Two regular n-gons in a plane

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Two regular n-gons in a plane

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#1 h Jul 10, 2018, 12:25 PM • 6 Y

Y by Amir Hossein, AlastorMoody, Siddharth03, Adventure10, Mango247, sabkx

Let $n\ge3$ be an integer. Two regular $n$ -gons $\mathcal{A}$ and $\mathcal{B}$ are given in the plane. Prove that the vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary are consecutive.

(That is, prove that there exists a line separating those vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary from the other vertices of $\mathcal{A}$ .)

This post has been edited 1 time. Last edited by Amir Hossein, Jul 10, 2018, 4:21 PM
Reason: Fixed.

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#2 h Jul 10, 2018, 12:26 PM • 3 Y

Y by Imayormaynotknowcalculus, Adventure10, sabkx

Wow, sounds so simple yet isn't trivial..

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#3 h Jul 11, 2018, 8:02 AM • 2 Y

Y by Adventure10, Mango247

Any solution?

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#4 h Jul 13, 2018, 7:59 AM • 4 Y

Y by CyclicISLscelesTrapezoid, Adventure10, Mango247, ohiorizzler1434

We solved this problem during our training.

Here, we refer to a polygon as both the interior and its boundary.

We try to find a regular $n$ -gon $\mathcal{C}$ inscribed in $\mathcal{B}$ which is either a translation or a positive dilation of $\mathcal{A}$ . We construct it as follows:
Let $O_A$ and $O_B$ be the centres of $\mathcal{A}$ and $\mathcal{B}$ respectively, and let $A$ be an arbitrary vertex of $\mathcal{A}$ . Let the ray originating from $O_B$ in the same direction to the vector $O_A A$ intersect $\mathcal{B}$ at $C$ . Rotating $C$ about $O_B$ with angle $\dfrac{2\pi}{n}$ yields a regular $n$ -gon $\mathcal{C}$ , which is a dilation/translation of $\mathcal{A}$ , so we're done.

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Let $d$ denote the mapping of $\mathcal{C}$ to $\mathcal{A}$ , and let the centre of this dilation be $O$ with factor $\lambda > 0$ . Note that if the dilation was a translation of $\mathcal{A}$ , then by applying a dilation on $\mathcal{A}$ with centre $O_B$ and factor of $1-\epsilon$ for sufficiently small positive $\epsilon$ will create a smaller polygon $\mathcal{A}'$ , and preserve the vertices already inside or outside $\mathcal{B}$ , so we can then assume the polygons are $\mathcal{A}'$ and $\mathcal{C}$ instead.

We will show the vertices of $\mathcal{C}$ that were already in $\mathcal{B}$ are consecutive upon $d$ being applied.
If $O \in \mathcal{B}$ , then if $\lambda < 1$ , then for each of the vertices $A$ of $\mathcal{A}$ lie on the segment $OC$ , where $C$ is the corresponding vertex to $A$ on $\mathcal{C}$ , which lie in $\mathcal{B}$ , so we're done in this case. If $\lambda > 1$ , $A$ lies on the extension of $OC$ , which lie outside $\mathcal{B}$ , so we're done in this case.

Thus, we can assume $O \not \in \mathcal{B}$ .
Case 1: $\lambda > 1$
Let $X$ and $Y$ be two vertices of $\mathcal{B}$ such that $XOY$ contains $\mathcal{B}$ . If there are multiple choices for this, choose the two vertices farthest from $O$ . For ease of understanding, consider the plane to be the Cartesian plane, with $XY$ parallel to the $y$ -axis. Also, WLOG $O$ lies to the right of $XY$ . Let $\mathcal{B}_R$ denote the set of perimeter points of $\mathcal{B}$ to the right of $XY$ , and $\mathcal{B}_L$ denote the set of perimeter points of $\mathcal{B}$ to the left of $XY$ . Thus, the vertices of $\mathcal{C}$ are also split into set $\mathcal{C}_R \subset \mathcal{B}_R$ and $\mathcal{C}_L \subset \mathcal{B}_L$ .

We have all the points from $\mathcal{B}_L$ (including $\mathcal{C}_L$ ) move out from $\mathcal{B}$ under $d$ since they are the farthest points of $\mathcal{B}$ on the corresponding rays from $O$ . Thus, it suffices to show the vertices of $\mathcal{C}_R$ which remain in $\mathcal{B}$ are consecutive. Let $C_1$ , $C_2$ , $C_3$ be 3 vertices in $\mathcal{C}_R$ such that $C_2$ lies in between $C_1$ and $C_3$ and both $d(C_1)$ and $d(C_3)$ lie in $\mathcal{B}$ . Let $A_i = d(C_i)$ . Thus, the ray $OC_2$ intersects $C_1C_3$ beyond $C_2$ , so intersects $A_1A_3$ beyond $A_2$ . Therefore, $A_2$ lies in triangle $C_2 A_1 A_3$ , so $A_2$ lies in $\mathcal{B}$ . Thus, proven in this case.

Case 2: $\lambda < 1$
We will apply the same logic as before, except have $X$ and $Y$ as the closest satisfying points to $O$ instead of furthest. All the points in $\mathcal{B}_R$ move out from $\mathcal{B}$ under $d$ since they are the closest points of $\mathcal{B}$ on the corresponding rays from $O$ . Thus, it suffices to show the vertices of $\mathcal{C}_L$ remain in $\mathcal{B}$ . Again, let $C_1$ , $C_2$ , $C_3$ be points in $\mathcal{C}_L$ such that $C_2$ lies in betwen $C_1$ and $C_3$ and both $h(C_1)$ and $h(C_3)$ lie in $\mathcal{B}$ , with $A_i = h(C_i)$ . Then $A_2$ lies ons egment $OC_2$ , and since the intersection of $OC_2$ and $C_1C_3$ lies on $OC_2$ , we must have $A_2$ inside $C_2A_1A_3$ , so $A_2$ is inside $\mathcal{B}$ . Thus, proven in this case.

Therefore, the vertices of $\mathcal{A}$ that remain inside $\mathcal{B}$ are consecutive.

This post has been edited 1 time. Last edited by SHARKYKESA, Jul 13, 2018, 8:00 AM

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#5 h Dec 29, 2021, 3:59 PM

Y by

What's the motivation behind considering the polygon which is a homothetic image of $\mathcal{A}$ whose vertices are in $\mathcal{B}$ ?

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#6 h May 18, 2023, 9:39 PM • 4 Y

Y by DevilDoesMath, L567, math_comb01, Om245

Hehe, mass point geo and convexity go brr

A very nice problem! Lemme share a completely different solution I came up with :

Note

Let $R(\mathcal{A})$ denote the region covered by $\mathcal{A}$ , i.e. boundary and interior. Let $D_{\mathcal{A}}(A,B)$ denote the number of sides between $A$ to $B$ while going counterclockwise on $\mathcal{A}$

Lemma 1: Let $ABC$ be a triangle formed by 3 vertices of $\mathcal{A}$ . Let $A',B'$ be vertices of $\mathcal{A}$ on the other side of $B$ w.r.t. $AC$ such that the vertices $A,A',C',C$ are in order ( $A$ can be the same as $A'$ and $C$ can be the same as $C'$ ). Consider the point $B'$ so that $\triangle ABC \sim \triangle A'B'C'$ . Then $B' \in R(\mathcal{A})$ .
Proof

In what follows, we shall mainly use the following 2 well-known results about mass points:

Theorem 1: Let $\Delta_1,\Delta_2,\dots,\Delta_k$ be $k$ similar triangles and let $\lambda_1,\lambda_2,\dots,\lambda_k$ be $k$ real numbers summing up to $1$ . Then $\lambda_1\Delta_1+\lambda_2\Delta_2+\dots+\lambda_k\Delta_k$ is also similar to each $\Delta_i$

Theorem 2: Let $A_1,A_2,\dots,A_k$ be any $k$ points. Then, for any point $X$ ; $X$ can be represented as $\lambda_1A_1+\lambda_2A_2+\dots+\lambda_kA_k$ where the $\lambda_i$ 's are non-negative reals summing up to $1$ iff $X$ is contained in the convex hull of $A_1,A_2,\dots,A_n$

Now, returning to the main problem, observe that it suffices to show that there do not exist distinct vertices $A,B,C,D$ of $\mathcal{B}$ (in order) so that $A,C\in R(\mathcal{A})$ but $B,D\not\in R(\mathcal{A})$ . FTSOC, assume that such 4 points exist.

Lemma 2: We can WLOG assume that $A,C$ are on the boundary of $\mathcal{A}$ .
Proof:

WLOG, let $\square ABCD$ be oriented in counterclockwise sense.
Let $b_1 = D_{\mathcal{B}}(A,C), b_2 = D_{\mathcal{B}}(C,A)$ .
Let the halfplanes determined by $\overleftrightarrow{AC}$ containing $B,D$ be $\mathcal{L},\mathcal{R}$ respectively.
Let $A$ be on side $A_0A_1$ and let $C$ be on side $C_0C_1$ in $\mathcal{C}$ (It's fine if either of $A,C$ are the endpoints of these sides) such that $A_0,C_0\in L\cup\{A,C\}$ and $A_1,C_1\in R\cup\{A,C\}$
Let $a_1 = D_{\mathcal{A}}(A_0,C_0), a_2 = D_{\mathcal{A}}(C_1,A_1)$ .
As $b_1 + b_2 = n = a_1+a_2+2$ we can now make the following two cases :
1. $a_1\geq b_1$ or $a_2\geq b_2$
2. $b_1 = a_1+1, b_2 = a_2+1$
WLOG, let the side length of $\mathcal{A}$ be $1$ and let $A_0A = a, C_0C = c$ .

Case 1

Case 2

(lmk if there are any typos)

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awesomeming327.

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#7 h Jun 14, 2024, 6:22 PM

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Yet another different proof...

Let $\mathcal{A}$ and $\mathcal{B}$ denote closed sets defining the interior of the polygon as their boundaries. Let $\mathcal{A}'$ and $\mathcal{B}'$ be open sets that are $\mathcal{A}$ and $\mathcal{B}$ with boundaries removed, respectively. Throughout the rest of this proof, we will take indices $\pmod n$ .

After translating $\mathcal{B}=B_1B_2\dots B_n$ so that its center $B$ coincides with the center $A$ of $\mathcal{A}=A_1A_2\dots A_n$ , we see that by rotational symmetry, we can positively dilate the translated polygon at $A$ so that it is inscribed in $\mathcal{A}$ . Call this polygon $\mathcal{C}=C_1C_2\dots=C_n$ where there exists either a positive homothety centered at $O$ , or a translation, that takes $C_i$ to $B_i$ for each $i$ . Let the factor of this homothety be $k$ . WLOG, let $C_i\in A_iA_{i+1}$ .

Note that there are three different cases, and we will look at the case of translation first. We define a function $f_T(X)$ for any point $X$ on $\mathcal{A}$ 's boundary that returns an interval that is defined as follows: draw a line $\ell_X$ through $X$ parallel to $AB$ . Note that $\ell_X\cap \mathcal{A}$ is a line segment, so represent this as a closed interval $[a_T(X),b_T(x)]$ where $X$ is zero and $A\to B$ is the positive direction. Note that $B_i\in \mathcal{A}$ if and only if $AB\in f_T(B_i)$ .

Now we look at the case of a homothety with factor $k\neq 1$ . Define $f_H(X)$ for any point $X$ on $\mathcal{A}$ 's boundary in a similar fashion: draw line $OX$ which intersects $\mathcal{A}$ in a line segment $X_1X_2$ where $X_1$ is closer to $O$ than is $X_2$ . Now, let our interval be
$[a_H(x),b_H(x)]=\left[\frac{OX_1}{OX},\frac{OX_2}{OX}\right]$ where $B_i\in\mathcal{A}$ if and only if $k\in f_H(B_i)$ .

We have created two indicator functions and these functions have the properties necessary for us to prove what we needed. We claim that $a_T(X)\le 0\le b_T(X)$ and $a_H(X)\le 1\le n_H(x)$ . These are obvious since $X$ is on $X$ is on $X_1X_2$ and $\ell_X\cap\mathcal{A}$ . We also claim that when we take each function as $X$ moves around the polygon counterclockwise, there is one contiguous subdomain in which the function is non-decreasing and one contiguous subdomain in which the function is non-increasing. This is obvious by convexity.

Now that we have proved properties of the functions, note that if $AB>0$ then $AB>a_T(X)$ . Therefore, $AB\in f_T(X)$ if and only if $AB\le b_T(X)$ . By the nature of $b_T$ , this is one contiguous segment of the $\mathcal{A}$ 's boundary. The other four proceeds similarly, and so we are done.

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