2nd USAMO 1973

Show that if two points lie inside a regular tetrahedron the angle they subtend at a vertex is less than π/3.

Solution

Let the tetrahedron be ABCD and the points be P and Q. Note that we are asked to prove the result for any vertex, not just some. So consider angle PAQ. Let the rays AP, AQ meet the plane BCD at P', Q' respectively. So we have to show that angle P'AQ' < 60^o for P' and Q' interior points of the triangle BCD. Extend P'Q' to meet the sides of the triangle at X and Y. Without loss of generality, X lies BC and Y lies on CD. Obviously if is sufficient to show that angle XAY < 60^o.

X and Y cannot both be vertices (or P', Q' would not have been interior points of the triangle and hence P and Q would not have been strictly inside the tetrahedron). So suppose X is not a vertex. We show that XY <= XD. Consider triangle XYD. ∠XDY < ∠BDC = 60^o, but ∠XYD = ∠XCD + ∠CXY ≥ 60^o, so ∠XDY < ∠XYD. Hence XY ≤ XD. But XD = AX (consider, for example, the congruent triangles AXB and DXB). Hence XY < AX. Similarly, XY ≤ AY. Hence angle XAY < 60^o.

2nd USAMO 1973

© John Scholes
jscholes@kalva.demon.co.uk
19 Aug 2002