cone.
The red line will form the circumference of the base. We can compute its length and use it to determine the radius.
The length of the red line is $\dfrac{252^{\circ}}{360^{\circ}... | import graph; unitsize(1.5cm); defaultpen(fontsize(8pt)); draw(Arc((0,0),1,-72,180),linewidth(.8pt) + red); draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt) + blue); label("$10$",(-0.5,0),S); draw(Arc((0,0),0.1,-72,180)); label("$252^{\circ}$",(0.05,0.05),NE); | [] |
564 | The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to
$\textbf{(A) }50 \qquad \textbf{(B) }52 \qquad \textbf{(C) }54 \qquad \textbf{(D) }56 \qquad \textbf{(E) }58$ | 2001 AMC 10 Problem 18 | Consider any single tile:
If the side of the small square is $a$, then the area of the tile is $\left(3a\right)^2 = 9a^2$, with $4a^2$ covered by squares, and therefore $9a^2-4a^2 = 5a^2$ by pentagons.
Hence exactly $\frac{5}{9}$ of any tile is covered by pentagons, and therefore pentagons cover $\frac{5}{9}$ of the... | unitsize(1cm); defaultpen(linewidth(0.8pt)); path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0); path p2=(0,1)--(1,1)--(1,0); path p3=(2,0)--(2,1)--(3,1); path p4=(3,2)--(2,2)--(2,3); path p5=(1,3)--(1,2)--(0,2); path p6=(1,1)--(2,2); path p7=(2,1)--(1,2); path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7; draw(p); | [] |
565 | A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
$\textbf{(A)}\ \frac{8}3\qquad\textbf{(B)}\ \frac{30}{11}\qquad\textbf{(C)}\ 3\qquad\tex... | 2001 AMC 10 Problem 21 | Let the diameter of the cylinder be $2r$. Examining the cross section of the cone and cylinder, we find two similar triangles. Hence, $\frac{12-2r}{12}=\frac{2r}{10}$ which we solve to find $r=\frac{30}{11}$. Our answer is $\boxed{\textbf{(B)}\ \frac{30}{11}}$. | // Block 1
draw((5,0)--(-5,0)--(0,12)--cycle);
unitsize(.75cm);
draw((-30/11,0)--(-30/11,60/11));
draw((-30/11,60/11)--(30/11,60/11));
draw((30/11,60/11)--(30/11,0));
draw((0,0)--(0,12));
label("$2r$",(0,30/11),E);
label("$12-2r$",(0,80/11),E);
label("$2r$",(0,60/11),S);
label("$10$",(0,0),S);
label("$A$",(0,12),N);
la... | [] |
565 | A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter $10$ and altitude $12$, and the axes of the cylinder and cone coincide. Find the radius of the cylinder.
$\textbf{(A)}\ \frac{8}3\qquad\textbf{(B)}\ \frac{30}{11}\qquad\textbf{(C)}\ 3\qquad\tex... | 2001 AMC 10 Problem 21 | Like in Solution 2, we draw a diagram.
It is known that $\overline{AH}$ has length $12$ and $\overline{BC}$ has length $10$, so triangle $\triangle ABC$ has area $60$. Also, let $x$ be equal to the radius of the cylinder.
Triangles $\triangle DBG$ and $\triangle ECF$ can be combined into one triangle with base $1... | // Block 1
draw((5,0)--(-5,0)--(0,12)--cycle);
unitsize(.75cm);
draw((-30/11,0)--(-30/11,60/11));
draw((-30/11,60/11)--(30/11,60/11));
draw((30/11,60/11)--(30/11,0));
draw((0,0)--(0,12));
label("$2x$",(0,30/11),E);
label("$2x$",(0,60/11),S);
label("$H$",(0,0),S);
label("$A$",(0,12),N);
label("$B$",(-5,0),SW);
label("$C... | [] |
566 | In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$, $w$, $x$, $y$, and $z$. Find $y + z$.
$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$ | 2001 AMC 10 Problem 22 | We know that $y+z=2v$, so we could find one variable rather than two.
$v+24+w=43+v$
$24+w=43$
$w=19$
$44+x=24+x+z \implies z=20$
The sum per row is $25+21+20=66$.
Thus $66-18-25=66-43=v=23$.
Since we needed $2v$ and we know $v=23$, $23 \times 2 = \boxed{\textbf{(D)}\ 46}$. | // Block 1
unitsize(10mm);
defaultpen(linewidth(1pt));
for(int i=0; i<=3; ++i)
{
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}
label("$25$",(0.5,0.5));
label("$z$",(1.5,0.5));
label("$21$",(2.5,0.5));
label("$18$",(0.5,1.5));
label("$x$",(1.5,1.5));
label("$y$",(2.5,1.5));
label("$v$",(0.5,2.5));
label("$24$",(1.5,2.5));
l... | [] |
566 | In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by $v$, $w$, $x$, $y$, and $z$. Find $y + z$.
$\textbf{(A)}\ 43 \qquad \textbf{(B)}\ 44 \qquad \textbf{(C)}\ 45 \qquad \textbf{(D)}\ 46 \qquad \textbf{(E)}\ 47$ | 2001 AMC 10 Problem 22 | $v+24+w=43+v$
$24+w=43$
$w=19$
$44+x=24+x+z \implies z=20$
The magic sum is determined by the bottom row. $25+20+21=66$.
Solving for $y$:
$y=66-19-21=66-40=26$.
To find our answer, we need to find $y+z$. $y+z=20+26 = \boxed{\textbf{(D)}\ 46}$. | // Block 1
unitsize(10mm);
defaultpen(linewidth(1pt));
for(int i=0; i<=3; ++i)
{
draw((0,i)--(3,i));
draw((i,0)--(i,3));
}
label("$25$",(0.5,0.5));
label("$z$",(1.5,0.5));
label("$21$",(2.5,0.5));
label("$18$",(0.5,1.5));
label("$x$",(1.5,1.5));
label("$y$",(2.5,1.5));
label("$v$",(0.5,2.5));
label("$24$",(1.5,2.5));
l... | [] |
567 | In trapezoid $ABCD$, $\overline{AB}$ and $\overline{CD}$ are perpendicular to $\overline{AD}$, with $AB+CD=BC$, $AB<CD$, and $AD=7$. What is $AB\cdot CD$?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 12.25 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 12.75 \qquad \textbf{(E)}\ 13$ | 2001 AMC 10 Problem 24 | If $AB=x$ and $CD=y$, then $BC=x+y$. By the Pythagorean theorem, we have $(x+y)^2=(y-x)^2+49.$
Solving the equation, we get $4xy=49 \implies xy = \boxed{\textbf{(B)}\ 12.25}$. | // Block 1
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */
import graph; size(7cm);
real labelscalefactor = 0.5; /* changes label-to-point distance */
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */
pen dotstyle = black... | [] |
568 | How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$?
$\textbf{(A) }768 \qquad \textbf{(B) }801 \qquad \textbf{(C) }934 \qquad \textbf{(D) }1067 \qquad \textbf{(E) }1167$ | 2001 AMC 10 Problem 25 | We have the following Venn diagram (sorry for the bad asymptote). To calculate this Venn diagram we can just use PIE to find the numbers or just work our way out from the inside. We just need to find the numbers that are divisible by $3$ or $4$ but not by $5$, and looking back at the diagram we just have the answer of ... | size(10cm); // Draw circles draw(circle((0,0),2), blue); // Circle for multiples of 3 draw(circle((2,0),2), red); // Circle for multiples of 4 draw(circle((1,-1.5),2), green); // Circle for multiples of 5 // Labels for the sets label("$3$", (-2.5,1.5), blue); label("$4$", (4.5,1.5), red); label("$5$", (1,-3.8), green... | [] |
569 | Which of the cones listed below can be formed from a $252^\circ$ sector of a circle of radius $10$ by aligning the two straight sides?
$\textbf{(A)} \text{ A cone with slant height of } 10 \text{ and radius } 6$
$\textbf{(B)} \text{ A cone with height of } 10 \text{ and radius } 6$
$\textbf{(C)} \text{ A cone wit... | 2001 AMC 12 Problem 8 | The blue lines will be joined together to form a single blue line on the surface of the cone, so $10$ will be the slant height of the cone.
The red line will form the circumference of the base. We can compute its length and use it to determine the radius.
The length of the red line is $\dfrac{252^{\circ}}{360^{\circ}... | // Block 1
import graph;
unitsize(1.5cm);
defaultpen(fontsize(8pt));
draw(Arc((0,0),1,-72,180),linewidth(.8pt) + red);
draw(dir(288)--(0,0)--(-1,0),linewidth(.8pt) + blue);
label("$10$",(-0.5,0),S);
draw(Arc((0,0),0.1,-72,180));
label("$252^{\circ}$",(0.05,0.05),NE);
// Block 2
import graph; unitsize(1.5cm); defaultpe... | [] |
570 | The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest to
$\textbf{(A) }50 \qquad \textbf{(B) }52 \qquad \textbf{(C) }54 \qquad \textbf{(D) }56 \qquad \textbf{(E) }58$ | 2001 AMC 12 Problem 10 | Consider any single tile:
If the side of the small square is $a$, then the area of the tile is $\left(3a\right)^2 = 9a^2$, with $4a^2$ covered by squares, and therefore $9a^2-4a^2 = 5a^2$ by pentagons.
Hence exactly $\frac{5}{9}$ of any tile is covered by pentagons, and therefore pentagons cover $\frac{5}{9}$ of the... | // Block 1
unitsize(1cm);
defaultpen(linewidth(0.8pt));
path p1=(0,0)--(3,0)--(3,3)--(0,3)--(0,0);
path p2=(0,1)--(1,1)--(1,0);
path p3=(2,0)--(2,1)--(3,1);
path p4=(3,2)--(2,2)--(2,3);
path p5=(1,3)--(1,2)--(0,2);
path p6=(1,1)--(2,2);
path p7=(2,1)--(1,2);
path[] p=p1^^p2^^p3^^p4^^p5^^p6^^p7;
draw(p);
// Block 2
uni... | [] |
571 | How many positive integers not exceeding $2001$ are multiples of $3$ or $4$ but not $5$?
$\textbf{(A) }768 \qquad \textbf{(B) }801 \qquad \textbf{(C) }934 \qquad \textbf{(D) }1067 \qquad \textbf{(E) }1167$ | 2001 AMC 12 Problem 12 | We have the following Venn diagram (sorry for the bad asymptote). To calculate this Venn diagram we can just use PIE to find the numbers or just work our way out from the inside. We just need to find the numbers that are divisible by $3$ or $4$ but not by $5$, and looking back at the diagram we just have the answer of ... | // Block 1
size(10cm);
// Draw circles
draw(circle((0,0),2), blue); // Circle for multiples of 3
draw(circle((2,0),2), red); // Circle for multiples of 4
draw(circle((1,-1.5),2), green); // Circle for multiples of 5
// Labels for the sets
label("$3$", (-2.5,1.5), blue);
label("$4$", (4.5,1.5), red);
label("$5$", (1,-... | [] |
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