Problem 11871
(American Mathematical Monthly, Vol.122, November 2015) Proposed by Cezar Lupu and Stefan Spataru (USA).
Let ABC be a triangle in the Cartesian plane with vertices in Z2. Show that, if P is an interior lattice point ofABC, then at least one of the angles P AB, P BC, and P CA has a radian measure that is not a rational multiple ofπ.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Preliminary remarks:
- if θ is the radian measure of an angle of a triangle with vertices in Z2then θ = π/2 or tan(θ) ∈ Q;
- the only rational values of tan(qπ) for q ∈ Q are 0 and ±1.
Assume that P AB = q1π, P BC = q2π, and P CA = q3π where q1, q2, q3∈ Q.
Moreeover q1, q2, q3> 0 and q1+ q2+ q3< 1 because P is an interior point.
Since the triangles P AB, P BC, and P CA have vertices in Z2, by the above remarks it follows that tan(q1π) = tan(q2π) = tan(q3π) = 1, that is q1= q2= q3= 1/4. Hence
CAP + ABP + BCP = π − (P AB + P BC + P CA) = π/4 which implies
sin(CAP ) sin(ABP ) sin(BCP ) < sin(π/4)3= 1 2√
2. On the other hand, by Ceva’s theorem,
sin(CAP ) sin(ABP ) sin(BCP ) = sin(P AB) sin(P BC) sin(P CA) = 1 2√
2
and we have a contradiction.
