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Math Magazine Problem 1560 Solution
by Peter Y. Woo, Biola University
4/18/99
Problem. (by Wu Wei Chao, Guang Zhou Normal Univ., China) Let BPTACQ be an arc Z of a circle such that BT = CT and AP = AQ = Ö(AB×AC). Let [K] denote the area of any polygon K.
(a) If ÐBAC ³ 90°, prove that [ABC] > [APQ].
(b) If AP £ BC, prove that [TBC] > [APQ].
Proof. Let O be the center of the circle. Let P", Q" be two points on the circle such that TP" = TQ" = AP = AQ = Ö (AB × AC). We need two lemmas.
Lemma 1: If X is a variable point on the arc BTC, then BX + XC is a monotonic decreasing function of ÐTOX.
Proof: Let O be the center and R the radius of the circle. For any X on the arc BTC, Let b = (½)ÐBOX = ÐC and g = (½)ÐCOX = ÐB. Then
BX + XC = 2 R (sin b + sin g) = R sin ((b+g)/2) cos ((b-g)/2) . Here (b+g)/2 is constant, and cos ((b - g)/2) = cos (C - B) = cos ÐTOX is a monotonic decreasing function of ÐTOX. QED
Corollary 1. TP" = TQ" = AP = AQ < (½)(AB + AC) < TB.
Proof. The geometric mean of AB and AC < arithmetic mean of them.
Lemma 2. Let ABC be a variable isosceles triangle inscribed in a fixed circle with radius 1, with AB = AC. Then as ÐA shrinks from p to 0, the area [ABC] increases monotonically from 0 to a maximum when ÐA = p/3, then decreases monotonically back to 0.
Proof. Let ÐA = 2q, then [ABC] = f(q) = (½)AB.AC sin 2q
= (½) (2 cos q)2 2 sin q cos q = 4 sin q cos3 q. A bit of calculus would yield the monotonicity, and the fact that maximum of f(q) occurs when q = p/6. QED
Proof of (a). Now assume ÐBAC > 90°. then by Cor. 1, TP" < TB and P"Q", BC are parallel chords with P"Q" < BC.
Hence [APQ] = [TP"Q"] = (½)TP×TQ sin ÐP"TQ" = (½)AP×AQ sin ÐP"TQ"
< (½)AP×AQ sin ÐBTC (due to ÐP"TQ" > ÐBTC > 90°)
= (½)AP×AQ sin ÐBAC = [ABC]. QED
Proof of (b). We now do not assume ÐBAC > 90°, but assume AP £ BC. We have two cases:
Case 1: ÐBTC ³ p/3. Then TB £ BC, so that 2 TP" < AB + BC < TB + TC = 2 TB, So TP" < TB, so by Lemma 2, [TBC] > [TP"Q"] = [APQ].
Case 2: ÐBTC < p/3. Then TB = TC > BC. Let P' be on the circle, on the same side of BC as T, such that BC = CP'. Then [TBC] > [P'BC] because P' is closer to line BC than T is. [P'BC] ³ [P"TQ"] because TP" £ BC, using Lemma 2. Hence [TBC] > [TP"Q"] also. QED
