# Interior Angle Problem | AIME I, 1990 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on Interior Angle.

## Interior Angle Problem - AIME I, 1990

Let $P_1$ be a regular r gon and $P_2$ be a regular s gon $(r \geq s \geq 3)$ such that each interior angle of $P_1$ is $\frac{59}{58}$ as large as each interior angle of $P_2$, find the largest possible value of s.

• is 107
• is 117
• is 840
• cannot be determined from the given information

### Key Concepts

Integers

Polygons

Algebra

AIME I, 1990, Question 3

Elementary Algebra by Hall and Knight

## Try with Hints

First hint

Interior angle of a regular sided polygon=$\frac{(n-2)180}{n}$

or, $\frac{\frac{(r-2)180}{r}}{\frac{(s-2)180}{s}}=\frac{59}{58}$

or, $\frac{58(r-2)}{r}=\frac{59(s-2)}{s}$

Second Hint

or, 58rs-58(2s)=59rs-59(2r)

or, 118r-116s=rs

or, r=$\frac{116s}{118-s}$

Final Step

for 118-s>0, s<118

or, s=117

or, r=(116)(117)

or, s=117.

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