# 1993 IMO Problems/Problem 4

For three points in the plane, we define to be the smallest length of the three heights of the triangle , where in the case , , are collinear, we set . Let , , be given points in the plane. Prove that for any point in the plane,

.

## Solution

First we prove the claim for all points within or on the triangle : In this specific case, suppose without loss of generality that . Then the length of the smallest height of is that of the height from vertex . This has a value of . Similarly, for triangles we have that the length of the smallest height is twice the area of the respective triangle divided by the longest side of that triangle. It is clear that since is the longest side of , no two points within have distance exceeding . Thus, since is within , the longest side of any of the triangles does not exceed . So, we have

as desired.

Now on to prove the assertion for outside the triangle . We shall assume without loss of generality that out of the points point is that farthest from .

If is concave or a degenerate quadrilateral, assume without loss of generality that inside or on triangle . We shall prove that to prove the main claim. If the shortest height of was from vertex then it is clear that the height of from is smaller than that since ray is closer to than ray . The case of the height from being the smallest of triangle is analogous, so we move on to the case of the height from is the smallest. If this is the case, then it is clear that the height of from is smaller than since . Thus the claim is proved.

If, instead, is convex, we can assign the letter to represent to point of intersection of and . Before proving the main claim, we shall prove that for triangle we have . We prove this by considering each of the vertices that the shortest height of is on. If the shortest height is that from , then it is obvious that the height from to side is smaller than that from since is closer to than is and so . If the shortest height of was from , then since the height from to is equal to the height from to , we have . If instead the shortest height of was from , the it is clear that . Thus, the projection of onto is on the same side of as . Now it is obvious that the height of from is smaller than that of from since the ray is closer to than the ray . Thus, we have in all cases. Notice that the case for triangle is analogous. Therefore, we have . But, this is the case of on the triangle , and this case was shown in the first part of this proof.

So, we have that is true in all cases, as desired.