ABCD is a cyclic quadrilateral. A circle whose center is on the side AB touches the other three sides. Show that AB = AD + BC. What is the maximum possible area of ABCD in terms of |AB| and |CD|?
Answer
(h/2 + k/2) √(hk/2 - h2/4), where h = |CD|, k = |AB|
Solution
Let the circle have center O on AB and radius r. Let ∠OAD = θ, ∠OBC = φ. Since ABCD is cyclic, ∠ADC = 180o-φ, so ∠ODA = 90o-φ/2. If AD touches the circle at X, then AD = AX + XD = r cot θ + r tan(φ/2). Similarly, BC = r cot φ + r tan(θ/2). Put t = tan(θ/2). Then cot θ = (1-t2)/2t, so cot θ + tan(θ/2) = (1+t2)/2t = 1/sin θ. Similarly for φ, so AD + BC = r/sin θ + r/sin φ = AO + OB = AB.
Suppose AD and BC meet at H (we deal below with the case where they are parallel). Then HCD and HAB are similar, so area HCD = (CD2/AB2) area HAB and area ABCD = (1 - CD2/AB2) area HAB. Also AB/CD = HA/HC = HB/HD = (HA+HB)/(HC+HD) = (HA+HB)/(HB-BC+HA-DA) = (HA+HB)/(HA+HB-AB). Hence HA+HB = AB2/(AB-CD), which is fixed. Now for fixed HA+HB we maximise the area of HAB by taking HA = HB and hence AD = BC.
Put h = CD, k = AB. So k cos θ + h = k. Hence cos θ = (1-h/k). Hence sin θ = √(2h/k - h2/k2). So area ABCD = ½(h+k) ½ k sin θ = (h/2 + k/2) √(hk/2 - h2/4) (*).
If AD and BC are parallel then A and B must lie on the circle, so that ∠DAB = ∠ABC = 90o. But ABCD is cyclic, so it must be a rectangle. Hence AB = CD and area ABCD = k2/2. In this case (*) still gives the correct answer.
Thanks to Vivek Kumar Mehra
© John Scholes
jscholes@kalva.demon.co.uk
3 February 2004
Last corrected/updated 3 Feb 04