Difference between revisions of "2019 AIME I Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | We know the area of the hexagon <math>ABCDEF</math> to be <math>\triangle PQR- \triangle PAF- \triangle BCQ- \triangle RED </math>. Since <math>PR^2+RQ^2=PQ^2</math>, we know that <math>\triangle PRQ</math> is a right triangle. Thus the area of <math>\triangle PQR</math> is <math>150</math>. Another way to compute the area is <cmath>\frac12 \cdot PQ\cdot RQ \sin \angle | + | We know the area of the hexagon <math>ABCDEF</math> to be <math>\triangle PQR- \triangle PAF- \triangle BCQ- \triangle RED </math>. Since <math>PR^2+RQ^2=PQ^2</math>, we know that <math>\triangle PRQ</math> is a right triangle. Thus the area of <math>\triangle PQR</math> is <math>150</math>. Another way to compute the area is <cmath>\frac12 \cdot PQ\cdot RQ \sin \angle PQR = \frac12 500 \cdot \sin \angle PQR=150 \implies \sin \angle PQR = \frac35.</cmath> Then the area of <math>\triangle BCQ = \frac12 \cdot BQ \cdot CQ \cdot \sin \angle PQR= \frac{25}{2}\cdot \frac{3}{5}=\frac{15}{2}</math>. Preceding in a similar fashion for <math>\triangle PAF</math>, the area of <math>\triangle PAF</math> is <math>10</math>. Since <math>\angle ERD = 90^{\circ}</math>, the area of <math>\triangle RED=\frac{25}{2}</math>. Thus our desired answer is <math>150-\frac{15}{2}-10-\frac{25}{2}=\boxed{120}</math> |
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=2|num-a=4}} | {{AIME box|year=2019|n=I|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:40, 14 March 2019
The 2019 AIME I takes place on March 13, 2019.
Problem 3
In , , , and . Points and lie on , points and lie on , and points and lie on , with . Find the area of hexagon .
Solution
We know the area of the hexagon to be . Since , we know that is a right triangle. Thus the area of is . Another way to compute the area is Then the area of . Preceding in a similar fashion for , the area of is . Since , the area of . Thus our desired answer is
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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