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

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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==
 
[[image:2018_AIME_I-8.png|center|500px]]
 
[[image:2018_AIME_I-8.png|center|500px]]
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- cooljoseph
 
  
 
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>.
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~novus677
 
~novus677
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==Note:==
 
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This is because the altitude of our equilateral triangle with side length <math>10</math> is perpendicular to the tangent line to the circle, which implies they are all <math>90</math> degrees (two <math>90</math> degree angles from altitude, two <math>90</math> degree angles from tangent lines). This allows us to calculate further. Tilt your head <math>120</math> degrees clockwise if you can't see what is being done. ~IronicNinja
 
  
 
==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 09:44, 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 (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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