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Difference between revisions of "2021 JMPSC Sprint Problems/Problem 18"

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==Solution==
 
==Solution==
 
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>.
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~tigerzhang
 
~tigerzhang
 +
 +
==See also==
 +
#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
 +
#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Latest revision as of 16:15, 11 July 2021

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

See also

  1. Other 2021 JMPSC Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

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