analytic geometrycircleseuclidean-geometrygeometrytriangles
An equilateral triangle is inscribed in a circle of radius $r$. If $P$ is any point on the circumference find the value of $PA^2 + PB^2 + PC^2$.
I have managed to solve this problem using co-ordinate geometry by taking a triangle with its centroid at (0,0) and assuming a point at $(r\cos\theta , r\sin \theta)$.
A simpler proof using geometry however eludes me. Is it possible to prove this using Euclidean geometry or without a calculation intensive approach?
The answer is
$6r^2$
Two circles + five lines = 7 steps:
Select an arbitrary point $P_0$ on a given circle $C_0$ and draw a circle $C_1$ with an arbitrary radius $r$, which is small enough to intersect $C_0$.
Draw a circle $C_2$ centered at the intersection point $P_1$ with the same radius $r$ to get the intersection points $P_2$ and $P_3$.
Draw a line through points $P_2,P_1$, intersecting $C_0$ at point $P_4$.
Draw a line through points $P_1,P_3$, intersecting $C_0$ at point $P_5$.
Draw a side $P_5 P_4$ of the inscribed equilateral triangle.
Draw a side $P_4 P_0$ of the inscribed equilateral triangle.
Draw a side $P_0 P_5$ of the inscribed equilateral triangle.
using the cosine rule as you state the third side is $140.73$.
then, since the sides are tangents to the incircle, we may write: $$ 119 = x + y \\ 202 = x+z \\ 140.73 = y+z $$ these equations give $x=90.136$
then $$ \begin{align} r &= x \tan 21.5^{\circ}\\ & = 35.5 \end{align} $$ (apologies if my calculations are in error, but i think the method suggested will work)
Best Answer
$PA^2+PB^2+PC^2$ is the moment of inertia of $\{A,B,C\}$ with respect to $P$, which by the parallel axis theorem only depends on the distance of $P$ from the centroid of $ABC$. In particular, if $ABC$ is equilateral and $P$ belongs to the circumcircle of $ABC$, $PA^2+PB^2+PC^2 = AA^2+AB^2+AC^2=6r^2. $