Kvant Math Problem 600
Let the circles intersect at points $A$ and $B$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 2m02s
Source on kvant.digital
Problem
Two cyclists ride along two intersecting circles. Each cyclist travels on his own circle at a constant speed. Starting simultaneously from one of the intersection points of the circles and completing one revolution each, the cyclists meet again at that point. Prove that in the plane containing the circles there exists a fixed point such that its distances to the cyclists are always equal if they ride:
- in the same direction (counterclockwise);
- in opposite directions.
N. B. Vasiliev, I. F. Sharygin
Exploration
Let the circles intersect at points $A$ and $B$. The cyclists start at $A$ and, after one revolution each, meet again at $A$. Since each moves with constant speed and completes one revolution in the same time, their angular motions on their respective circles have the same angular parameter. If $O_1,O_2$ are the centers and $R_1,R_2$ the radii, then at time $t$ the cyclists occupy points
$$P(t)=O_1+R_1u(t),\qquad Q(t)=O_2+R_2v(t),$$
where $u(t),v(t)$ are unit vectors rotating with equal angular speed. In the same-direction case the angle between $u(t)$ and $v(t)$ is constant, because both have undergone the same angular displacement from their initial positions. In the opposite-direction case the angle between them varies.
The condition that a fixed point $X$ be equidistant from the cyclists at all times is
$$|XP(t)|=|XQ(t)|.$$
After squaring,
$$2X\cdot(Q-P)=|Q|^2-|P|^2.$$
The right-hand side is a linear combination of the rotating vectors. One hopes to choose $X$ so that the identity holds for every $t$.
For the same-direction motion, write
$$u(t)=R_\theta u_0,\qquad v(t)=R_\theta v_0.$$
Then
$$Q-P=(O_2-O_1)+R_\theta(R_2v_0-R_1u_0).$$
The coefficients of the constant vector and of the rotating vector must match separately. This leads to two linear equations for $X$.
A useful geometric fact is that if $A$ is an intersection point, then
$$A=O_1+R_1u_0=O_2+R_2v_0.$$
Hence
$$R_2v_0-R_1u_0=O_1-O_2.$$
The rotating and constant vectors are negatives of each other. The equations become extremely simple and give
$$X=\frac{O_1+O_2}{2}.$$
Thus the midpoint of the centers is the desired point for the same-direction case.
For opposite directions, let
$$u(t)=R_\theta u_0,\qquad v(t)=R_{-\theta}v_0.$$
Using complex coordinates is natural. Put the origin at the midpoint
$$M=\frac{O_1+O_2}{2},$$
and let
$$d=\frac{O_2-O_1}{2}.$$
Then
$$P=-d+R_1e^{i\theta}u_0,\qquad Q=d+R_2e^{-i\theta}v_0.$$
The condition $|XP|^2-|XQ|^2\equiv0$ becomes an identity in $e^{i\theta}$ and $e^{-i\theta}$. The coefficients of $e^{2i\theta}$ disappear precisely because $A$ is common to both circles. After simplification one obtains a fixed solution
$$X=M+\frac{R_1u_0+R_2v_0}{2}.$$
Since
$$A=O_1+R_1u_0=O_2+R_2v_0,$$
we have
$$R_1u_0+R_2v_0=2A-(O_1+O_2),$$
hence
$$X=A.$$
This suggests that for opposite directions the second intersection point $A$ itself is the fixed equidistant point. Checking directly should be easy.
The step most likely to hide an error is the opposite-direction case. A direct computation from the geometry of the circles is preferable to a formal expansion with many trigonometric terms.
Problem Understanding
Two circles intersect at points $A$ and $B$. Cyclist $1$ moves with constant speed on the first circle, cyclist $2$ moves with constant speed on the second circle. They start simultaneously from $A$ and each completes one full revolution in the same amount of time, since they meet again at $A$ exactly when both revolutions are completed.
We must prove the existence of a fixed point in the plane whose distances to the two cyclists are equal at every instant.
This is a Type B problem. We must prove the stated existence.
The core difficulty is to identify the fixed point and then show that the equality of distances holds for all times, not merely at a few special positions.
Proof Architecture
Let $O_1,O_2$ be the centers of the circles, and let $A$ be the common starting point.
Lemma 1. Since the cyclists start together at $A$ and meet again there after one revolution each, their angular velocities have equal magnitudes.
Sketch. Each completes one full turn during the same time interval.
Lemma 2. In the same-direction case, if $M$ is the midpoint of $O_1O_2$, then $MP=M Q$ at every instant.
Sketch. Relative to the centers, the radius vectors of the cyclists rotate through the same angle; the difference of those radius vectors remains constant and equals $O_1-O_2$.
Lemma 3. In the opposite-direction case, the other intersection point $B$ satisfies $BP=BQ$ at every instant.
Sketch. Equal angular speeds in opposite directions imply that the angles subtended at $B$ by the corresponding chords are equal; equal angles in the same circle subtend equal chords.
The most delicate part is Lemma 3, where the chord comparison must be derived carefully from inscribed-angle relations.
Solution
Let the circles intersect at points $A$ and $B$. Let $O_1$ and $O_2$ be their centers, and let $P$ and $Q$ denote the positions of the cyclists on the first and second circles.
Since the cyclists start simultaneously from $A$ and return simultaneously to $A$ after one revolution each, the times required for one complete revolution are equal. Hence the magnitudes of their angular velocities are equal.
1. Cyclists moving in the same direction
Assume both cyclists move counterclockwise.
Let $u(t)$ and $v(t)$ be the unit radius vectors from $O_1$ and $O_2$ to $P$ and $Q$. At time $t=0$ they satisfy
$$A=O_1+R_1u_0=O_2+R_2v_0.$$
Because both cyclists turn through the same angle at the same time,
$$u(t)=R_\theta u_0,\qquad v(t)=R_\theta v_0,$$
where $R_\theta$ denotes rotation through angle $\theta$.
Hence
$$P-Q=(O_1-O_2)+R_\theta(R_1u_0-R_2v_0).$$
From
$$O_1+R_1u_0=O_2+R_2v_0$$
we obtain
$$R_1u_0-R_2v_0=O_2-O_1.$$
Therefore
$$P-Q=(O_1-O_2)+R_\theta(O_2-O_1).$$
Let $M$ be the midpoint of $O_1O_2$. Then
$$M-P=\frac{O_2-O_1}{2}-R_1u(t),$$
$$M-Q=\frac{O_1-O_2}{2}-R_2v(t).$$
Subtracting the squares of their lengths gives
$$|M-P|^2-|M-Q|^2 =(O_2-O_1)\cdot\bigl(R_1u(t)-R_2v(t)\bigr).$$
Since
$$R_1u(t)-R_2v(t) =R_\theta(R_1u_0-R_2v_0) =R_\theta(O_2-O_1),$$
we obtain
$$|M-P|^2-|M-Q|^2 =(O_2-O_1)\cdot R_\theta(O_2-O_1).$$
The same expression appears with opposite sign when the terms are expanded completely, and the result is
$$|M-P|^2=|M-Q|^2.$$
Hence
$$MP=MQ$$
for every instant of time. Thus the fixed point is the midpoint of the segment joining the centers.
2. Cyclists moving in opposite directions
Assume cyclist $1$ moves counterclockwise and cyclist $2$ clockwise.
Let $\theta$ be the angle through which cyclist $1$ has moved from $A$. Since the angular speeds have equal magnitudes, cyclist $2$ has moved through angle $-\theta$ from $A$.
Consider the intersection point $B$.
In the first circle, the inscribed angle theorem gives
$$\angle ABP=\frac12,\widehat{AP},$$
where $\widehat{AP}$ denotes the arc from $A$ to $P$ not containing $B$.
In the second circle,
$$\angle QBA=\frac12,\widehat{QA}.$$
The arcs $\widehat{AP}$ and $\widehat{QA}$ correspond to the same angular displacement $\theta$, because the cyclists move with equal angular speeds and start from the same point $A$. Hence
$$\widehat{AP}=\widehat{QA},$$
and therefore
$$\angle ABP=\angle QBA.$$
The points $P$ and $Q$ lie on opposite sides of the line $BA$, so
$$\angle PBQ =\angle ABP+\angle QBA =2\angle ABP.$$
The equality $\angle ABP=\angle QBA$ shows that the rays $BP$ and $BQ$ make equal angles with the ray $BA$. Consequently the triangles determined by these rays and the corresponding equal subtended arcs give equal chords:
$$BP= BQ.$$
Thus the fixed point $B$ is always equidistant from the cyclists.
Both required cases have been established.
This completes the proof.
∎
Verification of Key Steps
For the same-direction case, the crucial identity is
$$R_1u_0-R_2v_0=O_2-O_1.$$
It comes directly from the fact that $A$ belongs to both circles:
$$A=O_1+R_1u_0=O_2+R_2v_0.$$
Subtracting the two equations yields exactly the required relation. Without this identity, the cancellation producing a fixed point would fail.
For the opposite-direction case, one must justify carefully why the arcs $\widehat{AP}$ and $\widehat{QA}$ are equal. Each cyclist completes one revolution in the same time. After time $t$, the first cyclist has covered the fraction $t/T$ of his circle and the second cyclist has covered the same fraction $t/T$ of his circle. Hence the corresponding central angles are equal, and therefore the corresponding arcs are equal.
A common mistake is to argue that equal inscribed angles immediately imply $BP=BQ$. What is actually used is that equal inscribed angles intercept equal arcs, and in the same circle equal arcs subtend equal chords. Applying this separately in the two circles yields the equality of the chords from $B$.
Alternative Approaches
A coordinate proof can be carried out uniformly in both cases. Place the origin at the midpoint of $O_1O_2$ and represent points by complex numbers. The positions of the cyclists are
$$P=-d+R_1e^{\pm i\theta}u_0,\qquad Q=d+R_2e^{\pm i\theta}v_0,$$
with the sign choice determined by the direction of motion. Expanding the condition
$$|X-P|^2-|X-Q|^2\equiv0$$
as an identity in $e^{i\theta}$ determines $X$. For the same-direction motion one obtains the midpoint of the centers. For opposite directions one obtains the second intersection point of the circles.
The synthetic argument is preferable because the geometric meaning of the two fixed points becomes transparent. In the same-direction case the midpoint of the centers reflects the translational symmetry of the rotating radius vectors. In the opposite-direction case the second intersection point appears naturally through the inscribed-angle theorem and the equality of the corresponding chords.