Difference between revisions of "2004 AIME I Problems/Problem 11"
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Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</math>. | Our original solid <math>V</math> has [[surface area]] <math>A_v = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]] or Pythagorean Triple knowledge, we obtain <math>\ell = 5</math> and lateral area <math>A_\ell = 15\pi</math>. The area of the base is <math>A_B = 3^2\pi = 9\pi</math>. | ||
| − | Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>A_c | + | <math>V</math> and <math>C</math> are similar cones, because the plane that cut out <math>C</math> was parallel to the base of <math>V</math>. Let <math>x</math> be the scale factor between the original cone and the small cone <math>C</math> in one dimension. Because the scale factor is uniform in all dimensions, <math>x^2</math> relates corresponding areas of <math>C</math> and <math>V</math>, and <math>x^3</math> relates corresponding volumes. Then, the ratio of the painted areas <math>\frac{A_c}{A_f}</math> is <math>\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8 - 5 x^2} = k</math> and the ratio of the volumes <math>\frac{V_c}{V_f}</math> is <math>\frac{x^3}{1 - x^3} = k</math>. Since both ratios are equal to <math>k</math>, they are equal to each other. Therefore, <math>\frac{5 x^2}{8 - 5 x^2} = \frac{x^3}{1 - x^3}</math>. |
| − | Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{ | + | Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives <math>5 x^2(1 - x^3) = x^3(8 - 5 x^2)</math>. Dividing both sides by <math>x^2</math> and distributing the <math>x</math> on the right, we have <math>5 - 5 x^3 = 8 x - 5 x^3</math>, and so <math>8 x = 5</math> and <math>x = \frac{5}{8}</math>. Substituting back into the easier ratio, we have <math>\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}</math>. And so we have <math>m + n = 125 + 387 = \boxed{512}</math>. |
== See also == | == See also == | ||
Latest revision as of 21:12, 16 December 2015
Problem
A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid 
and a frustum-shaped solid 
in such a way that the ratio between the areas of the painted surfaces of and 
and the ratio between the volumes of and are both equal to 
. Given that 
where 
and 
are relatively prime positive integers, find 
Solution
Solution 1
Our original solid has volume equal to 
and has surface area 
, where 
is the slant height of the cone. Using the Pythagorean Theorem, we get 
and 
.
Let 
denote the radius of the small cone. Let 
and 
denote the area of the painted surface on cone and frustum , respectively, and let 
and 
denote the volume of cone and frustum , respectively. Because the plane cut is parallel to the base of our solid, is similar to the uncut solid and so the height and slant height of cone are 
and 
, respectively. Using the formula for lateral surface area of a cone, we find that 
. By subtracting from the surface area of the original solid, we find that 
.
Next, we can calculate 
. Finally, we subtract from the volume of the original cone to find that 
. We know that 
Plugging in our values for , , , and , we obtain the equation 
. We can take reciprocals of both sides to simplify this equation to 
and so 
. Then 
so the answer is 
.
Solution 2
Our original solid 
has surface area 
, where is the slant height of the cone. Using the Pythagorean Theorem or Pythagorean Triple knowledge, we obtain and lateral area 
. The area of the base is 
.
and are similar cones, because the plane that cut out was parallel to the base of . Let be the scale factor between the original cone and the small cone in one dimension. Because the scale factor is uniform in all dimensions, 
relates corresponding areas of and , and 
relates corresponding volumes. Then, the ratio of the painted areas 
is 
and the ratio of the volumes 
is 
. Since both ratios are equal to , they are equal to each other. Therefore, 
.
Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives 
. Dividing both sides by and distributing the on the right, we have 
, and so 
and 
. Substituting back into the easier ratio, we have 
. And so we have 
.
See also
| 2004 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
