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 real | + | Given that <math> x, y, </math> and <math>z</math> are [[real number]]s that satisfy: |
<center><math> x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}} </math> </center> | <center><math> x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}} </math> </center> | ||
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== Solution == | == Solution == | ||
− | + | 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>. | ||
− | + | 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 = \boxed{009}</math>. | |
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+ | 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. | |
− | + | {{incomplete|solution}} | |
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== See also == | == See also == |
Revision as of 21:15, 25 April 2008
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
Let be a triangle with sides of length and , and suppose this triangle is acute (so all altitudes are on 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 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 .
Since the area of the triangle must be the same no matter how we measure, and so and and . The semiperimeter of the triangle is so by Heron's formula we have . Thus and and the answer is .
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.
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 |