Difference between revisions of "2019 AIME I Problems/Problem 3"
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==Solution 2== | ==Solution 2== | ||
Let <math>R</math> be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that <math>A=(4,12), B=(16,3), C=(15,0), D=(5,0), E=(0,5)</math>, and <math>F=(0,10)</math>. Using the shoelace theorem, the area is <math>\boxed{120}</math>. | Let <math>R</math> be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that <math>A=(4,12), B=(16,3), C=(15,0), D=(5,0), E=(0,5)</math>, and <math>F=(0,10)</math>. Using the shoelace theorem, the area is <math>\boxed{120}</math>. | ||
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+ | ==Video Solution== | ||
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+ | https://www.youtube.com/watch?v=4jOfXNiQ6WM | ||
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==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 01:09, 17 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 1
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
Solution 2
Let be the origin. Noticing that the triangle is a 3-4-5 right triangle, we can see that , and . Using the shoelace theorem, the area is .
Video Solution
https://www.youtube.com/watch?v=4jOfXNiQ6WM
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.