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

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==Problem==
 
==Problem==
  
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Let  <math>T  =  TNFTPP</math>. Triangle  <math>ABC</math> has integer side lengths, including  <math>BC  =  696</math>, and a right angle,  <math>\angle ABC</math>. Let  <math>r</math> and  <math>s</math> denote the inradius and semiperimeter of  <math>ABC</math> respectively. Find the perimeter of the triangle ABC which minimizes  <math>\frac{s}{r}</math>.
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Triangle  <math>ABC</math> has integer side lengths, including  <math>BC  =  696</math>, and a right angle,  <math>\angle ABC</math>. Let  <math>r</math> and  <math>s</math> denote the inradius and semiperimeter of  <math>ABC</math> respectively. Find the perimeter of the triangle ABC which minimizes  <math>\frac{s}{r}</math>.
  
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==Solution==
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==Solutions==
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First, label the other leg <math>x</math> and the hypotenuse <math>y</math>. To minimize
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 +
===Solution 1 (credit to NikoIsLife)===
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 +
Let <math>AB = x</math> and <math>AC = y</math>.  By the Pythagorean Theorem, <math>y^2 - x^2 = 696^2</math>, and applying difference of squares yields <math>(y-x)(y+x)=696^2</math>.  Because <math>x</math> and <math>y</math> have the same parity (due to being integers), both <math>y+x</math> and <math>y-x</math> are even. 
 +
 
 +
<br>
 +
Let <math>y+x = 2b</math> abd <math>y-x = 2a</math>; then <math>ab = 348^2</math>.  Additionally,
 +
<cmath>r = \frac{x+696-y}{2} = 348 - a</cmath>
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<cmath>s = \frac{x+696+y}{2} = 348+b.</cmath>
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Therefore,
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<cmath>\begin{align*}
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\frac{s}{r} &= \frac{348+b}{348-a} \\
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&= \frac{348 + b}{348 - \frac{348^2}{b}} \\
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&= \frac{b^2 + 348b}{348b - 348^2} \\
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&= \frac{b}{348} + 2 + \frac{2 \cdot 348^2}{348b - 348^2} \\
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&= \frac{b}{348} + 2 + \frac{696}{b-348}.
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\end{align*}</cmath>
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Note that because <math>y > 696</math>, we must have <math>b > 696</math>.  We can do some optimization by using the [[derivative]] -- if we let <math>f(b) = \frac{b}{348} + 2 + \frac{696}{b-348}</math>, then
 +
<cmath>f'(b) = \frac{1}{348} + \frac{-696}{(b-348)^2},</cmath>
 +
which equals <math>0</math> if <math>b = 348 + 348\sqrt{2} \approx 840.146</math>.  Since <math>f'(b) < 0</math> if <math>696 < b < 348+348\sqrt{2}</math> and <math>f'(b) > 0</math> if <math>b > 348 + 348\sqrt{2}</math>, we can confirm that <math>b = 348+348\sqrt{2}</math> results in the absolute minimum of <math>\frac{s}{r}</math>.  However, the case where <math>b = 348+348\sqrt{2}</math> does not happen if <math>x, y</math> are integers, and since <math>b</math> is a factor of <math>348^2</math>, we need to test the largest factor of <math>348^2</math> less than <math>348 + 348\sqrt{2}</math> and the smallest factor of <math>348^2</math> greater than <math>348 + 348\sqrt{2}</math>.
 +
 
 +
<br>
 +
The largest factor of <math>348^2</math> less than <math>348 + 348\sqrt{2}</math> is <math>696</math> (which does not work), and the smallest factor of <math>348^2</math> greater than <math>348+348\sqrt{2}</math> is <math>841</math>.  Therefore, <math>b = 841</math>, which means that <math>a = 144</math>, <math>y = 985</math>, and <math>x = 697</math>.  Our wanted perimeter  is <math>696+697+985=\boxed{2378}</math>.
 +
 
 +
===Solution 2===
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 +
As before, label the other leg <math>x</math> and the hypotenuse <math>y</math>. Let the opposite angle to <math>x</math> be <math>\theta</math>, and let <math>t:=\tan\frac{\theta}{2}</math>; let the [[area]] be <math>A</math> and the [[semiperimeter]] <math>s</math>. Then we have <math>x = 696*\frac{2t}{1-t^2}, y = 696*\frac{1+t^2}{1-t^2}, A = 696^2*\frac{t}{1-t^2}, s = 696*\frac{1+t}{1-t^2} = 696*\frac{1}{1-t}</math>. This means that <math>\frac{s}{r} = \frac{s^2}{sr} = \frac{(\frac{1}{1-t})^2}{\frac{t}{1-t^2}} = \frac{1+t}{t(1-t)}</math>. By [[calculus]], we know that this function is minimized at <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 in terms of <math>\theta</math>, is symmetric around this point.
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Then we proceed as before, searching for [[Diophantine]] solutions of <math>y^2 - x^2 = 696^2</math> with <math>x</math> closest to <math>696</math>, and we find that <math>x = 697, y = 985</math> is the closest. (We can do so by noting that we would want <math>y - x \approx 696*(\sqrt{2}-1) = 288.29</math>.) Then the perimeter is <math>696+697+985=\boxed{2378}</math> as before, and we are done.
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 +
~duck_master
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{{alternate solutions}}
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==See Also==
 +
{{iTest box|year=2006|num-b=U6|num-a=U8|ver=[[2006 iTest Problems/Problem U1|U1]] '''•''' [[2006 iTest Problems/Problem U2|U2]] '''•''' [[2006 iTest Problems/Problem U3|U3]] '''•''' [[2006 iTest Problems/Problem U4|U4]] '''•''' [[2006 iTest Problems/Problem U5|U5]] '''•''' [[2006 iTest Problems/Problem U6|U6]] '''•''' [[2006 iTest Problems/Problem U7|U7]] '''•''' [[2006 iTest Problems/Problem U8|U8]] '''•''' [[2006 iTest Problems/Problem U9|U9]] '''•''' [[2006 iTest Problems/Problem U10|U10]]}}
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 +
[[Category:Intermediate Geometry Problems]]

Latest revision as of 15:03, 25 February 2021

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 $\frac{s}{r}$.

Solutions

Solution 1 (credit to NikoIsLife)

Let $AB = x$ and $AC = y$. By the Pythagorean Theorem, $y^2 - x^2 = 696^2$, and applying difference of squares yields $(y-x)(y+x)=696^2$. Because $x$ and $y$ have the same parity (due to being integers), both $y+x$ and $y-x$ are even.


Let $y+x = 2b$ abd $y-x = 2a$; then $ab = 348^2$. Additionally, \[r = \frac{x+696-y}{2} = 348 - a\] \[s = \frac{x+696+y}{2} = 348+b.\] Therefore, \begin{align*} \frac{s}{r} &= \frac{348+b}{348-a} \\ &= \frac{348 + b}{348 - \frac{348^2}{b}} \\ &= \frac{b^2 + 348b}{348b - 348^2} \\ &= \frac{b}{348} + 2 + \frac{2 \cdot 348^2}{348b - 348^2} \\ &= \frac{b}{348} + 2 + \frac{696}{b-348}. \end{align*} Note that because $y > 696$, we must have $b > 696$. We can do some optimization by using the derivative -- if we let $f(b) = \frac{b}{348} + 2 + \frac{696}{b-348}$, then \[f'(b) = \frac{1}{348} + \frac{-696}{(b-348)^2},\] which equals $0$ if $b = 348 + 348\sqrt{2} \approx 840.146$. Since $f'(b) < 0$ if $696 < b < 348+348\sqrt{2}$ and $f'(b) > 0$ if $b > 348 + 348\sqrt{2}$, we can confirm that $b = 348+348\sqrt{2}$ results in the absolute minimum of $\frac{s}{r}$. However, the case where $b = 348+348\sqrt{2}$ does not happen if $x, y$ are integers, and since $b$ is a factor of $348^2$, we need to test the largest factor of $348^2$ less than $348 + 348\sqrt{2}$ and the smallest factor of $348^2$ greater than $348 + 348\sqrt{2}$.


The largest factor of $348^2$ less than $348 + 348\sqrt{2}$ is $696$ (which does not work), and the smallest factor of $348^2$ greater than $348+348\sqrt{2}$ is $841$. Therefore, $b = 841$, which means that $a = 144$, $y = 985$, and $x = 697$. Our wanted perimeter is $696+697+985=\boxed{2378}$.

Solution 2

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

Then we proceed as before, searching for Diophantine solutions of $y^2 - x^2 = 696^2$ with $x$ closest to $696$, and we find that $x = 697, y = 985$ is the closest. (We can do so by noting that we would want $y - x \approx 696*(\sqrt{2}-1) = 288.29$.) Then the perimeter is $696+697+985=\boxed{2378}$ as before, and we are done.

~duck_master

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See Also

2006 iTest (Problems, Answer Key)
Preceded by:
Problem U6 		Followed by:
Problem U8
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