Equilateral triangle separating two homothetic circles

Question

This question is a follow-on of a question asked today that has been erased by its author an hour after being asked, for unknown reasons. I found it interesting ; I have a solution, mainly based on analytic geometry. I would appreciate to see other solutions (using synthetic geometry for example), hopefuly less computational.

Let me reformulate it in a slightly different way compared to the original question :

Given : a fixed line with two points $B$ and $C$ on it. Let us construct $A$ such that $ABC$ is equilateral. Let us take an arbitrary point $F$ on line $BC$ outside line segment $[BC]$. Let us denote by $D$, resp. $E$, the center of the incircle of triangle $FAC$, resp. the center of the excircle of triangle $FAB$ opposed to $F$.

Prove that

$$BE+CD=BC\tag{1}.$$

Solution :

Let us define $A'$ as being the symmetrical point of $A$ with respect to line $BC$. $A',C,D$ are aligned. Indeed, slope(A'C)=slope(C'D) for the following reason : as $\angle ACF = 180°-60°=120°$, we have $\angle DCF=\angle DCA=\angle ACB = 60°$.

For a similar reason, $A',B,E$ are aligned.

Let us call $D',E'$ the projections of $D,E$ resp. on line $AB$ ; (1) amounts to say that :

$$BE'+CD'=\tfrac12 BC\tag{2}.$$

Let us take coordinate axes such that :

$$B(-1,0),C(1,0),A(0,\sqrt{3}),F(a,0) \ \text{with} \ a>1$$

as one can see on the figure.

Let $2 \alpha = \angle AFB$. We have

$$\tan(2 \alpha)=\frac{\sqrt{3}}{a}$$

Using relationship $$\tan(2 \alpha)=\frac{2 t}{1-t^2} \ with \ t:=\tan(\alpha)\tag{3}$$

we get :

$$t:=\tan(\alpha)=\frac{\sqrt{3+a^2}-a}{\sqrt{3}}$$

(but in fact, we will not use this relationship).

Point $D$ being at the intersection of :

$$\begin{cases}\text{Line FD : } & y=-t(x-a)\\ \text{Line A'C : } & y=\sqrt{3}(x-1)\end{cases}$$

its abscissa, which is the same as the abscissa of $D'$, is

$$x_{D'}=\frac{at+\sqrt{3}}{t+\sqrt{3}}$$

For a similar reason :

$$x_{E'}=\frac{at+\sqrt{3}}{t-\sqrt{3}}$$

(2) will be established if we show that :

$$\left(-1-\frac{at+\sqrt{3}}{t-\sqrt{3}}\right)+\left(\frac{at+\sqrt{3}}{t+\sqrt{3}}-1\right)=1\tag{4}$$

With some simplifications, the LHS of (4) becomes :

$$(at+\sqrt{3})\frac{-2 \sqrt{3}}{t^2-3}-2\tag{5}$$

But, using (3),

$$t^2-1=\frac{-2ta}{\sqrt{3}}\tag{6}$$

Plugging (6) into (5) gives this LHS equal to $1$ as awaited.

Edit : I hadn't understood at first the track followed by the asker of the initial question, based on the sine law ; it can be completed by checking that for all $\theta$ :

$$\frac{\sin(\theta)}{\sin(2\pi/3-\theta)}+\frac{\sin(\pi/3-\theta)}{\sin(\pi/3+\theta)}=1.$$

Answer 1

Changing notation a bit ... Let $\triangle ABC$ be equilateral, and let $\ell$ be some line through $C$. Let $\bigcirc P$ and $\bigcirc Q$ be tangent to $\ell$ and $\overleftrightarrow{AB}$ (so that they're homothetic with respect to the point of intersection of those lines), and also tangent, respectively, to $\overline{AC}$ and $\overline{BC}$. Define $a:=|AP|$, $b:=|BQ|$, $c:=|AB|=|BC|=|CA|$.

Note that $\overline{PC}$ and $\overline{QC}$ bisect angles at $C$; let $\alpha$ and $\beta$ be the respective half-angles, and observe that $\alpha+\beta = 60^\circ$.

Let $P'$ and $Q'$ complete equilateral triangles $\triangle APP'$ and $\triangle BQQ'$ as shown. A little angle-chasing guarantees that $\triangle CPP'$ and $\triangle QCQ'$ are similar $\alpha$-$\beta$-$120^\circ$ triangles, giving us a proportion that readily yields the result: $$\frac{a}{c-a} = \frac{c-b}{b} \quad\to\quad c = a+b$$ Done! $\square$

Answer 2

$\triangle ABC$ is equilateral, $DE$ and $FG$ are tangent to the circles, $O_2M\perp O_1D$ and $O_2N\perp O_1F$

Let $AB=AC=BC=a$, $O_1D=R\sqrt3$ and $O_2E=r\sqrt3$

Since $BO_1$ and $CO_2$ are the bisectors, there will be $30^{\circ}- 60^{\circ} - 90^{\circ}$ right triangles then $O_1B=2R$, $O_2C=2r$ and $BD=BK=R$, $CE=CH=r$ and $DE=O_2M=a+R+r$, $O_1M=\sqrt3(R-r)$

Since $AB$ is tangent to circle $O_1$ and $AC$ is tangent to circle $O_2$ then $AF=AK=a-R$ , $AG=AH=a-r$ and $O_2N=FC=2a-R-r$ , $O_1N=\sqrt3(R-r)$

$O_1O_2M \cong O_1O_2N$ (RHS congruency theorem)

$O_2M=a+R+r = O_2N=2a-R-r $

$a=2(R+r)=O_1B +O_2C$

Answer 3

For our proof we need couple of line segments which are not present in OP’s sketch. So, we add $EA$, the angle bisector of $\angle BAM$, and $AH$, the angle bisector of $\angle FAC$. For brevity, we also let $\measuredangle BAM = \omega$, $BE=a$, $CD=b$, and $CH=x$. The length of the sides of the equilateral triangle $ABC$ is denoted as $c$. With all these additions, we obtained the configuration shown in the above diagram.

Since the line segment $CD$ is parallel to the line segment $AB$, the two triangles $BHA$ and $CHD$ are similar. Therefore, we shall write, $$\dfrac{c}{b}=\dfrac{c+x}{x}\quad\rightarrow\quad \dfrac{c-b}{b}=\dfrac{c}{x}.\tag{1} $$

By considering the two known angles of $\triangle BHA$, we can show that $\measuredangle BHA=\omega$. With that, we have another pair of similar triangles, namely $AEB$ and $DCH$. Hence, we have, $$\dfrac{c}{x}=\dfrac{a}{b}.\tag{2}$$

Using (1) and (2), we can deduce, $$\dfrac{c-b}{b }=\dfrac{a}{b}\quad\rightarrow\quad c-b=a\quad\rightarrow\quad c=a+b.$$