2012 AIME I Problems/Problem 13


Three concentric circles have radii $3,$ $4,$ and $5.$ An equilateral triangle with one vertex on each circle has side length $s.$ The largest possible area of the triangle can be written as $a + \tfrac{b}{c} \sqrt{d},$ where $a,$ $b,$ $c,$ and $d$ are positive integers, $b$ and $c$ are relatively prime, and $d$ is not divisible by the square of any prime. Find $a+b+c+d.$

Solution 1

Reinterpret the problem in the following manner. Equilateral triangle $ABC$ has a point $X$ on the interior such that $AX = 5,$ $BX = 4,$ and $CX = 3.$ A $60^\circ$ counter-clockwise rotation about vertex $A$ maps $X$ to $X'$ and $B$ to $C.$

[asy]import cse5; size(200); defaultpen(linewidth(0.4)+fontsize(8));  pair O = origin; pair A,B,C,Op,Bp,Cp; path c3,c4,c5; c3 = CR(O,3); c4 = CR(O,4); c5 = CR(O,5); draw(c3^^c4^^c5, gray+0.25); A = 5*dir(96.25); Op = rotate(60,A)*O; B = OP(CR(Op,4),c3); Bp = IP(CR(Op,4),c3); C = rotate(-60,A)*B; Cp = rotate(-60,A)*Bp; draw(A--B--C--A, black+0.8); draw(A--Bp--Cp--A, royalblue+0.8); draw(CR(Op,4), heavygreen+0.25); dot("$A$",A,N); dot("$X$",O,E); dot("$X'$",Op,E); dot("$C$",B,SE); dot("$C'$",Bp,NE); dot("$B$",C,2*SW); dot("$B'$",Cp,2*S); [/asy]

Note that angle $XAX'$ is $60$ and $XA = X'A = 5$ which tells us that triangle $XAX'$ is equilateral and that $XX' = 5.$ We now notice that $XC = 3$ and $X'C = 4$ which tells us that angle $XCX'$ is $90$ because there is a $3$-$4$-$5$ Pythagorean triple. Now note that $\angle ABC + \angle ACB = 120^\circ$ and $\angle XCA + \angle XBA = 90^\circ,$ so $\angle XCB+\angle XBC = 30^\circ$ and $\angle BXC = 150^\circ.$ Applying the law of cosines on triangle $BXC$ yields

\[BC^2 = BX^2+CX^2 - 2 \cdot BX \cdot CX \cdot \cos150^\circ = 4^2+3^2-24 \cdot \frac{-\sqrt{3}}{2} = 25+12\sqrt{3}\]

and thus the area of $ABC$ equals \[\frac{\sqrt{3}}{4}\cdot BC^2 = 25\frac{\sqrt{3}}{4}+9.\]

so our final answer is $3+4+25+9 = \boxed{041}.$

Solution 2

Here is a proof that shows that there are 2 distinct equilateral triangles (up to congruence) that have the given properties.

We claim that there are 2 distinct equilateral triangles (up to congruence) that have the given properties; one of which has largest area. We have two cases to consider; either the center $O$ of the circles lies in the interior of triangle $ABC$ or it does not (and we shall show that both can happen). To see that the first case can occur proceed as follows. Using the notation from Solution 1, let $\angle XAC = \theta$ so that $\angle BAX=60-\theta$. Let $AB=BC=AC=x$. The law of cosines on $\triangle BAX$ and $\triangle CAX$ yields \[\cos(60-\theta)=\frac{x^2+9}{10x}\quad \text{ and }\quad  \cos\theta=\frac{x^2+16}{10x}.\] Solving this system will yield the value of $x$. Since $\cos\theta=\frac{x^2+16}{10x}$ we have that \[\sin\theta=\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}.\] Substituting these into the equation \[\frac{x^2+9}{10x}=\cos(60-\theta)=\tfrac{1}{2}(\cos\theta+ \sqrt{3}\sin\theta)\] we obtain \[\frac{x^2+9}{10x}=\frac{1}{2}\frac{x^2+16}{10x}+\frac{\sqrt{3}}{2}\frac{\sqrt{100x^2-(x^2+16)^2}}{10x}.\] After clearing denominators, combining like terms, isolating the square root, squaring, and expanding, we obtain \[x^4-50x^2+193=0\] so that by the quadratic formula $x^2=25\pm12\sqrt{3}$. Under the hypothesis that $O$ lies in the interior of triangle $ABC$, $x^2$ must be $25+12\sqrt{3}$. To see this, note that the other value for $x^2$ is roughly $4.2$ so that $x\approx 2.05$, but since $AX=5$ and $AX\leq x$ we have a contradiction. We then obtain the area as in Solution 1.

Now, suppose $O$ does not lie in the interior of triangle $ABC$. We then obtain convex quadrilateral $OBAC$ with diagonals $CB$ and $OA$ intersecting at $X$. Here $AX=AB=AC=x$. We may let $\alpha$ denote the measure of angle $CAX$ so that angle $XAB$ measures $60-\alpha$. Note that the law of cosines on triangles $CXA$ and $BXA$ yield the same equations as in the first case with $\theta$ replaced with $\alpha$. Thus we obtain again $x^2=25\pm12\sqrt{3}$. If $x^2=25+12\sqrt{3}$ then $x\approx 6.8$, but this is impossible since $AX\leq 5$ but the shortest possible distance from $A$ to $X$ is the height of equilateral triangle $ABC$ which is $\approx6.8\sqrt{3}\approx5.8$; a contradiction. Hence in this case $x^2=25-12\sqrt{3}$. But, the area of this triangle is clearly less than that in the first case, so we are done. Hence the phrasing of the question (the triangle with maximal area) is absolutely necessary since there are 2 possible triangles (up to congruence).

Solution 3

[asy] import olympiad; import cse5; import graph;   dotfactor = 2;  unitsize(0.3inch);  pair B = (0,0), C= (5,0), A = (sqrt(9-2.4*2.4),2.4); pair D = rotate(60,B)*A, E=rotate(60,A)*C, F=rotate(60,C)*B; pair X = extension(A,F,D,C); pair L = (-1.5,2), M = (6.2,3), N = rotate(-60,L)*M;  dot("$C$", C, dir(0));  dot("$A$", A, dir(90));dot("$B$", B, dir(180)); dot("$D$", D, NE);  dot("$E$", E, dir(90));dot("$F$", F, dir(270)); dot("$M$", M, NE);  dot("$N$", N, dir(270));dot("$L$", L, NW); dot("$X$", X, dir(250)); draw(L--X); draw(M--X); draw(N--X);   draw(A--B--C--cycle); draw(A--D--B); draw(B--F--C); draw(A--E--C); draw(A--F,dashed); draw(D--C,dashed); draw(B--E,dashed); draw(L--M--N--cycle);   [/asy]

Let's call the circle center $X$. It has a distance of 3, 4, 5 to an equilateral triangle $LMN$. Consider $X$’s pedal triangle $ABC$. Since $X$’s antipedal triangle is equilateral, $X$ must be the one of the isogonic centers of $\triangle{ABC}$. We’ll take the one inside $ABC$, i.e., the Fermat point, because it leads to larger $\triangle LMN$. Now we construct the three equilateral triangles $ABD$, $ACE$, and $BCF$, the same way the Fermat point is constructed. Then we have $\angle DXE = \angle EXF = \angle FXE = 120$. Since $AEMCX$ is concyclic with $XM$=4 as diameter, we have $AC=4\sin(60)$. Similarly, $AB=3\sin(60)$, and $BC=5\sin(60)$. So $\triangle ABC$ is a 3-4-5 right triangle with $\angle BAC=90$. With some more angle chasing we get \[\angle MXC+\angle LXB = \angle MAC + \angle LAB = 180 – \angle BAC = 90\] \[\angle LXM = 360 – (\angle MXC + \angle LXB + \angle BXC) = 360 –(90+120)=150\] By Law of Cosines, we have \[LM^2 = 3^2+4^2-2*3*4\cos(150)=25+12\sqrt 3\] And the area follows. \[[LMN] = \frac{25}{4}\sqrt{3} + 9; \boxed{041}.\] By Mathdummy

Solution 4

Let $ABC$ be the equilateral triangle with $AB=BC=CA=x.$ Assume the coordinates of the vertices are $A(-\tfrac{x}{2},0), B(\tfrac{x}{2},0)$ and $C(0,\tfrac{\sqrt{3}}{2}x).$ Let $P(a,b)$ be such that $PA=3, PB=4$ and $PC=5.$ Then \[\left (a+\dfrac{x}{2}\right )^2 +b^2=3^2,\] \[\left (a-\dfrac{x}{2}\right )^2 +b^2=4^2,\] \[a^2+\left (b-\dfrac{\sqrt{3}}{2}x\right )^2 =5^2.\] Subtraction and addition of the first two equations yield $2ax=-7, 2a^2+\dfrac{1}{2}x^2+2b^2=25.$ The third equation gives $2a^2+2b^2-2\sqrt{3}bx+\dfrac{3}{2}x^2=50.$ Then $x^2-2\sqrt{3}bx=25.$ We can then solve for $a, b$ in terms of $x$ and have a substitution. We have \[2\left (\dfrac{-7}{2x}\right )^2 + \dfrac{1}{2}x^2 + 2\left (\dfrac{x^2-25}{2\sqrt{3}x}\right )^2 =25.\] Simplify it we have a quadratic equation for $x^2: x^4-50x^2+193=0.$ So $x^2=25\pm 12\sqrt{3}.$ The larger one leads to the solution. The smaller one relates to another equilateral triangle, as indicated in Solution 2.


Solution 5

[asy] import cse5; size(350); defaultpen(linewidth(0.6)+fontsize(12));   pair O=origin; pair Op,Bp,Cp;  path c3 = CR(O,3), c4 = CR(O,4), c5 = CR(O,5);  var theta=55.75; pair A = 3*dir(theta), Ap = rotate(150,O)*A, F=IP(c4,O--2*Ap), C=rotate(60,A)*F, E=rotate(60,A)*C, B=IP(c5,O--E), D=foot(C,O,E);    filldraw(A--O--B--cycle, rgb(255,255,200)); draw(c3, cyan);  draw(c4, green);  draw(c5, purple); draw(A--F--C--A--E--C, red+0.8);  draw(D--B--C--D--O--C, purple+0.6); draw(A--O^^E--B, cyan+0.6); draw(A--B^^O--F, heavygreen+0.6); draw(A--D); dot("$A$",A,N); dot("$O$",O,dir(A-B)); dot("$F$",F,dir(F-E)); dot("$C$",C,SE); dot("$B$",B,dir(-45)); dot("$D$",D,dir(B-C)); dot("$E$",E,dir(E-F)); label("$s$",A--F,dir(A-E),red); label("$s$",F--C,dir(C-B),red); label("$s$",A--C,dir(C-E),red); label("$s$",A--E,dir(E-C),red); label("$s$",C--E,dir(C),red); label("$x$",B--E,down,royalblue); label("$x$",O--A,dir(B-A),royalblue); label("$y$",A--B,dir(F-A),heavygreen); label("$y$",O--F,dir(C),heavygreen); label("$z$",C--B,dir(E-A),purple); label("$z$",O--C,dir(F),purple); label("$m$",A--D,dir(0),blue+fontsize(9)); markscalefactor=0.03; draw(rightanglemark(C,D,B), gray); MA("\varphi",A,D,O,0.25,blue); [/asy] We have $x=3$, $y=4$, and $z=5$. Because $AD=m$ is the median of $\triangle AOB$, by Stewart's Theorem we have \[m^2=\frac 12 (x^2+y^2)-\frac 14 \cdot z^2\quad \Rightarrow \quad m = \frac 52.\] Because $CD$ is the altitude of equilateral triangle $OBC$, we have $CD=\frac{\sqrt{3}}2\cdot z$. Then in $\triangle ADC$, we have $\angle ADC=\varphi+90^\circ$, so $\cos(\angle ADC)=-\sin\varphi$, and the Law of Cosines gives\[s^2=m^2+\frac 34\cdot z^2 + mz\sqrt{3}\sin\varphi = 25 \left(1+\frac {\sqrt{3}}{2}\cdot\sin\varphi  \right)\]To calculate $\sin\phi$ we apply the Law of Cosines to $\triangle ADO$ to get\[mz\cos\varphi = m^2+\frac 14\cdot z^2 - x^2 \quad \Rightarrow \quad \cos\varphi = \frac 7{25}\quad \Rightarrow \quad \sin\varphi = \frac {24}{25}.\] Finally, we get $s^2=25+12\sqrt{3}$ and and thus the area of $\triangle ABC$ equals\[\frac{\sqrt{3}}{4}\cdot s^2 = 25\frac{\sqrt{3}}{4}+9.\] so our final answer is $3+4+25+9 = \boxed{041}.$

vladimir.shelomovskii@gmail.com, vvsss

Video Solution by Richard Rusczyk


~ dolphin7

See also

2020 AMC 12A Problems/Problem 24

2012 AIME I (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AIME Problems and Solutions

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