## IMO 1995 Day 1

I decided to test myself on solving some IMO problems. Here are my solutions for IMO 1995 Day 1. Apart of the last problem which seemd a little more tricky, the first two were pretty straightforward.

**Problem 1** Let be four distinct points on a line, in that order. The circles with diameters and intersect at and . The line meets at . Let be a point on the line other than . The line intersects the circle with diameter at and , and the line intersects the circle with diameter at and . Prove that the lines are concurrent.

**Solution:** Denote by the circles with respective diameters. It is immediate that is the radical axis of and and of . Moreover, is the radical axis of the circles and and is the radical axis of the circles and . The radical axis of two circles is characterized by the fact it is the locus of the points which have the same power with respect to the two circles.

Denote by the intersection of and . By the above facts we have that has the same power with respect to the circles , and as a consequence, it is on the radical axis of the circles and which is . This proves that the three lines are concurrent.

**Problem 2.** Let , , be positive real numbers such that . Prove that

**Solution:** Since the inequality is equivalent to

Applying Cauchy-Schwarz and AM-GM we get

**Problem 3.** Determine all integers for which there exist points in the plane, no three collinear, and real numbers such that for , the area of is .

**Solution:** We note right away that if (in this order) correspond to vertices of a convex quadrilateral then

For the answer is of course yes. Take a square and four equal numbers equal to half the area of the square divided by .

First note that if the statement of the problem is not true for , then it is not true for all . For my intuition said that we cannot find such points in the plane, and a well known problem came to mind:

*If we have 5 points in the plane such that no three of them lie on the same line then there are four of them which form a convex quadrilateral.*

The solution to this problem auxiliary problem led to the solution of the third problem in the following way: the convex hull of non-collinear points is a triangle, a convex quadrilateral or a convex pentagon.

- Case 1 The convex hull of the points is a convex pentagon. Then using relation it is easy to see that the numbers must be equal. But then the triangles have the same area and are collinear. Contradiction.
- Case 2 The convex hull is a convex quadrilateral . Without loss of generality suppose that . Using the same relation we find that and therefore the areas of the triangles and are the same, which is impossible, since one of them is strictly contain in the other one.
- Case 3 The convex hull is a triangle and are in the interior of this triangle. Without loss of generality suppose that the line intersects the lines and and is closer to . Since we have
by the same relation we get

which is equivalent to

which is again a contradiction.