Baltic Way 1997 · Problem 12

Geometry

Two circles $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ intersect in $P$ and $Q$ . A line through $P$ intersects $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ again in $A$ and $B$ , respectively, and $X$ is the midpoint of $A B$ . The line through $Q$ and $X$ intersects $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ again in $Y$ and $Z$ , respectively. Prove that $X$ is the midpoint of $Y Z$ .

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Topics

Cyclic geometry · Angles and distances

Solutions

Solution

Solution:

Depending on the radii of the circles, the distance between their centres and the choice of the line through $P$ we have several possible arrangements of the points $A$ , $B$ , $P$ and $Y$ , $Z$ , $Q$ . We shall show that in each case the triangles $A X Y$ and $B X Z$ are congruent, whence $|Y X| = |X Z|$ .

a. Point $P$ lies within segment $AB$ and point $Q$ lies within segment $YZ$ (see Figure 3). Then

\angle A Y X = \angle A Y Q = \pi - \angle A P Q = \angle B P Q = \angle B Z Q = \angle B Z X.

Since also $\angle A X Y = \angle B X Z$ and $|A X| = |X B|$ , triangles $A X Y$ and $B X Z$ are congruent.

b. Point $P$ lies outside of segment $AB$ and point $Q$ lies within segment $YZ$ (see Figure 4). Then

\angle A Y X = \angle A Y Q = \angle A P Q = \angle B P Q = \angle B Z Q = \angle B Z X.

c. Point $P$ lies outside of segment $AB$ and point $Q$ lies outside of segment $YZ$ (see Figure 5). Then

\angle A Y X = \pi - \angle A Y Q = \angle A P Q = \angle B P Q = \angle B Z Q = \angle B Z X.

d. Point $P$ lies within segment $AB$ and point $Q$ lies outside of segment $YZ$ . This case is similar to (b): exchange the roles of points $P$ and $Q$ , $A$ and $Y$ , $B$ and $Z$ .