5th Eötvös Competition Problems 1898

1.  For which positive integers n does 3 divide 2ⁿ + 1?

2.  Triangles ABC, PQR satisfy (1) ∠A = ∠P, (2) |∠B - ∠C| < |∠Q - ∠R|. Show that sin A + sin B + sin C > sin P + sin Q + sin R. What angles A, B, C maximise sin A + sin B + sin C?

3.  The line L contains the distinct points A, B, C, D in that order. Construct a square such that two opposite sides (or their extensions) intersect L at A, B, and the other two sides (or their extensions) intersect L at C, D. 

Solutions

Problem 1

For which positive integers n does 3 divide 2ⁿ + 1?

Solution

2 = 3-1, so 2ⁿ + 1 = (-1)ⁿ + 1 mod 3. So it is divisible by 3 iff n odd. 

Problem 2

Triangles ABC, PQR satisfy (1) ∠A = ∠P, (2) |∠B - ∠C| < |∠Q - ∠R|. Show that sin A + sin B + sin C > sin P + sin Q + sin R. What angles A, B, C maximise sin A + sin B + sin C?

Solution

sin B + sin C = sin( (B+C)/2 + (B-C)/2) + sin( (B+C)/2 - (B-C)/2) = 2 sin((B+C)/2) cos((B-C)/2). Now B+C = Q+R, and |∠B - ∠C| < |∠Q - ∠R| so cos((B-C)/2) > cos((Q-R)/2). Hence sin B + sin C > sin Q + sin R.

If A, B, C are unequal, then we can increase the sum by moving to A, (B+C)/2, |B-C|/2. So the sum is maximised for A = B = C = 60^o. 

Problem 3

The line L contains the distinct points A, B, C, D in that order. Construct a square such that two opposite sides (or their extensions) intersect L at A, B, and the other two sides (or their extensions) intersect L at C, D.

Solution

Let the perpendicular from C to meet the far square side at X (as shown). Take the perpendicular to L through B to meet the square side through A at Y. Take the perpendicular through Y to the opposite square side to meet it at Z. Then triangles BYZ and DCX are congruent. Hence BY = CD.

So we can construct Y without knowing the position of the square. The rest of the construction is then trivial. (Join AY, take perpendiculars to it through C and D and a line through B parallel to AY).