# 1992 Canadian Mathematical Olympiad, Problem 2

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 =  ¼ a²
```