Difference between revisions of "2018 AIME I Problems/Problem 8"
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==Problem== | ==Problem== | ||
Let <math>ABCDEF</math> be an equiangular hexagon such that <math>AB=6, BC=8, CD=10</math>, and <math>DE=12</math>. Denote by <math>d</math> the diameter of the largest circle that fits inside the hexagon. Find <math>d^2</math>. | Let <math>ABCDEF</math> be an equiangular hexagon such that <math>AB=6, BC=8, CD=10</math>, and <math>DE=12</math>. Denote by <math>d</math> the diameter of the largest circle that fits inside the hexagon. Find <math>d^2</math>. | ||
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+ | ==Video Solution by Punxsutawney Phil== | ||
+ | https://www.youtube.com/watch?v=oc-cDRIEzoo | ||
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+ | ==Video Solution by Walt S== | ||
+ | https://www.youtube.com/watch?v=wGP9bjkdh1M | ||
==Solution 1== | ==Solution 1== | ||
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~novus677 | ~novus677 | ||
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==See Also== | ==See Also== | ||
{{AIME box|year=2018|n=I|num-b=7|num-a=9}} | {{AIME box|year=2018|n=I|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:44, 5 July 2022
Contents
Problem
Let be an equiangular hexagon such that , and . Denote by the diameter of the largest circle that fits inside the hexagon. Find .
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=oc-cDRIEzoo
Video Solution by Walt S
https://www.youtube.com/watch?v=wGP9bjkdh1M
Solution 1
First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that . Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length . Then, if you drew it to scale, notice that the "widest" this circle can be according to is . And it will be obvious that the sides won't be inside the circle, so our answer is .
-expiLnCalc
Solution 2
Like solution 1, draw out the large equilateral triangle with side length . Let the tangent point of the circle at be G and the tangent point of the circle at be H. Clearly, GH is the diameter of our circle, and is also perpendicular to and .
The equilateral triangle of side length is similar to our large equilateral triangle of . And the height of the former equilateral triangle is . By our similarity condition,
Solving this equation gives , and
~novus677
See Also
2018 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.