There is an equilateral triangle $ABC$ in the 2D plane. You don't know where it is, but have access to a magical fairy that, given a point $P$ in the same plane, will give you the value of $$f(P) = |PA|^2 + |PB|^2 + |PC|^2$$ i.e. the sum of squares of distances from $P$ to $A,B,C$ (henceforth we will drop the absolute sign)
i) Can you find out the area of triangle $ABC$ using only a finite number of queries to the fairy?
ii) Can you find the minimum number of queries you need to (and are sufficient) to guess the area?
iii) Can you find the locations of the three vertices?
Solution
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Part i)
In a previous post we showed the following:
For any triangle $ABC$ with centroid $M$ and any point $P$ in the same plane
$$ PA^2 + PB^2 + PC^2 = 3 PM^2 + AM^2 + BM^2 + CM^2$$
(Can easily be proven using complex numbers too, if required).
i.e we have $$f(P) = 3 PM^2 + S$$
where $S$ is a constant, independent of $P$ (and is actually a constant times the area of $ABC$, for equilateral $ABC$).
Now pick points $X, Y$ (say $(0,0), (1,0)$) and query the fairy for $X$ and $Y$.
We have that $$f(X) - f(Y) = 3(XM^2 - YM^2)$$
i.e $M$ lies on the locus of points $Q$ such that $$QX^2 - QY^2 = k$$
$k$ is a constant.
By simple co-ordinate geometry, this actually can easily be seen to be a line perpendicular to $XY$ (extended, if needed).
Thus using two points, we narrow down $M$, the centroid of $ABC$ to a straight line. With another point $Z$ (say $(0, 1)$), we get another line, and take the intersection of the two lines.
Once we have the location of $M$, we can compute $XM^2$ directly, and then use the already obtained $f(X)$ to compute the area of the equilateral triangle $ABC$, using
$$ AM^2 + BM^2 + CM^2 = f(X) - 3 XM^2$$
Thus three queries are sufficient.
Part ii)
Three queries are also necessary.
Suppose two were enough, (say X and Y), then as before $M$ lies on a line perpendicular to $XY$. Now if $M$ were the actual point, then the reflection $M'$ of $M$ on $XY$ gives the same $f(X)$ and $f(Y)$. Thus two queries are not enough.
Part iii)
Pinpointing the vertices is impossible! Any rotation of the $ABC$ around the centroid gives the same $f(P)$ values.
