3

This problem is from 'Geometry Revisited', Exercise number 4 in section 1.1. The question and answer below are from the textbook. Copy of book found here Question:-

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

Answer from the book :-

$$c = \frac{2p}{\sin B} = \frac{pb}{R}, b = \frac{2q}{\sin C} = \frac{qc}{R}$$ 

Multiply and simplify.

My understanding :-

  1. $R$ is the circumradius of triangle $ABC$. By Law Of Sines, $a/\sin A = b/\sin B = c/\sin C = 2R$.
  2. The side $AB$ is common to the 2 triangles having radii p and R, just as $AC$ is common to other 2 triangles having radii q and R.
  3. By the Law Of Sines, $2p = c/\sin C'$ ,if we call C' as the side which is the 3rd side of triangle whose radius is p and shares AB with ABC.
  4. Similarly $2q = a/\sin A'$ if we call A' as the side which is not shared by the triangle whose radius is q.

Have I understood correctly so far ? If then, how is $c=2p\sin B$?

Sawarnik
  • 7,284
  • 7
  • 33
  • 74
Chakra
  • 269

2 Answers2

3

Let $AB'$ a diameter of the circle whose radius is $p$ and $AC'$ a diameter of the circle whose radius is $q$.

See the following figure:

enter image description here

From $\triangle ABB'$ we get: $$\sin B'= \frac{c}{2p}, \quad (1)$$ and from $\triangle ACC'$: $$\sin C'= \frac{b}{2q}. \quad (2)$$ But $B'=B$ and $C'=C$ (Why?).

Recall that: $$\sin B= \frac{b}{2R}, \quad (3)$$ and $$\sin C= \frac{c}{2R}. \quad (4)$$

From $(1)$ and $(3)$ we get: $$c=\frac{pb}{R}, \quad(5)$$ and and from $(2)$ and $(4)$ we get: $$b=\frac{qc}{R}. \quad(6)$$ Finally from $(5)$ and $(6)$ we get: $$pq=R^2$$

RicardoCruz
  • 3,693
  • How is this true ? -----> "But B′=B and C′=C" (Why?). – Chakra May 25 '14 at 07:31
  • @Chakra Recall that $B'$ and $B$ are inscribed angles, and hence they have the same measure. The same occurs to $C'$ and $C$. See more information here: http://en.wikipedia.org/wiki/Inscribed_angle – RicardoCruz May 25 '14 at 11:48
  • Unless the question is stated in a manner having something assumed, my interpretation of the question is “…two circles (passing) through A”. However, your drawing is assuming “…two circles ([seem like] touching each other at only one point) A". – Mick Nov 04 '14 at 05:01
  • @Mick I agree the two circles seem like touching each other at only one point, but in fact they aren't. – RicardoCruz Nov 04 '14 at 16:59
  • Agreed but a drawing with two points of intersection would be more general. There is nothing wrong with the proof except a minor typo - should be b = qc/R instead. – Mick Nov 04 '14 at 17:07
  • @Mick Thanks for pointing out the typo. I'll fix it. – RicardoCruz Nov 04 '14 at 17:15
  • Since I have mentioned your name in post #100535, I should have you informed at the same time. Sorry for not having that done. I am glad you have found that out by yourself. – Mick Nov 05 '14 at 16:39
2

I have a slightly different solution that doesn't require the inscribed angle theorem.

First, I actually think this question is a bit ambiguous and confusing. For more on that see: Problem 1.1.4 from "Geometry Revisited" In short, the statement is true if "$BC$" is interpreted as "line $BC$" and false if interpreted as "line segment $BC$".

Let $P$ be the midpoint of the circle with radius $p$. Since this circle touches the line $BC$ in exactly one place, which is $B$, the angle $CBP$ is a right angle. Thus the angle $ABP$ measures $\pi/2 - B$ radians. Since the triangle $ABP$ is isosceles, the rest of the angles can be computed. Applying law of sines to the triangle and using $\sin(2B) = 2 \sin(B)\cos(B)$ gives the desired result $c = 2p \sin(B)$.

Edit: The above argument holds only if angle $ABC$ is acute. If it is obtuse, then a similar argument shows that the angle $ABP$ measures $B-\pi/2$ and the rest of the argument is similar. If angle $ABC$ is a right angle, then $c=2p \sin(B)$ follows directly.

Community
  • 1
AlexS
  • 53