2021 April MIMC 10 Problems/Problem 22
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 .
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 .