Kvant Math Problem 1163

Let the position of the first turtle at time $t$ be $P(t)$ and the position of the second turtle be $Q(t)$.

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Problem

A turtle started from the point $A$ and arrived at the point $B$, moving along an arbitrary trajectory with arbitrary speed. Following it, a second turtle started from the point $A$. At every moment of time, it moved in the direction of the first turtle (with arbitrary speed) and eventually also arrived at the point $B$. Prove that the distance traveled by the second turtle (by the time both have arrived at $B$) does not exceed the distance traveled by the first turtle.

A. Kh. Shen

Exploration

Let the position of the first turtle at time $t$ be $P(t)$ and the position of the second turtle be $Q(t)$. The condition says that whenever the second turtle moves, its instantaneous direction is toward the current position of the first turtle. Thus the velocity vector of the second turtle is always collinear with the vector $\overrightarrow{Q(t)P(t)}$ and points from $Q(t)$ toward $P(t)$.

The quantity that naturally compares the two motions is the distance between the turtles,

$$d(t)=|P(t)-Q(t)|.$$

At the beginning and at the end, both turtles are at the same point, so

$$d(0)=d(T)=0,$$

where $T$ is the moment when both have arrived at $B$.

Suppose during a small time interval $dt$, the first turtle moves by a vector $dP$ and the second by a vector $dQ$. Since $dQ$ is directed toward $P$, its projection onto the line $QP$ equals its whole length. The change of the distance $d(t)$ should then be

$$dd=(\text{projection of }dP\text{ on }QP)-|dQ|.$$

The projection of $dP$ onto any line does not exceed $|dP|$. Hence

$$dd \le |dP|-|dQ|.$$

Integrating this inequality from the beginning to the end gives

$$0=d(T)-d(0) \le L_1-L_2,$$

where $L_1$ and $L_2$ are the lengths traveled by the first and second turtles. Thus $L_2\le L_1$.

The delicate point is the formula for $dd$. One must justify carefully that the rate of change of the distance between two moving points equals the difference of the projections of their velocity vectors onto the line joining them.

Problem Understanding

We are given two turtles. The first moves from $A$ to $B$ along an arbitrary path with arbitrary speed. The second starts from $A$ later and always moves directly toward the current position of the first turtle, again with arbitrary speed. Eventually it also reaches $B$.

We must prove that the total distance traveled by the second turtle does not exceed the total distance traveled by the first turtle.

This is a Type B problem. The statement is already given and must be proved.

The core difficulty is relating the lengths of two completely arbitrary motions. The crucial observation is that the distance between the turtles begins and ends at $0$, while its rate of change is controlled by the projections of the two velocity vectors onto the line joining the turtles.

Proof Architecture

Let $P(t)$ and $Q(t)$ denote the positions of the first and second turtles at time $t$, and let $d(t)=|P(t)-Q(t)|$.

Lemma 1. At every moment when the velocities exist, the derivative of $d(t)$ equals the projection of the first turtle's velocity onto the line $QP$ minus the projection of the second turtle's velocity onto the same line. This is the standard formula for the derivative of the distance between two moving points.

Lemma 2. Because the second turtle moves exactly toward the first, its projection onto the line $QP$ equals the magnitude of its velocity. This follows from the definition of its motion.

Lemma 3. The projection of the first turtle's velocity onto the line $QP$ does not exceed the magnitude of that velocity. This is the elementary inequality between a vector and any of its projections.

From Lemmas 1-3 we obtain

$$d'(t)\le v_1(t)-v_2(t),$$

where $v_1(t)$ and $v_2(t)$ are the speeds of the first and second turtles.

Integrating over time and using $d(0)=d(T)=0$ yields

$$0\le L_1-L_2,$$

hence $L_2\le L_1$.

The lemma most likely to fail under insufficient scrutiny is Lemma 1, since it contains the whole geometric content of the argument.

Solution

Let $P(t)$ and $Q(t)$ be the positions of the first and second turtles at time $t$. Let

$$d(t)=|P(t)-Q(t)|$$

be the distance between them.

Denote by $v_1(t)$ and $v_2(t)$ the speeds of the first and second turtles. Let $T$ be a moment by which both turtles have arrived at $B$. Then

$$d(0)=0,\qquad d(T)=0.$$

Consider a moment $t$ at which both velocity vectors exist. Let

$$u(t)=\frac{P(t)-Q(t)}{|P(t)-Q(t)|}$$

be the unit vector directed from the second turtle toward the first. During a small time interval $dt$, the first turtle is displaced by

$$dP=\dot P(t),dt,$$

and the second by

$$dQ=\dot Q(t),dt.$$

The first-order change of the distance between the turtles is the projection of the relative displacement $dP-dQ$ onto the direction $u(t)$. Hence

$$d'(t)=u(t)\cdot\dot P(t)-u(t)\cdot\dot Q(t).$$

Since the second turtle always moves directly toward the first, its velocity vector is directed along $u(t)$. Therefore

$$u(t)\cdot\dot Q(t)=|\dot Q(t)|=v_2(t).$$

For the first turtle,

$$u(t)\cdot\dot P(t)\le |\dot P(t)|=v_1(t),$$

because the projection of a vector onto any direction does not exceed the vector's magnitude.

Substituting these relations into the expression for $d'(t)$ gives

$$d'(t)\le v_1(t)-v_2(t).$$

Integrating from $0$ to $T$, we obtain

$$d(T)-d(0) \le \int_0^T v_1(t),dt-\int_0^T v_2(t),dt.$$

The integrals are exactly the lengths of the paths traveled by the first and second turtles; denote them by $L_1$ and $L_2$. Since $d(0)=d(T)=0$, the left-hand side is $0$. Thus

$$0\le L_1-L_2.$$

Hence

$$L_2\le L_1.$$

This completes the proof.

∎

Verification of Key Steps

The first delicate step is the formula

$$d'(t)=u(t)\cdot\dot P(t)-u(t)\cdot\dot Q(t).$$

To rederive it, write

$$R(t)=P(t)-Q(t), \qquad d(t)=|R(t)|.$$

Then

$$d'(t) = \frac{R(t)\cdot R'(t)}{|R(t)|} = u(t)\cdot(\dot P(t)-\dot Q(t)),$$

which is exactly the claimed formula. No geometric intuition is needed.

The second delicate step is the equality

$$u(t)\cdot\dot Q(t)=v_2(t).$$

This uses the special condition of the problem. Since the second turtle moves toward the first, its velocity vector has the form

$$\dot Q(t)=v_2(t),u(t),$$

with $v_2(t)\ge0$. Taking the scalar product with $u(t)$ gives the equality.

The third delicate step is the passage from the differential inequality to the comparison of lengths. Integrating yields

$$d(T)-d(0)\le L_1-L_2.$$

Since both endpoints equal $0$, the left-hand side vanishes. If one forgot to use the fact that the turtles start and end together, the argument would not produce the desired inequality.

Alternative Approaches

One may avoid derivatives entirely. During a small time interval, let the first turtle travel a distance $ds_1$ and the second a distance $ds_2$. The increase of the distance between the turtles is at most the component of the first displacement along the line joining them, minus $ds_2$. Hence

$$dd\le ds_1-ds_2.$$

Summing these inequalities over a partition of the time interval and passing to the limit gives

$$d(T)-d(0)\le L_1-L_2.$$

The rest of the argument is unchanged.

The differential formulation is preferable because it expresses the geometric fact in a single exact identity for the derivative of the inter-turtle distance, after which the proof becomes a one-line integration.