2021 April MIMC 10 Problems/Problem 22

Revision as of 14:01, 26 April 2021 by Cellsecret (talk | contribs) (Solution)

In the diagram, $ABCD$ is a square with area $6+4\sqrt{2}$. $AC$ is a diagonal of square $ABCD$. Square $IGED$ has area $11-6\sqrt{2}$. Given that point $J$ bisects line segment $HE$, and $AE$ is a line segment. Extend $EG$ to meet diagonal $AC$ and mark the intersection point $H$. In addition, $K$ is drawn so that $JK//EC$. $FH^2$ can be represented as $\frac{a+b\sqrt{c}}{{d}}$ where $a,b,c,d$ are not necessarily distinct integers. Given that $gcd(a,b,d)=1$, and $c$ does not have a perfect square factor. Find $a+b+c+d$.


$\textbf{(A)} ~5 \qquad\textbf{(B)} ~15 \qquad\textbf{(C)} ~61 \qquad\textbf{(D)} ~349 \qquad\textbf{(E)} ~2009 \qquad$


To start this problem, we can first observe. Notice that $FGH$ is a right triangle because angle $FGH$ is supplementary to angle $IGE$ which is a right angle. Therefore, we just have to solve for the length of side $FG$ and $HG$.

Solve for $FG$: Triangles $AIF$ and $EGF$ are similar triangles, therefore, we can solve for length $AI$. $AI=AD-ID$. Use the technique of sum of squares and square root disintegration, $AD=2+\sqrt{2}$. Using the same technique, $ID=3-\sqrt{2}$. $AI=2+\sqrt{2}-3+\sqrt{2}=2\sqrt{2}-1$. Now, we can set up a ratio.

We can set $FG=x$, so $IF=3-\sqrt{2}-x$. Using the similar triangle, $\frac{GE}{AI}=\frac{FG}{IF}$. Plugging the numbers into the ratio, we can get $\frac{3-\sqrt{2}}{2\sqrt{2}-1}=\frac{x}{3-\sqrt{2}-x}$.







Solve for $HG$: Since angle $HEC$ is $90\degree$ (Error compiling LaTeX. ! Undefined control sequence.) and angle $HCE$ is $45\degree$ (Error compiling LaTeX. ! Undefined control sequence.), $EC=HE=2\sqrt{2}-1$. Since $GE=3-\sqrt{2}$, $HG=2\sqrt{2}-1-(3-\sqrt{2})=3\sqrt{2}-4$. Finally, we can solve for $FH^2$, that is, $FG^2+HG^2=(\frac{34-23\sqrt{2}}{2})^2+(3\sqrt{2}-4)^2=\frac{1175-830\sqrt{2}}{2}$. Therefore, our answer would be $1175-830+2+2=\fbox{\textbf{(D)} 349}$.

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