Difference between revisions of "2018 AIME I Problems/Problem 8"
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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, | 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, | ||
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<math>\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}</math> | <math>\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}</math> | ||
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Solving this equation gives <math>d=7\sqrt{3}</math>, and <math>d^2=\boxed{147}</math> | Solving this equation gives <math>d=7\sqrt{3}</math>, and <math>d^2=\boxed{147}</math> | ||
Revision as of 05:52, 8 March 2018
Let 
be an equiangular hexagon such that 
, and 
. Denote 
the diameter of the largest circle that fits inside the hexagon. Find 
.
Solution 1
- cooljoseph
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 
~novus 677
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. 
