For t(0,2), if ABC is an equilateral triangle with vertices A( t,- t), B( t, t) and C(a,b)…

Question

For $t\in(0,2\pi)$, if $ABC$ is an equilateral triangle with vertices $A(\sin t,-\cos t),\ B(\cos t,\sin t)$ and $C(a,b)$ such that its orthocentre lies on a circle with centre $\left(1,\dfrac13\right)$, then $(a^2-b^2)$ is equal to:

Accepted Answer

For an equilateral triangle, orthocentre = centroid: $H(h,k) = \left(\dfrac{\sin t+\cos t+a}3,\dfrac{-\cos t+\sin t+b}3\right)$.

So $(3h-a)^2+(3k-b)^2 = (\sin t+\cos t)^2+(\sin t-\cos t)^2 = 2$ (constant, since $A,B$ lie on the unit circle at fixed angular separation), i.e.
$$\left(h-\frac a3\right)^2+\left(k-\frac b3\right)^2 = \frac29$$

This is a circle centered at $\left(\dfrac a3,\dfrac b3\right)$; matching to the given centre $\left(1,\dfrac13\right)$: $a=3,\ b=1$.

$a^2-b^2 = 9-1=8$

Solutions · 1