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Difference between revisions of "Mock Geometry AIME 2011 Problems/Problem 1"
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Now we have <math> BE=BA\cdot\tan\angle EAB=1\cdot\tan30^\circ=\frac{\sqrt{3}}{3} </math>. Finally, <math> [ABE]=\frac{1}{2}\cdot1\cdot\frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{6} </math>, and our answer is <math> 2-2\left(\frac{\sqrt{3}}{6}\right)=\frac{6-\sqrt{3}}{3} </math>, and <math> a+b+c=\boxed{012} </math>.
 
Now we have <math> BE=BA\cdot\tan\angle EAB=1\cdot\tan30^\circ=\frac{\sqrt{3}}{3} </math>. Finally, <math> [ABE]=\frac{1}{2}\cdot1\cdot\frac{\sqrt{3}}{3}=\frac{\sqrt{3}}{6} </math>, and our answer is <math> 2-2\left(\frac{\sqrt{3}}{6}\right)=\frac{6-\sqrt{3}}{3} </math>, and <math> a+b+c=\boxed{012} </math>.
 +
 +
==Video Solution by Punxsutawney Phil==
 +
https://youtube.com/OEkt8HbUVbQ

Revision as of 21:04, 23 December 2022

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 $a+b+c=\boxed{012}$.

Video Solution by Punxsutawney Phil

https://youtube.com/OEkt8HbUVbQ

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