Categories

## Series and sum | AIME I, 1999 | Question 11

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.

• is 107
• is 177
• is 840
• cannot be determined from the given information

Angles

Triangles

Side Length

## Check the Answer

AIME I, 2009, Question 5

Plane Trigonometry by Loney

## Try with Hints

s=$\displaystyle\sum_{k=1}^{35}sin5k$

s(sin5)=$\displaystyle\sum_{k=1}^{35}sin5ksin5=\displaystyle\sum_{k=1}^{35}(0.5)[cos(5k-5)-cos(5k+5)]$=$\frac{1+cos5}{sin5}$

$=\frac{1-cos(175)}{sin175}$=$tan\frac{175}{2}$ then m+n=175+2=177.

Categories

## Triangles and sides | AIME I, 2009 | Question 5

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

Angles

Triangles

Side Length

## Check the Answer

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 $\frac{AM}{LP}=\frac{AL}{LB}+1=\frac{5}{2}$

$\frac{180}{LP}=\frac{5}{2}$ then LP=72.

Categories

## 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

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.