CEMC Cayley

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 $QT^2$. Since $RS$ is directly above $PQ$, then $RP$ is perpendicular to $PQ$. This means that $\triangle TPQ$ is right-angled at $P$. Since $PQ$ is a diameter, then $PQ = 2r$. By the Pythagorean Theorem, $QT^2 = PT^2 + PQ^2 = n^2 + (2r)^2 = n^2 + 4r^2$. So we need to determine the values of $n$ and $r$. We will use the information about $QU$ and $UT$ to determine these values. Join $U$ to $O$. Since $U$ is halfway between $R$ and $S$, then the arcs $RU$ and $US$ are each one-quarter of the circle that bounds the top face of the cylinder. This means that $\angle UOR = \angle UOS = 90^{\circ}$. We can use the Pythagorean Theorem in $\triangle UOR$ and $\triangle UOS$, which are both right-angled at $O$, to obtain $UR^2 = UO^2 + OR^2 = r^2 + r^2 = 2r^2$ and $US^2 = 2r^2$. Since $RP$ and $QS$ are both perpendicular to the top face of the cylinder, we can use the Pythagorean Theorem in $\triangle TRU$ and in $\triangle QSU$ to obtain $QU^2 = QS^2 + US^2 = m^2 + 2r^2$ and $UT^2 = TR^2 + UR^2 = (PR - PT)^2 + 2r^2 = (QS - n)^2 + 2r^2 = (m - n)^2 + 2r^2$. Since $QU = 9 \sqrt{33}$, then $QU^2 = 9^2 \cdot 33 = 2673$. Since $UT = 40$, then $UT^2 = 1600$. Therefore, $m^2 + 2r^2 = 2673$ and $(m - n)^2 + 2r^2 = 1600$. Subtracting the second equation from the first, we obtain the equivalent equations $m^2 - (m - n)^2 = 1073$ and $m^2 - (m^2 - 2mn + n^2) = 1073$ and $2mn - n^2 = 29 \cdot 37$ and $n(2m - n) = 29 \cdot 37$. Since $m$ and $n$ are integers, then $2m - n$ is an integer. Thus, $n$ and $2m - n$ 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: $n = 1, 2m - n = 1073, m = 537$; $n = 29, 2m - n = 37, m = 33$; $n = 37, 2m - n = 29, m = 33$; $n = 1073, 2m - n = 1, m = 537$. Since $m > n$, then $n$ cannot be 37 or 1073. Since $QU > QS$, then $m < 9 \sqrt{33} \approx 51.7$. This means that $n = 29$ and $m = 33$. Since $(m - n)^2 + 2r^2 = 1600$, we obtain $2r^2 = 1600 - (m - n)^2 = 1600 - 4^2 = 1584$ and so $QT^2 = n^2 + 4r^2 = 29^2 + 2(2r^2) = 841 + 3168 = 4009$. The remainder when $QT^2$ is divided by 100 is 9.