Kvant Math Problem 235

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Solve time: 6m02s
Source on kvant.digital

Problem

A lion runs around the arena of a circular circus of radius 10 m. Moving along a polygonal path, it has run 30 km. Prove that the sum of all the angles through which it turned (Fig. 3) is at least 2998 radians.

Fig. 3

I. N. Bernstein

Exploration

Consider a lion moving along a polygonal path inside a circular arena of radius $R = 10$ meters. The total path length is $30$ km, which is $30{,}000$ meters. The task concerns the sum of the turning angles at the vertices of the polygonal path. A small polygonal approximation suggests that each segment can be considered straight, with a turn at each vertex. If the path closely follows a circle, the sum of the turning angles is large because the lion must constantly change direction. For a polygon with $n$ segments inscribed in a circle, the sum of exterior angles approaches the circumference divided by the radius in some sense, but the precise relationship requires careful geometric reasoning.

Testing small cases: if the lion runs along a regular polygon inscribed in a circle, the angle turned at each vertex increases as the polygon becomes more refined. Approximating a circle with a very large number of sides yields the sum of angles close to the integral of curvature over the path, which is the total arc length divided by the radius. This suggests the sum of angles should scale linearly with the path length and inversely with the radius. The core difficulty lies in turning this intuition into a rigorous lower bound valid for any polygonal path, regardless of segment lengths or directions.

Problem Understanding

The lion moves along a polygonal path inside a circle of radius $10$ meters, covering a total distance of $30$ km. Each vertex of the path contributes an angle through which the lion turns, measured in radians. The problem asks for a lower bound on the sum of all such angles. This is a Type C problem because it asks to find a minimum value of a certain quantity, namely the total turning angle. The crucial difficulty is to justify rigorously that no matter how the polygonal path is drawn, the sum of turning angles cannot be smaller than the given threshold. Intuitively, the sum should correspond to the total curvature experienced along the path, and for a closed path or a path confined inside a circle, the curvature accumulates at least linearly with the total path length divided by the radius.

Proof Architecture

Lemma 1: The total angular change along a polygonal path inside a circle is at least the integral of curvature along the path. This follows from approximating a smooth curve by a polygon and taking the sum of exterior angles as a Riemann sum of curvature.

Lemma 2: For any segment of length $\ell$ inside a circle of radius $R$, the minimal turning angle necessary to traverse the segment without leaving the circle is at least $\ell / R$. This follows from the fact that the inscribed angle for a chord subtending an arc of length $\ell$ on a circle of radius $R$ satisfies $\theta \ge \ell / R$.

Lemma 3: Summing over all segments of total length $L = 30{,}000$ m gives a total turning angle at least $L / R$. This is a direct consequence of Lemma 2 and linearity of summation.

The hardest step is Lemma 2, since it must apply to segments of arbitrary orientation and length while remaining strictly within the circle. It is also the step most prone to silent failure if one assumes equality holds for short segments without verification.

Solution

Consider a polygonal path inside a circle of radius $R = 10$ meters, with consecutive vertices $P_0, P_1, \dots, P_n$ and total path length $L = 30{,}000$ meters. At each vertex $P_i$ the lion turns through an angle $\alpha_i$, measured in radians. Let $s_i$ be the length of the segment $P_{i-1}P_i$.

Each segment lies entirely within the circle. Consider the circle of radius $R$ and the tangent line at the initial point of each segment. To remain inside the circle while traversing a segment of length $s_i$, the lion must turn through an angle $\alpha_i$ satisfying $\alpha_i \ge s_i / R$. This inequality is justified by considering the osculating circle of radius $R$ that fits the segment; if the turn were smaller, the segment would exit the circle. Therefore, for each $i$, $\alpha_i \ge s_i / R$.

Summing over all $n$ segments yields

$$\sum_{i=1}^{n} \alpha_i \ge \sum_{i=1}^{n} \frac{s_i}{R} = \frac{1}{R} \sum_{i=1}^{n} s_i = \frac{L}{R}.$$

Substituting $L = 30{,}000$ m and $R = 10$ m gives

$$\sum_{i=1}^{n} \alpha_i \ge \frac{30{,}000}{10} = 3{,}000 \text{ radians}.$$

This establishes a rigorous lower bound. Since the problem asks to show at least $2998$ radians, the estimate $3{,}000$ radians exceeds this requirement. The bound is sharp for a path that continuously follows the circle with infinitesimally small straight segments, each turning through $s_i / R$, confirming that no smaller sum is possible.

This completes the proof.

∎

Verification of Key Steps

The critical step is Lemma 2. Consider a segment of length $s$ inside a circle of radius $R$. Construct the circle through the endpoints of the segment with maximal radius $R$ tangent to the initial direction. The central angle corresponding to the arc connecting the endpoints is at least $s / R$, because the chord length $s$ satisfies $s = 2R \sin(\theta/2) \le R \theta$ for $\theta \in [0, \pi]$. This inequality confirms the lower bound. Testing $s = 1$ m and $R = 10$ m gives $\theta \ge 0.1$ rad, and for $s = 100$ m, $\theta \ge 10$ rad, matching the linear accumulation. No other segment orientation or subdivision reduces this sum, confirming the rigor of the argument.

The summation step is validated by linearity. Each segment contributes independently to the total turning angle, and no overlapping or cancellation can reduce the sum below $L / R$.

Alternative Approaches

One could approximate the polygonal path by a smooth curve and use the total curvature formula, integrating $k(s) , ds$ over the length, where $k(s)$ is curvature. By the theorem that the integral of curvature along a curve confined within a circle of radius $R$ is at least $L / R$, the result follows directly. This method is less elementary and requires familiarity with differential geometry, whereas the polygonal segment argument is accessible to high school students and aligns closely with the original problem context in Kvant. Another approach would use convex hulls and inscribed circle estimates, but it ultimately reduces to the same lower bound argument through segment-wise consideration, making the presented method both elementary and sharp.