Difference between revisions of "2018 AIME I Problems/Problem 8"
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==Video Solution by Walt S== | ==Video Solution by Walt S== | ||
https://www.youtube.com/watch?v=wGP9bjkdh1M | https://www.youtube.com/watch?v=wGP9bjkdh1M | ||
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| + | ==Solution 2== | ||
| + | |||
| + | Like solution 1, draw out the large equilateral triangle with side length <math>24</math>. Let the tangent point of the circle at <math>\overline{CD}</math> be G and the tangent point of the circle at <math>\overline{AF}</math> be H. Clearly, GH is the diameter of our circle, and is also perpendicular to <math>\overline{CD}</math> and <math>\overline{AF}</math>. | ||
| + | |||
| + | The equilateral triangle of side length <math>10</math> is similar to our large equilateral triangle of <math>24</math>. And the height of the former equilateral triangle is <math>\sqrt{10^2-5^2}=5\sqrt{3}</math>. By our similarity condition, | ||
| + | <math>\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}</math> | ||
| + | |||
| + | Solving this equation gives <math>d=7\sqrt{3}</math>, and <math>d^2=\boxed{147}</math> | ||
| + | |||
| + | ~novus677 | ||
==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}} | ||
Revision as of 20:58, 30 October 2025
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 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
