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This problem is driving me crazy. It's from Andreescu's Mathematical Olympiad Challenges:

Let $AB$ be a chord in a circle and $P$ a point on the circle. Let $Q$ be the projection of $P$ onto $AB$ and $R$ and $S$ the projections of $P$ onto the tangents to the circle at $A$ and $B$. Prove that $PQ$ is the geometric mean of $PR$ and $PS$.

The solution given is as follows:

We will prove that the triangles $PRQ$ and $PQS$ are similar. This will imply $PR/PQ = PQ/PS$; hence $PQ^2 = PR\cdot PS$.

The quadrilaterals $PRAQ$ and $PQBS$ are cyclic, since each of them has two opposite right angles. In the first quadrilateral $\angle PRQ=\angle PAQ$ and in the second $\angle PQS = \angle PBS$. By inscribed angles, $\angle PAQ$ and $\angle PBS$ are equal. It follows that $\angle PRQ = \angle PQS$. A similar argument shows that $\angle PQR=\angle PSQ$. This implies that the triangles $PRQ$ and $PQS$ are similar, and the conclusion follows.

Now, just HOW by inscribed angles is $\angle PAQ=\angle PBS$? Those two do not subtend chords in the same circle, and I tried using angle chasing to find their values, but even if I consider the larger cyclic quadrilateral with vertices $P,R,S$ and the intersection of the tangents of $A$ and $B$, the two don't subtend any chords in common.

Can someone enlighten this tortured soul? I've been chasing angles for the past month now and it even haunts my dreams.

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In many countries "inscribed angles" subsumes a group of theorems, including

  1. two angles above the same chord are equal
  2. two angles above and below the same chord sum to $180^\circ$.
  3. an inscribed angle is half the angle at the center
  4. an angle above a chord equals the angles between the chord and each tangent at its endpoints.

It is the last point which is relevant for this step.

You can find it at the end of the wikipedia article as a "corollary": http://en.wikipedia.org/wiki/Inscribed_angle

There are many proofs possible, but you might want to use the fact that the endpoints of the chord, the center of the circle and the intersection of the two tangents also form a cyclic quadrilateral and the ordinary inscribed angle theorem gives the desired relation.

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  • $\begingroup$ Oh, I see. Let $T$ be the intersection point of the tangents. Then by point 4, $\angle TAQ = \angle TBA$, which implies that $\angle SBQ = \angle RAQ$, and hence the two cyclic quadrilaterals are similar, which then allows us to establish the similarity of the desired. triangles. Beautiful. Thank you! $\endgroup$ – laughing_man May 12 '14 at 3:57
  • $\begingroup$ Actually, we can use even less than point 4. The only property (that point 4 relies on) that is really needed is that the two tangents from the same point outside a circle have equal lengths. From this, we can derive that $\angle TAQ = \angle TBA$, and the rest follows. $\endgroup$ – laughing_man May 12 '14 at 4:18

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