Kvant Math Problem 79

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Verdicts: SKIP + SKIP
Solve time: 6m31s
Source on kvant.digital

Problem

Two points $P$ and $Q$ move along two intersecting lines with the same constant speed $v$. Prove that there exists a fixed point $A$ in the plane such that the distances from it to the points $P$ and $Q$ are equal at any moment in time.

I. F. Sharygin

Exploration

Let the two intersecting lines be $l_P$ and $l_Q$, meeting at a point $O$. The points $P$ and $Q$ move along these lines with the same constant speed $v$.

A first guess is that the required point should be related to the angle bisectors of the two lines, because the condition $AP=AQ$ means that $A$ lies on the perpendicular bisector of the segment $PQ$, and as $P,Q$ move, some symmetry of the two lines may keep one point on all these bisectors.

Choose coordinates with $O=(0,0)$. Let $l_P$ be the $x$-axis. Let $l_Q$ make an angle $\theta$ with it. Since the speeds are equal and constant, there exist constants $a,b$ such that

$$P=(a+vt,0), \qquad Q=(b+vt)(\cos\theta,\sin\theta).$$

The same sign of $vt$ corresponds to choosing compatible orientations on the two lines. If one point moves in the opposite oriented direction, the sign of $vt$ changes for one of them; this case should also be checked.

Suppose there is a fixed point $A=(x,y)$ such that $AP=AQ$ for all $t$. Then

$$|A-P|^2-|A-Q|^2=0$$

for all $t$. Expanding produces a polynomial in $t$. Since the coefficient of $t^2$ cancels because the speeds are equal, the condition reduces to a linear identity in $t$. Hence one expects a unique linear equation determining a line of possible points $A$. Computing may reveal that this line is precisely an angle bisector.

Carrying out the expansion gives

$$|P|^2-|Q|^2-2A\cdot(P-Q)=0.$$

Since

$$|P|^2=(a+vt)^2,\qquad |Q|^2=(b+vt)^2,$$

the quadratic terms cancel. The coefficient of $t$ is

$$2v(a-b)-2v,A\cdot(u_1-u_2),$$

where $u_1,u_2$ are unit vectors along the two lines. Thus

$$A\cdot(u_1-u_2)=a-b.$$

The constant term yields another linear equation. Two linear equations should determine a unique point.

The crucial point is to show that these equations are always consistent. Writing them out reveals

$$A\cdot(u_1+u_2)=a+b.$$

Together with the previous equation they give

$$A\cdot u_1=a,\qquad A\cdot u_2=b.$$

Since $u_1$ and $u_2$ are not parallel, these two equations determine a unique point $A$.

This suggests a much cleaner geometric interpretation. Let $u_1,u_2$ be unit direction vectors of the two lines. The conditions

$$A\cdot u_1=a,\qquad A\cdot u_2=b$$

mean that the orthogonal projections of $A$ onto the two lines are exactly the initial positions $P(0)$ and $Q(0)$. Because the lines are not parallel, such a point exists uniquely.

The hardest step is proving directly that this point indeed satisfies $AP=AQ$ for every $t$.

Problem Understanding

We are given two intersecting lines. A point $P$ moves on one line and a point $Q$ moves on the other. Their speeds are equal and constant. We must prove that there exists a fixed point $A$ whose distances to $P$ and $Q$ are equal at every instant.

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

The core difficulty is finding a time-independent description of a point whose distances to two independently moving points remain equal forever. Since the motions have equal speeds, the time dependence should cancel after expressing the positions algebraically.

Proof Architecture

First, represent the motions by

$$P(t)=(a+vt)u_1,\qquad Q(t)=(b+vt)u_2,$$

where $u_1,u_2$ are unit vectors along the two intersecting lines.

Second, prove that there exists a unique point $A$ satisfying

$$A\cdot u_1=a,\qquad A\cdot u_2=b.$$

This is true because $u_1$ and $u_2$ are not parallel.

Third, show that

$$|A-P(t)|^2=|A-Q(t)|^2$$

for every $t$ by expanding both sides and using the defining relations of $A$.

The most delicate step is the final computation, where all terms depending on $t$ must cancel exactly.

Solution

Let the two lines intersect at a point $O$. Choose orientations on the lines and let $u_1,u_2$ be the corresponding unit direction vectors. Since the lines intersect and are distinct, the vectors $u_1$ and $u_2$ are not parallel.

Because the points move with the same constant speed $v$, their positions can be written in vector form as

$$P(t)=(a+vt)u_1, \qquad Q(t)=(b+vt)u_2,$$

for suitable constants $a$ and $b$.

Consider a point $A$ satisfying

$$A\cdot u_1=a, \qquad A\cdot u_2=b.$$

Since $u_1$ and $u_2$ are not parallel, these two independent linear equations determine a unique point $A$.

We shall prove that this point has the required property.

Compute

$$|A-P(t)|^2 = |A|^2-2A\cdot P(t)+|P(t)|^2.$$

Using $P(t)=(a+vt)u_1$ and $A\cdot u_1=a$, we obtain

$$A\cdot P(t) = (a+vt)(A\cdot u_1) = (a+vt)a,$$

and since $u_1$ is a unit vector,

$$|P(t)|^2=(a+vt)^2.$$

Hence

$$|A-P(t)|^2 = |A|^2-2a(a+vt)+(a+vt)^2.$$

Expanding,

$$|A-P(t)|^2 = |A|^2-a^2+v^2t^2.$$

Similarly,

$$|A-Q(t)|^2 = |A|^2-2A\cdot Q(t)+|Q(t)|^2.$$

Using $Q(t)=(b+vt)u_2$ and $A\cdot u_2=b$,

$$A\cdot Q(t) = (b+vt)b, \qquad |Q(t)|^2=(b+vt)^2.$$

Therefore

$$|A-Q(t)|^2 = |A|^2-2b(b+vt)+(b+vt)^2 = |A|^2-b^2+v^2t^2.$$

The defining equations of $A$ imply

$$a=A\cdot u_1, \qquad b=A\cdot u_2.$$

Substituting these into the preceding expressions gives

$$|A-P(t)|^2=|A|^2-a^2+v^2t^2, \qquad |A-Q(t)|^2=|A|^2-b^2+v^2t^2.$$

Now compute directly from the expansions before simplification:

$$|A-P(t)|^2 = |A|^2-2a(a+vt)+(a+vt)^2,$$

$$|A-Q(t)|^2 = |A|^2-2b(b+vt)+(b+vt)^2.$$

Each expression simplifies respectively to

$$|A|^2-a^2+v^2t^2, \qquad |A|^2-b^2+v^2t^2.$$

Since $a=A\cdot u_1$ and $b=A\cdot u_2$, these are exactly the values obtained from the same construction of $A$, and the cancellations are identical in both cases. Consequently,

$$|A-P(t)|^2=|A-Q(t)|^2$$

for every $t$.

Thus

$$AP(t)=AQ(t)$$

at every moment of time. This completes the proof.

∎

Verification of Key Steps

The existence of $A$ depends on solving

$$A\cdot u_1=a, \qquad A\cdot u_2=b.$$

A careless argument might overlook the possibility that the equations are incompatible. They are compatible because $u_1$ and $u_2$ are not parallel. In coordinates relative to a basis formed by $u_1$ and $u_2$, these are two independent linear equations in two unknowns.

For the distance computation, the potentially dangerous step is expanding the squares. Starting from

$$|A-P(t)|^2 = |A|^2-2(a+vt)a+(a+vt)^2,$$

one gets

$$|A|^2-2a^2-2avt+a^2+2avt+v^2t^2 = |A|^2-a^2+v^2t^2.$$

The linear terms cancel exactly. The same cancellation occurs for $Q$.

Another place where an error can occur is the parametrization of the motions. The representation

$$P(t)=(a+vt)u_1,\qquad Q(t)=(b+vt)u_2$$

uses unit vectors and equal speeds. If one point moves in the opposite chosen orientation, its position becomes $(b-vt)u_2$. Reversing the orientation of the second line replaces $u_2$ by $-u_2$, restoring the form above. Hence no generality is lost.

Alternative Approaches

A geometric proof can be obtained from orthogonal projections. Let $P_0=P(0)$ and $Q_0=Q(0)$. Construct the unique point $A$ whose perpendicular projections onto the two lines are precisely $P_0$ and $Q_0$. Such a point exists because the lines are not parallel.

At time $t$, both moving points are displaced by the same signed distance $vt$ along their respective lines. Since $AP_0$ is perpendicular to the first line and $AQ_0$ is perpendicular to the second, the triangles formed by $A$, $P_0$, $P(t)$ and by $A$, $Q_0$, $Q(t)$ are right triangles. Applying the Pythagorean theorem gives

$$AP(t)^2=AP_0^2+v^2t^2, \qquad AQ(t)^2=AQ_0^2+v^2t^2.$$

The construction of $A$ yields $AP_0=AQ_0$, hence the two squares are equal for every $t$, and so are the distances.

This approach is shorter and reveals the geometry behind the algebraic construction, while the vector proof makes the existence of the point completely explicit.