Difference between revisions of "2021 JMPSC Sprint Problems/Problem 18"
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− | Notice that since <math>\angle ACD=45^\circ</math> and <math>\angle MEC=90^\circ</math>, <math>\triangle MEC</math> is a 45-45-90 triangle. Thus, <cmath>EC=\frac{MC}{\sqrt{2}}=\frac{14}{\sqrt{2}}=7\sqrt{2}.</cmath> Also, we have <math>AC=AD\sqrt{2}=28\sqrt{2}</math>, so <cmath>AE=AC-EC=21\sqrt{2}</cmath> which gives the answer of <math>\boxed{21}</math>. | + | Notice that since <math>\angle ACD=45^\circ</math> and <math>\angle MEC=90^\circ</math>, <math>\triangle MEC</math> is a 45-45-90 triangle. Thus, <cmath>EC=\frac{MC}{\sqrt{2}}=\frac{14}{\sqrt{2}}=7\sqrt{2}.</cmath> Also, we have <math>AC=AD\sqrt{2}=28\sqrt{2}</math>, so <cmath>AE=AC-EC=21\sqrt{2}</cmath> which gives the answer of <math>\boxed{21}</math>. |
+ | ~tigerzhang |
Revision as of 12:24, 11 July 2021
Problem
On square with side length , is the midpoint of . Let be the foot of the altitude from to . If can be represented as for some integer find the value of
Solution
Notice that since and , is a 45-45-90 triangle. Thus, Also, we have , so which gives the answer of . ~tigerzhang