Difference between revisions of "2006 AIME II Problems/Problem 15"
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| − | + | == Problem == | |
| + | Given that <math> x, y, </math> and <math>z</math> are real numbers that satisfy: | ||
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| + | <center><math> x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}} </math> </center> | ||
| + | <center><math> y = \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}} </math></center> | ||
| + | <center><math> z = \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}</math></center> | ||
| + | |||
| + | 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 == | ||
| + | This solution requires you to disregard rigor. Additional solutions, or justification for the nonrigorous steps, would be appreciated. | ||
| + | |||
| + | 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>. | ||
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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]]: | ||
| + | Still needed. | ||
| + | |||
| + | == See also == | ||
| + | {{AIME box|year=2006|n=II|num-b=14|after=Last Question}} | ||
| + | |||
| + | [[Category:Intermediate Algebra Problems]] | ||
| + | [[Category:Intermediate Geometry Problems]] | ||
Revision as of 18:12, 25 September 2007
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
This solution requires you to disregard rigor. Additional solutions, or justification for the nonrigorous steps, would be appreciated.
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 a 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.
Justification that this triangle is an acute triangle:
Still needed.
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 | ||
