The following simple problem has been included in two of Ross Honsberger's books. In the Mathematical Morsels he gave a solution by C. M. Ingleby and later in Mathematical Gems III added two more solutions, one by I. van Yzeren and the other by Karel A. Post. The three solution are given below. More recently, M. Hajja published 3 short trigonometric proofs and pointed to the relation of the diagram in the first solution below to the Broken Chord Theorem. The fourth solution on this page exploits Hajja's observation.
Solution 1
With center O draw a circle through A and B and extend AO to AC and AP to AQ, with C and Q on the circle.
AO = BO = CO,
implying that
∠AOB = 2 ∠ACB (= 2 ∠OBC.)
On the other hand, ∠APB = ∠AOB and ∠AQB = ∠ACB. Thus,
∠APB = 2 ∠AQB.
But, as an exterior angle of ΔBPQ, ∠APB = ∠AQB + ∠OBQ, so that ∠OBQ = ∠AQB, making ΔBPQ isosceles:
BP = PQ.
It follows that AQ = AP + PQ and AC = AO + CO. Both AQ and AC are chords in the extra circle, with AC being a diameter, the longest chord: AQ < AC.
Solution 2
Let K be the center of the circle (with arc AOC). Draw KE, the bisector of ∠AKP and KF, the bisector of ∠PKB and add EL ⊥ KP and FM ⊥ KP. If W is the point of intersection of AP and KE, then KW ⊥ AP and AW = PW.
By symmetry, PW = EL. Hence, EL = AP/2. Similarly, FM = BP/2. Let N be the intersection of KP and EF. Triangles ELN and FMN are right. Their hypotenuses exceed their legs: EN > AL and FM > FN. Adding up
(AP + BP) / 2 = EL + FM > EN + FN = EF.
But ∠EKF = ½∠AKB = ∠AKO = ∠OKB, making the subtending chords equal:
EF = AO = BO.
Finally, (AP + BP) / 2 > EF = (AO + BO) / 2.
Solution 3
Let X be the intersection of AP and BO.
Triangles AOX and BPX are equiangular and, thus, similar. With P between O and B, ΔAOX is the larger of the two:
AO/BP = AX/BX = OX/PX = k > 1.
We are concerned with the difference (AO + BO) - (AP + BP):
AO + BO = 2·AO ≥ 2·AU = AP + BP.
