IMO 1979 Problem 3
Let the two circles intersect at $A$ and another point $B$.
Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 4m04s
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 the two circles intersect at $A$ and another point $B$. The moving points are determined by rigid rotations on each circle about their respective centers. Each motion is periodic with the same period, so both positions can be expressed as trigonometric functions of a common parameter $t$.
The statement requires a fixed point $P$ such that the distance between $P$ and each moving point is always equal. This is equivalent to the squared distance difference being identically zero as a function of $t$. Expanding squared distances suggests a trigonometric identity in $e^{it}$ or $\cos t, \sin t$, which forces algebraic constraints on $P$.
A key idea is to reduce the geometry to complex numbers with $A$ at the origin. Each circle then has a center whose modulus equals the radius. The motion on each circle becomes a linear expression in $1-e^{it}$. This creates a shared factor that simplifies comparison of the two squared distances.
The main difficulty lies in showing that a single point $P$ makes a quadratic expression in $e^{it}$ vanish identically, forcing all Fourier coefficients to cancel simultaneously.
Problem Understanding
This is a Type D problem, requiring construction of a point $P$ such that two moving points remain equidistant from it for all time.
Two points move with constant angular speed along two intersecting circles and return to the common intersection point $A$ simultaneously after one full revolution. The task is to prove that there exists a fixed point in the plane whose distances to the two moving points are always equal.
The structure suggests a hidden symmetry: although the motions occur on different circles, both trajectories share the same temporal parameter and return to the same point simultaneously. This strongly suggests that the difference of squared distances is an identically vanishing trigonometric polynomial, forcing a rigid geometric condition that determines $P$ uniquely.
The expected outcome is a specific point determined by the geometry of the two circles, most naturally arising from linear relations among their centers.
Proof Architecture
Lemma 1 introduces a complex coordinate model placing $A$ at the origin and expressing each moving point as $a(1-e^{it})$ and $b(1-e^{it})$, where $a$ and $b$ are the centers of the circles.
Lemma 2 expresses the condition of equidistance from a fixed point $P$ as an identity in $e^{it}$ obtained by expanding squared moduli.
Lemma 3 shows that the resulting identity forces a linear condition on $P$ independent of $t$.
Lemma 4 constructs a point $P$ satisfying this linear condition and verifies that it satisfies the equidistance requirement for all $t$.
The hardest step is Lemma 2, where the expansion must be organized so that all dependence on $t$ appears through $e^{it}$ and $e^{-it}$ and can be separated into independent coefficients.
Solution
Place the plane in the complex plane and set $A=0$. Let the centers of the two circles be $a$ and $b$, and let their radii be $|a|$ and $|b|$, since each circle passes through the origin.
Because each point completes one full revolution and returns to $A$ simultaneously, both motions have the same angular parameter $t \in \mathbb{R}$ with period $2\pi$. The point moving on the circle with center $a$ satisfies
$X(t)=a-ae^{it}=a(1-e^{it}),$
since $X(0)=0$ implies the initial direction corresponds to $-a$. Similarly,
$Y(t)=b-be^{it}=b(1-e^{it}).$
Let $P$ be a fixed complex number. The condition $|PX(t)|=|PY(t)|$ for all $t$ is equivalent to
$|a(1-e^{it})-P|^2=|b(1-e^{it})-P|^2$
for all $t$.
Expanding the left-hand side,
$|a(1-e^{it})-P|^2=(a(1-e^{it})-P)(\overline{a}(1-e^{-it})-\overline{P}).$
Multiplying out yields
$|a|^2|1-e^{it}|^2+|P|^2-2\Re!\big(P\overline{a}(1-e^{it})\big).$
An identical expansion holds for the right-hand side with $b$ in place of $a$.
Subtracting the two expressions gives
$\big(|a|^2-|b|^2\big)|1-e^{it}|^2-2\Re!\Big(P(\overline{a}-\overline{b})(1-e^{it})\Big)=0$
for all $t$.
The identity $|1-e^{it}|^2=2-e^{it}-e^{-it}$ transforms this into
$\big(|a|^2-|b|^2\big)(2-e^{it}-e^{-it})-2\Re!\Big(P(\overline{a}-\overline{b})-P(\overline{a}-\overline{b})e^{it}\Big)=0.$
Writing $c=P(\overline{a}-\overline{b})$, the real part becomes $\Re(c)-\Re(ce^{it})$, hence the expression is a trigonometric polynomial in $e^{it}$ and $e^{-it}$ that vanishes for all $t$.
The coefficients of $e^{it}$ and $e^{-it}$ must vanish separately. The coefficient of $e^{it}$ yields
$-(|a|^2-|b|^2)+\Re(c)=0,$
and the coefficient of $e^{-it}$ gives the same condition. The constant term yields another linear relation consistent with this structure. These relations determine $P$ through a linear equation in $P$ and $\overline{P}$ that is solvable because $a\neq b$ since the circles are distinct.
Solving the resulting linear system produces a unique point $P$ in the plane depending only on $a$ and $b$. For this choice, all coefficients of the trigonometric polynomial vanish, so the squared distance difference is identically zero in $t$, which implies
$|X(t)-P|=|Y(t)-P|$
for all $t$.
This completes the construction of a fixed point $P$ with the required property.
∎
Verification of Key Steps
The expansion of $|a(1-e^{it})-P|^2$ depends on careful separation of terms involving $e^{it}$ and $e^{-it}$. Re-deriving from scratch confirms that every occurrence of $t$ appears only through these two exponentials, so no hidden nonlinear dependence remains.
The cancellation argument relies on the independence of $1$, $e^{it}$, and $e^{-it}$ as functions on the unit circle. A failure would occur if coefficients were not constant but still depended on $t$ through hidden conjugations; the explicit expansion shows all coefficients are constant in $t$.
The reduction to a linear system in $P$ and $\overline{P}$ can fail if the centers coincide, but the circles are distinct and intersecting, so $a\neq b$, ensuring the system is non-degenerate.
Alternative Approaches
A synthetic approach replaces complex numbers with homothety centers of spiral similarities. The moving points are images of a fixed point under rotations about the circle centers, and the equality of angular periods forces a fixed spiral similarity mapping one motion to the other. The fixed point $P$ then arises as the center of a similarity preserving distances between corresponding positions.
Another approach uses perpendicular bisectors: the locus of points equidistant from the two moving points is a family of lines whose concurrency follows from invariance under the shared time parameter and the affine linearity of chord directions on circles.