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=2068</math>

Revision as of 22:07, 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 minimized and $s$ must be maximized. 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 $676^2$, 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 $676+522+870=2068$

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