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Two circles of radii $4$ and $16$ units, respectively touch each other externally. Prove that the length of the chord intercepted on the transverse common tangent to the to circles, by a circle of diameter $D$ units which touches the given circle externally, is $\frac{4D}{5}$

enter image description here

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  • $\begingroup$ More generally (using slightly different notation): If $\bigcirc A$, $\bigcirc B$, $\bigcirc C$, of radius $a$, $b$, $c$, are pairwise externally tangent, then the line tangent to $\bigcirc A$ and $\bigcirc B$ at their intersection cuts $\bigcirc C$ in a chord of length $4c\sqrt{ab}/(a+b)$. I have a somewhat inelegant coordinate proof of this; I'm investigating a geometric approach. $\endgroup$ – Blue Oct 7 '17 at 5:15
  • $\begingroup$ Yes! Analytic geometry kills it, but it's very ugly. $\endgroup$ – Michael Rozenberg Oct 7 '17 at 5:39
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Here's an argument that could probably be streamlined a bit ...


Restating the problem:

Let $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ ---of radii $a$, $b$, $c$--- be pairwise externally tangent, with points of tangency $D$, $E$, $F$ as shown in the image. Let the line tangent to $\bigcirc A$ and $\bigcirc B$ at $F$ meet $\bigcirc C$ in chord $\overline{PQ}$ with midpoint $M$; define $p := |\overline{PM}| = |\overline{QM}|$. Then

$$p = \frac{2c\sqrt{ab}}{a+b} \tag{$\star$}$$

enter image description here

Extend $\overline{FE}$ and $\overline{FD}$ to meet $\bigcirc C$ again at $A^\prime$ and $B^\prime$. Note that $\triangle AEF$ and $\triangle CEA^\prime$ are isosceles, with congruent base angles at $E$ and therefore also at $F$ and $A^\prime$; thus, $\overline{A^\prime C}\parallel\overline{AF}$. Likewise, $\overline{B^\prime C}\parallel\overline{BF}$, making $A^\prime$, $B^\prime$, $C$ collinear on a diameter of $\bigcirc C$.

By Thales' Theorem, $\angle A^\prime PB^\prime$ is a right angle; because $\overline{PM}\perp\overline{A^\prime B^\prime}$, we have the classical construction of a geometric mean: $$|\overline{PM}|^2 = |\overline{A^\prime M}||\overline{B^\prime M}| \tag{1}$$

(This is why I chose $p$ to name the half-chord, instead of the full length of $\overline{PQ}$.) Now our task reduces to a matter of rewriting the right-hand side.

Observe that the perpendiculars at $D$, $E$, $F$ necessarily concur at a point equidistant from them; that is, they meet at center $I$ of the incircle (of radius $r$) of $\triangle ABC$. Of course, $\overline{AI}$ and $\overline{BI}$ are angle bisectors.

Since $\triangle A^\prime M F\sim \triangle IFA$ and $\triangle B^\prime MF\sim \triangle IFB$, we have $$\begin{align} \frac{|\overline{A^\prime M}|}{|\overline{MF}|} = \frac{r}{a} \quad\text{and}\quad \frac{|\overline{B^\prime M}|}{|\overline{MF}|} = \frac{r}{b} &\quad\implies\quad \frac{|\overline{B^\prime M}|}{|\overline{A^\prime M|}} = \frac{a}{b} \tag{2} \\ &\quad\implies\quad |\overline{A^\prime M}| = 2c\;\frac{b}{a+b} \quad\text{and}\quad |\overline{B^\prime M}| = 2c\;\frac{a}{a+b} \end{align}$$

(The above improves on my first pass at this proof. (See the revision history.) There may yet be a better argument that incorporates this proportionality relation: $$\frac{|\overline{A^\prime M}|}{b} = \frac{|\overline{A^\prime B^\prime}|}{|\overline{AB}|} = \frac{|\overline{B^\prime M}|}{a} \tag{2.5}$$ Be that as it may ...) Thus, $(1)$ becomes $$p^2 = \frac{4abc^2}{(a+b)^2} \qquad\to\qquad p = \frac{2c\sqrt{ab}}{a+b} \tag{3}$$ as claimed. $\square$

For the original question, we take $a = 4$, $b = 16$, $c = D/2$, so that the length of the chord is $$2p = \frac{4 \cdot D/2 \cdot \sqrt{4\cdot 16}}{4+16} = \frac{4D}{5}$$

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  • $\begingroup$ The "It is easy..." claim need to be PROVEN! $\endgroup$ – Moti Oct 7 '17 at 15:38
  • $\begingroup$ @Moti: $\overline{AB}\parallel\overline{A^\prime B^\prime}$, so isos. $\triangle AEF$ and $\triangle C E A^\prime$ have equal vertex angles, and thus also equal base angles. In particular, the angles at $E$ match, making $\overrightarrow{EF}$ and $\overrightarrow{EA^\prime}$ opposite rays. Done. $\square$ If you like, imagine that I constructed the converse: Extend $\overline{FE}$ to meet $\bigcirc C$ again at $A^\prime$. Then $\triangle AEF$ and $\triangle CEA^\prime$ have equal base angles, and thus also equal vertex angles, so $\overline{AB}\parallel\overline{A^\prime C}$. $\square$ OK? :) $\endgroup$ – Blue Oct 7 '17 at 18:14
  • $\begingroup$ @Moti: Actually, I like the converse construction better. I'm going to edit my answer. :) $\endgroup$ – Blue Oct 7 '17 at 18:16

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