Difference between revisions of "2006 AIME II Problems/Problem 15"
m |
|||
Line 18: | Line 18: | ||
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. | 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. | ||
+ | |||
+ | ==Video solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=M6sC26dzb_I | ||
== See also == | == See also == |
Revision as of 18:03, 21 July 2020
Contents
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.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.