Question

A point moves so that the sum of the squares of the perpendiculars let fall from it on the sides of an equilateral triangle is constant; prove that its locus is a circle.

Answer 1

Let $\Delta ABC$ be the given equilateral triangle of side $a$.

Let the point satisfying the given condition ${\left(d_{AB}\right)}^2+{\left(d_{BC}\right)}^2+{\left(d_{AC}\right)}^2=\lambda \to \left(1\right)$ be $(h,k)$.

The Equations of the sides are:

AB: $y=0$

AC: $y=\sqrt{3}x$

BC: $y+\sqrt{3}x=a\sqrt{3}$

Distance of a pt. $(h,k)$ from line $lx+my+n=0$ is:

$d=\frac{lh+mk+n}{\sqrt{l^2+m^2}}$

$d_{AB}$ can be directly found to be $k$.

For the other 2 sides, the distance can be found by the above formula:

$d_{AC}=\frac{k-\sqrt{3}h}{2}$

$d_{BC}=\frac{k+\sqrt{3}h-a\sqrt{3}}{2}$

To find the locus for $(h,k)$ we need to plug in the values above in equation $\left(1\right)$.

Plug in and simplify.

The final result would come out to be:

$\frac{3k^2}{2}+\frac{3h^2}{2}=\lambda -\frac{3a^2}{2}$

$⟹h^2+k^2=\mu$ (RHS was a constant)

This is a equation of a circle.