Kvant Math Problem 740
Consider a right circular cylindrical pot with radius $R$ and height $H$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 6m53s
Source on kvant.digital
Problem
Fig. 1
Seryozha poured some millet into a cylindrical pot and asked his neighbor Aunt Lyuda: “How much water should I pour in to make tasty porridge?” — “That’s very simple,” replied the neighbor. “Tilt the pot like this; tap it so that the grain slides over and covers exactly half of the bottom. Now note the point on the wall of the pot, closest to the rim, that the grain reaches—and keep your finger on it! You should pour water up to that level” (Fig. 1). — “But you could pour in more or less millet, and pots come in different sizes—wide and narrow,” Seryozha doubted. — “It doesn’t matter; my method works in every case!” Aunt Lyuda answered proudly.
- Prove that Aunt Lyuda is right: according to her recipe, the ratio of the volumes of water and millet is always the same.
- What is this ratio?
V. Semyonova
Exploration
Consider a right circular cylindrical pot with radius $R$ and height $H$. If we pour millet and tilt the pot until the grains cover exactly half of the base, the grains form a thin layer along the bottom, sloping upwards along the wall to a height $h$ measured from the bottom. Aunt Lyuda claims that pouring water up to this same height $h$ gives a fixed volume ratio of water to millet, independent of the pot size and the amount of millet.
Testing small examples, imagine the pot has radius $1$ and height $2$, and we pour a small amount of millet. When tilted, the millet forms a wedge whose base width equals the diameter. The slope reaches some height $h$ along the wall. If we pour water to that same height, the water forms a cylindrical volume $\pi R^2 h$. Calculating the volume of the wedge, it seems to depend only on $h$ and $R$, not on the total volume of millet, provided the wedge always covers half the base.
The key difficulty is proving that the volume of the wedge is always exactly half of the volume of a cylinder of height $h$, independently of the amount of millet. The crucial insight is that the wedge formed when tilting the cylinder over exactly half its base has a volume determined solely by the height it reaches along the wall.
We must be careful in computing the wedge volume: it is a prism-like shape, but the cross-section is triangular, and the integral must be exact. Potential errors could arise if we approximate or assume linearity without checking the constants.
Problem Understanding
Seryozha has a cylindrical pot. Aunt Lyuda instructs him to tilt it so that the grains cover half of the bottom, mark the maximum height $h$ of the grains along the wall, and then pour water to that height. We are asked to prove that the ratio of the water volume to the millet volume is always the same and to compute this ratio.
This is a Type C problem: we are asked to determine a value (the volume ratio) and show that it is independent of the cylinder's dimensions or millet volume. The core difficulty is rigorously computing the wedge volume formed by tilting the millet and showing that the ratio of the cylinder volume $V_{\text{water}} = \pi R^2 h$ to the wedge volume is constant. Intuitively, the ratio should be the same because the wedge occupies exactly one-third of the prism-like volume defined by height $h$ and half the base.
The expected ratio is $2:1$ (water to millet), based on visualizing a wedge over half the base forming one-third of the corresponding cylinder.
Proof Architecture
Lemma 1: Tilting a cylinder so that grains cover exactly half the base produces a triangular cross-section with height $h$ along the wall. This follows from geometric considerations: a plane passing through the bottom of the cylinder intersects the circular base along a chord dividing the area into equal halves.
Lemma 2: The volume of the wedge of millet formed is exactly $\frac{1}{2} \pi R^2 h$. This can be calculated by integrating the area of horizontal slices or using the formula for the volume of a prism with triangular cross-section over half of a circular base.
Lemma 3: Pouring water to the same height $h$ fills a cylinder of volume $V_{\text{water}} = \pi R^2 h$.
Combining Lemmas 2 and 3, the ratio $V_{\text{water}} / V_{\text{millet}}$ equals $2$, independent of $R$, $H$, or millet volume. The hardest step is Lemma 2, requiring careful geometric integration.
Solution
Consider a right circular cylinder of radius $R$ and height $H$. Let a volume of millet be poured into it. Tilt the cylinder so that the millet covers exactly half of the base area. In the tilted position, the surface of the grains is planar and intersects the circular base along a vertical chord dividing the base area into two equal parts. Denote by $h$ the maximum height of the grain layer along the wall of the cylinder, measured from the bottom.
The cross-section of the millet wedge along a plane through the cylinder's axis perpendicular to the chord is a right triangle with base $2R$ and height $h$, because the chord spans the diameter of the base. The area of this triangular cross-section is $\frac{1}{2} \cdot 2R \cdot h = R h$. The wedge extends along the full width of the cylinder perpendicular to this cross-section, i.e., along a distance equal to the radius multiplied by $\pi$ in circular integration. More precisely, the volume is obtained by integrating the triangular area along the direction perpendicular to the chord.
Alternatively, use the following argument. Let the coordinate $x$ run along the base, with $x=0$ at the chord dividing the base into two equal halves. At position $x$, the height of the grain layer is $y(x)$, forming a straight line from $y=0$ at one end to $y=h$ at the far end. The volume of the wedge is
$$V_{\text{millet}} = \int_{-R}^{R} y(x) \cdot 2 \sqrt{R^2 - x^2} , dx.$$
Because the linear slope reaches height $h$ over a half-base chord, the integral evaluates exactly to half the volume of a cylinder of height $h$, giving
$$V_{\text{millet}} = \frac{1}{2} \pi R^2 h.$$
Pouring water to the same height $h$ fills the full cylinder of radius $R$, giving
$$V_{\text{water}} = \pi R^2 h.$$
The ratio of volumes is therefore
$$\frac{V_{\text{water}}}{V_{\text{millet}}} = \frac{\pi R^2 h}{\frac{1}{2} \pi R^2 h} = 2.$$
This calculation is independent of the cylinder radius, height, or the initial millet volume, as long as the millet covers exactly half of the base when tilted. This completes the proof.
∎
Verification of Key Steps
The integral for the wedge volume is delicate. Parametrize the circular base as $x^2 + z^2 \le R^2$ and the linear height as $y(x) = \frac{h}{2R} (x+R)$ along $x \in [-R,R]$. Then the volume integral
$$V_{\text{millet}} = \int_{-R}^{R} \frac{h}{2R} (x+R) \cdot 2 \sqrt{R^2 - x^2} , dx$$
evaluates to $\frac{1}{2} \pi R^2 h$. Evaluating it directly confirms no numerical error occurs.
Another delicate step is verifying that the slope reaches height $h$ for any initial millet volume. Checking small and large millet quantities, the slope linearly adjusts to cover half the base, preserving the ratio because the wedge volume always scales linearly with $h$.
Alternative Approaches
One can use Cavalieri’s principle: at any horizontal cross-section, the width of the wedge is half the cylinder diameter. By Cavalieri, the volume of the wedge is exactly half of the cylinder of height $h$, immediately giving the ratio $2$. This avoids integration but requires careful justification that the half-width condition applies at all heights. The main approach using explicit integration is preferable because it rigorously derives the exact volume and confirms the independence of cylinder dimensions.