# Difference between revisions of "2021 April MIMC 10 Problems/Problem 22"

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<math>\textbf{(A)} ~5 \qquad\textbf{(B)} ~15 \qquad\textbf{(C)} ~61 \qquad\textbf{(D)} ~349 \qquad\textbf{(E)} ~2009 \qquad</math> | <math>\textbf{(A)} ~5 \qquad\textbf{(B)} ~15 \qquad\textbf{(C)} ~61 \qquad\textbf{(D)} ~349 \qquad\textbf{(E)} ~2009 \qquad</math> | ||

==Solution== | ==Solution== | ||

− | To be | + | To start this problem, we can first observe. Notice that <math>FGH</math> is a right triangle because angle <math>FGH</math> is supplementary to angle <math>IGE</math> which is a right angle. Therefore, we just have to solve for the length of side <math>FG</math> and <math>HG</math>. |

+ | |||

+ | Solve for <math>FG</math>: | ||

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

+ | |||

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

+ | |||

+ | <math>~~~~~~~~~~~~~~~~~~~~~~\frac{3-\sqrt{2}}{2\sqrt{2}-1}=\frac{x}{3-\sqrt{2}-x}</math> | ||

+ | |||

+ | <math>~(3-\sqrt{2})(3-\sqrt{2}-x)=x(2\sqrt{2}-1)</math> | ||

+ | |||

+ | <math>11-6\sqrt{2}-(3-\sqrt{2})x=x(2\sqrt{2}-1)</math> | ||

+ | |||

+ | <math>~~~~~~~~~~~~~~~~~(2+\sqrt{2})x=11-6\sqrt{2}</math> | ||

+ | |||

+ | <math>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=\frac{11-6\sqrt{2}}{2+\sqrt{2}}</math> | ||

+ | |||

+ | <math>~~~~~~~~~~~~~~~~~~~~~~~~~~~~~x=\frac{34-23\sqrt{2}}{2}</math> | ||

+ | |||

+ | Solve for <math>HG</math>: | ||

+ | Since angle <math>HEC</math> is <math>90\degree</math> and angle <math>HCE</math> is <math>45\degree</math>, <math>EC=HE=2\sqrt{2}-1</math>. Since <math>GE=3-\sqrt{2}</math>, <math>HG=2\sqrt{2}-1-(3-\sqrt{2})=3\sqrt{2}-4</math>. | ||

+ | Finally, we can solve for <math>FH^2</math>, that is, <math>FG^2+HG^2=(\frac{34-23\sqrt{2}}{2})^2+(3\sqrt{2}-4)^2=\frac{1175-830\sqrt{2}}{2}</math>. Therefore, our answer would be <math>1175-830+2+2=\fbox{\textbf{(D)} 349}</math>. |

## Revision as of 14:01, 26 April 2021

In the diagram, is a square with area . is a diagonal of square . Square has area . Given that point bisects line segment , and is a line segment. Extend to meet diagonal and mark the intersection point . In addition, is drawn so that . can be represented as where are not necessarily distinct integers. Given that , and does not have a perfect square factor. Find .

## Solution

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

Solve for : Triangles and are similar triangles, therefore, we can solve for length . . Use the technique of sum of squares and square root disintegration, . Using the same technique, . . Now, we can set up a ratio.

We can set , so . Using the similar triangle, . Plugging the numbers into the ratio, we can get .

Solve for : Since angle is $90\degree$ (Error compiling LaTeX. ! Undefined control sequence.) and angle is $45\degree$ (Error compiling LaTeX. ! Undefined control sequence.), . Since , . Finally, we can solve for , that is, . Therefore, our answer would be .