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

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==Solution==
 
==Solution==
  
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First, label the other leg <math>x</math> and the hypotenuse <math>y</math>. To minimize <math>\frac{s}{r}</math>, <math>r</math> must be maximize and <math>s</math> must be minimized. Through logic, it becomes clear that the triangle must be as close to equilateral as possible to maximize <math>r</math> and minimize <math>s</math> (Think about stretching one vertice of an equilateral triangle. The perimeter increases faster than the inradius).  
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First, label the other leg <math>x</math> and the hypotenuse <math>y</math>. To minimize <math>\frac{s}{r}</math>, <math>r</math> must be maximize and <math>s</math> must be minimized. Through logic, it becomes clear that the triangle must be as close to equilateral as possible to maximize <math>r</math> and minimize <math>s</math> (Think about stretching one vertice of an equilateral triangle. The perimeter increases “faster” than the inradius).  
  
 
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>696^2</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>696^2</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  

Revision as of 21:33, 17 November 2019

Problem

Triangle $ABC$ has integer side lengths, including $BC = 696$, and a right angle, $\angle ABC$. Let $r$ and $s$ denote the inradius and semiperimeter of $ABC$ respectively. Find the perimeter of the triangle ABC which minimizes $\frac{s}{r}$.

Solution

First, label the other leg $x$ and the hypotenuse $y$. To minimize $\frac{s}{r}$, $r$ must be maximize and $s$ must be minimized. Through logic, it becomes clear that the triangle must be as close to equilateral as possible to maximize $r$ and minimize $s$ (Think about stretching one vertice of an equilateral triangle. The perimeter increases “faster” than the inradius).

From the Pythagorean theorem, $y^2-x^2=696^2$, applying difference of squares yields $(y-x)(y+x)=696^2$. Since the question states $x$ and $y$ must be integers, we can find possible values of $x$ and $y$ by finding the prime factorization of $696^2$, which is $2^6 \cdot 3 \cdot 29$. The two values of $x$ and $y$ that are closest to each other are the values that satisfy $y-x=2^2 \cdot 3 \cdot 29$, and $y+x=2^4 \cdot 3 \cdot 29$. Solving the system yields $x = 522$ and $y = 870$. Thus, the perimeter is $676+522+870=\boxed{2068}$


Please check my solution as there may be errors.

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