Extraordinary Ellipse

We are looking at the Brocard inellipse of the triangle, which has foci at the first and second Brocard points. and semimajor and semiminor axes $A \; = \; \frac{a b c}{2\sqrt{a^2b^2+ b^2c^2 + c^2a^2}} \hspace{2cm} B \; = \; \frac{2a b c \Delta}{a^2b^2 + b^2c^2 + c^2a^2}$ and hence eccentricity $E \; = \; \sqrt{1 - \frac{B^2}{A^2}} \; = \; \sqrt{\frac{a^4 + b^4 + c^4}{a^2b^2 + b^2c^2 + c^2a^2} - 1}$ where $a,b,c$ are the sides of the triangle. Putting $a = \sin\tfrac{\pi}{7}$ , $b = \sin\tfrac{2\pi}{7}$ and $c = \sin\tfrac{4\pi}{7}$ into this formula gives $E = \boxed{\tfrac{1}{\sqrt{2}}}$ .

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Here is a detailed, but elementary, proof of the above.

First, let us show that the Brocard points exist. If there exists a point $P$ such that $\angle BAP = \angle CBP = \angle ACP$ , then the Alternate Segment Theorem tells us that the line $AB$ is tangential to the circumcircle of the triangle $ACP$ at $A$ , and that the line $BC$ is tangential to the circumcircle of the triangle $APB$ . Draw the circle that has $AC$ as a chord and which is tangential to $AB$ at $A$ , and draw the circle that has $AB$ as a chord and which is tangential to $AC$ at $B$ . Let $P$ be the point (other than $A$ ) where these circles intersect. Then $\angle ACP = \angle BAP = \angle CBP$ , and hence $P$ is a Brocard point (called the first Brocard point), and we note that $AC$ is also tangential to the circumcircle of the triangle $BCP$ at $C$ as well. Note that $\angle BPC = \pi - A$ , $\angle CPA = \pi - B$ and $\angle APB = \pi - C$ . Let the common angle $\angle PAB = \angle PBC = \angle PCA = \omega$ .

First note the following triangle identities: $\begin{aligned} \cot A \cot B \cot C + \csc \,A \,\csc \,B\, \csc \,C & = \; \cot A + \cot B + \cot C \\ \cot A \cot B + \cot A \cot C + \cot B \cot C & = \; 1 \end{aligned}$ The first follows because $2\cos^2\left(\tfrac12(u+v+w)\right) \; \equiv \; \sin u \sin v \sin w\big[ \cot u \cot v \cot w - \cot u - \cot v - \cot w + \csc\, u \,\csc\, v \,\csc\, w\big]$ for any $u,v,w$ , and hence the fact that $A+B+C=\pi$ gives the result. The second follows because $\cot C \; = \; \cot(\pi - A - B) \; = \; \frac{1 - \cot A \cot B}{\cot A + \cot B}$ Applying the Sine Rule to the triangles $ABP$ . $BPC$ and $CPA$ , we see that $\frac{y_a}{\sin(B-\omega)} = \frac{y_b}{\sin \omega} \hspace{1cm} \frac{y_b}{\sin(C-\omega)} = \frac{y_c}{\sin \omega} \hspace{1cm} \frac{y_c}{\sin(A-\omega)} = \frac{y_a}{\sin\omega}$ and hence $\begin{aligned} \sin(A-\omega)\sin(B-\omega)\sin(C-\omega) & = \; \sin^3\omega \\ (\cot\omega - \cot A)(\cot\omega - \cot B)(\cot\omega - \cot C) & = \; \csc\,A\,\csc\,B\,\csc\,C \\ \cot^3\omega - (\cot A + \cot B + \cot C)\cot^2\omega + \cot\omega - (\cot A +\cot B +\cot C) & = \; 0 \\ (\cot\omega - \cot A - \cot B - \cot C)(\omega^2 +1) & = \; 0 \end{aligned}$ and hence the Brocard angle $\omega$ is unique, and $\cot\omega \; = \; \cot A + \cot B + \cot C$ More applications of the Sine Rule tell us that $\frac{y_a}{\sin\omega} = \frac{b}{\sin A} \hspace{1cm} \frac{y_b}{\sin\omega} \; =\; \frac{c}{\sin B} \hspace{1cm} \frac{y_c}{\sin\omega} \; = \; \frac{a}{\sin C}$ and hence $y_a \; = \; \frac{2Rb\sin \omega}{a} \hspace{1cm} y_b \; = \; \frac{2Rc\sin\omega}{b} \hspace{1cm} y_c \; = \; \frac{2Ra\sin\omega}{c}$ where $R$ is the circumradius of $ABC$ . If we drop perpendiculars from $P$ to the sides $BC$ , $CA$ , $AB$ , with feet $D$ , $E$ , $F$ and lengths $x_a$ , $x_b$ , $x_c$ respectively, simple trigonometry tells us that $x_a \; = \; y_b\sin\omega \;=\; \frac{2Rc\sin^2 \omega}{b} \hspace{1cm} x_b \; = \; y_c\sin\omega \;=\; \frac{2Ra\sin^2\omega}{c} \hspace{1cm} x_c \; = \; y_a\sin\omega \;=\; \frac{2Rb\sin^2\omega}{a}$ Of course, $x_a\,:\,x_b\,:\,x_c$ are the absolute triangular coordinates of the point $P$ .

We can repeat these calculations for the second Brocard point $Q$ , where the vertices of $ABC$ are taken in the opposite order to locate the angles $\omega$ in slightly different places. We obtain the feet of the perpendiculars $G$ , $H$ , $I$ and lengths $p_a$ , $p_b$ , $p_c$ and $q_a$ , $q_b$ , $q_c$ in an analogous manner to that above, and we obtain that $p_a \; = \; \frac{2Rc\sin \omega}{a} \hspace{1cm} p_b \; = \; \frac{2Ra\sin\omega}{b} \hspace{1cm} p_c \; = \; \frac{2Rb\sin\omega}{c}$ and $q_a \; = \; \frac{2Rb\sin^2 \omega}{c} \hspace{1cm} q_b \;=\; \frac{2Rc\sin^2\omega}{a} \hspace{1cm} q_c \;=\; \frac{2Ra\sin^2\omega}{b}$

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It is easiest to do the later calculations in terms of the sides of the original triangle. We have $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ and $\sin A = \frac{2\Delta}{bc}$ , where $\Delta$ is the area of the triangle. Thus $\cot A = \frac{b^2 +c^2 - a^2}{4\Delta}$ , and hence $\cot\omega \; = \; \frac{a^2 + b^2 + c^2}{4\Delta}$ so that $\begin{aligned} \sin\omega & = \; \frac{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}{2\sqrt{a^2b^2 + a^2c^2 + b^2c^2}} \\ \cos\omega & = \; \frac{a^2 + b^2 + c^2}{2\sqrt{a^2b^2 + a^2c^2 + b^2c^2}} \end{aligned}$

We note that $PQ^2 \; =\; \big(c - y_a\cos\omega - p_b\cos\omega\big)^2 + (x_c - q_c)^2$ so that $\frac{PQ^2}{4R^2} \; = \; \left(\sin C - \big(\tfrac{a}{b} + \tfrac{b}{a}\big)\sin\omega\cos\omega\right)^2 + \left(\tfrac{a}{b} - \tfrac{b}{a}\right)^2\sin^4\omega \; = \; \sin^2\omega(1 - 4\sin^2\omega)$ after much simplification. Thus $PQ = 2R\sin\omega\sqrt{1 - 4\sin^2\omega}$ .

For any point $X$ on $BC$ , note that $PX + QX = P'X + QX$ , where $P'$ is the mirror image of $P$ in $BC$ . It is clear that $PX + QX \ge P'Q$ , and that this least value is achieved when $X = X_0$ . Now $P'Q^2 \; =\; \big(c - y_a\cos\omega - p_b\cos\omega\big)^2 + (x_c + q_c)^2$ so that $\frac{P'Q^2}{4R^2} \; = \; \left(\sin C - \big(\tfrac{a}{b} + \tfrac{b}{a}\big)\sin\omega\cos\omega\right)^2 + \left(\tfrac{a}{b} + \tfrac{b}{a}\right)^2\sin^4\omega \; = \; \sin^2\omega$ Thus $P'Q = 2R\sin\omega$ .

The locus of the set of points $X$ such that $PX + QX = 2R\sin\omega$ is an ellipse $\mathcal{E}$ . It touches the line $BC$ at the point $X_0$ only, and hence is tangential to the line $BC$ . Similar calculations can be done for the other two sides, and so we deduce that $\mathcal{E}$ is tangential to all three sides of the triangle, so is the Brocard ellipse desired in this question. Since the points $P$ , $Q$ are the foci of $\mathcal{E}$ , we see that the Brocard ellipse has semimajor axis $A$ and eccentricity $E$ where $A \; = \; R\sin\omega \hspace{2cm} AE \; = \; R\sin\omega\sqrt{1 -4\sin^2\omega}$ and hence the ellipse $\mathcal{E}$ has eccentricity $E = \sqrt{1 - 4\sin^2\omega}$ .

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For this particular problem, $\cot\omega \; = \; \cot\tfrac{\pi}{7} + \cot\tfrac{2\pi}{7} + \cot\tfrac{4\pi}{7} \; = \; \sqrt{7}$ so that $\sin\omega = \tfrac{1}{2\sqrt{2}}$ , and hence $E = \sqrt{1 - 4\sin^2\omega} = \boxed{\tfrac{1}{\sqrt{2}}}$ .