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My uncle gave me the following puzzle, hoping there was a mathematical proof to his conjecture.

Let $A$, $B$, and $C$ be points on a circle of some fixed radius. The radius for $C$ bisects the chord $AB$.

Let $P$ be any point on the segment $AC$. Construct a second circle of the same radius which passes through $P$ and $C$. Let $Q$ be the point of intersection of this new circle and the line $BC$.

(See image below.)

Problem Statement

My Uncle's Conjecture: $|AB| = |PQ|$.

I have created a proof which uses the Inscribed Angle Theorem a few times. The gist of it is below. Reference the following diagram.

Solution

Let $O$ be the center of the original circle, and let $\alpha = \angle COA$. Notice that also $\alpha = \angle COB$.

By the Inscribed Angle Theorem, we have $\angle CBA = \alpha/2$. By symmetry, $\angle CAB = \alpha/2$. Hence $\angle ACB = 180^\circ - \alpha$.

Next, let $R$ be the center of the constructed circle, and label $\beta = \angle CQP$ and $\gamma = \angle CPQ$. Then $\beta + \gamma = \alpha$.

Also, we have $\angle CRP = 2\beta$, and $\angle CRQ = 2\gamma$ by the same Inscribed Angle Theorem as before. Thus $\angle PRQ = 2\beta + 2\gamma = 2\alpha$.

Now, by SAS, $\Delta AOB \cong \Delta PRQ$. In particular, $|AB| = |PQ|$.


So, my question: Does this Theorem have a name or appear in the literature anywhere?

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In a somewhat more general form, this is the extended law of sines, which says that for a triangle $\triangle ABC$ with circumradius $r$, $\frac{BC}{\sin A} = \frac{AC}{\sin B} = \frac{AB}{\sin C} = 2r.$ Equivalently, this could be rearranged to say that $$AB = 2r \sin C.$$

Here, because both $2r$ and $\sin C$ are equal for $\triangle ABC$ and $\triangle PQC$, we get $AB = PQ$.

I admit that this straddles the line between "a more general form of your theorem" and "a short proof of your theorem using the extended law of sines", but I think this makes it unlikely that this theorem would show up somewhere with a name: it is too close to the law of sines to be properly its own thing.

On the other hand, if Euclid (who did not deal in trigonometric functions) has any propositions that come close to the law of sines, they would probably resemble your result very closely. But I haven't been able to find any.


Note: capital $R$ is conventional for the circumradius of a circle, but you already have a point with that name, so I use lowercase $r$ instead.

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  • $\begingroup$ I figured I was at least 2000 years late to the game with this one. $\endgroup$ – wckronholm Nov 7 '17 at 5:51
  • $\begingroup$ (And the alternate proof using Law of Sines is quite nice.) $\endgroup$ – wckronholm Nov 7 '17 at 5:53

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