A cylindrical tank has radius $10 \mathrm{~m}$ and length $30 \mathrm{~m}$. The tank is lying on its side on a flat surface and is filled with water to a depth of $5 \mathrm{~m}$. Water is added to the tank and the depth of the water increases from $5 \mathrm{~m}$ to $10+5 \sqrt{2} \mathrm{~m}$. If the volume of water added to the tank, in $\mathrm{m}^{3}$, can be written as $a \pi+b+c \sqrt{p}$ for

![](https://cdn.mathpix.com/cropped/2024_04_17_a167ac2f1051780ff733g-2.jpg?height=323&width=662&top_left_y=1134&top_left_x=1290)
some integers $a, b, c$ and prime number $p$, determine the quadruple $(a, b, c, p)$.
Consider the front face of the tank, which is a circle of radius $10 \mathrm{~m}$.

Suppose that when the water has depth $5 \mathrm{~m}$, its surface is along horizontal line $A B$.

Suppose that when the water has depth $(10+5 \sqrt{2}) \mathrm{m}$, its surface is along horizontal line $C D$. Let the area of the circle between the chords $A B$ and $C D$ be $x \mathrm{~m}^{2}$.

Since the tank is a cylinder which is lying on a flat surface, the volume of water added can be viewed as an irregular prism with base of area $x \mathrm{~m}^{2}$ and length $30 \mathrm{~m}$.

Thus, the volume of water equals $30 x \mathrm{~m}^{3}$. Therefore, we need to calculate the value of $x$.

Let $O$ be the centre of the circle, $N$ be the point where the circle touches the ground, $P$ the midpoint of $A B$, and $Q$ the midpoint of

$C D$.

Join $O$ to $A, B, C$, and $D$. Also, join $O$ to $N$ and to $Q$.

Since $Q$ is the midpoint of chord $C D$ and $O$ is the centre of the circle, then $O Q$ is perpendicular to $C D$.

Since $P$ is the midpoint of $A B$, then $O N$ passes through $P$ and is perpendicular to $A B$.

Since $A B$ and $C D$ are parallel and $O P$ and $O Q$ are perpendicular to these chords, then $Q O P N$ is a straight line segment.

![](https://cdn.mathpix.com/cropped/2024_04_30_be08db732f430ab57efag-05.jpg?height=471&width=441&top_left_y=201&top_left_x=1487)

Since the radius of the circle is $10 \mathrm{~m}$, then $O C=O D=O A=O N=O B=10 \mathrm{~m}$.

Since $A B$ is $5 \mathrm{~m}$ above the ground, then $N P=5 \mathrm{~m}$.

Since $O N=10 \mathrm{~m}$, then $O P=O N-N P=5 \mathrm{~m}$.

Since $C D$ is $(10+5 \sqrt{2}) \mathrm{m}$ above the ground and $O N=10 \mathrm{~m}$, then $Q O=5 \sqrt{2} \mathrm{~m}$.

Since $\triangle A O P$ is right-angled at $P$ and $O P: O A=1: 2$, then $\triangle A O P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with $\angle A O P=60^{\circ}$. Also, $A P=\sqrt{3} O P=5 \sqrt{3} \mathrm{~m}$.

Since $\triangle C Q O$ is right-angled at $Q$ and $O C: O Q=10: 5 \sqrt{2}=2: \sqrt{2}=\sqrt{2}: 1$, then $\triangle C Q O$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle with $\angle C O Q=45^{\circ}$. Also, $C Q=O Q=5 \sqrt{2} \mathrm{~m}$.

We are now ready to calculate the value of $x$.

The area between $A B$ and $C D$ is equal to the area of the circle minus the combined areas of the region under $A B$ and the region above $C D$.

The area of the region under $A B$ equals the area of sector $A O B$ minus the area of $\triangle A O B$.

The area of the region above $C D$ equals the area of sector $C O D$ minus the area of $\triangle C O D$.

Since $\angle A O P=60^{\circ}$, then $\angle A O B=2 \angle A O P=120^{\circ}$.

Since $\angle C O Q=45^{\circ}$, then $\angle C O D=2 \angle C O Q=90^{\circ}$.

Since the complete central angle of the circle is $360^{\circ}$, then sector $A O B$ is $\frac{120^{\circ}}{360^{\circ}}=\frac{1}{3}$ of the whole circle and sector $C O D$ is $\frac{90^{\circ}}{360^{\circ}}=\frac{1}{4}$ of the whole circle.

Since the area of the entire circle is $\pi(10 \mathrm{~m})^{2}=100 \pi \mathrm{m}^{2}$, then the area of sector $A O B$ is $\frac{100}{3} \pi \mathrm{m}^{2}$ and the area of sector $C O D$ is $25 \pi \mathrm{m}^{2}$.

Since $P$ and $Q$ are the midpoints of $A B$ and $C D$, respectively, then $A B=2 A P=10 \sqrt{3} \mathrm{~m}$ and $C D=2 C Q=10 \sqrt{2} \mathrm{~m}$.

Thus, the area of $\triangle A O B$ is $\frac{1}{2} \cdot A B \cdot O P=\frac{1}{2}(10 \sqrt{3} \mathrm{~m})(5 \mathrm{~m})=25 \sqrt{3} \mathrm{~m}^{2}$.

Also, the area of $\triangle C O D$ is $\frac{1}{2} \cdot C D \cdot O Q=\frac{1}{2}(10 \sqrt{2} \mathrm{~m})(5 \sqrt{2} \mathrm{~m})=50 \mathrm{~m}^{2}$.

This means that the area of the region below $A B$ is $\left(\frac{100}{3} \pi-25 \sqrt{3}\right) \mathrm{m}^{2}$ and the area of the region above $C D$ is $(25 \pi-50) \mathrm{m}^{2}$.

Finally, this means that the volume of water added, in $\mathrm{m}^{3}$, is

$$
\begin{aligned}
30 x & =30\left(100 \pi-\left(\frac{100}{3} \pi-25 \sqrt{3}\right)-(25 \pi-50)\right) \\
& =3000 \pi-1000 \pi+750 \sqrt{3}-750 \pi+1500 \\
& =1250 \pi+1500+750 \sqrt{3}
\end{aligned}
$$

Therefore, $a \pi+b+c \sqrt{p}=1250 \pi+1500+750 \sqrt{3}$ and so $(a, b, c, p)=(1250,1500,750,3)$.

## Team Problems