0
$\begingroup$

Let $ABCDE$ a regular pentagon inscribed in a circle of center $O$. Let $P$ an interior point of the pentagon from which we consider parallel line segments to all the sides of the pentagon. We know that $P$ is placed such that every segment has the endpoints on the sides of the pentagon. If $P$ divides all the segments drawn in subsegments with a difference in length less than $1$, prove that $OP \leq 1$.

Figure

Note that on my figure, the inner circle is the circle of radius $1$ centered at $O$. (Copyright note: The figure was made using GeoGebra Geometry tool, online version).

I'm thinking at vector geometry (probably scalar product applications). Maybe, by considering the midpoints of the segments drawn through $P$ and applying Steiner and by claculating scalar products, we may obtain $PO^2 \leq 1$, which is obviously equivalent to $PO \leq 1$.

$\endgroup$
11
  • $\begingroup$ What is meant by "If P divides all the segments drawn in subsegments with a difference in length less than 1"? Thanks. $\endgroup$
    – user502266
    Jan 29, 2022 at 12:15
  • 1
    $\begingroup$ @S.Dolan , I think that means, for example, a segment of length 5 can be divided into two segments 2.2 and 2.8 (difference-0.6<1) but not into 4 and 1 (difference-3>1). $\endgroup$
    – ACB
    Jan 29, 2022 at 12:22
  • $\begingroup$ @ACB Thanks that makes sense. $\endgroup$
    – user502266
    Jan 29, 2022 at 12:23
  • $\begingroup$ @ACB That's true, that is what I meant. Thanks for explaining for me! $\endgroup$ Jan 29, 2022 at 13:13
  • 1
    $\begingroup$ Consider the perpendicular bisector of a side of the pentagon. If $P$ divides the line through $P$ parallel to that side into two segments which differ in length by less than 1, then $P$ has to lie within distance $1/2$ of the bisector. Therefore point $P$ has to lie inside a regular decagon where the midpoint of each side is a distance 1/2 from $O$. This decagon can be inscribed inside a circle centered on $O$ with radius $\frac{1}{2\cos \pi/10} \approx 0.526$. $\endgroup$
    – Craig
    Feb 2, 2022 at 14:57

1 Answer 1

1
+50
$\begingroup$

In the figure below(geogebra link), the blue region represents $-0.5<x<0.5$ enter image description here

As long as $P$ lies in that region, the value of $|PQ-PR|$ cannot exceed $1$. Moreover if P lies outside this region, $|PQ-PR|$ is always greater than $1$. This can be easily proved in this case using the fact that $DE$ and $QR$ are parallel to the X axis. We know that the Y axis bisects QR.

So, $$QJ=JR$$

$$\implies PQ-PR = QJ+JP-PR$$ $$= JR+JP-PR$$ $$= JP+PR+JP-PR$$ $$= 2(JP)$$

By symmetry $|PQ-PR|=2|JP|=2(JP)$.

Given $|PQ-PR|<1$ implies $2(JP)<1 \implies JP<0.5$, which means $P$ lies in the blue region.

In other words, If $P$ divides $QR$ such that $|PQ-PR|<1$, then the distance from $P$ to the perpendicular bisector of $DE$ must be less than $0.5$.

This fact can be generalized for all the sides of the pentagon. If $P$ has to divide every one of those line segments into subsegments with difference less than 1, then the distance from $P$ to the perpendicular bisector of any of the pentagon's sides must be less than $0.5$. The intersection of all such regions can be seen below. enter image description here

It is a regular decagon. enter image description here

Clearly it lies inside the circle $r=1$. But let us prove it. From the figure $\angle HAI = \frac{2\pi}{10}$. That means $AX=AHcos(\frac{\pi}{10})=0.5$, which gives $AH=\frac{1}{2}sec(\frac{\pi}{10})<\frac{1}{2}sec(\frac{\pi}{3})=\frac{1}{2}(2)=1$.(We know that $sec(x)$ is an increasing function in $(0, \frac{\pi}{2})$ because $cos(x)$ is decreasing). Because of $AH$ being the hypotenuse, H is the farthest point from A in $\Delta AHX$. By symmetry we can extend this argument to claim that the farthest vertices from the center A in the regular decagon(for that matter, any regular polygon) are its vertices.

So, we proved that the distance of every point in the decagon from the center of the circle than 1, which proves that $AP<1$ since we have already proved that $P$ must lie inside the decagon.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .