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Difference between revisions of "2024 AMC 10 Problems/Problem 15"
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==Problem==
 
==Problem==
  

Revision as of 20:29, 14 November 2024

This page has been proposed for deletion. Reason: false problem, page does not say whether 10A or 10B

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Problem

Let $a$, $b$, and $c$ be positive integers such that $a^2 + b^2 = c^2$. What is the least possible value of $a + b + c$ such that $a$, $b$, and $c$ form a non-degenerate triangle?

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 25$

Solution

We know that $a^2 + b^2 = c^2$ represents a Pythagorean triple. The smallest Pythagorean triple is $(3, 4, 5)$.

To check if this forms a non-degenerate triangle, we verify the triangle inequality:

  • $3 + 4 > 5$
  • $3 + 5 > 4$
  • $4 + 5 > 3$

All inequalities hold, so $(3, 4, 5)$ is a valid solution.

Therefore, the least possible value of $a + b + c$ is $3 + 4 + 5 = \boxed{(D) 12}$.

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