21th Eötvös Competition Problems 1914

1.  Circles C and C' meet at A and B. The arc AB of C' divides the area inside C into two equal parts. Show that its length is greater than the diameter of C.

2.  a, b, c are reals such that |ax² + bx + c| ≤ 1 for all x ≤ |1|. Show that |2ax + b| ≤ 4 for all |x| ≤ 1.

3.  A circle meets the side BC of the triangle ABC at A₁, A₂. Similarly, it meets CA at B₁, B₂, and it meets AB at C₁, C₂. The perpendicular to BC at A₁, the perpendicular to CA at B₁ and the perpendicular to AB at C₁ meet at a point. Show that the perpendiculars at A₂, B₂, C₂ also meet at a point. 

Solutions

Problem 1

Circles C and C' meet at A and B. The arc AB of C' divides the area inside C into two equal parts. Show that its length is greater than the diameter of C.

Solution

 

The center O of C must lie inside C', because otherwise we take the tangent to C' at the point of C' between its center O' and O. Then the area inside C on the same side of the tangent as O is more than half its total area, but all outside C'. Contradiction.

Take the diameter AA' of C. It must intersect the arc AB, because otherwise the area inside C' is obviously smaller. So suppose they meet at X. Then the length of the arc AB is greater than AX + XB. But XB > XA' (for example, B will lie outside the circle center X radius XA'). So AX + XB > AX + XA' = AA'. 

Problem 2

a, b, c are reals such that |ax² + bx + c| ≤ 1 for all x ≤ |1|. Show that |2ax + b| ≤ 4 for all |x| ≤ 1.

Solution

The graph of 2ax + b is a straight line, so its max and min values are at x = ±1. So it is sufficient to show that |2a+b| ≤ 4 and |-2a+b| = |2a-b| ≤ 4. Putting x = 0, ±1 in the inequality given we get |c| ≤ 1, |a+b+c| ≤ 1 and |a-b+c| ≤ 1. Subtracting the first to the other two gives |a+b| ≤ 2, |a-b| ≤ 2. Adding gives |a| ≤ 2. Adding to the previous two gives: |2a+b| ≤ 4, |2a-b| ≤ 4, as required. 

Problem 3

A circle meets the side BC of the triangle ABC at A₁, A₂. Similarly, it meets CA at B₁, B₂, and it meets AB at C₁, C₂. The perpendicular to BC at A₁, the perpendicular to CA at B₁ and the perpendicular to AB at C₁ meet at a point. Show that the perpendiculars at A₂, B₂, C₂ also meet at a point.

Solution

Let the perpendiculars at A₁, B₁, C₁ meet at P and let O be the center of the circle. Take Q so that O is the midpoint of PQ. The perpendicular from O meets BC at the midpoint of A₁A₂, so Q must lie on the perpendicular to BC at A₂. Similarly since the perpendicular from O meets AC at the midpoint of B₁B₂, Q must lie on the perpendicular at B₂, and similarly on the perpendicular at C₂.