In the diagram, ABCD is a square, with U and V interior points of the sides AB and CD respectively. Determine all the possible ways of selecting U and V so as to maximize the area of the quadrilateral PUQV.

Solution by Steve Dinh, a.k.a. Vo Duc Dien (dedicated to my sister-in-law Dr. Giao Tran)

Let the side of the square be a. From P and Q draw perpendiculars to AD and BC, respectively, and let PE = p and QF = q. Let’s also denote (ABC) the area of shape ABC.

Note that the area of the quadrilateral PUQV is maximum when the total of the shaded areas is minimum. It’s easily seen that the total areas shaded with honey and bricks (AUD) + (BUC) = ½a ( AU + UB) = ½a² and is constant. So now the total areas shaded with squares (PDV) + (QVC) must be minimum.

But also note that (PDV) + (QVC) = (ADV) + (BCV) – (APD) – (BQC) = ½a² – (APD) – (BQC), so (PDV) + (QVC) is minimum when (APD) + (BQC) is maximum.

(It also follows from the Carpets Theorem that (PUQV) = (APD) + (BQC).)

(APD) + (BQC) = ½ a(p+ q) so the requirement now is for p + q to be maximum.

Since both EP and QF || with the vertical sides of the square, we have

p/ AU = DE /a = (a – AE) / a = 1 – AE /a = 1 – p /DV or p [(AU + DV)/(AU × DV)] = 1

or p = (AU × DV) / (AU + DV) Similarly, q = (BU × VC) / (BU + VC)

		p + q  =   (AU × DV)/(AU + DV)   +  (BU × VC)/(BU +VC)   =   

(AU ×DV ×BU + AU ×DV ×VC + AU ×BU ×VC + BU ×VC ×DV)/(AU × BU + AU × VC + DV × BU + DV × VC) =

[AU × BU (DV + VC) + DV ×VC (AU + BU)]/(AU × BU + AU × VC + DV × BU + DV × VC) = a (AU × BU + DV × VC)/(AU × BU + AU × VC + DV × BU + DV × VC)

Now divide both numerator and denominator by sum of products AU × BU + DV × VC, we have

p + q = a / [ 1 + (AU × VC + DV × BU )/(AU × BU + DV × VC)]

so now for p + q to be maximum, (AU × VC + DV × BU)/(AU × BU + DV × VC)

has to be minimum. Let it be k.

But AU = a – BU and DV = a – VC, and k = (AU × VC + DV × BU)/(AU × BU + DV × VC) becomes

k = [(a – BU) VC + (a – VC) BU]/[(a – BU) BU + (a – VC) VC] = [a (VC + BU) – 2 VC × BU ]/[a (VC + BU) – (VC² + BU²)] =

= [a (VC + BU) – 2 VC × BU]/[a (VC + BU) – 2 VC×BU – (VC – BU)²] = 1 / { 1 – (VC – BU)² /[a(VC+BU) – 2VCxBU]}

for k to be minimum the denominator 1 – (VC – BU)² / [ a (VC+BU) – 2VC×BU ] has to be maximum and (VC – BU)² / [ a (VC+BU) – 2VC×BU ] to be minimum. Note that the denominator is not zero, and the square (VC – BU)² is always greater than or equal to zero, and it’s a minimum when it’s zero or when VC = BU.

So to maximize the area of the quadrilateral PUQV, U and V has to be on a horizontal line between the top and bottom sides of the square ABCD. The maximal area of PUQV is then equal

			a²  -  ½ a² - ½ (a/2) × a =  Ό a²