Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution Diagram)
 
(17 intermediate revisions by 10 users not shown)
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>.
  
−
==Solutions==
+
==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==
 +
[[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>.
 
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
 
-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}}

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2018_AIME_I_Problems/Problem_8&oldid=175629"