37th Eötvös Competition Problems 1933

1.  If x² + y² = u² + v² = 1 and xu + yv = 0 for real x, y, u, v, find xy + uv.

2.  S is a set of 16 squares on an 8 x 8 chessboard such that there are just 2 squares of S in each row and column. Show that 8 black pawns and 8 white pawns can be placed on these squares so that there is just one white pawn and one black pawn in each row and column.

3.  A and B are points on the circle C, which touches a second circle at a third point P. The lines AP and BP meet the second circle again at A' and B' respectively. Show that triangles ABP and A'B'P are similar.

Solutions

Problem 1

If x² + y² = u² + v² = 1 and xu + yv = 0 for real x, y, u, v, find xy + uv.

Solution

If u = 0, then v = ±1, and yv = 0, so y = 0. Hence xy + uv = 0. So assume u ≠ 0. Then x = yv/u. So 1 = x² + y² = y²(v²/u² + 1) = y²/u², so y = ±u. If y = u, then x + v = 0, so x = -v and hence xy + uv = 0. If y = -u, then -x + v = 0, so x = v and xy + uv = 0. So in all cases xy + uv = 0. 

Problem 2

S is a set of 16 squares on an 8 x 8 chessboard such that there are just 2 squares of S in each row and column. Show that 8 black pawns and 8 white pawns can be placed on these squares so that there is just one white pawn and one black pawn in each row and column.

Solution

Take a graph with 8 points r₁, ..., r₈ and 8 points c₁, ... , c₈. Join r_i and c_j iff the square in row r_i and column c_j of the chessboard is in S. Now every point of the graph has degree 2. We can partition the edges into disjoint cycles. For given any point P, just trace out a path from P. Eventually the path must reach a point Q already on the path. Q must be P, because otherwise it would have degree > 2. So P belongs to a cycle. But any two cycles must be disjoint or we would have a point with degree > 2. Since we must alternate between r-points and c-points as we go around a cycle, each cycle must have an even number of points. So we can label the edges around each cycle alternately black and white. Now every point must have one black and one white edge, so there must be one black and one white pawn in each row and column. 

Problem 3

A and B are points on the circle C, which touches a second circle at a third point P. The lines AP and BP meet the second circle again at A' and B' respectively. Show that triangles ABP and A'B'P are similar.

Solution

Let the centers of the circles be O, O'. Consider triangles OPA and O'PA'. There are isosceles and ∠OPA = ∠O'PA', so they are similar. Hence PA/PA' = OA/OA'. Similarly, PB/PB' = OB/OB', so PA/PA' = PB/PB'. But ∠APB = ∠A'PB'. Hence APB and A'PB' are similar.