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Difference between revisions of "2006 AIME I Problems/Problem 15"

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== Solution ==
 
== Solution ==
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{{solution}}
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This solution requires you to disregard rigor.  Additional solutions, or justification for the nonrigorous steps, would be appreciated.
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Let <math>\triangle XYZ</math> be a [[triangle]] with sides of length <math>x, y</math> and <math>z</math>, and suppose this triangle is acute (so all [[altitude]]s are on the interior of the triangle).
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Let the altitude to the side of length <math>x</math> be of length <math>h_x</math>, and similarly for <math>y</math> and <math>z</math>.  Then we have by two applications of the [[Pythagorean Theorem]] that <math>x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}</math>.  As a [[function]] of <math>h_x</math>, the [[RHS]] of this [[equation]] is strictly decreasing, so it takes each value in its [[range]] exactly once.  Thus we must have that <math>h_x^2 = \frac1{16}</math> and so <math>h_x = \frac{1}4</math> and similarly <math>h_y = \frac15</math> and <math>h_z = \frac16</math>.
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Since the [[area]] of the triangle must be the same no matter how we measure, <math>x\cdot h_x = y\cdot h_y = z \cdot h_z</math> and so <math>\frac x4 = \frac y5 = \frac z6 = 2A</math> and <math>x = 8A, y = 10A</math> and <math>z = 12A</math>.  The [[semiperimeter]] of the triangle is <math>s = \frac{8A + 10A + 12A}{2} = 15A</math> so by [[Heron's formula]] we have <math>A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}</math>.  Thus <math>A = \frac{1}{15\sqrt{7}}</math> and <math>x + y + z = 30A = \frac2{\sqrt{7}}</math> and the answer is <math>2 + 7 = 009</math>.
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Justification that there is a triangle with sides of length <math>x, y</math> and <math>z</math>:
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Note that <math>x, y</math> and <math>z</math> are each the sum of two [[positive]] [[square root]]s of [[real number]]s, so <math>x, y, z \geq 0</math>.  (Recall that, by [[AIME]] [[mathematical convention| convention]], all numbers (including square roots) are taken to be real unless otherwise indicated.)  Also, <math>\sqrt{y^2-\frac{1}{16}} < \sqrt{y^2} = y</math>, so we have <math>x < y + z</math>, <math>y < z + x</math> and <math>z < x + y</math>.  But these conditions are exactly those of the [[triangle inequality]], so there does exist such a triangle. 
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Justification that this triangle is an [[acute triangle]]:
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Still needed.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2006|n=I|num-b=14|after=Last Question}}
 
{{AIME box|year=2006|n=I|num-b=14|after=Last Question}}

Revision as of 15:51, 25 September 2007

Problem

Given that $x, y,$ and $z$ are real numbers that satisfy:

$x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}$
$y = \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}$
$z = \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}$

and that $x+y+z = \frac{m}{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime, find $m+n.$

Solution

This solution requires you to disregard rigor. Additional solutions, or justification for the nonrigorous steps, would be appreciated.

Let $\triangle XYZ$ be a triangle with sides of length $x, y$ and $z$, and suppose this triangle is acute (so all altitudes are on the interior of the triangle). Let the altitude to the side of length $x$ be of length $h_x$, and similarly for $y$ and $z$. Then we have by two applications of the Pythagorean Theorem that $x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}$. As a function of $h_x$, the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once. Thus we must have that $h_x^2 = \frac1{16}$ and so $h_x = \frac{1}4$ and similarly $h_y = \frac15$ and $h_z = \frac16$.

Since the area of the triangle must be the same no matter how we measure, $x\cdot h_x = y\cdot h_y = z \cdot h_z$ and so $\frac x4 = \frac y5 = \frac z6 = 2A$ and $x = 8A, y = 10A$ and $z = 12A$. The semiperimeter of the triangle is $s = \frac{8A + 10A + 12A}{2} = 15A$ so by Heron's formula we have $A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}$. Thus $A = \frac{1}{15\sqrt{7}}$ and $x + y + z = 30A = \frac2{\sqrt{7}}$ and the answer is $2 + 7 = 009$.


Justification that there is a triangle with sides of length $x, y$ and $z$:

Note that $x, y$ and $z$ are each the sum of two positive square roots of real numbers, so $x, y, z \geq 0$. (Recall that, by AIME convention, all numbers (including square roots) are taken to be real unless otherwise indicated.) Also, $\sqrt{y^2-\frac{1}{16}} < \sqrt{y^2} = y$, so we have $x < y + z$, $y < z + x$ and $z < x + y$. But these conditions are exactly those of the triangle inequality, so there does exist such a triangle.


Justification that this triangle is an acute triangle: Still needed.

See also

2006 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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