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Difference between revisions of "2023 WSMO Tiebreaker Round Problems/Problem 3"

(Created page with "==Problem== Let <math>S</math> be the set of all values of <math>a</math> such that the area of a triangle with side lengths <math>5, 7, a</math> is a positive integer. Find...")
 
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
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Letting <math>5</math> be the base of the triangle, we have a height of <math>\frac{2h}{5}</math> for <math>0<\frac{2h}{5}\le 7\implies h\in\{1,2,\dots,17\}</math>. Note for each height, we have two posible values of <math>a</math>, if the triangle if acute or obtuse. We have
 +
\begin{align*}
 +
a_1^2 &= \left(5-\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\right)^2+\left(\frac{2h}{5}\right)^2\\
 +
&= 5^2+\left(7^2-\left(\frac{2h}{5}\right)^2\right)-10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}+\left(\frac{2h}{5}\right)^2\\
 +
&= 74-10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\\
 +
\end{align*}
 +
for when the triangle is acute and
 +
\begin{align*}
 +
a_2^2 &= \left(5+\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\right)^2+\left(\frac{2h}{5}\right)^2\\
 +
&= 5^2+\left(7^2-\left(\frac{2h}{5}\right)^2\right)+10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}+\left(\frac{2h}{5}\right)^2\\
 +
&= 74+10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\\
 +
\end{align*}
 +
for when the triangle is obtuse. We have <cmath>a_1^2+a_2^2=\left(74-10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\right)+\left(74+10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\right)=148.</cmath> So, <cmath>\sum_{x\in S}\left(x^2\right) = \sum_{h\in\{1,2,\dots,17\}}148 = 148\cdot17=\boxed{2516}.</cmath>
 +
 +
~pinkpig

Latest revision as of 11:14, 15 September 2025

Problem

Let $S$ be the set of all values of $a$ such that the area of a triangle with side lengths $5, 7, a$ is a positive integer. Find $\sum_{x\in S}\left(x^2\right).$

Solution

Letting $5$ be the base of the triangle, we have a height of $\frac{2h}{5}$ for $0<\frac{2h}{5}\le 7\implies h\in\{1,2,\dots,17\}$. Note for each height, we have two posible values of $a$, if the triangle if acute or obtuse. We have \begin{align*} a_1^2 &= \left(5-\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\right)^2+\left(\frac{2h}{5}\right)^2\\ &= 5^2+\left(7^2-\left(\frac{2h}{5}\right)^2\right)-10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}+\left(\frac{2h}{5}\right)^2\\ &= 74-10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\\ \end{align*} for when the triangle is acute and \begin{align*} a_2^2 &= \left(5+\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\right)^2+\left(\frac{2h}{5}\right)^2\\ &= 5^2+\left(7^2-\left(\frac{2h}{5}\right)^2\right)+10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}+\left(\frac{2h}{5}\right)^2\\ &= 74+10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\\ \end{align*} for when the triangle is obtuse. We have \[a_1^2+a_2^2=\left(74-10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\right)+\left(74+10\sqrt{7^2-\left(\frac{2h}{5}\right)^2}\right)=148.\] So, \[\sum_{x\in S}\left(x^2\right) = \sum_{h\in\{1,2,\dots,17\}}148 = 148\cdot17=\boxed{2516}.\]

~pinkpig

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