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

(Created page with "==Problem== There is a unique ordered triple of positive reals <math>(a,b,c)</math> satisfying the system of equations <cmath>\begin{align*} a^2 + 9 &= (b-8\sqrt{3})^2 + 4 \...")
 
m (→‎Solution)
 
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==Solution==
 
==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>\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>
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unitsize(0.8cm);
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import geometry;
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pair A,B,C,P,D,E,F;
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A=(0,0);
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C=(6,0);
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B=(3,5.196);
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P=(1.9,1.40);
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D=foot(P,A,B);
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E=foot(P,B,C);
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F=foot(P,C,A);
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draw(A--B--C--cycle);
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draw(P--D,red);
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draw(P--E,red);
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draw(P--F,red);
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draw(rightanglemark(P,D,B,6));
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draw(rightanglemark(P,E,C,6));
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draw(rightanglemark(P,F,A,6));
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draw(A--P,dashed+blue);
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draw(B--P,dashed+blue);
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draw(C--P,dashed+blue);
 +
 +
dot(P);
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dot(A);
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dot(B);
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dot(C);
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dot(D);
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dot(E);
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dot(F);
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label("$A$",A,dir(225));
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label("$B$",B,dir(90));
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label("$C$",C,dir(-45));
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label("$D$",D,dir(150));
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label("$E$",E,dir(30));
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label("$F$",F,dir(-90));
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label("$P$",P+(-0.05,-0.2),dir(-40));
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label("$a$",midpoint(A--F),dir(-90));
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label("$b$",midpoint(B--D),dir(150));
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label("$c$",midpoint(C--E),dir(30));
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label("$3$",midpoint(P--F),dir(180));
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label("$2$",midpoint(P--D),dir(60));
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label("$7$",midpoint(P--E),dir(-45));
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</asy>
 +
 +
We claim that <math>(a,b,c) = (AF,BD,CE)</math>. To prove this, we need to verify that the side length <math>s</math> of <math>\triangle ABC</math> is <math>8\sqrt{3}</math>. We determine the value of <math>s</math> by computing the area of <math>\triangle ABC</math> in two different ways. On one hand, <math>[ABC] = \tfrac{1}{4}s^2\sqrt{3}</math>. On the other hand, <math>[ABC] = [APB]+[BPC]+[CPA] = 6s</math>. Equating these two expressions for <math>[ABC]</math>, we find <math>s=8\sqrt{3}</math>, as desired.
 +
 +
Now, we use the Law of Cosines on <math>\triangle FAD</math> and <math>\triangle FPD</math> to obtain two expressions for the value of <math>DF^2</math> that we can set equal. Note that <math>\angle FAD = 60^\circ</math> and <math>\angle FPD = 120^\circ</math>. Thus, we have
 +
<cmath>DF^2 = a^2 + (8\sqrt{3}-b)^2 - a(8\sqrt{3}-b) = 19.</cmath>
 +
We also know that <math>(8\sqrt{3}-b)^2 = a^2+5</math>, and because <math>b=BD<8\sqrt{3}</math>, we have <math>8\sqrt{3}-b = \sqrt{a^2+5}</math>. Plugging in these expressions for <math>8\sqrt{3}-b</math> and <math>(8\sqrt{3}-b)^2</math> into the above equation and simplifying gives us
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<cmath>2a^2-14 = a\sqrt{a^2+5}.</cmath>
 +
Squaring and rearranging this equation yields the even quartic <math>3a^4 - 61a^2 + 196 = 0</math>. Solving and discarding negative solutions, we get <math>a = 2</math> or <math>a=\tfrac{7}{\sqrt{3}}</math>. However, <math>a=2</math> is extraneous as <math>2(2)^2 - 14 \ne 2\sqrt{2^2+5}</math>, so <math>a = \tfrac{7}{\sqrt{3}}</math>. It is now straightforward algebra to find that <math>(a,b,c) = (\tfrac{7}{\sqrt{3}}, \tfrac{16}{\sqrt{3}}, \tfrac{13}{\sqrt{3}})</math>. Therefore, <math>100a+10b+c = 291\sqrt{3}</math>, and we extract <math>291+3 = \boxed{294}</math>.
 +
 +
~Sedro

Latest revision as of 14:43, 17 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 $\triangle FAD$ and $\triangle FPD$ 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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