38th Eötvös Competition Problems 1934

1.  E is the product 2·4·6 ... 2n, and D is the product 1·3·5 ... (2n-1). Show that, for some m, D·2^m is a multiple of E.

2.  Given a circle, find the inscribed polygon with the largest sum of the squares of its sides.

3.  For i and j positive integers, let R_ij be the rectangle with vertices at (0, 0), (i, 0), (0, j), (i, j). Show that any infinite set of R_ij must have two rectangles one of which covers the other. 

Solutions

Problem 1

E is the product 2·4·6 ... 2n, and D is the product 1·3·5 ... (2n-1). Show that, for some m, D·2^m is a multiple of E.

Solution

We have D = (2n)!/(2ⁿ n!), so 2²ⁿD/E = 2nCn, which is an integer. 

Problem 2

Given a circle, find the inscribed polygon with the largest sum of the squares of its sides.

Solution

If there are more than 3 vertices, then the mean angle (between two adjacent sides) is ≥ 90^o, so there must be a vertex with angle ≥ 90^o. But now we can remove that vertex without increasing the sum of the squares.

So let us consider triangles.

Let M be the midpoint of BC. Applying the cosine formula to AMB and AMC and adding, we get AB² + AC² = 2AM² + 2BM². So if we keep B and C fixed and allow A to vary on the circle, then we maximize AB² + AC² by maximizing AM². But that is clearly achieved by taking AM to contain O the center of the circle (for AM² = AO² + OM² - 2·AO·OM cos AOM, which is maximized by taking ∠AOM = 180^o). Thus the triangle ABC is not optimal unless each pair of sides is equal, so the optimal triangle must be equilateral. Note that none of its sides are diameters, so the triangles we derive from any 4-gon must have smaller sum of squares, so no 4-gons are optimal. Since all n-gons are worse than 4-gons (since they have a vertex with angle > 90^o), no n-gon is optimal for n > 4. Thus the unique optimal polygons are the equilateral triangles. 

Problem 3

For i and j positive integers, let R_ij be the rectangle with vertices at (0, 0), (i, 0), (0, j), (i, j). Show that any infinite set of R_ij must have two rectangles one of which covers the other.

Solution

Take any rectangle R_ij. There are only finitely many rectangles R_mn with m ≤ i and n ≤ j. So there must be infinitely many with m > i or infinitely many with n > j. wlog we may assume there are infinitely many with m > i. They must all have n < j or they would contain R_ij. But there are only finitely many values for n available < j, so infinitely many must share some value n' < j. But now given any two of these, one contains the other.