# 2008 Austrian Mathematical Olympiad, Problem 3

The line g is given, and on it lie the four points P, Q, R, and S (in this order from left to right).

Construct all squares ABCD with the following properties: P lies on the line through A and D. Q lies on the line through B and C. R lies on the line through A and B. S lies on the line through C and D.

Solution by Steve Dinh, a.k.a. Vo Duc Dien It is easily recognized that A must lie on the circle with diameter PR, D on circle with diameter PS, and C on circle with diameter QS.

Let DS = d, CS = c, AP = a and DP = b. For ABCD to be a square, AD || BC, DC || AB and b  a = d  c. We then have

d / PS = c / QS = (d  c) / (PS  QS) = CD / (PS  QS) and b / PS = a / PR = (b  a) / (PS  PR) = AD / (PS  PR)

Now CD = AD gives us:

d (PS  QS) / PS = b (PS  PR) / PS or

d = b (PS  PR)/ (PS  QS)

We also have b² + d² = PS²

From those two equations, we have b² + b²(PS  PR)² / (PS  QS)² = PS² or

b² [(PS  QS)² + (PS  PR)²]/ (PS  QS)² = PS² or b = PS (PS  QS) / \sqrt {(PS  QS)² + (PS  PR )²}

The square is then defined. Its mirror image ABCD across g is also a solution.

The following construction is implied.

Construct circles with diameters PQ, RS, and PS. Using b as the radius and P the center, mark point D on the big circle. Join D and S to obtain C. Join D and P to obtain A. Let B be the intersection of AR and CQ. Since angles PDS, PAR, QCR are subtended by the diameters of the three circles, all three are right, and so is ∠ABC. The quadrilateral ABCD is then a reactangle. One needs to show that ist sides are equal. This is left as an exercise.