An unpainted cone has radius $3 \mathrm{~cm}$ and slant height $5 \mathrm{~cm}$. The cone is placed in a container of paint. With the cone's circular base resting flat on the bottom of the container, the depth of the paint in the container is $2 \mathrm{~cm}$. When the cone is removed, its circular base and the lower portion of its lateral surface are covered in paint. The fraction of the total surface area of the cone that is covered in paint can be written as $\frac{p}{q}$ where $p$ and $q$ are positive

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-017.jpg?height=312&width=209&top_left_y=234&top_left_x=1446)
integers with no common divisor larger than 1 . What is the value of $p+q$ ?

(The lateral surface of a cone is its external surface not including the circular base. A cone with radius $r$, height $h$, and slant height $s$ has lateral surface area equal to $\pi r s$.)
(A) 59
(B) 61
(C) 63
(D) 65
(E) 67
The total surface area of the cone includes the circular base and the lateral surface.

For the given unpainted cone, the base has area $\pi r^{2}=\pi(3 \mathrm{~cm})^{2}=9 \pi \mathrm{cm}^{2}$ and the lateral surface has area $\pi r s=\pi(3 \mathrm{~cm})(5 \mathrm{~cm})=15 \pi \mathrm{cm}^{2}$.

Thus, the total surface area of the unpainted cone is $9 \pi \mathrm{cm}^{2}+15 \pi \mathrm{cm}^{2}=24 \pi \mathrm{cm}^{2}$.

Since the height and base are perpendicular, the lengths $s, h$ and $r$ form a right-angled triangle with hypotenuse $s$.

For this cone, by the Pythagorean Theorem, $h^{2}=s^{2}-r^{2}=(5 \mathrm{~cm})^{2}-(3 \mathrm{~cm})^{2}=16 \mathrm{~cm}^{2}$ and so $h=4 \mathrm{~cm}$.

When the unpainted cone is placed in the container of paint so that the paint rises to a depth of $2 \mathrm{~cm}$, the base of the cone (area $9 \pi \mathrm{cm}^{2}$ ) is covered in paint.

Also, the bottom portion of the lateral surface is covered in paint.
![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-033.jpg?height=316&width=458&top_left_y=1262&top_left_x=889)

The unpainted portion of the cone is itself a cone with height 2.

When we take a vertical cross-section of the cone through its top vertex and a diameter of the base, the triangle formed above the paint is similar to the original triangle and has half its dimensions. The triangles are similar because both are right-angled and they share an equal angle at the top vertex. The ratio is $2: 1$ since their heights are $4 \mathrm{~cm}$ and $2 \mathrm{~cm}$.

Therefore, the unpainted cone has radius $1.5 \mathrm{~cm}$ (half of the original radius of $3 \mathrm{~cm}$ ) and slant height $2.5 \mathrm{~cm}$ (half of the original slant height of $5 \mathrm{~cm}$ ).

Thus, the unpainted lateral surface area is the lateral surface area of a cone with radius $1.5 \mathrm{~cm}$ and slant height $2.5 \mathrm{~cm}$, and so has area $\pi(1.5 \mathrm{~cm})(2.5 \mathrm{~cm})=3.75 \pi \mathrm{cm}^{2}$.

This means that the painted lateral surface area is $15 \pi \mathrm{cm}^{2}-3.75 \pi \mathrm{cm}^{2}=11.25 \pi \mathrm{cm}^{2}$. (This is in fact three-quarters of the total surface area. Can you explain why this is true in a different way?)

Thus, the fraction of the total surface area of the cone that is painted is

$$
\frac{9 \pi \mathrm{cm}^{2}+11.25 \pi \mathrm{cm}^{2}}{24 \pi \mathrm{cm}^{2}}=\frac{20.25}{24}=\frac{81}{96}=\frac{27}{32}
$$

Since $\frac{27}{32}$ is in lowest terms, then $p+q=27+32=59$.