Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 10"
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==Solution== | ==Solution== | ||
| − | + | We have the area of the trapezoid is <math>16 \cdot 16=256</math> since the height is <math>16</math>. Now, subtracting <math>144</math> we have <math>224=4x+28(16-x)</math> for <math>x</math> is the height of <math>\triangle PAB</math>. This means <math>x=\frac{28}{3}</math>, asserting the area of <math>\triangle PAB</math> is <math>\frac{56}{3} \implies 56+3=\boxed{59}.</math> | |
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| + | ~Geometry285, edited by CyclicISLscelesTrapezoid | ||
==See also== | ==See also== | ||
Latest revision as of 18:48, 23 September 2024
Problem
A point 
is chosen in isosceles trapezoid 
with 
, 
, 
, and 
. If the sum of the areas of 
and 
is 
, then the area of 
can be written as 
where 
and 
are relatively prime. Find 
Solution
We have the area of the trapezoid is 
since the height is 
. Now, subtracting we have 
for 
is the height of 
. This means 
, asserting the area of is 
~Geometry285, edited by CyclicISLscelesTrapezoid
See also
- Other 2021 JMPSC Invitationals Problems
- 2021 JMPSC Invitationals Answer Key
- All JMPSC Problems and Solutions
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. 
