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

Revision as of 17:30, 28 November 2021

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 $d^2$ .

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

Video Solution by Walt S

https://www.youtube.com/watch?v=wGP9bjkdh1M

@@ Line 21: / Line 21: @@
 ~novus677
-==Note:==
+==Video Solution by Walt S==
+https://www.youtube.com/watch?v=wGP9bjkdh1M
-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==
 {{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