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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

**Key Concepts**

Angles

Triangles

Side Length

## Check the Answer

But try the problem first…

Answer: is 177.

Source

Suggested Reading

AIME I, 2009, Question 5

Plane Trigonometry by Loney

## Try with Hints

First hint

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

Second Hint

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

Final Step

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

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequA

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