Points in the plane

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Points in the plane

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Source: Cono Sur 1996, Problem 6

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#1 h May 8, 2018, 10:22 PM • 2 Y

Y by Adventure10, Mango247

Find all integers $n \leq 3$ such that there is a set $S_n$ formed by $n$ points of the plane that satisfy the following two conditions:
Any three points are not collinear.
No point is found inside the circle whose diameter has ends at any two points of $S_n$ .

NOTE: The points on the circumference are not considered to be inside the circle.

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BSWJY

#2 h May 8, 2018, 11:05 PM • 2 Y

Y by Adventure10, Mango247

Let the convex polygon of $n$ points by $C$
Then first noticing that a point $P$ is inside the circle with diameter $AB$ $\Leftrightarrow$ $\angle APB>\frac{\pi}{2}$

So convex polygon $C$ satisfies condition so defining the number of vertices of $C$ by $m$
$\pi \cdot (m-2)\le m\cdot \frac{\pi}{2}$
Solving this equation $m\le 4$

Now suppose there exists a point $Q$ inside the polygon. Then making polygon into $n-2$ triangle, at least one area $\triangle ABC$ posseses $Q$
Then $\angle APB+\angle BPC+\angle CPA=2\pi$
Contradiction

So $n=3,4$
Equality holds for regular triangle and square

This post has been edited 2 times. Last edited by BSWJY, May 8, 2018, 11:08 PM

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suli

#3 h May 8, 2018, 11:07 PM • 2 Y

Y by Adventure10, Mango247

$n = 3, 4, 5$ . Constructions are obvious: take equilateral triangle, square, and square plus center.

Suppose $n \ge 6$ . If convex hull has at least 5 vertices, then its average angle is $\ge 108^\circ$ , so at least one angle is obtuse, clearly a contradiction. Thus convex hull has at most 4 vertices.

If convex hull is triangle $ABC$ , then there is a point $P$ inside it. At least one of $\angle APB, \angle BPC, \angle CPA$ is at least $120^\circ$ , or obtuse, contradiction.

If convex hull is quadrilateral $ABCD$ , then it has all four right angles (since none are obtuse). Any point $P$ inside $ABCD$ satisfies $\angle APB = \angle BPC = \angle CPD = \angle DPA = 90^\circ$ for same reason. There is at most one such point, the intersection of circles of diameters $AB$ and $BC$ that is not $B$ , so $n \le 5$ . This completes the casework and the proof that $n = 3, 4, 5$ .

This post has been edited 1 time. Last edited by suli, May 8, 2018, 11:09 PM

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#4 h May 8, 2018, 11:30 PM • 3 Y

Y by NiltonCesar, Adventure10, Mango247

suli wrote:

$n = 3, 4, 5$ . Constructions are obvious: take equilateral triangle, square, and square plus center.

But isn't the square plus center not true? Take the diameter that is a diagonal of the square, then the center is in this circle.

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#5 h Jul 14, 2018, 2:02 AM • 3 Y

Y by NiltonCesar, Adventure10, Mango247

also here

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