American Mathematical Monthly Problem 11433 by Marius Cavachi, Constant¸a, Romania.
Let n be a positive integer. Let A1, A2, ..., An, B1, B2, ..., Bn, C1, C2, ..., Cn be 3n points on the unit sphere S2. Prove that there exists a point P onS2 satisfying
n
X
k=1
|P −Ak|2 =
n
X
k=1
|P −Bk|2 =
n
X
k=1
|P −Ck|2.
Solution by Darij Grinberg.
We work with vectors in R3, identifying every pointT with the vector−→
0T, where 0 is the center of the sphere S2. We denote the scalar product of two vectorsv andw by v·w.
Let LAB be the locus of all points P ∈R3 such that
n
P
k=1
|P −Ak|2 =
n
P
k=1
|P −Bk|2. Since
n
X
k=1
|P −Ak|2−
n
X
k=1
|P −Bk|2 =
n
X
k=1
|P −Ak|2− |P −Bk|2
| {z }
=(P2−2P·Ak+A2k)−(P2−2P·Bk+Bk2)
=2P(Bk−Ak)+(A2k−Bk2)
= 2P
n
X
k=1
(Bk−Ak) +
n
X
k=1
A2k−B2k
depends linearly onP, this locusLAB is either a plane or the whole space or the empty set. Since 0∈LAB (because
n
P
k=1
|0−Ak|2 =
n
P
k=1
|0−Bk|2,as|0−Ak|2 =|0−Bk|2 = 1 for every k), the locus LAB cannot be the empty set, so that LAB is either a plane through 0 or the whole space. In either case, the locus LAB contains a plane through 0.
Similarly, denoting by LBC the locus of all pointsP ∈R3 such that
n
P
k=1
|P −Bk|2 =
n
P
k=1
|P −Ck|2, we can see that the locusLBC contains a plane through 0. Thus, the set LAB ∩LBC contains a line through 0 (since the intersection of any two sets, each of which contains a plane through 0, must contain a line through 0). Any line through 0 has nonzero intersection with S2. Thus, (LAB∩LBC)∩S2 6=∅. But this is exactly what the problem asks us to prove.
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