2021 JMPSC Sprint Problems/Problem 18

Revision as of 12:24, 11 July 2021 by Tigerzhang (talk | contribs) (Solution)

Problem

On square $ABCD$ with side length $28$, $M$ is the midpoint of $\overline{CD}$. Let $E$ be the foot of the altitude from $M$ to $\overline{AC}$. If $AE$ can be represented as $a\sqrt{2}$ for some integer $a,$ find the value of $a.$

Solution

Notice that since $\angle ACD=45^\circ$ and $\angle MEC=90^\circ$, $\triangle MEC$ is a 45-45-90 triangle. Thus, \[EC=\frac{MC}{\sqrt{2}}=\frac{14}{\sqrt{2}}=7\sqrt{2}.\] Also, we have $AC=AD\sqrt{2}=28\sqrt{2}$, so \[AE=AC-EC=21\sqrt{2}\] which gives the answer of $\boxed{21}$. ~tigerzhang