Let ABC be a triangle and X an interior point of ABC. Show that at least one of the angles XAB, XBC, XCA is less than or equal to 30o.
Solution
By Marcin Mazur, University of Illinois at Urbana-Champaign
Let P, Q, R be the feet of the perpendiculars from X to BC, CA, AB respectively. Use A, B, C to denote the interior angles of the triangle (BAC, CBA, ACB). We have PX = BX sin XBC = CX sin(C - XCA), QX = CX sin XCA = AX sin(A - XAB), RX = AX sin XAB = BX sin(B - XBC). Multiplying: sin(A - XAB) sin(B - XBC) sin(C - XCA) = sin A sin B sin C.
Now observe that sin(A - x)/sin x = sin A cot x - cos A is a strictly decreasing function of x (over the range 0 to π), so if XAB, XBC and XCA are all greater than 30, then sin(A - 30) sin(B - 30o) sin(C - 30o) > sin330o = 1/8.
But sin(A - 30o) sin(B - 30o) = (cos(A - B) - cos(A + B - 60o))/2 ≤ (1 - cos(A + B - 60o))/2 = (1 - sin(C - 30o))/2, since (A - 30o) + (B - 30o) + (C - 30o) = 90o. Hence sin(A - 30o) sin(B - 30o) sin(C - 30o) ≤ 1/2 (1 - sin(C - 30o)) sin(C - 30o) = 1/2 (1/4 - (sin(C - 30o) - 1/2)2) ≤ 1/8. So XAB, XBC, XCA cannot all be greater than 30o.
By Jean-Pierre Ehrmann
P, Q, R as above. Area ABX + area BCX + area CAX = area ABC, so AB·XR + BC·XP + CA·XQ = 2 area ABC ≤ BC·AP ≤ BC(AX + XP). Hence AB·XR/AX + CA·XQ/AX ≤ BC.
Squaring and using (λ + μ)2 ≥ 4 λμ, we have: BC2 ≥ 4 AB·CA. XR·XQ/AX2. Similarly: CA2 ≥ 4 BC·AB·XP·XR/BX2, and AB2 ≥ 4 AB·BC·XQ·XP/CX2.
Multiplying these three inequalities together gives: 1 ≥ 64 (XR/AX)2(XP/BX)2(XQ/CX)2, and hence: (XR/AX) (XP/BX) (XQ/CX) ≤ 1/8, or sin XAB sin XBC sin XCA ≤ 1/8. So not all XAB, XBC, XCA are greater than 30o.
Gerard Gjonej noted that the result follows almost immediately from the Erdos-Mordell inequality: XA + XB + XC ≥ 2(XP + XQ + XR). [For if all the angles are greater than 30, then XR/XA, XP/XB, XQ/XC are all greater than sin 30o = 1/2.]. This result was notoriously hard to prove - Erdos hawked it around a large number of mathematicians before Mordell found a proof - but the proof now appears fairly innocuous, at least if you do not have to rediscover it:
Let R1, Q1 be the feet of the perpendiculars from P to AB, CA respectively. Similarly, let P2, R2 be the feet of the perpendiculars from Q to BC, AB, and Q3, P3 the feet of the perpendiculars from R to CA, BC. Then P2P3 is the projection of QR onto BC, so P2P3/QR ≤ 1. Similarly, Q3Q1/RP ≤ 1, and R1R2/PQ ≤ 1. Hence XA + XB + XC ≥ XA.P2P3/QR + XB·Q3Q1/RP + XC·R1R2/PQ (*)
Now BPXR is cyclic, because BPX and XRB are both right angles. Hence angle BXR = angle BPR = angle RPP3, so triangles XBR and PRP3 are similar. Hence PP3 = PR.XR/XB.
Similarly, QQ1 = QP·XP/XC, RR2 = RQ·XQ/XA, and PP2 = PQ·XQ/XC, QQ3 = QR·XR/XA, RR1 = RP·XP/XB. Substituting into (*), we obtain:
XA + XB + XC ≥ XA( PQ/QR XQ/XC + PR/QR XR/XB ) + XB( QR/RP XR/XA + QP/RP XP/XC ) + XC( RP/PQ XP/XB + RQ/PQ XQ/XA ).
On the right hand side, the terms involving XP are: XP( QP/RP XB/XC + RP/PQ XC/XB ), which has the form XP (x + 1/x) and hence is at least 2 XP. Similarly for the terms involving XQ and XR.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
7 Sep 1999