AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

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




Check the Answer

Answer: is 117.

AIME I, 1990, Question 3

Elementary Algebra by Hall and Knight

Try with Hints

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

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

or, 118r-116s=rs

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

for 118-s>0, s<118

or, s=117

or, r=(116)(117)

or, s=117.

Subscribe to Cheenta at Youtube