Sorry I misreaded it

An equilateral triangle doesn’t have 60º rotational symmetry

This problem was proposed by me, and the flavortext is due to Michael Ren. (Maybe I should use the word “proposed” loosely, because I was heavily inspired by things I’d seen, as described in the remarks.) I'd like to imagine that the flavortext was an integral part of the problem being selected, so shoutouts to Michael Ren for coming up with it, and shoutouts to the RMM committee for not deleting it. Overall, I’m quite satisfied with how this problem turned out. Maybe I missed an unintended solution, but oh well. Here's my original submission. Looking back at it now, I find it... somewhat amusing (especially the part where I try to argue if this is used, it should be as problem 1 or 4).

CantonMathGuy wrote: This problem was proposed by me, and the flavortext is due to Michael Ren. (Maybe I should use the word “proposed” loosely, because I was heavily inspired by things I’d seen, as described in the remarks.) I'd like to imagine that the flavortext was an integral part of the problem being selected, so shoutouts to Michael Ren for coming up with it, and shoutouts to the RMM committee for not deleting it. Overall, I’m quite satisfied with how this problem turned out. Maybe I missed an unintended solution, but oh well. Here's my original submission. Looking back at it now, I find it... somewhat amusing (especially the part where I try to argue this should be 1 or 4). Nice problem

CantonMathGuy wrote: This problem was proposed by me, and the flavortext is due to Michael Ren. (Maybe I should use the word “proposed” loosely, because I was heavily inspired by things I’d seen, as described in the remarks.) I'd like to imagine that the flavortext was an integral part of the problem being selected, so shoutouts to Michael Ren for coming up with it, and shoutouts to the RMM committee for not deleting it. Overall, I’m quite satisfied with how this problem turned out. Maybe I missed an unintended solution, but oh well. Here's my original submission. Looking back at it now, I find it... somewhat amusing (especially the part where I try to argue if this is used, it should be as problem 1 or 4). How would you rate this in difficulty compare to imo or imo shortlist? Would it be less than regular C4?

scnwust wrote: How would you rate this in difficulty compare to IMO or IMO shortlist? Would it be less than regular C4? They did intend it as a P1/4 and statistically speaking, IMO P1/4s are less than C4. @below, you're right!

Statistically speaking, IMO P1/4s are not the same difficulty as RMM P1/4s

Wow this is extremely neat! My solution is essentially the same as the proposer's but I'll post it anyway because I liked this problem a lot. I will prove the problem by induction. For $n = 1$, its just a regular hexagon and put $7$ people on its vertices and center. Now suppose its true until $n-1$ Let $f(A)$ denote the rotation of point $A$ about $O$ $60^\circ$ clockwise. If Zoomtopia has a side of length $1$, use it, otherwise divide some other side into two parts, one containing a side of length $1$. Call this $AB$. Now, create equilateral triangles to the interior of Zoomtopia about sides $AB, f(A)f(B), f(f(A))f(f(B)), \cdots f^5(A), f^5(B)$. Now, to the left and right of these triangles, construct parallelograms with bases as the other sides. Since the rotations were of $60^\circ$ and the angle of the triangles constructed also had internal angles $60^\circ$, this is actually possible! Now place $6n$ people on the boundary (Possible since it has integral sides and its also easy to see every interior angle is $> 120^\circ$, so no two of these guys are at distance $< 1$ from each other). Observe that the points inside Zoomtopia that have been created also form a $60$ rotational symmetric polygon which has perimeter $6n -6 = 6(n-1)$ since every distance apart from the $6$ things lost in the triangles still remains. So, by the inductive hypothesis, we can place $$3(n-1)^2 + 3(n-1) + 1 + (6n) = 3n^2 - 6n + 3 + 3n - 3 + 1 + 6n = 3n^2 + 3n + 1$$people in the kingdom, while maintaining social distancing. The people on the perimeter are at least $1$ apart from anyone in the interior and they themselves are $1$ apart, so this indeed works. $\blacksquare$

oVlad wrote: They did intend it as a P1/4 and statistically speaking, IMO P1/4s are less than C4. Now the results are out and the average score suggests it's closer to P3/6 than to P1/4. I assume it is because there are a lot of possible ideas which look like they're worth trying but in fact lead to nowhere. It's a cute problem though, I enjoyed solving it (my solution is the same as the original one).

scnwust wrote: How would you rate this in difficulty compare to imo or imo shortlist? Would it be less than regular C4? I'm not sure about the first question, but this problem would not be less than a regular C4, because it is a geometry problem.

Consider any portion of perimeter of Zoomtopia that is a union of entire segments(i.e. each segment of this portion has integer length) and subtending an angle of $60$ degrees at O(the same point as in the problem).Call this portion $\Omega$. Let $A$ and $B$ be the endpoints of $\Omega$, that has length $n$.Let $X_0=A$ and $X_n=B$. Construct points $X_i$,$(0<i<n)$ on $\Omega$ such that $X_iX_{i+1}=1$ for all $i<n$. Due to $60$ degree rotational symmetry, $OA=OB$. Hence $\triangle AOB$ is equilateral. Let $Q$ be the center of $\triangle AOB$ .Rotate $\Omega$ about $Q$ by $120$ degrees in an anticlockwise manner to get $\Omega1$. Let $X_i$ go to $Y_i$ under this transformation. Note that $A$ goes to $B$ and $B$ goes to $O$. (A sample geometric diagram for $n=4$ is attached at the end of the solution for better understanding. All further definitions and proofs are according to the orientation of this diagram) Now let line $AB$ be the real axis of the Argand plane with $A$ as the origin. Let $B=b$ and let $X_i=x_i$ and $Y_i=y_i$ for all $(i<n+1)$. Denote $e^{\frac{i\pi}{3}}=t$.Then $O=bt$ Now using $Q=\frac{b(t+1)}{3}$ and rotation of complex numbers,$$y_i=t^2x_i+b$$. Having defined the notations we proceed to the problem; Lemma 1 Arg($\frac{x_n-x_{n-1}}{x_1-x_0}$)$\leq \frac{\pi}{3}.$ Proof Since Zoomtopia is given to be a convex polygon the interior angle at $B$ must be less than $180$ degrees. Due to rotational symmetry, this is just $\angle OAX_1+\angle OBX_{n-1}$. This is also equal to $120+\angle BAX_1+\angle ABX_{n-1}$.Since the LHS of our above inequality is just $\angle BAX_1+\angle ABX_{n-1}$, we have proved our lemma. Lemma_2 :Let $x_i$ and $x_j$ be two given complex numbers with $i<j$. For any two complex numbers $x_k$ and $x_l$ with $k\leq i$,$l\leq j$ and $k\neq l$ we have the following result: Arg$(\frac{x_l-x_k}{x_j-x_i})=\theta \in [\frac{-\pi}{3},0]\cap [\frac{2\pi}{3},\pi]$ Proof Case 1:($k>l$).In that case it can easily be seen geometrically that $\theta\geq$Arg$(\frac{x_0-x_1}{x_n-x_{n-1}})\geq \frac{2\pi}{3}$(by lemma 1) Case 2:($k<l$).This case is supplementary to Case 1. Lemma 3: $ |x_i-x_j|\geq 1$ for all $i\neq j$. Proof :For any $k$($i<k<j$),by lemma 2,$\angle X_iX_kX_{k+1}\geq 120$ therefore $X_iX_{k+1}>X_iX_k$ and consequently $X_iX_j\geq X_iX_{i+1}=1$. Now consider the set $S$ of complex numbers defined by: $S=(x_i+(y_j-b)|0\leq j\leq i\leq n)$ We claim any two points in $S$ are atleast $1$ unit apart. Since previously we found $y_j=t^2x_j+b$,set S is basically:$(x_i+t^2x_j|0\leq j\leq i\leq n)$ So for any two complex numbers $p=x_i+t^2x_j$ and $q=x_k+t^2x_l$,$|p-q|= |(x_i-x_k)+t^2(x_j-x_l)|$. If either $i=k$ or $j=l$,then $|p-q|\geq 1$ from Lemma 3. If neither holds true,then note that by Lemma 2,the magnitude of angle between $x_i-x_k$ and $t^2(x_j-x_l)$ is less than $120$ degrees. Therefore by Cosine rule,$$|p-q|\geq \sqrt {|x_i-x_k|^2+|x_j-x_l|^2-|x_i-x_k||x_j-x_l|}\geq \sqrt {|x_i-x_k||x_j-x_l|}\geq 1$$by Lemma 3. Consider the $60$ degree anticlockwise rotation of $\Omega$ about point $A$ and call it $\Omega2$.Then all points of $S$ are (enclosed by/on) $\Omega,\Omega1$ and $\Omega2$.Call this region $R$.Then succesive $60$ degree rotations of $R$ would create entire Zoomtopia. Then number of points inside $R$ or in the interior $\Omega$ is $\frac{(n-1)(n)}{2}$.The number of points on $\Omega1$(excluding O) is $n$. Creating such $5$ other copies of $R$ by $60$ degree rotations of $R$ about $O$ and including O, we have total $6(\frac{n(n-1)}{2}+n)+1=3n^2+3n+1$ points marked in Zoomtopia. We claim that this is our desired set of points. For any two points $M$ and $N$ in the same "region"(that is $R$ or some copy of $R$) we have already proved their distance is greater than $1$ previously. So it remains to show the condition is satisfied for $2$ points lying in different "regions". But this is trivially true as any point in the interior of a region is already at a distance of atleast 1 from its boundary, so it would be easily at a distance of atleast 1 from a point in the interior of the neighbouring region.$\blacksquare$ In the attached diagram the construction is also shown.

My favorite problem of the exam

Solved with YaoAOPS and blackbluecar. The construction we use is as follows. First, place a hexagon and draw its center. For $k \ge 1$, we draw a $6(k+1)$-gon of side length $1$ surrounding the given $6k$-gon construction by doing the following: At every $k$th side of the $6k$-gon, draw an outward rhombi with acute angle $\alpha_i$ for the $i$-th rhombus, so that the acute/obtuse orientation is the same in the clockwise direction. Next, draw an equilateral triangle outwards on the side of each square that is perpendicular to the corresponding side of the $6k$-gon and in the clockwise direction. Choose $(\alpha_i)$ such that the resulting convex hull is the desired $6(k+1)$-gon. Finally, fill the remaining region of the convex hull of the resulting figure with multiple disjoint rhombi outward each side when $k>1$, so that the acute/obtuse orientation is also the same in the clockwise direction. Apply this process for $k=1, \dots, n-1$. Claim: The above construction works. Proof. We induct on $n$, where the $n=1$ case is clear as it is a hexagon and its center. In the inductive step, suppose that $n$ works, and we want to show that $n+1$ also works. Since the pattern outside the $6n$-gon is periodic with period $n$, there is a rotational symmetry of $60^{\circ}$. It is clear that the distance between any two vertices of the construction is at least $1$, because the rhombi outside of the $6n$-gon for $n>1$ have an acute angle of at least $60^{\circ}$. The side length of all polygons appended is $1$, so the new perimeter is $6(n+1)$. Finally, since we are adding $6(n+1)$ new vertices, the number of vertices in the $n+1$ config is \[ (3n^2+3n+1)+6(n+1) = 3(n+1)^2+3(n+1)+1. \]Therefore the induction is complete.

We induct on $n$. The base case of $n=1$ is obvious, since Zoomtopia is just a regular hexagon. Now fix an $n>1$. WLOG let all sides of Zoomtopia have length $1$ by allowing $180^\circ$ internal angles. Let Zoomtopia have vertices $A_1A_2\dots A_nB_1\dots B_n\dots F_1\dots F_n$. Construct equilateral triangle $A'_1A_1A_2$ with $A'_1$ inside Zoomtopia. Then for each $2\le i\le n-1$, define $A'_i = A_{i+1} + \overrightarrow{A_2A_1'}$. Repeat this construction symmetrically in all $6$ parts of Zoomtopia to obtain a $6n-6-$gon $A_1'A_2'\dots A_{n-1}'B_1'\dots B_{n-1}'\dots F_1'\dots F_{n-1}'$, call it Webextopia. Clearly Webextopia is convex, $60^\circ$ rotationally symmetric, and has all sides of length $1$, since $A_1'A_2'\dots A_{n-1}'$ is a translation of the section $A_2A_3\dots A_n$ of Zoomtopia, and similarly for the other $5$ sections. Moreover, by convexity of Zoomtopia, any translation of a vertex $X$ of Zoomtopia to a vertex $Y$ of Webextopia leaves $\ge 60^\circ$ angles on both sides of $\angle X$, so all vertices of Webextopia are distance at least $1$ away from the closest vertex of Zoomtopia. Hence by the inductive hypothesis, we can relocate $3(n-1)^2 + 3(n-1) + 1 = 3n^2 - 3n+1$ citizens in Webextopia, and adding in the $6n$ vertices of Zoomtopia gets us a total of $3n^2+3n+1$, as desired.