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

(Added solution)
(→‎Solution Diagram)
Line 4: Line 4:
  
 
==Solution Diagram==
 
==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

Revision as of 19:30, 7 March 2018

Let $ABCDEF$ be an equiangular hexagon such that $AB=6, BC=8, CD=10$, and $DE=12$. Denote $d$ the diameter of the largest circle that fits inside the hexagon. Find $d^2$.

Solutions

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("$A$",(9,15.59),NW); label("$B$",(15,15.59),NE); label("$C$",(19,8.66),NE); label("$D$",(14,0),S); label("$E$",(2,0),S); label("$F$",(1,1.73),NW); pair O; O=(11.25,7.36); dot(O); label("$O$",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 $EF=2, FA=16$. 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

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