Difference between revisions of "2006 AIME II Problems/Problem 15"
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== Problem == | == Problem == | ||
− | Given that <math> x, y, </math> and <math>z</math> are [[ | + | 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 (Heron Bash)== | |
+ | We can rewrite the equations as follows: | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{1}{4}x &= \sqrt{\left(y+\frac{1}{4}\right) \left(y-\frac{1}{4}\right)\left(\frac{1}{4}\right) \left(\frac{1}{4}\right)}+\sqrt{\left(z+\frac{1}{4}\right) \left(z-\frac{1}{4}\right) \left(\frac{1}{4}\right) \left(\frac{1}{4}\right)} \ | ||
+ | \frac{1}{5} y &= \sqrt{\left(z+\frac{1}{5}\right) \left(z-\frac{1}{5}\right) \left(\frac{1}{5}\right) \left(\frac{1}{5}\right)}+\sqrt{\left(x+\frac{1}{5}\right) \left(x-\frac{1}{5}\right) \left(\frac{1}{5}\right) \left(\frac{1}{5}\right)} \ | ||
+ | \frac{1}{6} z &= \sqrt{\left(x+\frac{1}{6}\right) \left(x-\frac{1}{6}\right) \left(\frac{1}{6}\right) \left(\frac{1}{6}\right)}+\sqrt{\left(y+\frac{1}{6}\right) \left(y-\frac{1}{6}\right) \left(\frac{1}{6}\right) \left(\frac{1}{6}\right)} | ||
+ | \end{align*}</cmath> | ||
− | + | Take the first equation. The first square root is the area of a triangle with side lengths <math>y,y,\frac{1}{2}</math> by Heron’s Formula. Similarly, the second square root is the area of a triangle with side lengths <math>z,z,\frac{1}{2}</math>. If we connect these two triangles together at the <math>\frac{1}{2}</math> side, we obtain a kite. The area of the kite is <math>\frac{1}{4}x</math>, and since the first diagonal is <math>\frac{1}{2}</math>, the second diagonal is <math>\frac{2\cdot\frac{1}{4}x}{\frac{1}{2}}=x</math>. If we draw this diagonal, we obtain two triangles with side lengths <math>x,y,z</math>. Let this triangle have area <math>A</math>. Then <math>\frac{1}{4}x=2A</math>; extend this for the other two equations. We can substitute into the first equation to obtain a value for <math>x</math>, and the answer is <math>\frac{2}{\sqrt{7}}\Rightarrow\boxed{009}</math>. | |
− | + | ~eevee9406 | |
− | == Solution | + | == Solution 3 (Algebraic) == |
Note that none of <math>x,y,z</math> can be zero. | Note that none of <math>x,y,z</math> can be zero. | ||
Line 44: | Line 67: | ||
And thus the answer is <math>\boxed{009}</math> | And thus the answer is <math>\boxed{009}</math> | ||
+ | |||
+ | ~phoenixfire | ||
==Video solution== | ==Video solution== |
Latest revision as of 12:40, 3 December 2024
Contents
[hide]Problem
Given that and are real numbers that satisfy: and that where and are positive integers and is not divisible by the square of any prime, find
Solution 1 (Geometric Interpretation)
Let be a triangle with sides of length and , and suppose this triangle is acute (so all altitudes are in the interior of the triangle).
Let the altitude to the side of length be of length , and similarly for and . Then we have by two applications of the Pythagorean Theorem we that
As a function of , the RHS of this equation is strictly decreasing, so it takes each value in its range exactly once. Thus we must have that and so and similarly and .
The area of the triangle must be the same no matter how we measure it; therefore gives us and and .
The semiperimeter of the triangle is so by Heron's formula we have
Thus, and and the answer is .
The justification that there is an acute triangle with sides of length and :
Note that and are each the sum of two positive square roots of real numbers, so . (Recall that, by AIME convention, all numbers (including square roots) are taken to be real unless otherwise indicated.)
Also, , so we have , and . But these conditions are exactly those of the triangle inequality, so there does exist such a triangle.
Solution 2 (Heron Bash)
We can rewrite the equations as follows:
Take the first equation. The first square root is the area of a triangle with side lengths by Heron’s Formula. Similarly, the second square root is the area of a triangle with side lengths . If we connect these two triangles together at the side, we obtain a kite. The area of the kite is , and since the first diagonal is , the second diagonal is . If we draw this diagonal, we obtain two triangles with side lengths . Let this triangle have area . Then ; extend this for the other two equations. We can substitute into the first equation to obtain a value for , and the answer is .
~eevee9406
Solution 3 (Algebraic)
Note that none of can be zero.
Each of the equations is in the form
Isolate a radical and square the equation to get
Now cancel, and again isolate the radical, and square the equation to get
Rearranging gives
Now note that everything is cyclic but the last term (i.e. ), which implies
Or
Plug these values into the middle equation to get
Simplifying gives
Substituting the value of for and gives
And thus the answer is
~phoenixfire
Video solution
https://www.youtube.com/watch?v=M6sC26dzb_I
See also
2006 AIME II (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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