Mock Geometry AIME 2011 Problems/Problem 1

Problem

Let $ABCD$ be a unit square, and let $AB_1C_1D_1$ be its image after a $30$ degree rotation about point $A.$ The area of the region consisting of all points inside at least one of $ABCD$ and $AB_1C_1D_1$ can be expressed in the form $\frac{a-\sqrt{b}} {c},$ where $a,b,c$ are positive integers, and $b$ shares no perfect square common factor with $c$ . Find $a+b+c.$

Solution

$[asy] unitsize(5cm); draw(unitsquare); draw((0,1)--(1/2, (2-sqrt(3))/2)--((1+sqrt(3))/2,(3-sqrt(3))/2)--(sqrt(3)/2,3/2)--cycle); label("$A$",(0,1),NW); label("$B$",(1,1),E); label("$C$",(1,0),SE); label("$D$",(0,0),SW); label("$B_1$",(sqrt(3)/2,3/2),N); label("$C_1$",((1+sqrt(3))/2,(3-sqrt(3))/2),ENE); label("$D_1$",(1/2,(2-sqrt(3))/2),SSW); label("$E$",(1,(3-sqrt(3))/3),SE); draw((0,1)--(1,(3-sqrt(3))/3)); [/asy]$

Denote the intersection of $BC$ and $C_1D_1$ as $E$ . Also, denote a polygon enclosed in square brackets as the area as that polygon. (For example, $[ABCD]$ denotes the area of polygon $ABCD$ .)

The area of the union of the two squares is equal to $[ABCD]+[AB_1C_1D_1]-[ABED_1]$ . Note that $[ABCD]=[AB_1C_1D_1]=1$ , so we wish to find $2-[ABED_1]$ .

Draw in $AE$ . Notice that $\triangle AED_1\cong\triangle AEB$ , since $\angle ABE=\angle AD_1E=90^\circ$ , $AD_1=AB=1$ , and $AE=AE$ , so the triangles are congruent by HL. Thus, their areas are equal, and we need to find $2-2[ABE]$ .

Note that $\angle BAB_1=30^\circ$ , since that's the angle of rotation, so $\angle D_1AB=90^\circ-30^\circ=60^\circ$ . Also, since $\angle D_1AE=\angle EAB$ , we have $\angle EAB=\frac{60^\circ}{2}=30^\circ$ .

Now we have $BE=BA\cdot\tan\angle EAB=1\cdot\tan30^\circ=\frac{\sqrt{3}}{3}$ . Finally, $[ABE]=\frac{1}{2}\cdot1\cdot\frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{6}$ , and our answer is $2-2\left(\frac{\sqrt{3}}{6}\right)=\frac{6-\sqrt{3}}{3}$ , and $a+b+c=\boxed{012}$ .

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Mock Geometry AIME 2011 Problems/Problem 1

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