Difference between revisions of "2006 iTest Problems/Problem U7"
Duck master (talk | contribs) (fixed someonenumber011's sol and added own) |
Duck master (talk | contribs) (fixed someonenumber011's sol and added own) |
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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 | ||
− | <math>y-x=2^ | + | <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>696+697+985=\boxed{2378}</math> |
~Someonenumber011 | ~Someonenumber011 | ||
− | ==Solution 2== | + | ==Solution 2 (calculus)== |
− | As before, label the other leg <math>x</math> and the hypotenuse <math>y</math>. Let the angle | + | 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. |
− | Then we proceed as before | + | 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. |
+ | |||
+ | ~duck_master |
Revision as of 23:15, 9 May 2020
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
.
Solution
First, label the other leg and the hypotenuse
. To minimize
,
must be maximize and
must be minimized. Through logic, it becomes clear that the triangle must be as close to equilateral as possible to maximize
and minimize
(Think about stretching one vertice of an equilateral triangle. The perimeter increases “faster” than the inradius).
From the Pythagorean theorem, , applying difference of squares yields
. Since the question states
and
must be integers, we can find possible values of
and
by finding the prime factorization of
, which is
. The two values of
and
that are closest to each other are the values that satisfy
, and
. Solving the system yields
and
. Thus, the perimeter is
~Someonenumber011
Solution 2 (calculus)
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