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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.

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.

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

Key Concepts


Parallelograms

Rectangles

Side Length

Check the Answer


Answer: is 36.

AIME I, 2011, Question 2

Geometry Vol I to IV by Hall and Stevens

Try with Hints


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

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

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

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