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Difference between revisions of "2006 AIME II Problems/Problem 15"
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and that <math> x+y+z = \frac{m}{\sqrt{n}}, </math> where <math> m </math> and <math> n </math> are positive integers and <math> n </math> is not divisible by the square of any prime, find <math> m+n.</math>
 
and that <math> x+y+z = \frac{m}{\sqrt{n}}, </math> where <math> m </math> and <math> n </math> are positive integers and <math> n </math> is not divisible by the square of any prime, find <math> m+n.</math>
  
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== Solution ==
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== Solution 1 (Geometric Interpretation)==
 
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).
 
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).
 
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>.
 
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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Note that <math>x, y</math> and <math>z</math> are each the sum of two positive [[square root]]s of real numbers, 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.
 
Note that <math>x, y</math> and <math>z</math> are each the sum of two positive [[square root]]s of real numbers, 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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== Solution 2 (Algebraic) ==
  
 
==Video solution==
 
==Video solution==

Revision as of 00:09, 25 September 2020

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 1 (Geometric Interpretation)

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 $2 + 7 = \boxed{9}$.


Justification that there is an acute 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.

Solution 2 (Algebraic)

Video solution

https://www.youtube.com/watch?v=M6sC26dzb_I

See also

2006 AIME II (Problems • Answer Key • Resources)
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