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

Latest revision as of 16:18, 27 June 2021

Problem

Given that $x, y,$ and $z$ are real numbers that satisfy: $\begin{align*} x &= \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}, \\ y &= \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}, \\ z &= \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}, \end{align*}$ 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 in 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 we 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 $h_z = \frac16$ .

The area of the triangle must be the same no matter how we measure it; therefore $x\cdot h_x = y\cdot h_y = z \cdot h_z$ gives us $\frac x4 = \frac y5 = \frac z6 = 2A$ 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 $2 + 7 = \boxed{009}$ .

The 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)

Note that none of $x,y,z$ can be zero.

Each of the equations is in the form $a=\sqrt{b^2-d^2}+\sqrt{c^2-d^2}$

Isolate a radical and square the equation to get $b^2-d^2=a^2-2a\sqrt{c^2-d^2}+c^2-d^2$

Now cancel, and again isolate the radical, and square the equation to get $a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2$

Rearranging gives $a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2$

Now note that everything is cyclic but the last term (i.e. $-4a^2d^2$ ), which implies $-4x^2\cdot\frac1{16}=-4y^2\cdot\frac1{25}=-4z^2\cdot\frac1{36}$

Or

$x: y: z=4: 5: 6 \implies x=\frac{4y}5 \textrm{ and } z=\frac{6y}5$

Plug these values into the middle equation to get $\frac{256y^4+625y^4+1296y^4}{625}=\frac{800y^4}{625}+\frac{1800y^4}{625}+\frac{1152y^4}{625}-\frac{100y^2}{625}$

Simplifying gives $1575y^4=100y^2 \textrm{ but } y \neq 0 \implies y^2=\frac4{63} \textrm{ or } y=\frac2{3\sqrt7}$

Substituting the value of $y$ for $x$ and $z$ gives $x+y+z = \frac{4y+5y+6y}5 = 3y = 3 \cdot \frac{2}{3\sqrt7} = \frac{2}{\sqrt7}$

And thus the answer is $\boxed{009}$

~phoenixfire

Video solution

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

@@ Line 1: / Line 1: @@
-#REDIRECT [[2006 AIME A Problems/Problem 15]]
+== Problem ==
+Given that <math> x, y, </math> and <math>z</math> are real numbers that satisfy:
+<cmath>\begin{align*}
+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}},
+\end{align*}</cmath>
+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>
+== 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 in 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]] we that <cmath>x = \sqrt{y^2 - h_x^2} + \sqrt{z^2 - h_x^2}</cmath>
+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>.
+The area of the triangle must be the same no matter how we measure it; therefore <math>x\cdot h_x = y\cdot h_y = z \cdot h_z</math> gives us <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 <cmath>A = \sqrt{15A \cdot 7A \cdot 5A \cdot 3A} = 15A^2\sqrt{7}</cmath>
+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 = \boxed{009}</math>.
+-------------------------------------
+The justification that there is an acute triangle with sides of length <math>x, y</math> and <math>z</math>:
+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.
+== Solution 2 (Algebraic) ==
+Note that none of <math>x,y,z</math> can be zero.
+Each of the equations is in the form <cmath>a=\sqrt{b^2-d^2}+\sqrt{c^2-d^2}</cmath>
+Isolate a radical and square the equation to get <cmath>b^2-d^2=a^2-2a\sqrt{c^2-d^2}+c^2-d^2</cmath>
+Now cancel, and again isolate the radical, and square the equation to get <cmath>a^4+b^4+c^4+2a^2c^2-2a^2b^2-2b^2c^2=4a^2c^2-4a^2d^2</cmath>
+Rearranging gives <cmath>a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2-4a^2d^2</cmath>
+Now note that everything is cyclic but the last term (i.e. <math>-4a^2d^2</math>), which implies <cmath>-4x^2\cdot\frac1{16}=-4y^2\cdot\frac1{25}=-4z^2\cdot\frac1{36}</cmath>
+Or
+<cmath>x: y: z=4: 5: 6 \implies x=\frac{4y}5 \textrm{ and } z=\frac{6y}5</cmath>
+Plug these values into the middle equation to get <cmath>\frac{256y^4+625y^4+1296y^4}{625}=\frac{800y^4}{625}+\frac{1800y^4}{625}+\frac{1152y^4}{625}-\frac{100y^2}{625}</cmath>
+Simplifying gives <cmath>1575y^4=100y^2 \textrm{ but } y \neq 0 \implies y^2=\frac4{63} \textrm{ or } y=\frac2{3\sqrt7}</cmath>
+Substituting the value of <math>y</math> for <math>x</math> and <math>z</math> gives
+<cmath>x+y+z = \frac{4y+5y+6y}5 = 3y = 3 \cdot \frac{2}{3\sqrt7} = \frac{2}{\sqrt7}</cmath>
+And thus the answer is <math>\boxed{009}</math>
+~phoenixfire
+==Video solution==
+https://www.youtube.com/watch?v=M6sC26dzb_I
+== See also ==
+{{AIME box|year=2006|n=II|num-b=14|after=Last Question}}
+[[Category:Intermediate Algebra Problems]]
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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