Difference between revisions of "2014 AIME II Problems/Problem 3"
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==Solution== | ==Solution== | ||
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+ | When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below. | ||
+ | |||
+ | <asy> | ||
+ | pair R,S,T,X,Y,Z; | ||
+ | dotfactor = 2; | ||
+ | unitsize(.1cm); | ||
+ | R = (12*2.236 +22,0); | ||
+ | S = (12*2.236 + 22 - 13.4164,12); | ||
+ | T = (12*2.236 + 22,24); | ||
+ | X = (12*4.472+ 22,24); | ||
+ | Y = (12*4.472+ 22 + 13.4164,12); | ||
+ | Z = (12*4.472+ 22,0); | ||
+ | draw(R--S--T--X--Y--Z--cycle); | ||
+ | draw(T--R,red); | ||
+ | draw(X--Z,red); | ||
+ | dot(" ",R,NW); | ||
+ | dot(" ",S,NW); | ||
+ | dot(" ",T,NW); | ||
+ | dot(" ",X,NW); | ||
+ | dot(" ",Y,NW); | ||
+ | dot(" ",Z,NW); | ||
+ | // sqrt180 = 13.4164 | ||
+ | // sqrt5 = 2.236</asy> | ||
+ | |||
+ | By Heron's Formula, the area of each isosceles triangle is <math>\sqrt{(30)(12)(12)(6)}=\sqrt{180\times 12^2}=72\sqrt{5}</math>. So the area of both is <math>144\sqrt{5}</math>. From the rectangle, our original area is <math>36a</math>. The area of the rectangle in the hexagon is <math>24a</math>. So we have <cmath>24a+144\sqrt{5}=36a\implies 12a=144\sqrt{5}\implies a=12\sqrt{5}\implies a^2=\boxed{720}.</cmath> |
Revision as of 20:04, 27 March 2014
Problem
A rectangle has sides of length and 36. A hinge is installed at each vertex of the rectangle, and at the midpoint of each side of length 36. The sides of length can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length parallel and separated by a distance of 24, the hexagon has the same area as the original rectangle. Find .
Solution
When we squish the rectangle, the hexagon is composed of a rectangle and two isosceles triangles with side lengths 18, 18, and 24 as shown below.
By Heron's Formula, the area of each isosceles triangle is . So the area of both is . From the rectangle, our original area is . The area of the rectangle in the hexagon is . So we have