Symmedian Simplicity

The Symmmedian point $X_6$ is the isogonal conjugate of the centroid $G$ of the triangle. Suppose that $\angle BAC = \theta$ , so that $A,B,C$ have coordinates $A\; (-1,0) \hspace{1.5cm} B\; (1,0) \hspace{1.5cm} C\;(\cos2\theta,\sin2\theta)$

If $CF$ is the symmedian of $ABC$ corresponding to the median $CO$ then, since $\angle OCA = \theta$ , we deduce that $\angle FCB = \theta$ , which implies that triangle $CFB$ is right-angled, with $\angle CFB = 90^\circ$ .

Now let $AD$ be the symmedian corresponding to the median $AM_A$ , and let $\angle M_AAC = \alpha$ , so that $\tan\alpha = \tfrac{\sin\theta}{2\cos\theta} = \tfrac12\tan\theta$ . But then $\angle DAB = \alpha$ , and hence $X_6F \; = \; AF\tan\alpha \; = \; (\cos2\theta + 1) \times \tfrac12\tan\theta \; = \; \sin\theta\cos\theta \; = \; \tfrac12\sin2\theta$ which means that $X_6$ has coordinates $\big(\cos2\theta,\tfrac12\sin2\theta\big)$ , and hence the locus of $X_6$ is the ellipse $x^2 + 4y^2 \; = \; 1$ which has semimajor axis $a=1$ and semiminor axis $b=\tfrac12$ . Thus the area of the enclosed region is $\pi ab \; = \; \tfrac12\pi$ which makes the answer $1 + 2 = 3$ .

The Lemoine point of a right angled triangle is the midpoint of its altitude corresponding to the right angle.

Proof:-

Let $ACB$ be a right triangle with $\angle C= 90^{\circ}$ . Let O be the midpoint of AB and F the foot of $\perp$ from $C$ on AB. We have:

$\angle ACO = \angle A = \pi/2 - \angle B = \angle BCF$

$\Rightarrow CF$ and $CO$ are isogonals.

Let $M$ be the midpoint of $CF$ and $M_A$ the midpoint of $BC$ . We have:

$\Delta ACB \sim \Delta AFC$

$\Rightarrow \angle CAM_A = \angle FAM$

$\Rightarrow AM_A$ and $AM$ are isogonals.

$\therefore M$ is the isogonal conjugate of the centroid $G$ , so, $M \equiv X_6$ $\square$

Let $f(x)$ denote the locus of the point $C$ as $x \equiv F$ varies over path $\rho = [A-B-A]$ .

The required area = $\large{\int_\rho |\frac{f(x)}{2} |dx} = \dfrac{\large{\int_\rho} |f(x)|dx}{2} = \dfrac{\text{Area of Circle(r=1)}}{2} = \boxed{\dfrac{\pi}{2}}$ .