### 16th Putnam 1956

**Problem B3**

ABCD is an arbitrary tetrahedron. The inscribed sphere touches ABC at S, ABD at R, ACD at Q and BCD at P. Show that the four sets of angles {ASB, BSC, CSA}, {ARB, BRD, DRA}, {AQC, CQD, DQA}, {BPC, CPD, DPB} are the same.

**Solution**

The key is to notice that the two angles in the sets subtended by the same side of the tetrahedron are the same. For example ∠ASC = ∠AQC. Let O be the centre of the sphere and r its radius. Then AQ^{2} + r^{2 = }AO^{2} = AS^{2} + r^{2}, so AQ = AS. Similarly, CQ = CS. So the triangles AQC, ASC are similar.

Now the sum of the angles in each set is the same. But ∠ASB in the first set equals ∠ARB in the second set, so ∠BSC + ∠CSA = ∠BRD + ∠DRA. Similarly, ∠CQD in the third set equals ∠CPD in the fourth, so ∠AQC + ∠DQA = ∠BPC + ∠DPB. Adding these two equations and using ∠BSC = ∠BPC, ∠CSA = ∠AQC, ∠BRD = ∠DPB, ∠DRA = ∠DQA gives ∠CSA = ∠BRD. In other words, the angles in the set subscribed by opposite sides of the tetrahedron are also the same.

That gives us all we need: ∠ASC = ∠BRD = ∠AQC = ∠BPD; ∠ASB = ∠ARB = ∠CQD = ∠CPD; ∠BSC = ∠ARD = ∠AQD = ∠BPC.

16th Putnam 1956

© John Scholes

jscholes@kalva.demon.co.uk

4 Dec 1999