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

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

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

@@ Line 1: / Line 1: @@
-Let <math>ABCDEF</math> be an equiangular hexagon such that <math>AB=6, BC=8, CD=10</math>, and <math>DE=12</math>. Denote <math>d</math> the diameter of the largest circle that fits inside the hexagon. Find <math>d^2</math>.
+==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>.
-==Solution 2==
+==Video Solution by Punxsutawney Phil==
-Like solution 1, draw out the large equilateral triangle of side length <math>24</math>.
+https://www.youtube.com/watch?v=oc-cDRIEzoo
+==Video Solution by Walt S==
+https://www.youtube.com/watch?v=wGP9bjkdh1M
 ==Solution 1==
 [[image:2018_AIME_I-8.png|center|500px]]
-- 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>.
 -expiLnCalc
+==Solution 2==
+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,
+<math>\frac{10}{24}=\frac{5\sqrt{3}}{d+5\sqrt{3}}</math>
+Solving this equation gives <math>d=7\sqrt{3}</math>, and <math>d^2=\boxed{147}</math>
+~novus677
 ==See Also==
 {{AIME box|year=2018|n=I|num-b=7|num-a=9}}
 {{MAA Notice}}

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

Page

Toolbox

Search