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)
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
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=(\({{17}^{2}-{10}^{2}})^\frac{1}{2}\)=\(3({21})^\frac{1}{2}\)
A parallelogram is a quadrilateral with opposite sides parallel (and therefore opposite angles equal). A quadrilateral with equal sides is called a rhombus, and a parallelogram whose angles are all right angles is called a rectangle.
Try the problem
\begin{equation} \begin{array}{l} {\text { 4. Let } A B C D \text { be a parallelogram. Let } O \text { be a point in its interior such that } \angle A O B+} \\ {\angle D O C=180^{\circ} . \text { Show that } \angle O D C=\angle O B C \text { . }} \end{array} \end{equation}
CMI Entrance, 2019
Parallelogram
6 out of 10
Challenging Problems in Geometry
Use some hints
First construct AP and BP parallel to DO and CO respectively. Basically translate triangle DOC to triangle APB with DC and AB as parallel bases respectively. Notice that product of the fractions is 1. Can you use this fact to compute the geometric mean of the fractions?
Now using angle manipulations by spotting the required parallelograms and show that APBO is cyclic quadrilateral.
Now using Hint 2 and the fact that PBCO is a parallelogram arrive at the proof.
After implementing Hint 3 by spotting the parallelogram PBCO, notice that the the diagonal OB forms the the alternate interior angles i.e. angle POB and angle OBC which will be equal to each other as PB is parallel to OC since PBCO is a parallelogram. Since angle POB is equal to angle PAB (angle POB and angle PAB share the same arc AB and hence are equal to each other), which again itself is equal to angle ODC as AP is parallel to DO as we translated triangle DOC to triangle APB. Hence, angle ODC = angle PAB = angle POB = angle OBC. Hence we are done.
