Problem 924

Solution. If we tilt the cylindrical tank so that its base makes an angle of $30^{\circ}$ with its original position, then after the liquid has drained, it will occupy a horizontal position $D^{\prime \prime} C^{\prime} \| A B$ (see
diagram). The amount of liquid that has drained is equal to half the volume of the cylinder with cross-section $D^{\prime \prime} C^{\prime \prime} C^{\prime} D^{\prime}$. Let $x=\overline{C^{\prime \prime} C^{\prime}}$ be the height of this cylinder, which has the same base as the given cylinder, so $\overline{D^{\prime \prime} C^{\prime}}=2 R$, where $R$ is the radius of the base of the given cylinder. The right triangle $C^{\prime} C^{\prime \prime} D^{\prime \prime}$ is half of an equilateral triangle, because

$$\measuredangle C^{\prime} D^{\prime \prime} C^{\prime \prime}=\measuredangle A B A^{\prime}=30^{\circ}$$

as angles with mutually parallel sides, so $\overline{D^{\prime \prime} C^{\prime}}=2 x$. From the Pythagorean theorem we have

$${\overline{D^{\prime \prime} C^{\prime \prime}}}^{2}+{\overline{C^{\prime \prime} C^{\prime}}}^{2}={\overline{D^{\prime \prime} C^{\prime}}}^{2}$$

or $(2 R)^{2}+x^{2}=(2 x)^{2}$, from which we get $x=\frac{2 \sqrt{3} R}{3}$. The height of the given cylinder is $H=2 R$, so its volume is $V=2 \pi R^{3}$.

!
Then the volume of the drained water is

$$V_{1}=\frac{1}{2} \pi R^{2} x=\frac{\sqrt{3}}{6} \cdot 2 \pi R^{3}=\frac{\sqrt{3}}{6} V<\frac{1}{3} V$$

which means that less than one-third of the liquid will drain from the tank.

## 2nd Year

1AB. Find all natural numbers $n$ for which $n^{2}+25 n+19$ is a perfect square.