In the diagram, $P Q$ is a diameter of the circular base of the cylinder. $R S$ is a diameter of the top face of the cylinder and is directly above $P Q$, as shown. Point $U$ is on the circumference of the top face, halfway between $R$ and $S$. Point $T$ is on the cylinder and is directly above $P$. Suppose that $Q S=m$ and $P T=n$, where $m$ and $n$ are integers with $1<n<m$. If $Q U=9 \sqrt{33}$ and $U T=40$, what is the remainder when the integer equal to $Q T^{2}$ is divided by 100 ?
(A) 29
(B) 49
(C) 9
(D) 89
(E) 69

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-017.jpg?height=563&width=417&top_left_y=1231&top_left_x=1301)
Let $O$ be the centre of the top face of the cylinder and let $r$ be the radius of the cylinder.

We need to determine the value of $Q T^{2}$.

Since $R S$ is directly above $P Q$, then $R P$ is perpendicular to $P Q$.

This means that $\triangle T P Q$ is right-angled at $P$.

Since $P Q$ is a diameter, then $P Q=2 r$.

By the Pythagorean Theorem, $Q T^{2}=P T^{2}+P Q^{2}=n^{2}+(2 r)^{2}=n^{2}+4 r^{2}$.

So we need to determine the values of $n$ and $r$. We will use the information about $Q U$ and $U T$ to determine these values.

Join $U$ to $O$.

Since $U$ is halfway between $R$ and $S$, then the $\operatorname{arcs} R U$ and $U S$ are each one-quarter of the circle that bounds the top face of the cylinder.

This means that $\angle U O R=\angle U O S=90^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-034.jpg?height=261&width=285&top_left_y=791&top_left_x=974)

We can use the Pythagorean Theorem in $\triangle U O R$ and $\triangle U O S$, which are both right-angled at $O$, to obtain

$$
U R^{2}=U O^{2}+O R^{2}=r^{2}+r^{2}=2 r^{2} \quad \text { and } \quad U S^{2}=2 r^{2}
$$

Since $R P$ and $Q S$ are both perpendicular to the top face of the cylinder, we can use the Pythagorean Theorem in $\triangle T R U$ and in $\triangle Q S U$ to obtain

$$
\begin{aligned}
& Q U^{2}=Q S^{2}+U S^{2}=m^{2}+2 r^{2} \\
& U T^{2}=T R^{2}+U R^{2}=(P R-P T)^{2}+2 r^{2}=(Q S-n)^{2}+2 r^{2}=(m-n)^{2}+2 r^{2}
\end{aligned}
$$

Since $Q U=9 \sqrt{33}$, then $Q U^{2}=9^{2} \cdot 33=2673$.

Since $U T=40$, then $U T^{2}=1600$.

Therefore,

$$
\begin{aligned}
m^{2}+2 r^{2} & =2673 \\
(m-n)^{2}+2 r^{2} & =1600
\end{aligned}
$$

Subtracting the second equation from the first, we obtain the equivalent equations

$$
\begin{aligned}
m^{2}-(m-n)^{2} & =1073 \\
m^{2}-\left(m^{2}-2 m n+n^{2}\right) & =1073 \\
2 m n-n^{2} & =29 \cdot 37 \\
n(2 m-n) & =29 \cdot 37
\end{aligned}
$$

Since $m$ and $n$ are integers, then $2 m-n$ is an integer. Thus, $n$ and $2 m-n$ are a factor pair of $29 \cdot 37=1073$.

Since 29 and 37 are prime numbers, the integer 1073 has only four positive divisors: 1, 29, 37, 1073.

This gives the following possibilities:

| $n$ | $2 m-n$ | $m=\frac{1}{2}((2 m-n)+n)$ |
| :---: | :---: | :---: |
| 1 | 1073 | 537 |
| 29 | 37 | 33 |
| 37 | 29 | 33 |
| 1073 | 1 | 537 |

Since $m>n$, then $n$ cannot be 37 or 1073 .

Since $Q U>Q S$, then $m<9 \sqrt{33} \approx 51.7$.

This means that $n=29$ and $m=33$.

Since $(m-n)^{2}+2 r^{2}=1600$, we obtain $2 r^{2}=1600-(m-n)^{2}=1600-4^{2}=1584$ and so

$$
Q T^{2}=n^{2}+4 r^{2}=29^{2}+2\left(2 r^{2}\right)=841+3168=4009
$$

The remainder when $Q T^{2}$ is divided by 100 is 9 .

ANSWER: $(\mathrm{C})$