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 | ||
− | <math>y-x=2^ | + | <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 | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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}$, as before. |
Revision as of 23:05, 9 May 2020
Problem
Triangle has integer side lengths, including
, and a right angle,
. Let
and
denote the inradius and semiperimeter of
respectively. Find the perimeter of the triangle ABC which minimizes
.
Solution
First, label the other leg and the hypotenuse
. To minimize
,
must be maximize and
must be minimized. Through logic, it becomes clear that the triangle must be as close to equilateral as possible to maximize
and minimize
(Think about stretching one vertice of an equilateral triangle. The perimeter increases “faster” than the inradius).
From the Pythagorean theorem, , applying difference of squares yields
. Since the question states
and
must be integers, we can find possible values of
and
by finding the prime factorization of
, which is
. The two values of
and
that are closest to each other are the values that satisfy
, and
. Solving the system yields
and
. Thus, the perimeter is
~Someonenumber011
Solution 2
As before, label the other leg and the hypotenuse
. Let the angle opposite
be
, and let
. Then
, so that the area is
and the semiperimeter is
. Now we have
. By calculus, we know that this is minimized when
, which corresponds to
and
; by geometry, we know that this function, expressed as a function of
, is symmetric about this point.
Then we proceed as before. Searching the integer solutions of , we find that
and
is the closest integer approximation to
. (We can do so by noting that we should have
696+697+985=\boxed{2378}$, as before.