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Source: Brazilian Math Olympiad 2006, Problem 2

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#1 h Oct 30, 2006, 9:21 PM • 3 Y

Y by Davi-8191, Adventure10, Mango247

Let $n$ be an integer, $n \geq 3$ . Let $f(n)$ be the largest number of isosceles triangles whose vertices belong to some set of $n$ points in the plane without three colinear points. Prove that there exists positive real constants $a$ and $b$ such that $an^{2}< f(n) < bn^{2}$ for every integer $n$ , $n \geq 3$ .

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#2 h Oct 31, 2006, 9:36 AM • 2 Y

Y by Adventure10, Mango247

First consider a convex regular $2k$ -gon. It is clear that no three vertices are collinear, and for each vertex $A$ there are $k-1$ isosceles triangles at $A$ .
Thus, since an isosceles triangle is counted at most three times in this way, the total number of isosceles triangles is at least $\frac{2k(k-1)}3.$
We have the same kind of estimation for a regular convex $2k+1$ -gon.
It follows that $f(n) > an^{2}$ for some appropriate positive constant $a$ .

In another hand, let's consider any set of $n$ points in the plane, no three collinear. There are $\frac{n(n-1)}2$ choices for two distinct points among the $n$ . For each, say $A$ and $B$ , there are at most two points $M$ among the given ones such that $AMB$ is isosceles at $M$ , because such points $M$ belong to the perpendicular bissector of $[AB]$ and no three of the points can belong to this line. Thus the total number of isosceles triangles is at most $n(n-1) <n^{2}$ .
Therefore $f(n) < n^{2}.$

Pierre.

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Johann Peter Dirichlet

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Johann Peter Dirichlet

#3 h Nov 6, 2006, 4:06 AM • 2 Y

Y by Adventure10, Mango247

Well, what is the "best" constants $a$ and $b$ ?

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#4 h May 13, 2009, 8:48 PM • 1 Y

Y by Adventure10

It seems that "the best" constants $a$ and $b$ are 1.
$f(n)<n^2$ is already proved.

Lets consider an example. Regular $2n-1$ -gon with a point in its center. It is easy to see that we get $(2n-1)(2n-2)$ isosceles triangles for such configuration (every 2 points of $2n-1$ -gon forms two triangles - one with another point and one with the center point. These triangles have two points as a base, that is all triangles are different). So we have $2n$ points and $(2n-1)(2n-2)$ triangles, i.e.
$(2n-1)(2n-2) < f(2n) < (2n-1)(2n)$ .

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#5 h Mar 29, 2021, 12:30 AM

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For the lower bound we claim we can take $a=\frac{1}{20}$ . We find an explicit construction that yields $>an^2$ isosceles triangles for some set of $n$ points with no three collinear, which proves $f(n) > an^2$ .
-For $n=3, 4$ we can take an isosceles triangle, which yields $1 > \frac{4^2}{20}> \frac{3^2}{20}$ isosceles triangles.
-For $5 \le 10 \le 9$ we can take $5$ of the $n$ points to be a regular pentagon and appropriately place the other points, which yields at least $5 > \frac{9^2}{20}$ isosceles triangles.
Now suppose $n \ge 10$ . Let $k = 2\lfloor \frac{n}{2} \rfloor$ and consider the following construction on $k$ of the $n$ points. Take one of the $k$ points to be the center $O$ of a regular $k-1$ -gon and space the other $k-1$ points to be vertices of the regular $k-1$ gon. Assuming the other $n-k$ points are properly placed it follows that there are no three collinear points, since $k-1$ is odd. Every choice of 2 of the $k-1$ points and $O$ yields a different isosceles triangle, so letting $x=n/2$ , this construction gives
$\dbinom{k-1}{2} > \frac{1}{2}(2x-3)(2x-4) = \frac{(n-3)(n-4)}{2}$ isosceles triangles. But for $n \ge 5$ we can check that
$4 > \frac{35}{9} \implies (n-\frac{35}{9})^2 \ge \frac{35^2}{9^2}-\frac{120}{9} \implies n^2 - \frac{70}{9}n+\frac{120}{9} \ge 0 \implies 9n^2 - 70n+120 \ge 0 \implies \frac{(n-3)(n-4)}{2} = \frac{n^2-7n+12}{2} \ge \frac{n^2}{20}$ so this construction indeed works.

For the upper bound we claim we can take $b=1$ . Note for every pair of two points $p,q$ , there are at most 2 points that lie on the perpendicular bisector of $p,q$ ; else there would be three collinear points. So $f(n) \le 2\dbinom{n}{2}=n(n-1)<n^2$ .

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#6 h Jan 19, 2022, 9:30 PM

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For any set of $n$ points in general position on the plane, observe that for each pair of points there are at most $2$ distinct points lying on its perpendicular bisector. Hence, for each pair of points, it is the basis of at most $2$ isosceles triangles. Therefore, we have at most $2 \binom{n}{2}= n^2-n<n^2$ , so $b=1$ works.

Now, we claim that $a= \frac{1}{12}$ works. Observe that if $n$ is odd, then we can take a regular $n-$ agon. It will have at least $\binom{n}{2}-\frac{2n}{3}$ isosceles triangles (when $3|n$ we count $\frac{n}{3}$ equilateral triangles twice). Observe that $\binom{n}{2}-\frac{2n}{3}= \frac{3n^2-7n}{6}>\frac{n^2}{12} \iff 3n^2-7n > \frac{n^2}{2}$ , which is clearly true for $n \geq 3$ . If $n$ is even, we can take a regular $(n-1)-$ agon and its center. Clearly, the number of isosceles triangles in this set is greater than the number of isosceles triangles in a regular $(n-1)-$ agon, so $f(n)>\frac{n^2}{12}$ . Hence, $a=\frac{1}{12}, b=1$ works.

$\blacksquare$

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