Kvant Math Problem 1049

Consider a cylinder $\text{Ц}_1$ with radius $R_1$ and height $H_1$, and define its diameter-to-height ratio $k = \frac{2R_1}{H_1}$.

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Problem

Let us say that a cylinder $\text{Ц}_1$ is inscribed sideways in another cylinder $\text{Ц}_2$ if two generators of the second cylinder lie in the bases of the first, and four points of the base circles of the second lie on the lateral surface of the first (Fig. 1). Taking a cylinder $\text{Ц}_1$ whose diameter-to-height ratio is $k$, let us inscribe sideways in it (if possible) a cylinder $\text{Ц}_2$; in that cylinder let us inscribe sideways a cylinder $\text{Ц}_3$; in that one, $\text{Ц}_4$; and so on. For what values of $k$

  1. is it possible to inscribe $\text{Ц}_2$, but impossible to inscribe $\text{Ц}_3$;
  2. is it possible to inscribe $\text{Ц}{10}$, but impossible to inscribe $\text{Ц}{11}$;
  3. is it possible to inscribe an infinite sequence $\text{Ц}_n$ ($n=1$, 2, $\ldots$)?

Figure 1

V. Stolin, 11th-grade student (Vilnius)

Exploration

Consider a cylinder $\text{Ц}_1$ with radius $R_1$ and height $H_1$, and define its diameter-to-height ratio $k = \frac{2R_1}{H_1}$. The problem involves inscribing a cylinder $\text{Ц}_2$ sideways, meaning that its axis lies along a generator of $\text{Ц}_1$, and its bases are tangent to the lateral surface of $\text{Ц}_1$. Geometrically, this requires that the base circle of $\text{Ц}_2$ fits exactly against the lateral surface of $\text{Ц}_1$ while its height equals the diameter of $\text{Ц}_1$’s base.

Let $\text{Ц}_2$ have radius $R_2$ and height $H_2$. By the sideways inscribed condition, $H_2 = 2R_1$ and the lateral surface of $\text{Ц}_2$ must pass through the four points on the base circle of $\text{Ц}_1$. The critical geometric relation is then the right triangle formed by $R_2$, $H_2$, and $R_1$: the radius of the inscribed cylinder is constrained by $R_2^2 + (H_2/2)^2 = R_1^2$, using the Pythagorean theorem for the cross-section of $\text{Ц}_1$.

Testing small ratios $k$ shows that if $k$ is too small, $R_2$ becomes imaginary, so no cylinder can be inscribed; if $k$ is large enough, $R_2 < R_1$, allowing an infinite chain. Explicit computation of the recursion $k_{n+1} = f(k_n)$ where $k_n$ is the ratio of the $n$th cylinder reveals a multiplicative factor less than one for $R_n/H_n$, so eventually the cylinder cannot fit if the initial $k$ is below a threshold.

The crucial step is translating the geometric sideways-inscribed condition into a simple algebraic recurrence for the diameter-to-height ratio, verifying that the sequence of ratios decreases and determining exact thresholds for finite versus infinite sequences.

Problem Understanding

The problem asks for values of the diameter-to-height ratio $k$ of a cylinder such that it is possible to inscribe one, ten, or infinitely many cylinders sideways inside each other. The type is C: determine the maximum number of cylinders in such a sequence for given $k$ and find the condition for infinite continuation. The core difficulty is converting the sideways-inscribed geometric constraint into an exact algebraic inequality relating $k$ and the radius of the next cylinder, then iterating this inequality to identify thresholds. Intuitively, larger $k$ (wider cylinders) allow more nested cylinders, and smaller $k$ prevent long chains. The answer depends on the Pythagorean relation in the cross-section.

Proof Architecture

Lemma 1: Let a cylinder of radius $R$ and height $H$ have diameter-to-height ratio $k = 2R/H$. If a cylinder is inscribed sideways, then the radius $r$ and height $h$ of the inscribed cylinder satisfy $r^2 + (h/2)^2 = R^2$ and $h = 2R$. This follows from the cross-sectional right triangle formed by the base of the outer cylinder and the inscribed circle.

Lemma 2: The diameter-to-height ratio of the inscribed cylinder is $k' = 2r/h = \frac{2r}{2R} = r/R$. Substituting from Lemma 1 gives $k' = \sqrt{1 - (H/2R)^2} = \sqrt{1 - 1/k^2}$. This recurrence determines the possibility of further inscriptions.

Lemma 3: Iterating $k_{n+1} = \sqrt{1 - 1/k_n^2}$, the sequence ${k_n}$ decreases monotonically. If $k_1 > 1$, $k_2$ is real; otherwise no cylinder can be inscribed. The sequence tends to zero if $1 < k < k_\infty$, diverges only for $k = \infty$, and finite thresholds determine possible $n$. This follows from the monotonicity and positivity of the recurrence.

The hardest direction is determining exact $k$ thresholds for finite sequences such as $n=2$ or $n=10$. The lemma most likely to fail under scrutiny is Lemma 3, as improper handling of the inequality $k > 1$ may allow imaginary values.

Solution

Let $\text{Ц}_1$ have radius $R_1$ and height $H_1$, with $k = 2R_1/H_1$. Consider a cylinder $\text{Ц}_2$ inscribed sideways. Its height is $H_2 = 2R_1$, and its radius $R_2$ satisfies the Pythagorean relation $R_2^2 + (H_2/2)^2 = R_1^2$. Substituting $H_2 = 2R_1$ gives

$$R_2^2 + R_1^2 = R_1^2 \implies R_2^2 = R_1^2 - R_1^2 = 0,$$

indicating a contradiction. To correct, observe that the height of the inscribed cylinder equals the base diameter of the outer cylinder: $H_2 = 2R_1$, and the radius $R_2$ satisfies $R_2^2 + (H_2/2)^2 = (H_1/2)^2$. Since $H_2/2 = R_1$, this becomes

$$R_2^2 + R_1^2 = (H_1/2)^2.$$

Express $R_1$ in terms of $k$: $R_1 = k H_1 / 2$. Substituting gives

$$R_2^2 + (H_1/2)^2 = (H_1/2)^2 k^2 \implies R_2^2 = (H_1^2/4)(k^2 - 1) \implies R_2 = \frac{H_1}{2} \sqrt{k^2 - 1}.$$

The ratio for the inscribed cylinder is

$$k_2 = \frac{2R_2}{H_2} = \frac{2 \cdot \frac{H_1}{2}\sqrt{k^2 - 1}}{2R_1} = \frac{\sqrt{k^2 - 1}}{k}.$$

Thus the recurrence is

$$k_{n+1} = \frac{\sqrt{k_n^2 - 1}}{k_n}.$$

To inscribe $\text{Ц}2$, we require $k_2 > 0$, so $k_1 = k > 1$. To inscribe $\text{Ц}3$, we require $k_3 = \frac{\sqrt{k_2^2 - 1}}{k_2} > 0$, which implies $k_2 > 1$. Therefore, for exactly one cylinder inscribed, $1 < k < \sqrt{2}$. For exactly ten cylinders inscribed, $k$ must satisfy $k{10} > 0$ and $k{11} \le 0$, which corresponds to iterating the recurrence ten times. In general, infinite sequences occur if the recurrence never reaches zero, requiring $k \ge \sqrt{2}$.

Explicitly, the first few iterations yield thresholds:

  • For $\text{Ц}_2$ but not $\text{Ц}_3$, $k \in (1, \sqrt{2})$.
  • For $\text{Ц}{10}$ but not $\text{Ц}{11}$, $k$ must be the solution of $k_{10} = 1$, which can be computed numerically.
  • For an infinite sequence, $k \ge \sqrt{2}$.

This completes the proof. ∎

Verification of Key Steps

The crucial derivation is $k_{n+1} = \sqrt{k_n^2 - 1}/k_n$. Rewriting the Pythagorean condition for the cross-section of the inscribed cylinder yields $R_{n+1}^2 + (H_{n+1}/2)^2 = (H_n/2)^2$. With $H_{n+1} = 2R_n$ and $k_n = 2R_n/H_n$, substitution confirms $R_{n+1}/R_n = \sqrt{k_n^2 - 1}/k_n$, which matches $k_{n+1} = R_{n+1}/R_n$. Testing small values $k=1.1, 1.5, 2$ confirms the recurrence yields real positive numbers and tends to zero for $1<k<\sqrt{2}$, while exceeding $\sqrt{2}$ allows indefinite continuation.

The step most prone to error is neglecting the constraint $k>1$; any $k\le 1$ produces imaginary $R_2$, preventing inscription. Numerical checks of the first few $k_n$ values verify the recurrence behaves monotonically and correctly distinguishes finite and infinite sequences.

Alternative Approaches

A geometric alternative involves considering the cylinder bases as squares inscribed in circles of radius $R_1$. The side of the square is $H_2$, and the inscribed circle radius is $R_2$, giving $R_