Difference between revisions of "1965 IMO Problems/Problem 6"
| Line 47: | Line 47: | ||
== Solution 2 == | == Solution 2 == | ||
| − | For the purpose of this proof, let us define an <math>equilateral\ arc\ triangle</math> | + | <math>\mathbf{Definition:}</math> For the purpose of this proof, let us define an |
| − | as the shape we obtain when we take a triangle <math>\triangle ABC</math> in which we | + | <math>equilateral\ arc\ triangle</math> as the shape we obtain when we take a triangle |
| − | replace the side <math>AB</math> by the <math>\pi/6</math> arc of the circle centered at <math>C</math> with | + | <math>\triangle ABC</math> in which we replace the side <math>AB</math> by the <math>\pi/6</math> arc of the |
| − | radius <math>CA = CB</math>, going from <math>A</math> to <math>B</math>, and similarly for the other two sides | + | circle centered at <math>C</math> with radius <math>CA = CB</math>, going from <math>A</math> to <math>B</math>, and |
| − | <math>BC</math> and <math>AC</math>. | + | similarly for the other two sides <math>BC</math> and <math>AC</math>. |
| + | |||
| + | See the picture below. | ||
[[File:prob_1965_6_fig1.png|300px]] | [[File:prob_1965_6_fig1.png|300px]] | ||
Revision as of 19:31, 29 October 2024
Contents
Problem
In a plane a set of 
points (
) is given. Each pair of points is connected by a segment. Let 
be the length of the longest of these segments. We define a diameter of the set to be any connecting segment of length . Prove that the number of diameters of the given set is at most .
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.
4. I will give another solution below, in the section "Solution 2".
Solution (by user awe-sum)
Solution 2
For the purpose of this proof, let us define an
as the shape we obtain when we take a triangle
in which we replace the side 
by the 
arc of the
circle centered at 
with radius 
, going from 
to 
, and
similarly for the other two sides 
and 
.
See the picture below.
Note if the sides of the original equilateral triangle were of length ,
then the distance from a vertex to any point on the opposite arc is .
TO BE CONTINUED. SAVING MID WAY SO I DON'T LOOSE WORK DONE SO FAR.
See Also
| 1965 IMO (Problems) • Resources | ||
| Preceded by Problem 5 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Last Question |
| All IMO Problems and Solutions | ||
