Difference between revisions of "2006 AIME II Problems/Problem 15"
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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>. | ||
− | 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{ | + | 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{9}</math>. |
Revision as of 11:16, 8 October 2019
Problem
Given that and
are real numbers that satisfy:
![$x = \sqrt{y^2-\frac{1}{16}}+\sqrt{z^2-\frac{1}{16}}$](http://latex.artofproblemsolving.com/4/8/2/482ec834e100a5c445119023117da55b0a08dae6.png)
![$y = \sqrt{z^2-\frac{1}{25}}+\sqrt{x^2-\frac{1}{25}}$](http://latex.artofproblemsolving.com/0/7/e/07e3a5bfe68da11f9902c4b34d8381e913e31859.png)
![$z = \sqrt{x^2 - \frac 1{36}}+\sqrt{y^2-\frac 1{36}}$](http://latex.artofproblemsolving.com/0/0/b/00b6096d5f7c9f1584a2efea64e687039b0c28ff.png)
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.