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AIME I Math Olympiad USA Math Olympiad

A Parallelogram and a Line | AIME I, 1999 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on A Parallelogram and a Line.

A Parallelogram and a Line – AIME I, 1999


Consider the parallelogram with vertices (10,45),(10,114),(28,153) and (28,84). A line through the origin cuts this figure into two congruent polygons. The slope of the line is \(\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 118
  • is 840
  • cannot be determined from the given information

Key Concepts


Parallelogram

Slope of line

Integers

Check the Answer


Answer: is 118.

AIME I, 1999, Question 2

Geometry Vol I to IV by Hall and Stevens

Try with Hints


By construction here we see that a line makes the parallelogram into two congruent polygons gives line passes through the centre of the parallelogram

Centre of the parallogram is midpoint of (10,45) and (28,153)=(19,99)

Slope of line =\(\frac{99}{19}\) then m+n=118.

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AIME I Geometry Math Olympiad USA Math Olympiad

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.

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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