1965 IMO Problems/Problem 6

Problem

In a plane a set of $n$ points ( $n\geq 3$ ) is given. Each pair of points is connected by a segment. Let $d$ be the length of the longest of these segments. We define a diameter of the set to be any connecting segment of length $d$ . Prove that the number of diameters of the given set is at most $n$ .

Solution

Image of problem Solution. Credits to user awe-sum.

Remarks (added by pf02, October 2024)

1. As a public service, I will upload the "Image of problem Solution" to this web page. That way, a reader can see the "Solution" immediately, without having to go to another web site, and we are not subjected to the imgur.com website being taken down, or Imgur's parent company deciding to delete this particular image. Credits for the image are due to user awe-sum, as pointed out above.

2. This "Solution" is presented very badly, and edited very badly. Indeed, some terms are undefined, left to the reader to make sense of (e.g. "incident", "degree"). But let us be forgiving, and let us do our best to make sense of the "Solution".

3. The "Solution" is incomplete, to the point of not being a solution. Some serious questions are not addressed, and a reader can not be expected to fill in the details. These are:

a. The author says "... at least on point is incident to at least three diagonals." But, it could also happen that two points are "incident" to two diagonals each. The author does not address this possibility at all.

b. The author says "consider all points at distance exactly d from point Q (the green circle). We see that all of them are outside the locus, the exception being P." This is far from obvious. It assumes that all the other k-4 points (those points of the k given points which are not highlighted in the picture) are inside the "locus (the blue shaded region)". in fact, it seems to this reader that this is not necessarily true.

c. Another issue with this "Solution" is that it assumes $n \ge 4$ , while the statement of the problem has $n \ge 3$ . This shortcoming is easy to fix, unlike the previous two I mentioned.

4. I will give another solution below, in the section "Solution 2".

Solution (by user awe-sum)

Solution 2

$\mathbf{Definition:}$ For the purpose of this proof, let us define an $equilateral\ arc\ triangle$ as the shape we obtain when we take a triangle $\triangle ABC$ in which we replace the side $AB$ by the $\pi/6$ arc of the circle centered at $C$ with radius $CA = CB$ , going from $A$ to $B$ , and similarly for the other two sides $BC$ and $AC$ .

See the picture below.

Note that if the sides of the original equilateral triangle were of length $d$ , then the distance from a vertex to any point on the opposite arc is $d$ .

TO BE CONTINUED. SAVING MID WAY SO I DON'T LOOSE WORK DONE SO FAR.

1965 IMO (Problems) • Resources
Preceded by Problem 5	1 • 2 • 3 • 4 • 5 • 6	Followed by Last Question
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