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# Rectangles and sides | AIME I, 2011 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Rectangles and sides.

## Rectangles and sides - AIME I, 2011

In rectangle ABCD, AB=12 and BC=10 points E and F are inside rectangle ABCD so that BE=9 and DF=8, BE parallel to DF and EF parallel to AB and line BE intersects segment AD. The length EF can be expressed in theorem $m n^\frac{1}{2}-p$ where m , n and p are positive integers and n is not divisible by the square of any prime, find m+n+p.

• is 107
• is 36
• is 840
• cannot be determined from the given information

Parallelograms

Rectangles

Side Length

## Check the Answer

AIME I, 2011, Question 2

Geometry Vol I to IV by Hall and Stevens

## Try with Hints

First hint

here extending lines BE and CD meet at point G and drawing altitude GH from point G by line BA extended till H GE=DF=8 GB=17

Second Hint

In a right triangle GHB, GH=10 GB=17 by Pythagorus thorem, HB=(${{17}^{2}-{10}^{2}})^\frac{1}{2}$=$3({21})^\frac{1}{2}$

Final Step

HA=EF=$3({21})^\frac{1}{2}-12$ then 3+21+12=36.

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