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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>696^2</math>, which is <math>2^6 \cdot 3^2 \cdot 29^2</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 3^2 \cdot 29^2</math>. The two values of <math>x</math> and <math>y</math> that are closest to each other are the values that satisfy  
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<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>
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<math>y-x=2^5 \cdot 3^2</math>, and <math>y+x=2 \cdot 29^2</math>. Solving the system yields <math>x = 697</math> and <math>y = 985</math>. Thus, the perimeter is <math>696+697+985=\boxed{2378}</math>
  
 
~Someonenumber011
 
~Someonenumber011
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==Solution 2==
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As before, label the other leg <math>x</math> and the hypotenuse <math>y</math>. Let the angle opposite <math>x</math> be <math>\theta</math>, and let <math>t:=\tan\frac{\theta}{2}</math>. Then <math>x = 696*\frac{2t}{1-t^2}, y = 696*\frac{1+t^2}{1-t^2}</math>, so that the area is <math>A = 696^2*\frac{t}{1-t^2}</math> and the semiperimeter is <math>s = 696*\frac{1+t}{1-t^2} = 696*\frac{1}{1-t}</math>. Now we have <math>\frac{s}{r} = \frac{s^2}{sr} = \frac{s^2}{A} = \frac{(\frac{1}{1-t})^2}{\frac{t}{1-t^2}} = \frac{1+t}{t(1-t)}</math>. By calculus, we know that this is minimized when <math>t = \sqrt{2}-1</math>, which corresponds to <math>\theta = 45^\circ</math> and <math>x = 696</math>; by geometry, we know that this function, expressed as a function of <math>\theta</math>, is symmetric about this point.
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Then we proceed as before. Searching the integer solutions of <math>y^2 - x^2 = 696^2</math>, we find that <math>x = 697</math> and <math>y = 985</math> is the closest integer approximation to <math>x \approx 696</math>. (We can do so by noting that we should have <math>y - x \approx 696(\sqrt{2}-1) \approx 288.29.) This means that the perimeter is </math>696+697+985=\boxed{2378}$, as before.

Revision as of 00:05, 10 May 2020

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 $2^6 \cdot 3^2 \cdot 29^2$. The two values of $x$ and $y$ that are closest to each other are the values that satisfy $y-x=2^5 \cdot 3^2$, and $y+x=2 \cdot 29^2$. Solving the system yields $x = 697$ and $y = 985$. Thus, the perimeter is $696+697+985=\boxed{2378}$

~Someonenumber011

Solution 2

As before, label the other leg $x$ and the hypotenuse $y$. Let the angle opposite $x$ be $\theta$, and let $t:=\tan\frac{\theta}{2}$. Then $x = 696*\frac{2t}{1-t^2}, y = 696*\frac{1+t^2}{1-t^2}$, so that the area is $A = 696^2*\frac{t}{1-t^2}$ and the semiperimeter is $s = 696*\frac{1+t}{1-t^2} = 696*\frac{1}{1-t}$. Now we have $\frac{s}{r} = \frac{s^2}{sr} = \frac{s^2}{A} = \frac{(\frac{1}{1-t})^2}{\frac{t}{1-t^2}} = \frac{1+t}{t(1-t)}$. By calculus, we know that this is minimized when $t = \sqrt{2}-1$, which corresponds to $\theta = 45^\circ$ and $x = 696$; by geometry, we know that this function, expressed as a function of $\theta$, is symmetric about this point.

Then we proceed as before. Searching the integer solutions of $y^2 - x^2 = 696^2$, we find that $x = 697$ and $y = 985$ is the closest integer approximation to $x \approx 696$. (We can do so by noting that we should have $y - x \approx 696(\sqrt{2}-1) \approx 288.29.) This means that the perimeter is$696+697+985=\boxed{2378}$, as before.

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