1979 IMO Problems/Problem 3


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 $P.$

Solution 1

Let $O_1, O_2$ be the centers of the 2 circles and let the points B, C move on the circles $(O_1), (O_2)$, recpectively. Let A' be the other intersection of the 2 circles different from the point A. Since the angles $\angle BO_1A = \angle CO_2A$ are equal, the isosceles triangles $\triangle AO_1B \sim \triangle AO_2C$ are similar. Consequently, the angles $\angle BAO_1 = \angle CAO_2$ are equal and the angles

$\angle BAC = BAO_1 + \angle O_1AO_2 + \angle O_2AC =$

$= BAO_1 + \angle O_1AO_2 - \angle CAO_2 = \angle O_1AO_2$

are also equal. The angles $\angle A'BA = \frac{\angle A'O_1A}{2} = \angle O_2O_1A$ are equal and the angles $\angle A'CA = \frac{\angle A'O_2A}{2} = \angle O_1O_2A$ are also equal. As a result, the point A' is on the side BC of the triangle $\triangle ABC$, which is always similar to the fixed triangle $\triangle O_1AO_2$.

Lemma (Yaglom's Geometric Transformations II, pp 68-69). Assume that a figure F' (the triangle $\triangle ABC$) moves in such a way that it remains at all times similar to a fixed figure F (the triangle $\triangle O_1AO_2$) and in addition, some point of the figure F' does not move at all (the point A). If a point B of the figure F' describes a curve $\Gamma$ (the circle $(O_1)$), then any other point M of the figure F' describes a curve $\Gamma'$ similar to the curve $\Gamma$.

Consequently, the midpoint M of the segment BC decribes a circle (Q). When the points B, C, become identical with the point A, the midpoint M of BC also becomes identical with A. Hence, the circle (Q) passes through the point A. The line BC always passes through A'. Just after the start of motion counter-clockwise, the points B, C, A' follow on the line BC in this order. The points B, C both pass through the point A' one after another and just before the end of one period of motion, the points A', B, C follow on the line BC In this order. From the continuity principle, the midpoint M of BC also passes through the point A' and consequently, the circle (Q), the locus of the midpoints M, also passes through the point A'. Thus the circle (Q) belongs to the same pencil with the circles $(O_1), (O_2)$ and it is centered on their center line $O_1O_2$. Let p be the perpendicular bisector of the segment BC intersectong the circle (Q) at a point P different from the point M. The point A' and the circle (Q) passing through this point are fixed and the angle $\angle A'MP = 90^\circ$ is right. Therefore, A'P is a fixed diameter of the circle (Q), which means that the point P is also fixed. Q.E.D.

When the points B, C are both diametrally opposite to the point A on their respective circles, the segment $O_1O_2$ is the midline of the triangle $\triangle ABC$, i.e., $BC \parallel O_1O_2$, $BC \perp AA'$. The cyclic quadrilateral AA'MP inscribed in the circle (Q) is then a rectangle. Its circumcenter Q is the intersection of its diagonals AM, A'P intersecting at their midpoint and the midpoint of AM is identical with the midpoint of $O_1O_2$. Thus the fixed point P is the reflection of the intersection A' of the circles $(O_1), (O_2)$ in the midpoint Q of the segment $O_1O_2$.

This solution was posted and copyrighted by yetti. The original thread for this problem can be found here: [1]

Solution 2

Let $B$ and $C$ be the antipodes of $A$ on the two circles and let $D$ be the foot of the altitude from $A$, which is the other intersection point of the two circles. Also, let $M$ be the midpoint of $\overline{BC}$, and construct rectangle $ADMZ$. Our claim is that $Z$ is the fixed point. We let $X$ and $Y$ be the two points; by the condition the angles are the same. So we have a spiral similarity \[\triangle AXY \sim \triangle ABC.\] [asy] size(10cm); pair A = dir(110); pair B = dir(210); pair C = dir(330); pair D = foot(A, B, C); pair M = midpoint(B--C); pair Z = A+M-D; pair X = midpoint(A--B)+dir(200)*abs(A-B)/2; pair Y = -D+2*foot(midpoint(A--C), X, D); draw(circumcircle(A, B, D), blue); draw(circumcircle(A, C, D), blue); draw(circumcircle(A, D, M), dashed+deepgreen); pair N = midpoint(X--Y); filldraw(A--X--Y--cycle, invisible, purple); filldraw(A--B--C--cycle, invisible, blue); draw(A--D, blue); draw(A--Z--M, deepgreen); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$D$", D, dir(D)); dot("$M$", M, dir(M)); dot("$Z$", Z, dir(Z)); dot("$X$", X, dir(X)); dot("$Y$", Y, dir(Y)); dot("$N$", N, dir(N)); /* TSQ Source: A = dir 110 B = dir 210 C = dir 330 D = foot A B C M = midpoint B--C Z = A+M-D X = midpoint(A--B)+dir(200)*abs(A-B)/2 Y = -D+2*foot midpoint A--C X D circumcircle A B D blue circumcircle A C D blue circumcircle A D M dashed deepgreen N = midpoint X--Y A--X--Y--cycle 0.1 red / purple A--B--C--cycle 0.1 yellow / blue A--D blue A--Z--M deepgreen */ [/asy] Now let $N$ be the midpoint of $\overline{XY}$. By spiral similarity, since $N$ maps to $M$, it follows $M$, $N$, $A$, and $D$ are cyclic too. So actually $N$ lies on the circumcircle of rectangle $ADMZ$, meaning $\overline{ZN} \perp \overline{XY}$, hence $ZX = ZY$ as needed.

Remark: The special point $Z$ can be identified by selecting the special case $X \to A$, $Y \to A$ and $X=B$, $Y=C$.

This solution was posted and copyrighted by v_Enhance. The original thread for this problem can be found here: [2]

Solution 3

Let the points on the circles be $E$ and $F$. Then, let $B$ the the other intersection of the two circles, and $C$ and $D$ be the centers of the circles containing $E$ and $F$, respectively. WLOG, let us name $E$ and $F$ such that $CA>DA$. Let $O$ be the midpoint of $CD$, and the antipode of $B$ with respect to the circle centered at $O$ with radius $OA$ be $P$. Let $M$ be where this such circle hits $EF$. We first note that\[\angle ADF = 2\angle ABF = 2 \angle APM = \angle AOM\]and also\[2 \angle APM = 360^{\circ} - 2\angle MBA = 360^{\circ} - 2\angle EBA = \angle ACE\]meaning that since $AD=DF$, $AO=OM$, and $AC=CE$, then SAS similarity gets us that $\triangle ADF \sim \triangle AOM \sim \triangle ACE$. This means we have $\frac{AO}{AM} = \frac{AC}{AE}$ so $\frac{AO}{AC} = \frac{AM}{AE}$ and because $\angle MAE = \angle OAC$, then $\triangle AOC \sim \triangle AME$ and similarly, we also have $\triangle AOD \sim \triangle AMF$. This means that\[\frac{EM}{MA} = \frac{OC}{OA} = \frac{DO}{OA} = \frac{FM}{MA} \to EM=FM\]meaning that $M$ is the midpoint of $EF$. Now, since $P$ is the antipode of $B$, then $\angle BMP = 90^{\circ} = \angle EMP = \angle FMP$ so $EP=FP$ for all $E$ and $F$ for some fixed $P$ and we are done.

This solution was posted and copyrighted by kevinmathz. The original thread for this problem can be found here: [3]

See Also

1979 IMO (Problems) • Resources
Preceded by
Problem 2
1 2 3 4 5 6 Followed by
Problem 4
All IMO Problems and Solutions