# Difference between revisions of "2006 iTest Problems/Problem U7"

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From the Pythagorean theorem, <math>y^2-x^2=696^2</math>, applying difference of squares yields <math>(y-x)(y+x)=696^2</math>. Since the question states <math>x</math> and <math>y</math> must be integers, we can find possible values of <math>x</math> and <math>y</math> by finding the prime factorization of <math>676^2</math>, which is <math>2^6 \cdot 3 \cdot 29</math>. The two values of <math>x</math> and <math>y</math> that are closest to each other are the values that satisfy | From the Pythagorean theorem, <math>y^2-x^2=696^2</math>, applying difference of squares yields <math>(y-x)(y+x)=696^2</math>. Since the question states <math>x</math> and <math>y</math> must be integers, we can find possible values of <math>x</math> and <math>y</math> by finding the prime factorization of <math>676^2</math>, which is <math>2^6 \cdot 3 \cdot 29</math>. The two values of <math>x</math> and <math>y</math> that are closest to each other are the values that satisfy | ||

− | <math>y-x=2^2 \cdot 3 \cdot 29</math>, and <math>y+x=2^4 \cdot 3 \cdot 29</math>. Solving the system yields <math>x = 522 </math> and <math>y = 870</math>. Thus, the perimeter is <math>676+522+870=2068</math> | + | <math>y-x=2^2 \cdot 3 \cdot 29</math>, and <math>y+x=2^4 \cdot 3 \cdot 29</math>. Solving the system yields <math>x = 522 </math> and <math>y = 870</math>. Thus, the perimeter is <math>676+522+870=\boxed{2068}</math> |

## Revision as of 22:08, 17 November 2019

## Problem

Triangle has integer side lengths, including , and a right angle, . Let and denote the inradius and semiperimeter of respectively. Find the perimeter of the triangle ABC which minimizes .

## Solution

First, label the other leg and the hypotenuse . To minimize , must be minimized and must be maximized. Through logic, it becomes clear that the triangle must be as close to equilateral as possible to maximize and minimize (Think about stretching one vertice of an equilateral triangle. The perimeter increases faster than the inradius).

From the Pythagorean theorem, , applying difference of squares yields . Since the question states and must be integers, we can find possible values of and by finding the prime factorization of , which is . The two values of and that are closest to each other are the values that satisfy , and . Solving the system yields and . Thus, the perimeter is