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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Cones and circle | AIME I, 2008 | Question 5

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Cones and circle.

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2008 based on Cones and circle.

Cones and circle – AIME I, 2008


A right circular cone has base radius r and height h the cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cones base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making 17 complete rotations. The value of \(\frac{h}{r}\) can be written in the form \(m{n}^\frac{1}{2}\) where m and n are positive integers and n in not divisible by the square of any prime, find m+n.

  • is 107
  • is 14
  • is 840
  • cannot be determined from the given information

Key Concepts


Cones

Circles

Algebra

Check the Answer


Answer: is 14.

AIME I, 2008, Question 5

Geometry Vol I to IV by Hall and Stevens

Try with Hints


First hint

The path is circle with radius =\(({r}^{2}+{h}^{2})^\frac{1}{2}\) then length of path=\(2\frac{22}{7}({r}^{2}+{h}^{2})^\frac{1}{2}\)

Second Hint

length of path=17 times circumference of base then \(({r}^{2}+{h}^{2})^\frac{1}{2}\)=17r then \({h}^{2}=288{r}^{2}\)

Final Step

then \(\frac{h}{r}=12{(2)}^\frac{1}{2}\) then 12+2=14.

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