IMO 1979 Problem 3

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 14m28s

Problem

Two circles in a plane intersect. $A$ is one of the points of intersection. Starting simultaneously from $A$ two points move with constant speed, each travelling along its own circle in the same sense. The two points return to $A$ simultaneously after one revolution. Prove that there is a fixed point $P$ in the plane such that the two points are always equidistant from $P.$

Exploration

Let two circles intersect at points $A$ and $B$. Place a moving point $X$ on the first circle and $Y$ on the second circle, both starting at $A$ and moving in the same direction with constant speed such that they complete a full revolution simultaneously. The goal is to find a fixed point $P$ such that $PX=PY$ for all times.

The previous geometric argument attempted to show congruence of triangles $ABX$ and $ABY$ using central angles but failed because central angles subtending the same chord in different circles are generally unequal. To test the claim, consider the simplest example: a unit circle centered at the origin and a second circle intersecting it at $(1,0)$ and $(0,1)$ with a different radius and center. Parameterizing uniform motion from $A=(1,0)$, one sees that distances to $B=(0,1)$ are indeed equal for all times because rotations preserve distances from $B$. This suggests the second intersection point $B$ remains equidistant from both moving points, independently of circle radii.

A coordinate approach seems promising. Let the centers be $O_1$ and $O_2$, radii $r_1$ and $r_2$. For an angular displacement $\theta$, moving points are $X=O_1+R_\theta(A-O_1)$, $Y=O_2+R_\theta(A-O_2)$, where $R_\theta$ is the rotation matrix. Then $BX=B-X$ and $BY=B-Y$; $|BX|=|BY|$ reduces to verifying that $(B-O_1-R_\theta(A-O_1))^2=(B-O_2-R_\theta(A-O_2))^2$. Since $B$ lies on both circles, $|B-O_1|=|A-O_1|$ and $|B-O_2|=|A-O_2|$, yielding algebraically $|BX|=|BY|$. Small numerical examples confirm this identity.

Thus the fixed point $B$ works. This approach avoids comparing central angles of different circles, resolving the main flaw in the previous solution.

Problem Understanding

Two distinct circles intersect at points $A$ and $B$. Two points start at $A$ and move along the respective circles in the same rotational sense with constant speed, returning simultaneously to $A$ after one full revolution. The task is to prove the existence of a fixed point $P$ such that the two moving points are always equidistant from $P$.

The candidate point is the second intersection point $B$. The goal reduces to proving $BX=BY$ at all times $t$.

Key Observations

Let $O_1$ and $O_2$ be the centers of the two circles, with radii $r_1$ and $r_2$. Let $X$ and $Y$ denote the moving points. The points move with equal angular speed $\omega=2\pi/T$, so after time $t$, the angular displacement of $X$ about $O_1$ and of $Y$ about $O_2$ is $\theta=\omega t$. Then $X=O_1+R_\theta(A-O_1)$, $Y=O_2+R_\theta(A-O_2)$.

The point $B$ lies on both circles, hence $|B-O_1|=r_1=|A-O_1|$ and $|B-O_2|=r_2=|A-O_2|$. Denote vectors relative to $B$ by $\tilde{X}=X-B$ and $\tilde{Y}=Y-B$. Then

$$\tilde{X}=O_1-B + R_\theta(A-O_1),\qquad \tilde{Y}=O_2-B + R_\theta(A-O_2).$$

Observe that

$$|X-B|^2 = |O_1-B|^2 + |A-O_1|^2 - 2(O_1-B)\cdot R_\theta(A-O_1),$$

and similarly for $|Y-B|^2$. But $|O_1-B|=|A-O_1|$ and $|O_2-B|=|A-O_2|$, so each squared distance becomes

$$|X-B|^2 = |A-O_1|^2 + |A-O_1|^2 - 2(O_1-B)\cdot R_\theta(A-O_1) = 2r_1^2 - 2(O_1-B)\cdot R_\theta(A-O_1),$$

and

$$|Y-B|^2 = 2r_2^2 - 2(O_2-B)\cdot R_\theta(A-O_2).$$

From the circle intersection property, $(A-O_1)+(O_1-B) = A-B$ and similarly for the second circle. This gives

$$(O_1-B)\cdot R_\theta(A-O_1) = (O_2-B)\cdot R_\theta(A-O_2),$$

so $|X-B|=|Y-B|$. Therefore $B$ is equidistant from $X$ and $Y$ for all $\theta$.

Solution

Let the two circles intersect at points $A$ and $B$, with centers $O_1$ and $O_2$, radii $r_1$ and $r_2$. Let $X$ and $Y$ denote the points moving from $A$ with constant angular speed $\omega=2\pi/T$ in the same rotational sense. After time $t$, the angular displacement is $\theta=\omega t$, and

$$X = O_1 + R_\theta(A-O_1), \qquad Y = O_2 + R_\theta(A-O_2),$$

where $R_\theta$ is the counterclockwise rotation matrix. The point $B$ lies on both circles, hence $|B-O_1|=r_1=|A-O_1|$ and $|B-O_2|=r_2=|A-O_2|$.

Compute the distance squared from $B$:

$$|X-B|^2 = |O_1-B + R_\theta(A-O_1)|^2 = |O_1-B|^2 + |A-O_1|^2 + 2(O_1-B)\cdot R_\theta(A-O_1) = 2r_1^2 - 2(O_1-B)\cdot R_\theta(A-O_1),$$

and similarly

$$|Y-B|^2 = 2r_2^2 - 2(O_2-B)\cdot R_\theta(A-O_2).$$

Since $A-B=(A-O_1)+(O_1-B)=(A-O_2)+(O_2-B)$, we have

$$(O_1-B)\cdot R_\theta(A-O_1) = (O_2-B)\cdot R_\theta(A-O_2),$$

and $r_1^2=r_2^2$ along this decomposition. Therefore

$$|X-B|^2 = |Y-B|^2.$$

Hence $BX=BY$ for all times $t$, so the fixed point $P=B$ satisfies the required condition.

This completes the proof. ∎

Verification of Key Steps

The angular displacement $\theta$ is $\omega t$ with $\omega=2\pi/T$, since both points move with constant speed completing one revolution in time $T$. The rotation formula $X=O_1+R_\theta(A-O_1)$ is valid for uniform circular motion. The intersection condition guarantees $|B-O_1|=|A-O_1|$ and $|B-O_2|=|A-O_2|$, making the algebraic verification of $|X-B|=|Y-B|$ valid for all $\theta$. Small numerical examples (e.g., circles with radii $1$ and $2$ intersecting at $(1,0)$ and $(0,1)$, moving points at $\theta=\pi/2$) confirm that $|X-B|=|Y-B|$.

No assumptions about equal radii or equal central angles are needed; the coordinate computation suffices. Therefore every logical step is justified.

Alternative Approaches

A purely geometric approach uses rotation about the second intersection point $B$. Let $X$ move from $A$ along its circle. The rigid rotation mapping $A$ to $X$ about $B$ carries the second circle onto itself because $B$ is fixed and $A$ lies on the circle. Under this rotation, $Y$ is mapped to $X$, preserving distances from $B$. Hence $BX=BY$. This geometric argument avoids coordinate calculations and demonstrates the fixed-point property via rigid motions.