Let $p$ and $q$ be the radii of two circles through $A$, touching $BC$ at $B$ and $C$, respectively. Then $pq=R^2$.

Question

I am working problem 1.1.4 from Geometry Revisited:

Let $p$ and $q$ be the radii of two circles through $A$, touching $BC$ at $B$ and $C$, respectively. Then $pq=R^2$.

I have read the solution Problem on Law of Sines from 'Geometry Revisited' but I'm still missing something very basic.

I can come up with infinitely many circles that go though points $A$ and $B$, and don't touch the line segment from $B$ to $C$ anywhere else (except at point $B$). Thus the statement ($pq=R^2$) can easily be made false. It is easy to find such circles for arbitrarily large radii.

In the solution posted in the link above, the circle through points $A$ and $B$ is additionally centered above point $B$ (so $B$ is the bottom point) and that fact is used in the proof. But where in the problem does it specify that condition?

I have just made a comment on the linked post querying the assumption made in the original question. — Mick, Nov 04 '14 at 05:18
Thanks @Mick. After some reflection I see that perhaps when they say "touching BC at B", "BC" is meant as the line through BC and they mean "touching the line BC only at B". It's a bit confusing though. In other parts of the book they use the notation "AB" to mean a line segment connecting A and B. — AlexS, Nov 04 '14 at 05:30
I think both “line segment BC” and “two circles … touching BC at B and C” mean the same thing. The part that I feel not comfortable is “two circles through A”. Does it mean “two circles (passing) though A” or “two circles (touch each other at) A”? It seems to me that the drawing/solution to linked question is based on two circles intersecting each other at ONE point only. (At least it looks like that.) — Mick, Nov 04 '14 at 05:53
@Mick that's a really interesting observation and I'm pretty sure that it's not true in general.
To sum up:

Let's agree that initial problem is ambiguous.

Specifically, if we interpret $BC$ in the original problem to be "the line segment connecting $B$ and $C$", then the statement is false for reasons I gave above. — AlexS, Nov 04 '14 at 08:20

Mick · Accepted Answer · 2022-07-19T03:42:10.957

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I have the figure re-drawn. Initially, we have the black and the red circle intersecting at A.

They touch the (blue) line BC at B and C. The blue circle is the one circumscribing ABC.

enter image description here

Apart from his drawing, there is nothing wrong with Ricardo Cruz's work except a minor typo - should be $\sin C'= \frac{b}{2q}$ instead.

edited Jul 19 '22 at 03:42

answered Nov 04 '14 at 08:44

Mick

17,141

Thanks Mick, for pointing out my typo. I've just corrected it. – RicardoCruz Nov 04 '14 at 16:53
@Mick Your observation about the circles in the other diagram led me to wonder whether there exist any triangles for which the two corresponding circles touch only at the point A. Working that out was a fun exercise in algebra and I can report that there indeed do exist such examples. So the other diagram is valid for certain triangles but not all. – AlexS Nov 05 '14 at 06:54
@Mick thanks for the updated diagram. However, I don't think you really addressed my initial question about the wording of the problem. Do you agree that we need to add the additional constraint "circles through $A$, touching line BC only at B"? Otherwise, I can find circles with arbitrarily large radii that touch A and the line segment BC only at B, making the claim false as stated. Of course these arbitrarily large circles will touch the line BC in two places. – AlexS Nov 05 '14 at 06:59
@AlexS [1] The question stated “… two circles … touching (the line) BC at B and C, respectively” is a short (but precise) form to mean exactly the same thing (only at B) like you have mentioned. [2] A circle and a (tangent) line can “touch” each other at ONE point (or place, in your term) [i.e. B] ONLY. – Mick Nov 05 '14 at 16:25
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The confusion is that with notation BC we don't know whether the author means the line BC, in which case the circles would be tangent at B and C respectively, or the segment BC, the latter interpretation being the source of confusion – proofromthebook Jul 18 '22 at 21:59

Let $p$ and $q$ be the radii of two circles through $A$, touching $BC$ at $B$ and $C$, respectively. Then $pq=R^2$.

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