Difference between revisions of "2006 iTest Problems/Problem U7"
Duck master (talk | contribs) (fixed someonenumber011's sol and added own) |
Rockmanex3 (talk | contribs) (Solution to U7 -- first solution writeup of 2021 and finally got that problem down!) |
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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>. | 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>. | ||
− | == | + | ==Solutions== |
− | + | ===Solution 1 (credit to NikoIsLife)=== | |
− | + | 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. | |
− | <math>y-x | ||
− | + | <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> | ||
+ | <cmath>s = \frac{x+696+y}{2} = 348+b.</cmath> | ||
+ | Therefore, | ||
+ | <cmath>\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*}</cmath> | ||
+ | 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>. | ||
− | ==Solution 2 | + | <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=== | ||
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. | 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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~duck_master | ~duck_master | ||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | ==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]]}} | ||
+ | |||
+ | [[Category:Intermediate Geometry Problems]] |
Latest revision as of 16:03, 25 February 2021
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 .
Solutions
Solution 1 (credit to NikoIsLife)
Let and . By the Pythagorean Theorem, , and applying difference of squares yields . Because and have the same parity (due to being integers), both and are even.
Let abd ; then . Additionally,
Therefore,
Note that because , we must have . We can do some optimization by using the derivative -- if we let , then
which equals if . Since if and if , we can confirm that results in the absolute minimum of . However, the case where does not happen if are integers, and since is a factor of , we need to test the largest factor of less than and the smallest factor of greater than .
The largest factor of less than is (which does not work), and the smallest factor of greater than is . Therefore, , which means that , , and . Our wanted perimeter is .
Solution 2
As before, label the other leg and the hypotenuse . Let the opposite angle to be , and let ; let the area be and the semiperimeter . Then we have . This means that . By calculus, we know that this function is minimized at , which corresponds to and ; by geometry, we know that this function, expressed in terms of , is symmetric around this point.
Then we proceed as before, searching for Diophantine solutions of with closest to , and we find that is the closest. (We can do so by noting that we would want .) Then the perimeter is 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) | ||
Preceded by: Problem U6 |
Followed by: Problem U8 | |
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