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

Revision as of 20:20, 16 March 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

Note:

This is because the altitude of our equilateral triangle with side length $10$ is perpendicular to the tangent line to the circle, which implies they are all $90$ degrees (two $90$ degree angles from altitude, two $90$ degree angles from tangent lines). This allows us to calculate further. Tilt your head $120$ degrees clockwise if you can't see what is being done. ~IronicNinja

@@ 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>.
-==Solutions==
+==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 1==
+[[image:2018_AIME_I-8.png|center|500px]]
-==Solution Diagram==
-[asy]
-draw((0,0)--(12,20.78)--(24,0)--cycle);
-draw((1,1.73)--(2,0));
-draw((9,15.59)--(15,15.59));
-draw((14,0)--(19,8.66));
-label("<math>A</math>",(9,15.59),NW);
-label("<math>B</math>",(15,15.59),NE);
-label("<math>C</math>",(19,8.66),NE);
-label("<math>D</math>",(14,0),S);
-label("<math>E</math>",(2,0),S);
-label("<math>F</math>",(1,1.73),NW);
-pair O;
-O=(11.25,7.36);
-dot(O);
-label("<math>O</math>",O,SW);
-draw(Circle(O,6.06));
-[/asy]
-asymptote code for a picture
-- 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
+==Note:==
+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)
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