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

(Video Solution by Walt S:)
 
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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==
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https://www.youtube.com/watch?v=oc-cDRIEzoo
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==Video Solution by Walt S==
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https://www.youtube.com/watch?v=wGP9bjkdh1M
  
 
==Solution 1==
 
==Solution 1==
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~novus677
 
~novus677
 
==Video Solution by Walt S==
 
https://www.youtube.com/watch?v=wGP9bjkdh1M
 
  
 
==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

Problem

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

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

2018 AIME I-8.png

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, $\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}$

Solving this equation gives $d=7\sqrt{3}$, and $d^2=\boxed{147}$

~novus677

See Also

2018 AIME I (ProblemsAnswer KeyResources)
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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