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>. |
| + | |||
| + | ~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 
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, ![\[EC=\frac{MC}{\sqrt{2}}=\frac{14}{\sqrt{2}}=7\sqrt{2}.\]](http://latex.artofproblemsolving.com/7/1/6/716d2ea385605dbb21615c829ef3f8cbc0dae24a.png)
Also, we have 
, so ![\[AE=AC-EC=21\sqrt{2}\]](http://latex.artofproblemsolving.com/2/8/b/28bb35f5a0698db8af9babd6aaebfce1036eb741.png)
which gives the answer of 
.
~tigerzhang
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. 
