AoPS - Problem

Processing math: 100%

Problem

Source: JBMO 2005, Problem 3

Tags: combinatorics proposed, combinatorics

Valentin Vornicu

30.10.2005 01:33

Prove that there exist (a) 5 points in the plane so that among all the triangles with vertices among these points there are 8 right-angled ones; (b) 64 points in the plane so that among all the triangles with vertices among these points there are at least 2005 right-angled ones.

pbornsztein

30.10.2005 08:07

a) Choose the four vertices of a square and its center b) Just consider a

$8 \times 8$ grid. For each point, say

$M$ of the grid, there are

$7$ other points in the same column and

$7$ in the same row. Thus, there are at least

$49$ right-angled triangles with right-angle in

$M$ . Thus, there are at least

$49 \times 64 =3136$ right-angled triangles. And much more, if we consider the triangles whose sides re not parallel to the sides of the grid...

Pierre.

Valentin Vornicu

30.10.2005 09:17

Well it is easy. I don't think you missed anything Pierre, but let's remember that the competition is for students up to 15.5 years old

Marinchoo

02.07.2020 09:37

Another approach for b) because a) is almost trivial: Set a bipartite graph with $32$ points on a line so that we form $31$ squares in a line as such $: : : :$ and so on. If we start counting the right triangles with one side, not the hypotenuse , on the one line we have $(32)(31^{2})(2)(2)=1984$ . Now we have another $60$ triangles with the form $. : .$ so overall we get more than $2005$ .