Irodov 1.12: A Shrinking Triangle That Never Turns Ninety

Problem Irodov 1.12

Three points sit at the vertices of an equilateral triangle of side $a$ . Starting at the same instant, all three move with the same constant speed $v$ : point 1 always heads straight at the current position of point 2, point 2 always heads straight at the current position of point 3, and point 3 always heads straight at the current position of point 1. Find the time after which the three points meet.

The starting configuration: an equilateral triangle of side a with a point at each vertex. Each velocity arrow has length v and runs along a side toward the point it chases, so the three circulate in one sense. The dot is the centroid. — The starting configuration: an equilateral triangle of side $a$ with a point at each vertex. Each velocity arrow has length $v$ and runs along a side toward the point it chases, so the three circulate in one sense. The dot is the centroid.

Given

side $a$
speed $v$

Find

the time

t

for the three points to converge

Solution

Step 1The configuration stays an equilateral triangle

By the threefold symmetry of the setup, the triangle formed by the three points remains equilateral at every instant, shrinking in side length $r(t)$ (with $r(0)=a$ ) while rotating about the fixed centroid.

Step 2Closing rate of one side

Consider the side from point 1 to point 2, of instantaneous length $r$ , with unit vector $\hat{\vect n}$ pointing from point 1 to point 2. Point 1’s velocity is exactly $v\hat{\vect n}$ , since it is aimed straight at point 2. Point 2’s velocity is $v$ directed along the side toward point 3; the triangle’s interior angle at vertex 2, between its incoming side (from point 1) and outgoing side (to point 3), is $60^\circ$ , but $\hat{\vect n}$ points the other way along the incoming side, from point 1 toward point 2, so the angle it makes with the outgoing side toward point 3 is the supplement, $120^\circ$ . The rate of change of $r$ is the component of the relative velocity $\vect v_2-\vect v_1$ along $\hat{\vect n}$ :

\frac{dr}{dt} = (\vect v_2-\vect v_1)\cdot\hat{\vect n} = v\cos120^\circ - v

Evaluating the cosine,

\frac{dr}{dt} = -\frac{v}{2}-v .

\dfrac{dr}{dt} = -\dfrac{3v}{2}

(1)

Step 3Integrate the constant closing rate

The rate (1) does not depend on $r$ or $t$ , so it integrates directly; the points converge when $r$ reaches zero.

\int_a^0 dr = \int_0^t \left(-\frac{3v}{2}\right)dt'

Both sides are elementary: the left is the total change in $r$ , the right a constant times the elapsed time.

0 - a = -\frac{3v}{2}\,(t-0)

Cancelling the common minus sign leaves the initial side written as the closing rate times the elapsed time.

a = \frac{3v}{2}\,t

t = \dfrac{2a}{3v}

(2)

Check

Dimensionally, $[a]/[v]=\text{m}/(\text{m/s})=\text{s}$ , a time, as required. As an independent geometric check, each point travels a total path length $v t=v\cdot 2a/(3v)=2a/3$ before meeting at the centroid; the straight line distance from a vertex to the centroid is $a/\sqrt3\approx0.577a$ , and $2a/3\approx0.667a$ exceeds it, consistent with each point following a curved (spiral) path rather than a straight one.

Remark

The closing speed $3v/2$ is faster than a head-on approach at speed $v$ even though each point’s own velocity has magnitude only $v$ : the target itself is fleeing to the side, but the fleeing target’s velocity still has a component toward the chaser (here $v\cos60^\circ=v/2$ , the same $120^\circ$ read from the other point), and that component adds to, rather than subtracts from, the closing rate. Each point’s actual path is a logarithmic spiral into the centroid, even though the algebra above never had to construct it.