A geometry problem by Omar Ortiz

Geometry Level 1

ABC is a right triangle with measures 3n, 3n + 3 and 5n. If n is a positive integer, greater than 1. What is the smallest value of n?


The answer is 3.

Omar Ortiz by Omar Ortiz , 19, Mexico

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2 solutions

Caleb Townsend
Feb 26, 2015

Since n ≥ 2 , n \geq 2, 5 n > 3 n + 3 > 3 n . 5n > 3n+3 > 3n. So by applying pythagorean theorem, ( 3 n ) 2 + ( 3 n + 3 ) 2 = ( 5 n ) 2 (3n)^2 + (3n+3)^2 = (5n)^2 9 n 2 + ( 3 n + 3 ) 2 = 25 n 2 9n^2 + (3n+3)^2 = 25n^2 ( 3 n + 3 ) 2 = 16 n 2 = ( 4 n ) 2 (3n+3)^2 = 16n^2 = (4n)^2 3 n + 3 = 4 n 3n + 3 = 4n n = 3 n = \boxed{3}

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( 3 n ) 2 + ( 3 + n ) 2 = ( 5 n ) 2 ⟹ 9 ( n 2 + n 2 + 2 n + 1 ) = 25 n 2 ∴ 7 n 2 − 18 n − 9 = 0 ⟹ n = − 3 / 7 o r n = 3. O n l y + t i v e v a l u e o f n i s 3 (3n)^2+(3+n)^2=(5n)^2 \implies~9(n^2+n^2+2n+1)=25n^2 \\\therefore~7n^2-18n-9=0~~\implies n=- 3/7~~or~~~ n=3.\\Only~+tive~value~of~n~is~\boxed{\color{#3D99F6}{\Large 3}}

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