Difference between revisions of "2021 AIME II Problems/Problem 2"
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Now we can make an equation: | Now we can make an equation: | ||
− | + | <cmath>\frac{\triangle AFG}{\triangle BED} = \frac{8}{9}</cmath> | |
− | \frac{\triangle AFG}{\triangle BED} | + | <cmath>\frac{\frac{1}{2}(\sin 120)(x^2)}{\frac{1}{2}(\sin 30)(420\sqrt{3} - \frac{\sqrt{3}}{2}x)(840-x)} = \frac{8}{9}</cmath> |
− | \frac{\frac{1}{2}(\sin 120)(x^2)}{\frac{1}{2}(\sin 30)(420\sqrt{3} - \frac{\sqrt{3}}{2}x)(840-x)} | + | <cmath>\frac{x^2}{(420 - \frac{x}{2})(840-x)} = \frac{8}{9}</cmath> |
− | \frac{x^2}{(420 - \frac{x}{2})(840-x)} | ||
− | |||
To make further calculations easier, we scale everything down by <math>420</math> (while keeping the same variable names, so keep that in mind). | To make further calculations easier, we scale everything down by <math>420</math> (while keeping the same variable names, so keep that in mind). | ||
− | + | <cmath>\frac{x^2}{(1-\frac{x}{2})(2-x)} = \frac{8}{9}</cmath> | |
− | + | <cmath>8(1-\frac{x}{2})(2-x) = 9x^2</cmath> | |
− | + | <cmath>16-16x + 4x^2 = 9x^2</cmath> | |
− | \frac{x^2}{(1-\frac{x}{2})(2-x)} | + | <cmath>5x^2 + 16x -16 = 0</cmath> |
− | 8(1-\frac{x}{2})(2-x) | + | <cmath>(5x-4)(x+4) = 0</cmath> |
− | 16-16x + 4x^2 | ||
− | 5x^2 + 16x -16 | ||
− | (5x-4)(x+4) | ||
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− | |||
Revision as of 20:15, 22 March 2021
Contents
Problem
Equilateral triangle has side length . Point lies on the same side of line as such that . The line through parallel to line intersects sides and at points and , respectively. Point lies on such that is between and , is isosceles, and the ratio of the area of to the area of is . Find .
Solution
SOMEBODY DELETED A SOLUTION - PLEASE WAIT FOR THE ORIGINAL AUTHOR TO COME BACK ~ARCTICTURN
Solution
We express the ares of and in terms of in order to solve for
We let Because is isoceles and is equilateral,
Let the height of be and the height of be Then we have that and
Now we can find and in terms of Because we are given that This allows us to use the sin formula for triangle area: the area of is Similarly, because the area of
Now we can make an equation:
To make further calculations easier, we scale everything down by (while keeping the same variable names, so keep that in mind).
Thus Because we scaled down everything by the actual value of is
~JimY
See also
2021 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
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.