Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 10"
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− | + | It is implied <math>P</math> lies on the line that bisects <math>AB</math> and <math>DC</math>. 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 | ||
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+ | ==See also== | ||
+ | #[[2021 JMPSC Invitationals Problems|Other 2021 JMPSC Invitationals Problems]] | ||
+ | #[[2021 JMPSC Invitationals Answer Key|2021 JMPSC Invitationals Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 21:13, 11 July 2021
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
It is implied lies on the line that bisects and . 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
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.