Kvant Math Problem 794

Let the first circle be $\omega_1$ and the second circle be $\omega_2$.

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Verified: yes
Verdicts: PASS + PASS
Solve time: 24m15s
Source on kvant.digital

Problem

Two circles intersect at the points $A$ and $B$. Through the point $K$ on the first circle, the lines $KA$ and $KB$ are drawn; they intersect the second circle at the points $P$ and $Q$, respectively (Fig. 2). Prove that the chord $PQ$ of the second circle is perpendicular to the diameter $KM$ of the first circle.

Figure 2

Alla Ivchenko, 9th-grade student (Mohyliv-Podilskyi, School No. 1)

Exploration

Let the first circle be $\omega_1$ and the second circle be $\omega_2$. The points $A$ and $B$ are their common points. The point $K$ lies on $\omega_1$. The line $KA$ meets $\omega_2$ again at $P$, and the line $KB$ meets $\omega_2$ again at $Q$.

The statement to prove is that the chord $PQ$ of $\omega_2$ is perpendicular to the diameter through $K$ of $\omega_1$.

A natural idea is to compare angles on the two circles. Since $A,B,K$ lie on $\omega_1$, the angle $\angle AKB$ subtends the chord $AB$. Since $A,P,B,Q$ lie on $\omega_2$, the same chord $AB$ determines several equal inscribed angles there.

Compute the angle between $PQ$ and one of the lines $PA$ or $PB$. Since $P,A,K$ are collinear and $Q,B,K$ are collinear, angle relations on $\omega_2$ may convert into angles involving $KA$ and $KB$.

Let

$$\alpha=\angle AKB.$$

On $\omega_2$,

$$\angle APQ=\angle ABQ.$$

To avoid ambiguity arising from opposite rays, all subsequent angle equalities are interpreted as directed angles modulo $180^\circ$, equivalently as angles between lines. Since $B,Q,K$ are collinear, the lines $BQ$ and $BK$ coincide. Hence

$$\angle(AB,BQ)=\angle(AB,BK).$$

In the usual notation for directed angles between lines, this is

$$\angle ABQ=\angle ABK.$$

Therefore

$$\angle(AP,PQ)=\angle ABK.$$

Since $AP$ is the same line as $AK$, the angle between $AK$ and $PQ$ equals $\angle ABK$.

Now compare this with the tangent at $K$ to $\omega_1$. By the tangent-chord theorem, the angle between the tangent at $K$ and $KA$ is also $\angle ABK$. Hence $PQ$ is parallel to the tangent at $K$.

Once $PQ$ is parallel to the tangent at $K$ to $\omega_1$, the desired result follows because the tangent at a point of a circle is perpendicular to the radius through that point, hence also perpendicular to the diameter $KM$.

Problem Understanding

Two circles intersect at $A$ and $B$. A point $K$ lies on the first circle. The lines $KA$ and $KB$ meet the second circle again at $P$ and $Q$, respectively. We must prove that the chord $PQ$ is perpendicular to the diameter $KM$ of the first circle.

The connection between the two circles comes from the common points $A$ and $B$. The plan is to show that $PQ$ is parallel to the tangent to the first circle at $K$. The required perpendicularity then follows from the tangent-radius theorem.

Proof Architecture

Lemma 1. The angle between the lines $AK$ and $PQ$ equals $\angle ABK$.

Sketch. Since $A,P,K$ are collinear, the angle between $AK$ and $PQ$ is $\angle APQ$. On the second circle, $\angle APQ=\angle ABQ$. Interpreting angles as directed angles modulo $180^\circ$, the collinearity of $B,Q,K$ gives $\angle ABQ=\angle ABK$.

Lemma 2. The angle between the tangent to the first circle at $K$ and the chord $KA$ equals $\angle ABK$.

Sketch. Apply the tangent-chord theorem in the first circle.

Lemma 3. The line $PQ$ is parallel to the tangent to the first circle at $K$.

Sketch. By Lemmas 1 and 2, both lines make the same directed angle with the line $AK$.

The only delicate point is the passage from $\angle ABQ$ to $\angle ABK$. Because $Q$ generally lies on the extension of $KB$, this step must be interpreted as an equality of directed angles between lines rather than ordinary vertex angles.

Solution

Let $t$ be the tangent to the first circle at the point $K$.

All angle equalities below are interpreted as directed angles modulo $180^\circ$.

Since $A,P,K$ are collinear, the angle between the lines $AK$ and $PQ$ equals the angle between the lines $AP$ and $PQ$, namely

$$\angle(AK,PQ)=\angle APQ.$$

The points $A,P,B,Q$ lie on the second circle. The inscribed angles $\angle APQ$ and $\angle ABQ$ subtend the same chord $AQ$. Hence

$$\angle APQ=\angle ABQ.$$

Since $B,Q,K$ are collinear, the lines $BQ$ and $BK$ are the same geometric line. Directed angles between lines depend only on the underlying lines, so replacing $BQ$ by the coincident line $BK$ does not change the angle with $AB$. Thus

$$\angle ABQ=\angle(AB,BQ)=\angle(AB,BK)=\angle ABK.$$

Combining the previous equalities gives

$$\angle(AK,PQ)=\angle ABK.$$

Now consider the first circle. By the tangent-chord theorem, the directed angle between the tangent $t$ at $K$ and the chord $KA$ equals the directed inscribed angle subtending the chord $KA$. Since $B$ lies on the first circle, that inscribed angle is $\angle ABK$. Therefore

$$\angle(t,KA)=\angle ABK.$$

Since $AK$ and $KA$ are the same line,

$$\angle(AK,PQ)=\angle(t,KA).$$

Both $PQ$ and $t$ make the same directed angle with the same reference line $AK$. Hence

$$PQ\parallel t.$$

Let $O$ be the center of the first circle. Since $KM$ is a diameter, the points $K,O,M$ are collinear.

The tangent at $K$ is perpendicular to the radius $OK$, so

$$t\perp OK.$$

Because $OK$ and $KM$ are the same line,

$$t\perp KM.$$

Since $PQ\parallel t$, a line parallel to $t$ is also perpendicular to $KM$. Therefore

$$PQ\perp KM.$$

This completes the proof.

$$\boxed{PQ\perp KM}.$$

Verification of Key Steps

The equality

$$\angle APQ=\angle ABQ$$

follows from the inscribed-angle theorem on $\omega_2$. Both angles subtend the chord $AQ$.

The equality

$$\angle ABQ=\angle ABK$$

must be interpreted in the directed-angle sense. Because $B,Q,K$ are collinear, the lines $BQ$ and $BK$ coincide. Directed angles between lines depend only on the lines themselves, not on the choice of opposite rays along those lines. Hence

$$\angle(AB,BQ)=\angle(AB,BK).$$

The conclusion

$$PQ\parallel t$$

comes from the equality of directed angles with the same reference line $AK$:

$$\angle(AK,PQ)=\angle(t,KA).$$

Since $AK$ and $KA$ represent the same line, the two lines $PQ$ and $t$ have the same direction modulo $180^\circ$.

The tangent-chord theorem yields

$$\angle(t,KA)=\angle ABK,$$

because $\angle ABK$ is the inscribed angle subtending the chord $KA$ of the first circle.

Alternative Approaches

A coordinate proof can be obtained by placing the first circle in a convenient coordinate system and computing the second intersections $P$ and $Q$ explicitly. One then checks that the slope of $PQ$ is the negative reciprocal of the slope of the diameter through $K$. The calculations are substantially longer than the synthetic argument.

Another synthetic approach uses properties of the common chord $AB$ and projective geometry on the pencil of lines through $K$. Such methods also lead to the fact that $PQ$ is parallel to the tangent at $K$, but they introduce machinery that is unnecessary for this problem. The angle argument above isolates the essential geometric relation directly.