Antipodal pair
Named by 6 essays across one field — each of them below, with the objects they name alongside it.
One line that halves them both
Two shapes lying anywhere on a page, of any sizes and any shapes at all. There is always a single straight line that cuts both of them into two equal halves at once — and finding it needs no cleverness, only the observation that a quantity which reverses sign has to pass through zero.
Two opposite points that agree twice
At any moment there are two points on opposite sides of the Earth with the same temperature and the same pressure. On a seeded globe they sit at 11.9°N 44.6°E and 11.9°S 135.4°W. The reason is the circle argument that halved two shapes, run one dimension up: the differences between opposite readings, walked round the equator, wind round zero an odd number of times — and an odd number cannot be zero.
As many cuts as colours
Two thieves steal a necklace and want half of every colour of bead each. However the beads are strung, they never need more cuts than there are colours — three cuts for three colours, four for four — and sometimes they need every one. The guarantee is the Borsuk–Ulam theorem again, with a point on a sphere read as a way of cutting the necklace, and every necklace of several small kinds has been checked against it.
Opposite labels that have to meet
Cut a square into triangles, label every corner +1, −1, +2 or −2, and insist only that opposite points of the edge get opposite labels. Somewhere inside, an edge must join a label to its negative. The proof counts quarter-turns round a diamond — an odd number on the boundary, zero in any triangle that avoids opposites — and making the triangles smaller turns the count back into the theorem about opposite points on the Earth.
The colours a circle forces
Take every pair from five things and join two pairs when they share nothing. Three colours are enough to colour the result so joined pairs differ, and two are not — but no triangle, no dense cluster and no counting argument explains why. The reason is five points on a circle and a direction that cannot be told apart from its opposite, and the same reason, one sphere at a time, settles Kneser's question for every size.
Several colours on every vertex
Give every pair from six points three colours, so that pairs with nothing in common share no colour. Counting says nine colours might do; ten are needed. Stahl conjectured in 1976 exactly how many colours every such problem needs — a formula that meets Lovász's topological answer at one colour a vertex and the obvious answer at k — and a search over stars and triangles confirms it in every case small enough to run.
Named alongside it
The objects these essays reach for when they reach for this one.
Existence proofContinuityExhaustive searchNonconstructiveWinding numberBrouwerChromatic numberComplexityDegreeFair divisionGraph colouringIntermediate value theorem