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

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{{delete|false problem, page does not say whether 10A or 10B}}
 
==Problem==
 
==Problem==
  
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<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 25</math>
 
<math>\textbf{(A)}\ 6\qquad\textbf{(B)}\ 9\qquad\textbf{(C)}\ 10\qquad\textbf{(D)}\ 12\qquad\textbf{(E)}\ 25</math>
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==Solution==
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We know that <math>a^2 + b^2 = c^2</math> represents a Pythagorean triple. The smallest Pythagorean triple is <math>(3, 4, 5)</math>.
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To check if this forms a non-degenerate triangle, we verify the triangle inequality:
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* <math>3 + 4 > 5</math>
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* <math>3 + 5 > 4</math>
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* <math>4 + 5 > 3</math>
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All inequalities hold, so <math>(3, 4, 5)</math> is a valid solution.
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Therefore, the least possible value of <math>a + b + c</math> is <math>3 + 4 + 5 = \boxed{(D) 12}</math>.

Latest revision as of 19:29, 14 November 2024

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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 $3 + 4 + 5 = \boxed{(D) 12}$.