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 |
|---|---|---|---|---|---|
226 | 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 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);
... | [] |
226 | 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 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);
... | [] |
226 | 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 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);
... | [] |
226 | 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 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... | [] |
226 | 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 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... | [] |
227 | 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 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... | [] |
227 | 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 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);... | [] |
227 | 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 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... | [] |
228 | 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 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... | [] |
229 | 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 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... | [] |
230 | 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 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)... | [] |
230 | 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 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... | [] |
230 | 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 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... | [] |
231 | 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 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 =... | [] |
232 | 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 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... | [] |
233 | Find the number of integers $c$ such that the equation \[\left||20|x|-x^2|-c\right|=21\]has $12$ distinct real solutions. | 2021 AIME I Problem 8 | We take cases for the outermost absolute value, then rearrange: \[\left|20|x|-x^2\right|=c\pm21.\]
Let $f(x)=\left|20|x|-x^2\right|.$ We rewrite $f(x)$ as a piecewise function without using absolute values:
\[f(x) = \begin{cases} \left|-20x-x^2\right| & \mathrm{if} \ x \le 0 \begin{cases} 20x+x^2 & \mathrm{if} \ x\le-... | // Block 1
/* Made by MRENTHUSIASM */
size(1200,300);
real xMin = -65;
real xMax = 65;
real yMin = -50;
real yMax = 125;
draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
label("$x$",(xMax,0),(2,0));
label("$y$",(0,yMax),(0,2));
real f(real x) { re... | [] |
234 | 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 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... | [] |
234 | 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 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... | [] |
235 | Let $ABCD$ be a cyclic quadrilateral with $AB=4,BC=5,CD=6,$ and $DA=7.$ Let $A_1$ and $C_1$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to line $BD,$ and let $B_1$ and $D_1$ be the feet of the perpendiculars from $B$ and $D,$ respectively, to line $AC.$ The perimeter of $A_1B_1C_1D_1$ is $\frac mn... | 2021 AIME I Problem 11 | This solution refers to the Diagram section.
By the Converse of the Inscribed Angle Theorem, if distinct points $X$ and $Y$ lie on the same side of $\overline{PQ}$ (but not on $\overline{PQ}$ itself) for which $\angle PXQ=\angle PYQ,$ then $P,Q,X,$ and $Y$ are cyclic. From the Converse of the Inscribed Angle Theorem, ... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, C, D, A1, B1, C1, D1, P, M1, M2;
A = origin;
C = (sqrt(53041)/31,0);
B = intersectionpoints(Circle(A,4),Circle(C,5))[0];
D = intersectionpoints(Circle(A,7),Circle(C,6))[1];
A1 = foot(A,B,D);
C1 = foot(C,B,D);
B1 = foot(B,A,C);
D1 = foot(D,A,C);
P = intersecti... | [] |
236 | 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 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... | [] |
236 | 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 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... | [] |
236 | 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 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... | [] |
237 | For positive real numbers $s$, let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$. The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$. Find $a^2+b^2$. | 2021 AIME II Problem 5 | We have the diagram below.
We proceed by taking cases on the angles that can be obtuse, and finding the ranges for $s$ that they yield .
If angle $\theta$ is obtuse, then we have that $s \in (0,20)$. This is because $s=20$ is attained at $\theta = 90^{\circ}$, and the area of the triangle is strictly decreasing as ... | // Block 1
draw((0,0)--(1,2*sqrt(3)));
draw((1,2*sqrt(3))--(10,0));
draw((10,0)--(0,0));
label("$A$",(0,0),SW);
label("$B$",(1,2*sqrt(3)),N);
label("$C$",(10,0),SE);
label("$\theta$",(0,0),NE);
label("$\alpha$",(1,2*sqrt(3)),SSE);
label("$4$",(0,0)--(1,2*sqrt(3)),WNW);
label("$10$",(0,0)--(10,0),S);
// Block 2
draw((0,... | [] |
237 | For positive real numbers $s$, let $\tau(s)$ denote the set of all obtuse triangles that have area $s$ and two sides with lengths $4$ and $10$. The set of all $s$ for which $\tau(s)$ is nonempty, but all triangles in $\tau(s)$ are congruent, is an interval $[a,b)$. Find $a^2+b^2$. | 2021 AIME II Problem 5 | For $\triangle ABC,$ we fix $AB=10$ and $BC=4.$ Without the loss of generality, we consider $C$ on only one side of $\overline{AB}.$
As shown below, all locations for $C$ at which $\triangle ABC$ is an obtuse triangle are indicated in red, excluding the endpoints.
Note that:
The region in which $\angle B$ is obtuse... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
pair A, B, O, P, Q, C1, C2, D;
A = origin;
B = (10,0);
O = midpoint(A--B);
P = B - (4,0);
Q = B + (4,0);
C1 = intersectionpoints(D--D+(100,0),Arc(B,Q,P))[1];
C2 = B + (0,4);
D = intersectionpoint(Arc(O,B,A),Arc(B,Q,P));
draw(Arc(O,B,A)^^Arc(B,C2,D)^^A--Q);
draw(Arc(B,Q,... | [] |
238 | An ant makes a sequence of moves on a cube where a move consists of walking from one vertex to an adjacent vertex along an edge of the cube. Initially the ant is at a vertex of the bottom face of the cube and chooses one of the three adjacent vertices to move to as its first move. For all moves after the first move, th... | 2021 AIME II Problem 8 | For all positive integers $k,$ let
$N(k,\mathrm{BB})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the bottom face to the bottom face.
$N(k,\mathrm{BT})$ be the number of ways to make a sequence of exactly $k$ moves, where the last move is from the bottom face to the top f... | // Block 1
/* Made by MRENTHUSIASM */
size(9cm);
pair A, B, C, D, E, F, G, H, X, Y;
A=(0,6);
B=(0,4);
C=(0,2);
D=(0,0);
E=(10,6);
F=(10,4);
G=(10,2);
H=(10,0);
X=(-1,8);
Y=(11,8);
label("BB", A, (-2,0));
label("BT", B, (-2,0));
label("TB", C, (-2,0));
label("TT", D, (-2,0));
label("BB", E, (2,0));
lab... | [] |
239 | Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$. The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$. The distance from line $\ell$ to the point where the sphere with ra... | 2021 AIME II Problem 10 | This solution refers to the Diagram section.
As shown below, let $O_1,O_2,O_3$ be the centers of the spheres (where sphere $O_3$ has radius $13$) and $T_1,T_2,T_3$ be their respective points of tangency to plane $\mathcal{P}.$ Let $\mathcal{R}$ be the plane that is determined by $O_1,O_2,$ and $O_3.$ Suppose $A$ is th... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
import graph3;
import solids;
currentprojection=orthographic((1,1/2,0));
triple O1, O2, O3, T1, T2, T3, A, L1, L2;
O1 = (0,-36,0);
O2 = (0,36,0);
O3 = (0,0,-sqrt(1105));
T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105);
T2 = (864*sqrt(1105)/1105,36,-828*sqrt(1105)/11... | [] |
239 | Two spheres with radii $36$ and one sphere with radius $13$ are each externally tangent to the other two spheres and to two different planes $\mathcal{P}$ and $\mathcal{Q}$. The intersection of planes $\mathcal{P}$ and $\mathcal{Q}$ is the line $\ell$. The distance from line $\ell$ to the point where the sphere with ra... | 2021 AIME II Problem 10 | This solution refers to the Diagram section.
The isosceles triangle of centers $O_1 O_2 O$ ($O$ is the center of sphere of radii $13$) has sides $O_1 O = O_2 O = 36 + 13 = 49,$ and $O_1 O_2 = 36 + 36 = 72.$
Let $N$ be the midpoint $O_1 O_2$.
The isosceles triangle of points of tangency $T_1 T_2 T$ has sides $T_1... | // Block 1
/* Made by MRENTHUSIASM */
size(300);
import graph3;
import solids;
currentprojection=orthographic((10,-3,-40));
triple O1, O2, O3, T1, T2, T3, A, L1, L2, M;
O1 = (0,-36,0);
O2 = (0,36,0);
O3 = (0,0,-sqrt(1105));
T1 = (864*sqrt(1105)/1105,-36,-828*sqrt(1105)/1105);
T2 = (864*sqrt(1105)/1105,36,-828*sqrt(11... | [] |
240 | A teacher was leading a class of four perfectly logical students. The teacher chose a set $S$ of four integers and gave a different number in $S$ to each student. Then the teacher announced to the class that the numbers in $S$ were four consecutive two-digit positive integers, that some number in $S$ was divisible by $... | 2021 AIME II Problem 11 | Note that $\operatorname{lcm}(6,7)=42.$ It is clear that $42\not\in S$ and $84\not\in S,$ otherwise the three other elements in $S$ are divisible by neither $6$ nor $7.$
In the table below, the multiples of $6$ are colored in yellow, and the multiples of $7$ are colored in green. By the least common multiple, we obtai... | // Block 1
/* Made by MRENTHUSIASM */
size(20cm);
fill((5,0)--(6,0)--(6,2)--(5,2)--cycle,yellow);
fill((11,0)--(12,0)--(12,3)--(11,3)--cycle,yellow);
fill((17,1)--(18,1)--(18,3)--(17,3)--cycle,yellow);
fill((23,1)--(24,1)--(24,3)--(23,3)--cycle,yellow);
fill((29,1)--(30,1)--(30,3)--(29,3)--cycle,yellow);
fill((35,1)--... | [] |
241 | A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2021 AIME II Problem 12 | Since we are asked to find $\tan \theta$, we can find $\sin \theta$ and $\cos \theta$ separately and use their values to get $\tan \theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$... | // Block 1
unitsize(4cm);
pair A,B,C,D,X;
A = (0,0);
B = (1,0);
C = (1.25,-1);
D = (-0.75,-0.75);
draw(A--B--C--D--cycle,black+1bp);
X = intersectionpoint(A--C,B--D);
draw(A--C);
draw(B--D);
label("$A$",A,NW);
label("$B$",B,NE);
label("$C$",C,SE);
label("$D$",D,SW);
dot(X);
label("$X$",X,S);
label("$5$",(A+B)/2,N);
lab... | [] |
241 | A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2021 AIME II Problem 12 | In convex quadrilateral $ABCD,$ let $AB=5,BC=6,CD=9,$ and $DA=7.$ Let $A'$ and $C'$ be the feet of the perpendiculars from $A$ and $C,$ respectively, to $\overline{BD}.$ We obtain the following diagram:
Let $BC'=p,C'E=q,EA'=r,A'D=s,AA'=h_1,$ and $CC'=h_2.$ We apply the Pythagorean Theorem to right triangles $\triangle... | // Block 1
/* Made by MRENTHUSIASM */
size(500);
pair A, B, C, D, P, A1, C1;
B = origin;
D = (3*sqrt(32498*(29400*sqrt(47)+312523))/32498,0);
A = intersectionpoints(Circle(B,5),Circle(D,7))[0];
C = intersectionpoints(Circle(B,6),Circle(D,9))[1];
P = intersectionpoint(A--C,B--D);
A1 = foot(A,B,D);
C1 = foot(C,B,D);
mar... | [] |
241 | A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. | 2021 AIME II Problem 12 | Bretschneider's Formula
Given quadrilateral $ABCD$, let, $a, b, c, d$, be the sides, $s$ the semiperimeter, and $u, v$, the diagonals. Then the area, $K$, is given by \[K = \tfrac 14 \sqrt {4u^2v^2-(b^2+d^2-a^2-c^2)^2}\]
Solution
By Bretschneider's Formula, \[30=\tfrac{1}{4}\sqrt{4u^2v^2-(b^2+d^2-a^2-c^2)^2}=\tfrac{1... | size(200); import olympiad; defaultpen(linewidth(0.8)+fontsize(10)); pair A,B,C,D; A=origin; B=(1.25,0); D=dir(65); C=D+0.85*dir(90)*(A-D); draw(A--B--C--D--cycle); draw(A--C^^B--D, gray+0.5); dot("$A$", A, SW); dot("$B$", B, SE); dot("$C$", C, NE); dot("$D$",D,NW); label("$a$", A--B, S); label("$b$", B--C, E); label... | [] |
242 | Let $\Delta ABC$ be an acute triangle with circumcenter $O$ and centroid $G$. Let $X$ be the intersection of the line tangent to the circumcircle of $\Delta ABC$ at $A$ and the line perpendicular to $GO$ at $G$. Let $Y$ be the intersection of lines $XG$ and $BC$. Given that the measures of $\angle ABC, \angle BCA,$ and... | 2021 AIME II Problem 14 | $\angle OAX = \angle OGX = 90^\circ \implies$ quadrilateral $XAGO$ is cyclic $\implies$
$\angle GXO = \angle GAO,$ as they share the same intersept $\overset{\Large\frown} {GO}.$
$\angle OGY = \angle OMY = 90^\circ \implies$ quadrilateral $OGYM$ is cyclic $\implies$
$\angle GYO = \angle OMG,$ as they share the sa... | // Block 1
/* Made by MRENTHUSIASM */
size(375);
pair A, B, C, O, G, X, Y;
A = origin;
B = (1,0);
C = extension(A,A+10*dir(585/7),B,B+10*dir(180-585/7));
O = circumcenter(A,B,C);
G = centroid(A,B,C);
Y = intersectionpoint(G--G+(100,0),B--C);
X = intersectionpoint(G--G-(100,0),A--scale(100)*rotate(90)*dir(O-A));
pair O... | ["https://artofproblemsolving.com/wiki/images/thumb/6/6c/AIME-II-2021-14.png/230px-AIME-II-2021-14.png"] |
243 | 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 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... | [] |
243 | 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 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... | [] |
243 | 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 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... | [] |
243 | 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 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... | [] |
243 | 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 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... | [] |
243 | 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 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... | [] |
244 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per ... | 2022 AIME I Problem 5 | Define $m$ as the number of minutes they swim for.
Let their meeting point be $A$. Melanie is swimming against the current, so she must aim upstream from point $A$, to compensate for this; in particular, since she is swimming for $m$ minutes, the current will push her $14m$ meters downstream in that time, so she must ... | // Block 1
unitsize(0.02cm);
draw((0,0)--(0,264)--(550,264)--(550,0)--cycle);
pair B = (198,264);
dot(B^^(0,0)^^(550,0),linewidth(5));
draw((0,0)--B,dashed);
draw((550,0)--B,dashed);
label("$60m$", (0,0)--B, E);
label("$80m$", (550,0)--B, W);
label("$264$", (0,0)--(0,264), W);
label("$\frac{D}{2} - 14m$", (0,264)--B, ... | [] |
244 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per ... | 2022 AIME I Problem 5 | We have the following diagram:
Since Melanie and Sherry swim for the same distance and the same amount of time, they swim at the same net speed.
Let $x$ and $y$ be some positive numbers. We have the following table:
\[\begin{array}{c||c|c|c} & \textbf{Net Velocity Vector (m/min)} & \textbf{Natural Velocity Vector... | // Block 1
/* Made by MRENTHUSIASM */
size(350);
pair A, B, C;
A = (0,264);
B = (-275,0);
C = (275,0);
draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2));
dot("Finish",A,1.75*N,linewidth(5));
dot("Sherry",B,1.75*S,linewidth(5));
dot("Melanie",C,1.75*S,linewidth(5));
Label L1 = Label("$D$", align=(0... | [] |
244 | A straight river that is $264$ meters wide flows from west to east at a rate of $14$ meters per minute. Melanie and Sherry sit on the south bank of the river with Melanie a distance of $D$ meters downstream from Sherry. Relative to the water, Melanie swims at $80$ meters per minute, and Sherry swims at $60$ meters per ... | 2022 AIME I Problem 5 | We can break down movement into two components: the $x$-component and the $y$-component. Suppose that Melanie travels a distance of $a$ in the $x$-direction and a distance of $c$ in the $y$-direction in one minute when there is no current. Similarly, suppose that Sherry travels a distance of $a$ in the $x$-direction bu... | // Block 1
/* Ruthlessly plagiarized from MRENTHUSIASM by Curious_crow */
size(350);
pair A, B, C;
A = (0,264);
B = (-275,0);
C = (275,0);
draw((-300,0)--(300,0)^^(-300,264)--(300,264)^^A--B^^A--C,linewidth(2));
dot("Finish",A,1.75*N,linewidth(5));
dot("Sherry",B,1.75*S,linewidth(5));
dot("Melanie",C,1.75*S,linewidth(5... | [] |
245 | 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 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 =... | [] |
245 | 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 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 = ... | [] |
246 | 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 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)... | [] |
247 | 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 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... | [] |
248 | A circle with radius $6$ is externally tangent to a circle with radius $24$. Find the area of the triangular region bounded by the three common tangent lines of these two circles. | 2022 AIME II Problem 7 | $r_1 = O_1A = 24$, $r_2 = O_2B = 6$, $AG = BO_2 = r_2 = 6$, $O_1G = r_1 - r_2 = 24 - 6 = 18$, $O_1O_2 = r_1 + r_2 = 30$
$\triangle O_2BD \sim \triangle O_1GO_2$, $\frac{O_2D}{O_1O_2} = \frac{BO_2}{GO_1}$, $\frac{O_2D}{30} = \frac{6}{18}$, $O_2D = 10$
$CD = O_2D + r_2 = 10 + 6 = 16$,
$EF = 2EC = EA + EB = AB = GO_2 ... | // Block 1
//Created by isabelchen
size(12cm, 12cm);
draw(circle((0,0),24));
draw(circle((30,0),6));
draw((72/5, 96/5) -- (40,0));
draw((72/5, -96/5) -- (40,0));
draw((24, 12) -- (24, -12));
draw((0, 0) -- (40, 0));
draw((72/5, 96/5) -- (0,0));
draw((168/5, 24/5) -- (30,0));
draw((54/5, 72/5) -- (30,0));
dot((72/5, ... | [] |
248 | A circle with radius $6$ is externally tangent to a circle with radius $24$. Find the area of the triangular region bounded by the three common tangent lines of these two circles. | 2022 AIME II Problem 7 | First, we want to find $O_2D$. We know that $\angle O_1AD = \angle O_2BD = 90^{\circ}$, so by AA similarity, $\triangle O_1AD \sim \triangle O_2BD$. We want to find the length of $x$, and using the similar triangles, we write an equation: $\frac{30 + x}{4} = x$. Solving, we get $x=10$. Therefore, $CD = 10 + 6 = 16$. Ne... | // Block 1
//Created by isabelchen and edited by afly
size(12cm, 12cm);
draw(circle((0,0),24));
draw(circle((30,0),6));
draw((72/5, 96/5) -- (40,0));
draw((72/5, -96/5) -- (40,0));
draw((24, 12) -- (24, -12));
draw((0, 0) -- (40, 0));
draw((72/5, 96/5) -- (0,0));
draw((168/5, 24/5) -- (30,0));
draw((54/5, 72/5) -- (3... | [] |
249 | Let $ABCD$ be a convex quadrilateral with $AB=2, AD=7,$ and $CD=3$ such that the bisectors of acute angles $\angle{DAB}$ and $\angle{ADC}$ intersect at the midpoint of $\overline{BC}.$ Find the square of the area of $ABCD.$ | 2022 AIME II Problem 11 | According to the problem, we have $AB=AB'=2$, $DC=DC'=3$, $MB=MB'$, $MC=MC'$, and $B'C'=7-2-3=2$
Because $M$ is the midpoint of $BC$, we have $BM=MC$, so: \[MB=MB'=MC'=MC.\]
Then, we can see that $\bigtriangleup{MB'C'}$ is an isosceles triangle with $MB'=MC'$
Therefore, we could start our angle chasing: $\angle{MB'C... | // Block 1
defaultpen(fontsize(12)+0.6); size(300);
pair A,B,C,D,M,H; real xb=71, xd=121;
A=origin; D=(7,0); B=2*dir(xb); C=3*dir(xd)+D; M=(B+C)/2; H=foot(M,A,D); path c=CR(D,3); pair A1=bisectorpoint(D,A,B), D1=bisectorpoint(C,D,A), Bp=IP(CR(A,2),A--H), Cp=IP(CR(D,3),D--H);
draw(B--A--D--C--B); draw(A--M--D^^M--H^^B... | [] |
250 | 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 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"] |
250 | 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 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... | [] |
251 | 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 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,... | [] |
251 | 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 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,... | [] |
251 | 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 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... | [] |
251 | 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 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... | [] |
252 | 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 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 = ... | [] |
252 | 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 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... | [] |
253 | 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 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... | [] |
253 | 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 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,... | [] |
254 | 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 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--... | [] |
254 | 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 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... | [] |
255 | The following analog clock has two hands that can move independently of each other.
Initially, both hands point to the number $12$. The clock performs a sequence of hand movements so that on each movement, one of the two hands moves clockwise to the next number on the clock face while the other hand does not move.
Le... | 2023 AIME I Problem 14 | This problem is, in essence, the following: A $12\times12$ coordinate grid is placed on a (flat) torus; how many loops are there that pass through each point while only moving up or right? In other words, Felix the frog starts his journey at $(0,0)$ in the coordinate plane. Every second, he jumps either to the right... | // Block 1
size(7cm);
for (int x=0; x<12; ++x){
for (int y=0; y<12; ++y){
fill(circle((x,y),0.05));}}
for (int i=0; i<12; ++i){
fill(circle((i,11-i),0.1),red);}
pen p=green+dashed;
path u=(3,8)--(4,8)--(4,9)--(4,10)--(4,11)--(5,11)--(5,11.5);
path v=(5,-0.5)--(5,0)--(5,1)--(6,1)--(6,2)--(6,3)--(6,4)--(7,4);
draw(u,p);
... | [] |
256 | 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 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... | [] |
257 | 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 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... | [] |
258 | Rectangles $ABCD$ and $EFGH$ are drawn such that $D,E,C,F$ are collinear. Also, $A,D,H,G$ all lie on a circle. If $BC=16$,$AB=107$,$FG=17$, and $EF=184$, what is the length of $CE$? | 2024 AIME I Problem 5 | We use simple geometry to solve this problem.
We are given that $A$, $D$, $H$, and $G$ are concyclic; call the circle that they all pass through circle $\omega$ with center $O$. We know that, given any chord on a circle, the perpendicular bisector to the chord passes through the center; thus, given two chords, taki... | // Block 1
import graph;
unitsize(0.1cm);
pair A = (0,0);pair B = (107,0);pair C = (107,16);pair D = (0,16);pair E = (3,16);pair F = (187,16);pair G = (187,33);pair H = (3,33);
label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, NW);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);labe... | [] |
258 | Rectangles $ABCD$ and $EFGH$ are drawn such that $D,E,C,F$ are collinear. Also, $A,D,H,G$ all lie on a circle. If $BC=16$,$AB=107$,$FG=17$, and $EF=184$, what is the length of $CE$? | 2024 AIME I Problem 5 | Suppose $DE=x$. Extend $AD$ and $GH$ until they meet at $P$. From the Power of a Point Theorem, we have $(PH)(PG)=(PD)(PA)$. Substituting in these values, we get $(x)(x+184)=(17)(33)=561$. We can use guess and check to find that $x=3$, so $EC=\boxed{104}$.
~alexanderruan
~diagram by Technodoggo
rabbit47 - quadratic... | // Block 1
import graph;
unitsize(0.1cm);
pair A = (0,0);pair B = (107,0);pair C = (107,16);pair D = (0,16);pair E = (3,16);pair F = (187,16);pair G = (187,33);pair H = (3,33);pair P = (0,33);
label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, W);label("$E$", E, S);label("$F$", F, SE);label("$G... | [] |
259 | Find the largest possible real part of\[(75+117i)z+\frac{96+144i}{z}\]where $z$ is a complex number with $|z|=4$. Here $i = \sqrt{-1}$. | 2024 AIME I Problem 7 | Suppose $z = 4e^{i\theta}$
\begin{align}
& \phantom{\text{ = }}\mathrm{Re}\left((75 + 117i)z + \dfrac{96 + 144i}{z}\right)\\
& = \mathrm{Re}\left((300 + 468i)e^{i\theta} + (24 + 36i)e^{-i\theta}\right)\\
& = \mathrm{Re}\left((300 + 468i)e^{i\theta} + (24 - 36i)e^{i\theta}\right)\\
& = \mathrm{Re}\left((324 + 432i)e^{i... | unitsize(25); draw((-1, 0) -- (5, 0), Arrows(TeXHead)); draw((0, -3) -- (0, 5), Arrows(TeXHead)); draw((0, 0) -- (3, 3), EndArrow); draw((0, 0) -- (1, 2), EndArrow); draw((0, 0) -- (1, -2), dashed, EndArrow); draw((3, 3) -- (4, 1) -- (1, -2), dashed); draw((0, 0) -- (4, 1), red, EndArrow); draw(arc((0, 0), (3, 3), rota... | [] |
260 | Each vertex of a regular octagon is independently colored either red or blue with equal probability. The probability that the octagon can then be rotated so that all of the blue vertices end up at positions where there had been red vertices is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Fi... | 2024 AIME I Problem 11 | Notice that the question's condition mandates all blues to go to reds, but reds do not necessarily have to go to blue. Let us do casework on how many blues there are.
If there are no blues whatsoever, there is only one case. This case is valid, as all of the (zero) blues have gone to reds. (One could also view it as:... | // Block 1
import graph;
unitsize(1cm);
filldraw(circle((0.5,1.207),2.5),green,black);
void oct11(int[] pts) {
pair[] vertices = {(0,0),(1,0),(1.707,0.707),(1.707,1.707),(1,2.414),(0,2.414),(-0.707,1.707),(-0.707,0.707)};
draw((0,0)--(1,0)--(1.707,0.707)--(1.707,1.707)--(1,2.414)--(0,2.414)--(-0.707,1.707)--(-0.707,0.7... | [] |
261 | 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 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-... | [] |
262 | 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 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 ... | [] |
263 | Let $\triangle$$ABC$ have incenter $I$, circumcenter $O$, inradius $6$, and circumradius $13$. Suppose that $\overline{IA} \perp \overline{OI}$. Find $AB \cdot AC$. | 2024 AIME II Problem 10 | Start off by (of course) drawing a diagram! Let $I$ and $O$ be the incenter and circumcenters of triangle $ABC$, respectively. Furthermore, extend $AI$ to meet $BC$ at $L$ and the circumcircle of triangle $ABC$ at $D$.
We'll tackle the initial steps of the problem in two different manners, both leading us to the s... | // Block 1
size(300);
import olympiad;
real c=8.1,a=5*(c+sqrt(c^2-64))/6,b=5*(c-sqrt(c^2-64))/6;
pair B=(0,0),C=(c,0), D = (c/2-0.01, -2.26);
pair A = (c/3,8.65*c/10);
draw(circumcircle(A,B,C));
pair I=incenter(A,B,C);
pair O=circumcenter(A,B,C);
pair L=extension(A,I,C,B);
dot(I^^O^^A^^B^^C^^D^^L);
draw(A--L);
draw(A--... | [] |
263 | Let $\triangle$$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 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... | [] |
264 | 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 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("... | [] |
265 | Find the number of rectangles that can be formed inside a fixed regular dodecagon ($12$-gon) where each side of the rectangle lies on either a side or a diagonal of the dodecagon. The diagram below shows three of those rectangles. | 2024 AIME II Problem 15 | By Furaken
There are two kinds of such rectangles: those whose sides are parallel to some edges of the regular 12-gon (Case 1), and those whose sides are not (Case 2).
For Case 1, WLOG assume that the rectangle's sides are horizontal and vertical (don't forget to multiply by 3 at the end of Case 1). Then the rectangl... | // Block 1
real r = pi/6;
pair A1 = (cos(r),sin(r));
pair A2 = (cos(2r),sin(2r));
pair A3 = (cos(3r),sin(3r));
pair A4 = (cos(4r),sin(4r));
pair A5 = (cos(5r),sin(5r));
pair A6 = (cos(6r),sin(6r));
pair A7 = (cos(7r),sin(7r));
pair A8 = (cos(8r),sin(8r));
pair A9 = (cos(9r),sin(9r));
pair A10 = (cos(10r),sin(10r));
pai... | [] |
266 | An isosceles trapezoid has an inscribed circle tangent to each of its four sides. The radius of the circle is $3$, and the area of the trapezoid is $72$. Let the parallel sides of the trapezoid have lengths $r$ and $s$, with $r \neq s$. Find $r^2+s^2$ | 2025 AIME I Problem 6 | Draw angle bisectors from the bottom left vertex to the center of the circle. Call the angle formed $\theta$. Drawing a line from the center of the circle to the midway point of the bottom base of the trapezoid makes a right angle, and the other angle has to be $90^{\circ} - \theta$. Then draw a line segment from the c... | // Block 1
size(15cm);
draw(circle((0,0), 3));
draw((-0.5 * (12 + 6sqrt(3)), -3) -- (0.5 * (12 + 6sqrt(3)), -3) -- (0.5 * (12 - 6sqrt(3)), 3) -- (-0.5 * (12 - 6sqrt(3)), 3) -- cycle);
draw((0, 0) -- (-0.5 * (12 + 6*sqrt(3)),-3) -- cycle);
draw((0, 0) -- (0.5 * (12 + 6*sqrt(3)),-3) -- cycle);
draw((0, 0) -- (-0.5 * (12... | [] |
267 | Let $k$ be a real number such that the system
\begin{align*}
&|25 + 20i - z| = 5 \\
&|z - 4 - k| = |z - 3i - k|
\end{align*}
has exactly one complex solution $z$. The sum of all possible values of $k$ can be written as $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$. Here $i = \sqr... | 2025 AIME I Problem 8 | The complex number $z$ must satisfy the following conditions on the complex plane:
$1.$ The magnitude between $z$ and $(25,20)$ is $5.$ This can be represented by drawing a circle with center $(25,20)$ and radius $5.$
$2.$ It is equidistant from the points $(4+k,0)$ and $(k,3).$ Hence it must lie on the perpendicula... | // Block 1
size(300);
draw((0, 0) -- (0, 20), EndArrow(10));
label("$y$", (0, 20), NW);
dot((25,20));
draw((0, 0) -- (25, 0), EndArrow(10));
label("$x$", (25, 0), SE);
draw(circle((25,20),5));
label(scale(0.7)*"$(25,20)$", (25,20), S);
draw((7,0) -- (3,3), blue);
draw((5,3/2) -- (21,23), dashed);
label("$(4+k,0)$", (7,... | [] |
268 | The parabola with equation $y = x^2 - 4$ is rotated $60^\circ$ counterclockwise around the origin. The unique point in the fourth quadrant where the original parabola and its image intersect has $y$-coordinate $\frac{a - \sqrt{b}}{c}$, where $a$, $b$, and $c$ are positive integers, and $a$ and $c$ are relatively prime.... | 2025 AIME I Problem 9 | We need to relate the rotation to something simpler because substituting the equation of the rotated parabola into the original will give us a quartic.
Notice that the vertices of the original parabola and its image are symmetrical about the angle bisector of the 60 degree rotation (shown in green).
The equation of... | // Block 1
size(300);
import graph;
// View window
real L = 6;
// Original parabola y = x^2 - 4
real f(real x){ return x^2 - 4; }
// Rotation by +60 degrees about the origin
pair rot60(pair P){
real ang = pi/3;
return (P.x*cos(ang) - P.y*sin(ang), P.x*sin(ang) + P.y*cos(ang));
}
// Axes
draw((-L,0)--(L,0), gray... | [] |
269 | The set of points in $3$-dimensional coordinate space that lie in the plane $x+y+z=75$ whose coordinates satisfy the inequalities \[x-yz<y-zx<z-xy\]forms three disjoint convex regions. Exactly one of those regions has finite area. The area of this finite region can be expressed in the form $a\sqrt{b},$ where $a$ and $b... | 2025 AIME I Problem 12 | Rewriting we have $z=75-x-y.$
From the inequality $x-yz<y-zx$ we can rewrite to get,
\[x-y(75-x-y)< y-x(75-x-y).\] \[76x-76y+y^2-x^2<0.\] \[(76-x-y)(x-y)<0.\]
Similarly from the inequality $y-zx<z-xy$ we rewrite to get,
\[y-x(75-x-y)<(75-x-y)-xy.\]
\[2y + 2xy + x^2 - 74x - 75 < 0.\]
\[(x + 1)(2y + x - 75) < 0.\]
Our... | // Block 1
import graph;
size(400, 400);
real xmin = -250, xmax = 250;
real ymin = -150, ymax = 150;
draw((xmin,0)--(xmax,0), black+0.8bp, Arrow);
draw((0,ymin)--(0,ymax), black+0.8bp, Arrow);
pair A = (38, 38),
B = (-10, -10),
C = (-227, 151),
D = (-10, 85/2),
E = (25, 25);
path L1 = (xmin, xmi... | [] |
270 | Alex divides a disk into four quadrants with two perpendicular diameters intersecting at the center of the disk. He draws $25$ more lines segments through the disk, drawing each segment by selecting two points at random on the perimeter of the disk in different quadrants and connecting these two points. Find the expect... | 2025 AIME I Problem 13 | First, we calculate the probability that two segments intersect each other. Let the quadrants be numbered $1$ through $4$ in the normal labeling of quadrants, let the two perpendicular diameters be labeled the $x$-axis and $y$-axis, and let the two segments be $A$ and $B.$
$\textbf{Case 1:}$ Segment $A$ has endpoints ... | // Block 1
pair A,B,C,D,E,F,O;
A=(1,0);B=(0,1);C=(-1,0);D=(0,-1);E=(0.5,0.86602540);F=(-0.707106,-0.707106);O=(0,0);
draw(A--C);draw(B--D);draw(circle((0,0),1));draw(E--F,blue);dot(E,blue);dot(F,blue);
// Block 2
pair A,B,C,D,E,F,O;
A=(1,0);B=(0,1);C=(-1,0);D=(0,-1);E=(0.5,0.86602540);F=(-0.707106,0.707106);O=(0,0);
dr... | [] |
271 | 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 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... | [] |
272 | Six points $A, B, C, D, E,$ and $F$ lie in a straight line in that order. Suppose that $G$ is a point not on the line and that $AC=26, BD=22, CE=31, DF=33, AF=73, CG=40,$ and $DG=30.$ Find the area of $\triangle BGE.$ | 2025 AIME II Problem 1 | Let $AB=a$, $BC=b$, $CD=c$, $DE=d$ and $EF=e$. Then we know that $a+b+c+d+e=73$, $a+b=26$, $b+c=22$, $c+d=31$ and $d+e=33$. From this we can easily deduce $c=14$ and $a+e=34$ thus $b+c+d=39$. Using Heron's formula we can calculate the area of $\triangle{CGD}$ to be $\sqrt{(42)(28)(12)(2)}=168$, and since the base of $\... | // Block 1
pair A,B,C,D,E,F,G;
A=(0,0);
label("$A$", A, S);
B=(1.5,0);
label("$B$", B, S);
C=(2.9,0);
label("$C$", C, S);
D=(4.2,0);
label("$D$", D, S);
E=(5.3,0);
label("$E$", E, S);
F=(6.5,0);
label("$F$", F, S);
G=(3.7,3);
label("$G$", G, N);
draw(A--B--C--D--E--F);
draw(C--G--D);
draw(B--G--E);
// Block 2
pair A,B... | [] |
273 | Four unit squares form a $2 \times 2$ grid. Each of the $12$ unit line segments forming the sides of the squares is colored either red or blue in such a way that each unit square has $2$ red sides and $2$ blue sides. One example is shown below (red is solid, blue is dashed). Find the number of such colorings. | 2025 AIME II Problem 3 | Let the red edges be "used" edges. In the digarams below, dashed lines are uncolored lines yet to be decided. Since all four center edges are common to both squares, we consider five distinct cases:
$\textbf{Case 1:}$ All center edges are used. There is only one way to do this.
$\textbf{Case 2:}$ Three center edges... | // Block 1
pair A,B,C,D,E,F,G,H,I;
A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1);
draw(A--C--I--G--cycle,dashed); draw(B--H,red); draw(D--F,red);
// Block 2
pair A,B,C,D,E,F,G,H,I;
A=(-1,-1);B=(-1,0);C=(-1,1);D=(0,-1);E=(0,0);F=(0,1);G=(1,-1);H=(1,0);I=(1,1);
draw(A--C--I--G--cycle,dashe... | [] |
274 | Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$ | 2025 AIME II Problem 14 | Let $O$ be the midpoint of $BC$. Take the diagram and rotate it $180^{\circ}$ around $O$ to get the diagram shown. Notice that we have $\angle ABC+\angle ACB=90^{\circ}$. Because $\triangle AKL$ is equilateral, then $\angle KAL=60^{\circ}$, so $\angle BAK+\angle CAL=30^{\circ}$. Because of isosceles triangles $\triangl... | // Block 1
import math; import geometry; import olympiad;
point A,C,B,L,K,D,F,G,O; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13); D=(16sqrt(3),26); F=(13sqrt(3),13); G=(8sqrt(3),24); O=(8sqrt(3),13);
draw(A--B--D--C--A--L--C--F--L--K--A--D); draw(K--B--G--D--F--G--K--F); draw(B--O--L); draw(C--O--... | [] |
274 | Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$ | 2025 AIME II Problem 14 | Immediately we should see that $\triangle{AKL}$ is equilateral, so $\angle{KAL}=60$.
We assume $\angle{LCA}=x$, and it is easily derived that $\angle{KBA}=30-x$. Using trigonometry, we can say that $AC=28\cos{x}$ (imagine extending $\overrightarrow{CL}$ to intersect $\overline{AB}$, forming a right triangle with hypot... | // Block 1
import math; import geometry; import olympiad;
point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13);
draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L); draw(B--L);
label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,W); label("L",L,NE);
markscalefac... | [] |
274 | Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$ | 2025 AIME II Problem 14 | Consider a diagram to the original problem (credit to solution 4):
Now, let us simplify the problem further. We know that $K$ and $L$ must lie on the perpendicular bisectors of $AB$ and $AC$, respectively. The real problem here is the equilateral triangle in the middle, inscribed in a rectangle with diagonal length ... | // Block 1
import math; import geometry; import olympiad;
point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13);
draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L); draw(B--L);
label("A",A,SW); label("B",B,NW); label("C",C,SE); label("K",K,W); label("L",L,NE);
markscalefac... | [] |
274 | Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$ | 2025 AIME II Problem 14 | This diagram is modified from the solution 4 diagram. Let $M$ be the midpoint of $AC$, and let $N$ be the midpoint of $AB$.
We place the diagram onto the Cartesian coordinate grid. Let $A = (0, 0)$, $M = (a, 0)$, $C = (2a, 0)$, $N = (0, b)$, and $B = (0, 2b)$. We are given $AL = CL$, so $\triangle ACL$ is isosceles. T... | // Block 1
import math; import geometry; import olympiad;
point A,B,C,L,K; A=(0,0); C=(16sqrt(3),0); B=(0,26); L=(8sqrt(3),2); K=(3sqrt(3),13);
point M,N; M=(8sqrt(3), 0); N=(0,13);
draw(A--B--C--cycle); draw(A--K--L--cycle); draw(B--K); draw(C--L);
draw(N--K); draw(L--M);
label("A",A,SW); label("B",B,NW); label("C",C,... | [] |
274 | Let ${\triangle ABC}$ be a right triangle with $\angle A = 90^\circ$ and $BC = 38.$ There exist points $K$ and $L$ inside the triangle such\[AK = AL = BK = CL = KL = 14.\]The area of the quadrilateral $BKLC$ can be expressed as $n\sqrt3$ for some positive integer $n.$ Find $n.$ | 2025 AIME II Problem 14 | We observe that the desired area can be found by subtracting the areas of the three smaller triangles $\triangle BAK$, $\triangle KAL$, and $\triangle LAC$ from the area of the large triangle $\triangle ABC$. This can be expressed as \([ABC]-[BAK]-[KAL]-[LAC]\). Thus, our goal is to compute these four areas.
Since $\t... | // Block 1
import olympiad;
import geometry;
size(300);
defaultpen(linewidth(0.7)+fontsize(10pt));
pair A = (0,0);
pair B = (0,26);
pair C = (16*sqrt(3),0); // approx 27.71
pair K = (sqrt(14^2 - 13^2), 13);
pair L = (8*sqrt(3), sqrt(14^2 - (8*sqrt(3))^2));
draw(A--B--C--cycle);
draw(B--K--L--C--cycle);
draw(A--K, lin... | [] |
275 | Let $\triangle ABC$ have side lengths $AB=13$, $BC=14$, and $CA=15$. Triangle $\triangle A'B'C'$ is obtained by rotating $\triangle ABC$ about its circumcenter so that $\overline{A'C'}$ is perpendicular to $\overline{BC}$, with $A'$ and $B$ not on the same side of line $B'C'$. Find the integer closest to the area of he... | 2026 AIME I Problem 10 | Since a $13$-$14$-$15$ triangle is composed of a $5$-$12$-$13$ triangle connected to a $9$-$12$-$15$ triangle, we set up coordinates
\[A = (5,12),\quad B = (0,0),\quad C = (14,0).\]
From this we have
\[[ABC] = \frac{1}{2} \cdot 14 \cdot 12 = 84,\]
and thus
\[R = \frac{abc}{4[ABC]} = \frac{65}{8}.\]
We compute the cir... | // Block 1
import geometry;
size(10cm);
// the diagram was all done by gemini
// 1. Define the coordinates of Triangle ABC (13-14-15 triangle)
// Base BC = 14 on the x-axis. Height to A is 12.
pair B = (0,0);
pair C = (14,0);
pair A = (5,12);
// 2. Calculate Circumcenter O and Radius R
pair O = (7, 33/8);
real R = 65... | [] |
276 | For each nonnegative integer $r$ less than $502$, define\[S_r=\sum_{m\geq 0}\binom{10,000}{502m+r},\]where $\binom{10,000}{n}$ is defined to be $0$ when $n>10,000$. That is, $S_r$ is the sum of all the binomial coefficients of the form $\binom{10,000}{k}$ for which $0\leq k\leq 10,000$ and $k-r$ is a multiple of $502$.... | 2026 AIME I Problem 13 | Let $p = 503$. Since $p$ is a prime number, the field $\mathbb{F}_p$ is a finite field of order $p$. Let $N = 10,000$ and $d = 502$. Observe that $d = p - 1$.
The problem asks for the number of indices $r \in \{0, 1, \dots, d-1\}$ such that $S_r \equiv 0 \pmod{p}$.
The sum $S_r$ corresponds to the sum of every $d$-th ... | // Block 1
unitsize(1cm);
// Colors
pen accentBlue = blue + linewidth(1.5pt);
pen accentRed = red + linewidth(1.5pt);
pen shadeRed = lightred + opacity(0.2);
// Draw background shading for zero region
fill((7.5, -1)--(11.5, -1)--(11.5, 2)--(7.5, 2)--cycle, shadeRed);
// Main Axis
draw((-1,0)--(12,0), black+linewidth... | [] |
277 | In an equiangular pentagon, the sum of the squares of the side lengths equals $308$, and the sum of the squares of the diagonal lengths equals $800$. The square of the perimeter of the pentagon can be expressed as $m\sqrt{n}$, where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. ... | 2026 AIME I Problem 14 | Consider an equiangular pentagon in the complex plane with vertices $V_1, V_2, V_3, V_4, V_5$ in counterclockwise order. Let the side lengths be $a_k = |V_{k+1} - V_k|$ for $k=1, \dots, 5$ (indices modulo 5), and let the diagonal lengths be $d_k = |V_{k+2} - V_k|$.
Since the pentagon is equiangular, each interior ang... | // Block 1
size(7.5cm); // Roughly equivalent to scale=2.5
// Define the vertices using polar coordinates (r=1.5)
pair[] V = new pair[6];
for(int i=1; i<=5; ++i) {
V[i] = 1.5 * dir(90 - (i-1)*72);
}
// Pens (colors and styles)
pen sidePen = blue + linewidth(1.2pt);
pen mainDiagPen = deepgreen + dashed + linewidth... | [] |
278 | The figure below shows a grid of $10$ squares in a row. Each square has a diagonal connecting its lower left vertex to its upper right vertex. A bug moves along the line segments from vertex to vertex, never traversing the same segment twice and never moving from right to left along a horizontal or diagonal segment. Le... | 2026 AIME II Problem 2 | Consider each of the ten instances of
Let the number of ways to start at the top-left corner be $a$ and the number of ways to start at the bottom-left corner be $b$. Then, there are $a+b$ ways to get to the top-right corner without passing through the top-left corner, and there are $a$ ways to get to the bottom-right... | size(1cm); draw((0, 0)--(0, 1)); for (int i = 0; i <= 0; i += 1) { draw((i, 1)--(i + 1, 1)--(i + 1, 0)--(i, 0)--(i + 1, 1)); } | [] |
279 | Isosceles triangle $\triangle ABC$ has $AB=BC$. Let $I$ be the incenter of $\triangle ABC$. The perimeters of $\triangle ABC$ and $\triangle AIC$ are in the ratio $125:6$, and all the sides of both triangles have integer lengths. Find the minimum possible value of $AB$. | 2026 AIME II Problem 8 | Since $\triangle ABC$ is isosceles with $AB = BC$, the angle bisector, median, and altitude from $B$ to $AC$ coincide, which is the perpendicular bisector of $AC$. The incenter $I$ lies on this common line. Let $D$ be the intersection of this line with $AC$, so $D$ is the midpoint of $AC$ and $ID \perp AC$.
We set $AC ... | // Block 1
size(200,0);
pair A=(-5,0);
pair C=(5,0);
pair D=(0,0);
pair B=(0,12);
pair I=(0,3);
dot(A,red);
dot(B,red);
dot(C,red);
dot(D,red);
dot(I,red);
label("$A$",A,SW);
label("$B$",B,N);
label("$C$",C,SE);
label("$D$",D,S);
label("$I$",I,W);
draw(A--B);
draw(B--C);
draw(A--C);
draw(B--D);
draw(A--I);
draw(C--... | [] |
280 | Let $\triangle ABC{}$ be a triangle with $D$ on $\overline{BC}$ such that $\overline{AD}{}$ bisects $\angle BAC.$ Let $\omega{}$ be the circle that passes through $A$ and is tangent to segment $\overline{BC}{}$ at $D.$ Let $E \neq A$ and $F \neq A$ be the intersections of $\omega{}$ with segments $\overline{AB}{}$ and ... | 2026 AIME II Problem 10 | Let $BC=x$. By Angle Bisector Theorem, we have \[BD = \frac{200}{200+225}x=\frac{8x}{17}, CD = \frac{9x}{17}\] Since both of these integers, we write $x=17y$, for some $y\in \mathbb{Z}^+$. Then, $BD=8y$ and $CD=9y$.
Since $BD$ is tangent to $\omega$, it follows that $\angle BDE = \angle EAD$. Similarly, $\angle CDF = ... | // Block 1
size(300);
import olympiad;
import geometry;
pair B = (0,0);
pair C = (170,0);
pair D = (80,0);
pair A = (53.75, 192.64);
pair O = (80, 98.11);
real R = 98.11;
pair E = interp(B, A, 32/200);
pair F = interp(C, A, 36/225);
draw(circle(O, R), Black);
draw(A--B--C--cycle, Black);
draw(A--D, Black);
draw(F--D... | [] |
280 | Let $\triangle ABC{}$ be a triangle with $D$ on $\overline{BC}$ such that $\overline{AD}{}$ bisects $\angle BAC.$ Let $\omega{}$ be the circle that passes through $A$ and is tangent to segment $\overline{BC}{}$ at $D.$ Let $E \neq A$ and $F \neq A$ be the intersections of $\omega{}$ with segments $\overline{AB}{}$ and ... | 2026 AIME II Problem 10 | Let the side lengths of $\triangle ABC$ be denoted by $AB = c$, $AC = b$, and $BC = a$. We are given $c = 200$ and $b = 225$. Let $D$ be a point on $BC$ such that $AD$ is the internal angle bisector of $\angle BAC$. By the Angle Bisector Theorem, the ratio of the segments of the opposite side is proportional to the adj... | // Block 1
size(300);
// Basic Coordinates
pair B = (0,0);
pair C = (340,0);
pair D = (160,0);
pair A = (154.375, 127.17);
pair O = (160, 63.71);
real R = 63.71;
// Points E and F
pair E = interp(B, A, 128/200);
pair F = interp(C, A, 144/225);
// Main Drawing
draw(circle(O, R), cyan);
draw(A--B--C--cycle, orange);
d... | [] |
281 | Consider a tetrahedron with two isosceles triangle faces with side lengths $5\sqrt{10}, \,5\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\sqrt{10},\, 5\sqrt{10},$ and $18.$ The four vertices of the tetrahedron lie on a sphere with center $S,$ and the four faces of the tetrahedron are tangent... | 2026 AIME II Problem 12 | We place the tetrahedron in a coordinate system by symmetry:
\[A(-5, 0, 0),\ B(5, 0, 0),\ C(0, 9, h),\ D(0, -9, h).\]
Given $AC = 5\sqrt{10}$, we have
\begin{align*}
AC^2 &= 5^2 + 9^2 + h^2 = 250, \\
h^2 &= 144 \implies h = 12.
\end{align*}
So the vertices are
\[A(-5,0,0),\ B(5,0,0),\ C(0,9,12),\ D(0,-9,12).\]
By symme... | import three;size(200,0);currentprojection=orthographic(10,20,15);triple A=(-5,0,0);triple B=(5,0,0);triple C=(0,9,12);triple D=(0,-9,12);triple S=(0,0,25/3);triple R=(0,0,75/16);dot(A,red);dot(B,red);dot(C,red);dot(D,red);dot(S,purple+linewidth(3));dot(R,deepgreen+linewidth(3));label("$A$",A,SW);label("$B$",B,SE);labe... | [] |
281 | Consider a tetrahedron with two isosceles triangle faces with side lengths $5\sqrt{10}, \,5\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\sqrt{10},\, 5\sqrt{10},$ and $18.$ The four vertices of the tetrahedron lie on a sphere with center $S,$ and the four faces of the tetrahedron are tangent... | 2026 AIME II Problem 12 | ~Diagram by Confringo
Choose face $ABC$ as the base. Let $O$ be the midpoint of $AB$ as the origin, with $OA$ along the positive $x$-axis, $OC$ along the positive $y$-axis, and the line through $O$ perpendicular to plane $ABC$ as the $z$-axis.
First, find the coordinates of point $D$. Due to the symmetry of the f... | // Block 1
import three;size(200,0);currentprojection=orthographic(10,20,15);triple A=(-5,0,0);triple B=(5,0,0);triple C=(0,21.79,0);triple D=(0,14.35,16.39);dot(A,red);dot(B,red);dot(C,red);dot(D,red);label("$A$",A,SW);label("$B$",B,W);label("$C$",C,E);label("$D$",D,NE);draw(A--B,black);draw(A--C,black);draw(B--C,blac... | [] |
281 | Consider a tetrahedron with two isosceles triangle faces with side lengths $5\sqrt{10}, \,5\sqrt{10},$ and $10$ and two isosceles triangle faces with side lengths $5\sqrt{10},\, 5\sqrt{10},$ and $18.$ The four vertices of the tetrahedron lie on a sphere with center $S,$ and the four faces of the tetrahedron are tangent... | 2026 AIME II Problem 12 | We let $AB$ be the side of length $10$ and $CD$ of length $18$. Call midpoints of side $AB$ and $CD$ as $M$ and $N$. The faces of our tetrahedron are isosceles triangle so $AM=5$ and $CN=9$, along with $MN\perp AB$ and $MN\perp CD$. By symmetry we know both $R$ and $S$ locate on the segment $MN$.
For the position of... | // Block 1
import olympiad; size(150);pair M=(0,0),C=(15,0),D=(21/5,72/5),R=(15/4,45/16),E=(15/4,0),N=(C+D)/2;draw(M--C--D--cycle);draw(R--E);draw(M--N);dot(M);dot(C);dot(D);dot(R);dot(E);dot(N);label("$M$",M,SW);label("$R$",R,NW);label("$E$",E,S);label("$C$",C,SE);label("$D$",D,NW);label("$N$",N,W);label("$15$",(M+D)/... | [] |
282 | Find the number of seven-tuple $(a_1, a_2, a_3, a_4, a_5, a_6, a_7)$ that follows the conditions.
$\{a_1, a_2, a_3, a_4, a_5, a_6, a_7\} \subset \{1,2,3\}$
$a_1+a_2+a_3+a_4+a_5+a_6+a_7$ is a multiple of $3$.
$a_1 a_2 a_4 + a_2 a_3 a_5 + a_3 a_4 a_6 + a_4 a_5 a_7 + a_5 a_6 a_1 + a_6 a_7 a_2$ is a multiple of $3$. | 2026 AIME II Problem 15 | Let's model the problem over the finite field $\mathbb{F}_3$. We define a bijection between the set $\{1, 2, 3\}$ and the field elements $\{1, -1, 0\}$ (modulo $3$) via the mapping $1 \mapsto 1$, $2 \mapsto -1 \equiv 2$, and $3 \mapsto 0$. Let $x = (x_1, \dots, x_7) \in \mathbb{F}_3^7$ correspond to the tuple $(a_1, \d... | // Block 1
size(200);
// Coordinates for the 7 points
real R = 1.5;
pair P1 = R * dir(90);
pair P2 = R * dir(210);
pair P4 = R * dir(330);
pair P3 = interp(P1, P2, 0.5);
pair P5 = interp(P2, P4, 0.5);
pair P6 = interp(P4, P1, 0.5);
pair P7 = (0,0);
// Draw Triangle and Medians in Cyan
draw(P1--P2--P4--cycle, cyan+lin... | [] |
283 | In a given right triangle $ABC$, the hypotenuse $BC$, of length $a$, is divided into $n$ equal parts ($n$ an odd integer). Let $\alpha$ be the acute angle subtending, from $A$, that segment which contains the midpoint of the hypotenuse. Let $h$ be the length of the altitude to the hypotenuse of the triangle. Prove that... | 1960 IMO Problems/Problem 3 | Using coordinates, let $A=(0,0)$, $B=(b,0)$, and $C=(0,c)$. Also, let $PQ$ be the segment that contains the midpoint of the hypotenuse with $P$ closer to $B$.
Then, $P = \frac{n+1}{2}B+\frac{n-1}{2}C = \left(\frac{n+1}{2}b,\frac{n-1}{2}c\right)$, and $Q = \frac{n-1}{2}B+\frac{n+1}{2}C = \left(\frac{n-1}{2}b,\frac{n... | // Block 1
size(8cm);
pair A,B,C,P,Q;
A=(0,0);
B=(4,0);
C=(0,3);
P=(2.08,1.44);
Q=(1.92,1.56);
dot(A);
dot(B);
dot(C);
dot(P);
dot(Q);
label("A",A,SW);
label("B",B,SE);
label("C",C,NW);
label("P",P,ENE);
label("Q",Q,NNE);
draw(A--B--C--cycle);
draw(A--P);
draw(A--Q);
// Block 2
size(8cm); pair A,B,C,P,Q; A=(0,0); B=(... | [] |
284 | A given convex pentagon $ABCDE$ has the property that the area of each of the five triangles $ABC$, $BCD$, $CDE$, $DEA$, and $EAB$ is unity. Show that all pentagons with the above property have the same area, and calculate that area. Show, furthermore, that there are infinitely many non-congruent pentagons having the a... | 1972 USAMO Problems/Problem 5 | Lemma: Convex pentagon $A_0A_1A_2A_3A_4$ has the property that $[A_0A_1A_2] = [A_1A_2A_3] = [A_2A_3A_4] = [A_3A_4A_0] = [A_4A_0A_1]$ if and only if $\overline{A_{n - 1}A_{n + 1}}\parallel\overline{A_{n - 2}A_{n + 2}}$ for $n = 0, 1, 2, 3, 4$ (indices taken mod 5).
Proof: For the "only if" direction, since $[A_0A_1A_2]... | // Block 1
size(120);
defaultpen(fontsize(10));
pathpen = black;
pair A=MP("A",dir(90),dir(90)), B=MP("B",dir(90-72),dir(90-72)), C=MP("C",dir(90-2*72),dir(90-2*72)), D=MP("D",dir(90-3*72),dir(90-3*72)), E=MP("E",dir(90-4*72),dir(90-4*72));
D(A--B--C--D--E--cycle);
D(A--C--E--B--D--cycle);
pair Ap = MP("A'",IP(B--D,C--... | [] |
285 | Consider the two triangles $\triangle ABC$ and $\triangle PQR$ shown in Figure 1. In $\triangle ABC$, $\angle ADB = \angle BDC = \angle CDA = 120^\circ$. Prove that $x=u+v+w$. | 1974 USAMO Problems/Problem 5 | Solution 1
We rotate figure $PRQM$ by a clockwise angle of $\pi/3$ about $Q$ to obtain figure $RR'QM'$:
Evidently, $MM'Q$ is an equilateral triangle, so triangles $MRM'$ and $ABC$ are congruent. Also, triangles $PMQ$ and $RM'Q$ are congruent, since they are images of each other under rotations. Then
\[[ABC] + \frac... | // Block 1
size(300); defaultpen(1); pair P=(7,0), Q=(14,0), R=P+7expi(pi/3), M=(10,1.2); pair RR=R+Q-P, MM= rotate(-60,Q)*M; draw(P--R--RR--Q--P--M--MM--RR); draw(Q--R--M--Q--MM--R); label("$P$",P,W); label("$Q$",Q,E); label("$R$",R,W); label("$M$",M,NW); label("$R'$",RR,NE); label("$M'$",MM,ESE);
// Block 2
size(20... | [] |
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