# 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.