Difference between revisions of "2018 AIME I Problems/Problem 8"
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First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that <math>EF=2, FA=16</math>. Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length <math>6+8+10=24</math>. Then, if you drew it to scale, notice that the "widest" this circle can be according to <math>AF, CD</math> is <math>7\sqrt{3}</math>. And it will be obvious that the sides won't be inside the circle, so our answer is <math>\boxed{147}</math>. | First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that <math>EF=2, FA=16</math>. Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length <math>6+8+10=24</math>. Then, if you drew it to scale, notice that the "widest" this circle can be according to <math>AF, CD</math> is <math>7\sqrt{3}</math>. And it will be obvious that the sides won't be inside the circle, so our answer is <math>\boxed{147}</math>. | ||
Revision as of 19:20, 16 March 2021
Contents
Problem
Let 
be an equiangular hexagon such that 
, and 
. Denote by 
the diameter of the largest circle that fits inside the hexagon. Find 
.
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
Note:
This is because the altitude of our equilateral triangle with side length is perpendicular to the tangent line to the circle, which implies they are all 
degrees (two degree angles from altitude, two degree angles from tangent lines). This allows us to calculate further. Tilt your head 
degrees clockwise if you can't see what is being done. ~IronicNinja
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. 
