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Difference between revisions of "2018 AIME I Problems/Problem 8"

(→‎Solution 1)
(→‎Solution 2)
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==Solution 2==
 
==Solution 2==
  
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Like solution 1, draw out the large equilateral triangle with side length <math>24</math>.
+
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,
 +
\begin{center}
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<math>\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}</math>
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\end{center}
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Solving this equation gives <math>d=7\sqrt{3}</math>, and <math>d^2=\boxed{147}</math>
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 +
~novus 677
  
 
==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 05:51, 8 March 2018

Let $ABCDEF$ be an equiangular hexagon such that $AB=6, BC=8, CD=10$, and $DE=12$. Denote $d$ the diameter of the largest circle that fits inside the hexagon. Find $d^2$.

Solution 1

2018 AIME I-8.png

- cooljoseph

First of all, draw a good diagram! This is always the key to solving any geometry problem. Once you draw it, realize that $EF=2, FA=16$. Why? Because since the hexagon is equiangular, we can put an equilateral triangle around it, with side length $6+8+10=24$. Then, if you drew it to scale, notice that the "widest" this circle can be according to $AF, CD$ is $7\sqrt{3}$. And it will be obvious that the sides won't be inside the circle, so our answer is $\boxed{147}$.

-expiLnCalc

Solution 2

Like solution 1, draw out the large equilateral triangle with side length $24$. Let the tangent point of the circle at $\overline{CD}$ be G and the tangent point of the circle at $\overline{AF}$ be H. Clearly, GH is the diameter of our circle, and is also perpendicular to $\overline{CD}$ and $\overline{AF}$.

The equilateral triangle of side length $10$ is similar to our large equilateral triangle of $24$. And the height of the former equilateral triangle is $\sqrt{10^2-5^2}=5\sqrt{3}$. By our similarity condition, \begin{center} $\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}$ \end{center} Solving this equation gives $d=7\sqrt{3}$, and $d^2=\boxed{147}$

~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

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