Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.
Series and sum - AIME I, 1999
given that \(\displaystyle\sum_{k=1}^{35}sin5k=tan\frac{m}{n}\) where angles are measured in degrees, m and n are relatively prime positive integer that satisfy \(\frac{m}{n} \lt 90\), find m+n.
Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Triangles and sides.
Triangles and sides - AIME I, 2009
Triangle ABC AC=450 BC=300 points K and L are on AC and AB such that AK=CK and CL is angle bisectors of angle C. let P be the point of intersection of Bk and CL and let M be a point on line Bk for which K is the mid point of PM AM=180, find LP
is 107
is 72
is 840
cannot be determined from the given information
Key Concepts
Angles
Triangles
Side Length
Check the Answer
Answer: is 72.
AIME I, 2009, Question 5
Geometry Vol I to IV by Hall and Stevens
Try with Hints
since K is mid point of PM and AC quadrilateral AMCP is a parallelogram which implies AM parallel LP and triangle AMB is similar to triangle LPB
then \(\frac{AM}{LP}=\frac{AB}{LB}=\frac{AL+LB}{LB}=\frac{AL}{LB}+1\)
from angle bisector theorem, \(\frac{AL}{LB}=\frac{AC}{BC}=\frac{450}{300}=\frac{3}{2}\) then \(\frac{AM}{LP}=\frac{AL}{LB}+1=\frac{5}{2}\)
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
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=(\({{17}^{2}-{10}^{2}})^\frac{1}{2}\)=\(3({21})^\frac{1}{2}\)
