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Difference between revisions of "2025 SSMO Accuracy Round Problems/Problem 7"

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==Solution==
 
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Rewrite the given system of equations as\begin{align*} a^2 + 3^2 &= (8\sqrt{3}-b)^2 + 2^2 \\ b^2 + 2^2 &= (8\sqrt{3}-c)^2 + 7^2 \\ c^2 + 7^2 &= (8\sqrt{3}-a)^2 + 3^2. \end{align*}We provide a geometric interpretation for this system. Let <math>\triangle ABC</math> be equilateral, let <math>P</math> be a point in the interior of <math>ABC</math>, and let <math>D</math>, <math>E</math>, and <math>F</math> be the projections of <math>P</math> onto <math>AB</math>, <math>BC</math>, and <math>CA</math>, respectively. Furthermore, suppose that <math>PD=2</math>, <math>PE=7</math>, and <math>PF=3</math>.
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Rewrite the given system of equations as\begin{align*} a^2 + 3^2 &= (8\sqrt{3}-b)^2 + 2^2 \\ b^2 + 2^2 &= (8\sqrt{3}-c)^2 + 7^2 \\ c^2 + 7^2 &= (8\sqrt{3}-a)^2 + 3^2. \end{align*}We provide a geometric interpretation for this system. Let <math>\triangle ABC</math> be equilateral, let <math>P</math> be a point in the interior of <math>\triangle ABC</math>, and let <math>D</math>, <math>E</math>, and <math>F</math> be the projections of <math>P</math> onto <math>AB</math>, <math>BC</math>, and <math>CA</math>, respectively. Furthermore, suppose that <math>PD=2</math>, <math>PE=7</math>, and <math>PF=3</math>.
  
 
<asy>
 
<asy>

Revision as of 20:15, 15 September 2025

Problem

There is a unique ordered triple of positive reals $(a,b,c)$ satisfying the system of equations \begin{align*} a^2 + 9 &= (b-8\sqrt{3})^2 + 4 \\ b^2 + 4 &= (c-8\sqrt{3})^2 + 49 \\ c^2 + 49 &= (a-8\sqrt{3})^2 + 9. \end{align*} The value of $100a+10b+c$ can be expressed as $m\sqrt{n},$ where $m$ and $n$ are positive integers such that $n$ is square-free. Find $m+n$.

Solution

Rewrite the given system of equations as\begin{align*} a^2 + 3^2 &= (8\sqrt{3}-b)^2 + 2^2 \\ b^2 + 2^2 &= (8\sqrt{3}-c)^2 + 7^2 \\ c^2 + 7^2 &= (8\sqrt{3}-a)^2 + 3^2. \end{align*}We provide a geometric interpretation for this system. Let $\triangle ABC$ be equilateral, let $P$ be a point in the interior of $\triangle ABC$, and let $D$, $E$, and $F$ be the projections of $P$ onto $AB$, $BC$, and $CA$, respectively. Furthermore, suppose that $PD=2$, $PE=7$, and $PF=3$.

[asy] unitsize(0.8cm); import geometry; pair A,B,C,P,D,E,F; A=(0,0); C=(6,0); B=(3,5.196); P=(1.9,1.40); D=foot(P,A,B); E=foot(P,B,C); F=foot(P,C,A); draw(A--B--C--cycle); draw(P--D,red); draw(P--E,red); draw(P--F,red); draw(rightanglemark(P,D,B,6)); draw(rightanglemark(P,E,C,6)); draw(rightanglemark(P,F,A,6)); draw(A--P,dashed+blue); draw(B--P,dashed+blue); draw(C--P,dashed+blue); dot(P); dot(A); dot(B); dot(C); dot(D); dot(E); dot(F); label("$A$",A,dir(225)); label("$B$",B,dir(90)); label("$C$",C,dir(-45)); label("$D$",D,dir(150)); label("$E$",E,dir(30)); label("$F$",F,dir(-90)); label("$P$",P+(-0.05,-0.2),dir(-40)); label("$a$",midpoint(A--F),dir(-90)); label("$b$",midpoint(B--D),dir(150)); label("$c$",midpoint(C--E),dir(30)); label("$3$",midpoint(P--F),dir(180)); label("$2$",midpoint(P--D),dir(60)); label("$7$",midpoint(P--E),dir(-45)); [/asy]

We claim that $(a,b,c) = (AF,BD,CE)$. To prove this, we need to verify that the side length $s$ of $\triangle ABC$ is $8\sqrt{3}$. We determine the value of $s$ by computing the area of $\triangle ABC$ in two different ways. On one hand, $[ABC] = \tfrac{1}{4}s^2\sqrt{3}$. On the other hand, $[ABC] = [APB]+[BPC]+[CPA] = 6s$. Equating these two expressions for $[ABC]$, we find $s=8\sqrt{3}$, as desired.

Now, we use the Law of Cosines on triangles $FAD$ and $FPD$ to obtain two expressions for the value of $DF^2$ that we can set equal. Note that $\angle FAD = 60^\circ$ and $\angle FPD = 120^\circ$. Thus, we have \[DF^2 = a^2 + (8\sqrt{3}-b)^2 - a(8\sqrt{3}-b) = 19.\] We also know that $(8\sqrt{3}-b)^2 = a^2+5$, and because $b=BD<8\sqrt{3}$, we have $8\sqrt{3}-b = \sqrt{a^2+5}$. Plugging in these expressions for $8\sqrt{3}-b$ and $(8\sqrt{3}-b)^2$ into the above equation and simplifying gives us \[2a^2-14 = a\sqrt{a^2+5}.\] Squaring and rearranging this equation yields the even quartic $3a^4 - 61a^2 + 196 = 0$. Solving and discarding negative solutions, we get $a = 2$ or $a=\tfrac{7}{\sqrt{3}}$. However, $a=2$ is extraneous as $2(2)^2 - 14 \ne 2\sqrt{2^2+5}$, so $a = \tfrac{7}{\sqrt{3}}$. It is now straightforward algebra to find that $(a,b,c) = (\tfrac{7}{\sqrt{3}}, \tfrac{16}{\sqrt{3}}, \tfrac{13}{\sqrt{3}})$. Therefore, $100a+10b+c = 291\sqrt{3}$, and we extract $291+3 = \boxed{294}$.

~Sedro

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