Difference between revisions of "2006 AIME II Problems/Problem 15"
m (Fixed typos) |
Math101010 (talk | contribs) m (→Solution 1 (Geometric Interpretation)) |
||
Line 21: | Line 21: | ||
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> | 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{ | + | 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>: | The justification that there is an acute triangle with sides of length <math>x, y</math> and <math>z</math>: |
Latest revision as of 23:32, 28 February 2021
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 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 (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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.