Difference between revisions of "2004 AIME I Problems/Problem 11"
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== Solution == | == Solution == | ||
− | {{solution}} | + | Let <math>x</math> denote the radius of the small cone. Let <math>A_c</math> and <math>A_f</math> denote the area of the painted surface on cone <math>C</math> and frustum <math>F</math>, respectively. Let <math>V_c</math> and <math>V_f</math> denote the volume of cone <math>C</math> and frustum <math>F</math>, respectively. Using the Pythagorean Theorem, we find that the height and slant height of cone <math>C</math> are <math>\frac{4}{3}x</math> and <math>\frac{5}{3}x</math>, respectively. Using the formula for lateral surface area of a cone, we find that <math>A_c=\frac{1}{2}c\cdot \ell=\frac{1}{2}(2\pi x)\left(\frac{5}{3}x\right)=\frac{5}{3}\pi x^2</math>. By subtracting <math>A_c</math> from the lateral surface area of the original cone and adding to this the are of the base, we find that <math>A_f=12\pi - \frac{4}{9}\pi x^3</math>. Next, we find that <math>V_c=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi x^2 \left(\frac{4}{3}x\right)=\frac{4}{9}\pi x^3.</math> Finally, we subtract <math>V_c</math> from the volume of the original cone to find that <math>V_f=12\pi - \frac{4}{9}\pi x^3</math>. We know that <math>\frac{A_c}{A_f}=\frac{V_c}{V_f}=k.</math> Plugging in our values for <math>A_c</math>, <math>A_f</math>, <math>V_c</math>, and <math>V_f</math>, we obtain the equation <math>\frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}=\frac{\frac{4}{9}\pi x^3}{12\pi - \frac{4}{9}\pi x^3}.</math> We solve for <math>x</math> to obtain <math>x=\frac{15}{8}</math> (Note that we have disregarded the trivial solution <math>x=0</math>). We plug this value of <math>x</math> into our expression for <math>k</math> and simplify to obtain <math>k=\frac{125}{387}</math>. Hence <math>m+n=125+387=512</math>. |
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+ | Solution provided by 177h4x. | ||
== See also == | == See also == |
Revision as of 18:40, 30 November 2006
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
Let denote the radius of the small cone. Let
and
denote the area of the painted surface on cone
and frustum
, respectively. Let
and
denote the volume of cone
and frustum
, respectively. Using the Pythagorean Theorem, we find that 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 lateral surface area of the original cone and adding to this the are of the base, we find that
. Next, we find that
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 solve for
to obtain
(Note that we have disregarded the trivial solution
). We plug this value of
into our expression for
and simplify to obtain
. Hence
.
Solution provided by 177h4x.