Angle Tangled

As I have shown elsewhere , the locus of the points $P$ such that $\angle PBA = \angle PCA$ is a rectangular hyperbola $\mathcal{H}$ which is the isogonal conjugate of the perpendicular bisector $\mathcal{L}$ of the line $BC$ . The section of the hyperbola which is internal to the triangle $ABC$ is the isogonal conjugate of the segment $MN$ of $\mathcal{L}$ which is internal to $ABC$ . If $P$ is a point on $\mathcal{H}$ , let $Q \in \mathcal{L}$ be its isogonal conjugate. If we denote $\varphi = \angle APC$ and $\theta = \angle AQC$ , then we note that $180^\circ = \varphi + \alpha + \beta$ and $B + \alpha + \beta + 360^\circ - \theta = 360^\circ$ , so that $\theta = 180^\circ + B - \varphi$ .

Thus to minimize $\varphi$ we need to maximize $\theta$ . It is clear that the maximum value $\theta_0$ of $\theta$ will occur when we construct a circle that passes through both $A$ and $C$ , and which is tangential to the line $\mathcal{L}$ . The centre $X$ of this circle can be found as the point of intersection of the perpendicular bisector of $AC$ with the parabola $\mathcal{P}$ with $C$ as focus and $\mathcal{L}$ as directrix. Since $\angle BPC = 180^\circ - (B - \alpha) - (C - \alpha) = A + 2\alpha$ , this means that, in the case that $\theta$ is minimized, $\angle BPA = 90^\circ$ , and hence $\frac{\sin \varphi_0}{b} = \frac{\sin\alpha}{AP} \; = \; \frac{1}{c}$ so that $\sin\varphi_0 = \boxed{\tfrac{b}{c}}$ .

Thus, if $\varphi$ is minimized at some point internal to the triangle, then the minimum value of $\sin\varphi$ is $\tfrac{b}{c}$ . There are three conditions that the triangle has to satisfy for this minimum to be achieved in the desired manner, and they can most easily be expressed in terms of the angles of the triangle:

• $B < C$ ,
• $C$ is acute,
• $\angle CAM$ is acute ( $M$ is the midpoint of $BC$ ).

If $B > C$ then $A$ is on the same side of $\mathcal{L}$ as $B$ , and hence the maximum value of $\theta$ clearly occurs when $Q = N$ , which now lies on the edge $AC$ . For $\angle APB$ to be equal to $90^\circ$ (which is necessary if the maximum value of $\theta$ occurs inside $ABC$ ), we must have $C$ an acute angle, because $\angle ACB < \angle APB$ . Assume now that $B < C < 90^\circ$ . If $\angle CAM$ is obtuse, then $A$ lies inside the circle with diameter $\tfrac12a$ passing through $M$ and $C$ . Let $A'$ be the intersection of the extension of $AC$ with this circle. Then the perpendicular bisector of $A'C$ passes through the centre of this circle, which is the vertex of the parabola $\mathcal{P}$ . The perpendicular bisector of $AC$ is parallel to this other perpendicular bisector, but crosses the the $BC$ "below" the vertex of the parabola. Thus the perpendicular bisector of $AC$ meets the parabola at a point $X$ which is to the "left" of $BC$ , and hence $Q$ lies outside the line segment $MN$ . Conversely, if $\angle CAM$ is acute, then $A$ lies outside this circle, and $Q$ lies inside $MN$ .

Using the Cosine Rule, and the known formula for the length of a triangle median, the conditions that $B < C$ , $C < 90^\circ$ and $\angle CAM < 90^\circ$ can be written in terms of the sides of the triangle as $b < c$ , $c^2 < a^2 + b^2$ and $a^2 < 3b^2 + c^2$ , respectively.

Let $\angle PBA = \angle PCA = \theta , \> \angle BPC = x = \angle A + 2 \theta , \> \angle APB = \beta = 2 \pi - x - \alpha$ .

Applying the sine rule for $\Delta APB, \> \Delta APC$ , we get $\begin{aligned} \frac{\sin \beta}{c} &=\frac{\sin \theta}{AP} = \frac{\sin \alpha}{b} \\ c \sin \alpha &= b \sin (2 \pi -x -\alpha) \\ c \sin \alpha &= - b \sin (x +\alpha) \\ c \sin \alpha &= - b \sin x \cos \alpha -b \cos x \sin \alpha \\ \tan \alpha &= \frac{-b \sin x}{c+b \cos x} \\ \implies \alpha &= \tan^{-1} \Big(\frac{-b \sin x}{c+b \cos x} \Big) \\ \therefore \frac{ d\alpha}{dx} &= \frac{-b^2-bc \cos x}{(c + b \cos x)^2+(b \sin x)^2} \end{aligned}$

For an extremum, $\begin{aligned} \frac{ d\alpha}{dx} &= 0 \\ \frac{-b^2-bc \cos x}{(c + b \cos x)^2+(b \sin x)^2} &= 0 \\ -b^2-bc \cos x &= 0 \\ \implies \cos x &= - \frac{b}{c} \end{aligned}$

So, $\begin{aligned} \tan \alpha &= \frac{-b \sin x}{c+b \cos x} \\ &= \frac{-b (\frac{ \sqrt{c^2-b^2} }{c}) }{c+b (- \frac{b}{c})} \\ &= - \frac{b}{ \sqrt{c^2-b^2} } \\ \therefore \sin \alpha &= \frac{b}{c} \end{aligned}$

As we can see from the expression for $\tan \alpha$ , there is a radical $\sqrt{c^2-b^2}$ . So, for an extremum to exist, we need $c > b$ . Also, we need the point $P$ to stay inside $\Delta ABC$ . For that we need to satisfy $0 < \theta < \angle B$ . We can also write this in terms of $x$ to get $\angle A < x < \angle A + 2 \angle B$ . This inequality can be written in terms of side lengths as $c^2 -b^2 < a^2 < c^2 + 3b^2$ .

Let $\angle PBA = \angle PCA = \theta$ . Then $\angle BPC = 180^\circ - (B-\theta) - (C-\theta) = 180^\circ - B - C = 2\theta = A + 2\theta$ and $\angle APB = 360^\circ - A - 2\theta - \alpha$ . Let $AP = d$ . By sine rule ,

$\begin{aligned} \frac {\sin \alpha}b & = \frac {\sin \theta}d = \frac {\sin (360^\circ - A - 2\theta -\alpha)}c \\ \implies \frac {\sin \alpha}b & = \frac {-\sin (A + 2\theta + \alpha)}c \\ c \sin \alpha & = - b \sin (A+2\theta + \alpha) \end{aligned}$

To find the minimum $\alpha$ , we first find the $\theta$ such that $\dfrac {d\alpha}{d\theta} = 0$ .

$\begin{aligned} c \cos \alpha \frac {d\alpha}{d\theta} & = - b \cos (A + 2 \theta + \alpha) \left(2 + \frac {d\alpha}{d\theta} \right) \\ \implies \frac {d\alpha}{d \theta} & = \frac {-2b\cos (A+2\theta+\alpha)}{c \cos \alpha + b \cos(A+2\theta + \alpha)} \end{aligned}$

We note that $\dfrac {d\alpha}{d\theta} = 0$ , when $A+2\theta + \alpha = 270^\circ$ , since the angle is larger than $90^\circ$ . Since

$\begin{aligned} c \sin \alpha & = - b \sin (A+2\theta + \alpha) \\ \implies \sin \alpha_{\min} & = \frac {-b \sin 270^\circ}c = \boxed{\frac bc} \end{aligned}$

Note that as Point P moves toward Point B, angle alpha decreases. Also, side b is opposite angle alpha. For the previous considerations, side c eventually approximates the hypotenuse of the minimum angle alpha. Q.E.D.:, by definition, the sine of angle alpha is opposite over hypotenuse, or b / c.

BTW, I'm not sure I could have deduced an answer if b / a was a distractor ...