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Difference between revisions of "2021 JMPSC Invitationals Problems/Problem 10"

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==Solution==
 
==Solution==
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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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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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~Geometry285, edited by CyclicISLscelesTrapezoid
  
 
==See also==
 
==See also==

Latest revision as of 18:48, 23 September 2024

Problem

A point $P$ is chosen in isosceles trapezoid $ABCD$ with $AB=4$, $BC=20$, $CD=28$, and $DA=20$. If the sum of the areas of $PBC$ and $PDA$ is $144$, then the area of $PAB$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime. Find $m+n.$

Solution

We have the area of the trapezoid is $16 \cdot 16=256$ since the height is $16$. Now, subtracting $144$ we have $224=4x+28(16-x)$ for $x$ is the height of $\triangle PAB$. This means $x=\frac{28}{3}$, asserting the area of $\triangle PAB$ is $\frac{56}{3} \implies 56+3=\boxed{59}.$

~Geometry285, edited by CyclicISLscelesTrapezoid

See also

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

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