Difference between revisions of "2024 AMC 10 Problems/Problem 15"
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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> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | To check if this forms a non-degenerate triangle, we verify the triangle inequality: | ||
+ | * <math>3 + 4 > 5</math> | ||
+ | * <math>3 + 5 > 4</math> | ||
+ | * <math>4 + 5 > 3</math> | ||
+ | |||
+ | All inequalities hold, so <math>(3, 4, 5)</math> is a valid solution. | ||
+ | |||
+ | Therefore, the least possible value of <math>a + b + c</math> is <math>3 + 4 + 5 = \boxed{(D) 12}</math>. |
Revision as of 08:43, 31 October 2024
Problem
Let , , and be positive integers such that . What is the least possible value of such that , , and form a non-degenerate triangle?
Solution
We know that represents a Pythagorean triple. The smallest Pythagorean triple is .
To check if this forms a non-degenerate triangle, we verify the triangle inequality:
All inequalities hold, so is a valid solution.
Therefore, the least possible value of is .