Points $P$, $Q$, and $R$ lie on the same line. Three semi-circles with the diameters $PQ$, $QR$, and $PR$ are drawn on the same side of the line segment $PR$. (That is, suppose we have an arbelos.) The centers of the semi-circles are $A$, $B$, and $O$, respectively. A circle with center $C$ touches all three semi-circles. Show that the radius of this circle is $$c = \frac{ab(a+b)}{a^2+ab+b^2}$$ where $a :=|AQ|$ and $b :=|BQ|$ are the radii of the smaller two semi-circles.

I know that since this is a trigonometry question, I have to construct a triangle somewhere. However, I am unsure as to whether I should construct the triangle between points ACQ or points ACB.

  • $\begingroup$ You have enough reputation to post a diagram. Please do so. It makes it so much easier. What have you tried? Where are you stuck? $\endgroup$ – Ross Millikan May 1 '17 at 3:07
  • $\begingroup$ FYI: This is a case of Descartes' "Kissing Circles" Theorem. Our circles have radii $a$, $b$, $a+b$, $c$, and thus curvatures $k_1 = 1/a$, $k_2 = 1/b$, $k_3 = -1/(a+b)$, $k_4 = 1/c$. (We negate $k_3$ because we're "inside" the corresponding circle.) Descartes' formula $$(k_1 + k_2 + k_3 + k_4 )^2 = 2 ( k_1^2 +k_2^2 + k_3^2 + k_4^2 ) \tag{$\star$}$$ reduces to $$\frac{\left( a b (a+b) - c ( a^2 + a b + b^2) \right)^2}{a^2 b^2 (a + b)^2 c^2} = 0$$ and the result follows. Proving $(\star)$ is nice exercise, but is a bit beyond the scope here. $\endgroup$ – Blue May 1 '17 at 4:01

Hint: Draw the figure and you will have $AC=a+c$, $BC=b+c$, $OA=b$, $OB=a$ and $OC=a+b-c$. Apply the cosine formula to find $\cos\angle BAC$ in

(1) $\triangle ABC$

(ii) $\triangle OAC$

  • $\begingroup$ Hello, thank you for the hint! I was just wondering, why is OA = b and OB = a and OC= a + b -c? @CY Kwong $\endgroup$ – jenboo12138 May 2 '17 at 0:42
  • $\begingroup$ OP is a+b and AP is a. So OA is b. OB is a for a similar reason. Extend OC to meet the big semicircle at X. OX is a+b and CX is c. So OC = a+b-c $\endgroup$ – CY Aries May 2 '17 at 1:07

I will outline an approach that can be ultimately used to prove Descartes' (kissing circles) theorem too. The key idea is to perform a circle inversion and to keep track of some distances.

enter image description here

If we perform a circle inversion with respect to a circle centered at $Q$ through $P$ (first dotted circle) the $PQ$-circle is mapped into a line through $P$ (red line), the $QR$-circle is mapped into a parallel line (blue line) and the $PR$-circle is mapped into a circle tangent to both lines (green circle). If $PQ=2a$ and $QR=2b$ the radius of the green circle is $$ \frac{1}{2}\left(2a+\frac{4a^2}{2b}\right) = \frac{a^2+ab}{b}. $$ A circle $\Xi$ congruent to the green circle (the second dotted circle) lies above the green circle and is tangent to the green circle, the blue line and the red line. Its inverse (the orange circle) is the solution to the arbelos. It follows that it is enough to compute where the intersections of $\Xi$ with the red line, the blue line and the green circle lie (in terms of the previously computed radius) to get three points on the orange circle by inverting with respect to $Q$, and that also gives the radius of the orange circle.


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