Call a transformation good if it can be decomposed into a sequence of reflections upon the sides of $K$. Lemma 1: if the reflections on the sides of a polygon $L$ are good transformations, then so are the reflections upon the sides of $L^*$, where $L^*$ is the reflection of $L$ on one of its sides $b$. Proof: let $a^*$ be some side of $L^*$, where $a^*$ is the reflection of a side $a$ of $L$ upon line $b$. Then the reflecting on $a^*$ is identical to reflecting on $b$, then $a$, then $b$. Using Lemma 1, it is enough to cover $X$ using a sequence $K=L_0,L_1,\dots,L_k$ of polygons for which $L_{i+1}$ is the reflection of $L_i$ upon a side of $L_i$, because the reflection upon a side of $L_i$ is good. To do this, it's enough to prove Lemma 2: start from $K$, and reflect it on all of its sides. After that, take every new and old polygon, and reflect then on each of their sides (we might get a polygon twice, but that doesn't matter), and so on. Then iterating these steps, we can cover any point of the plane. Proof: Note that there exists a $d_0$ such that, iterating finitely many times, we can cover every point of distance at most $d_0$ from the perimeter of $K$. To prove this, take any side $AB$ of $K$. It is clear that there exists some $r$ such that iterating enough times, we can cover the disc around $A$ and around $B$ with radius $r$. Let $K^*$ be the reflection of $K$ upon $AB$, and let it intersect the disc around $A$ and $B$ at points $A_1$ and $B_1$. Then since $K$ is convex, we can be sure that every point with at most $d_{AB}=\min\{d(A_1,AB),d(B_1,AB)\}$ distance from $AB$ will be covered. Taking the minimal $d_{AB}$ over all sides of $K$ as $d_0$ works. Let $O\in K$ be some point. We claim that for any $n\in\mathbb{N}$, the disc centered at $O$ with radius $nd_0$ will be covered after enough iterations. This can be proven by induction. Indeed, suppose there is such a disc with radius $nd_0$, covered by some figure $F$ formed by copies of $K$, then since using enough iterations, we can cover any point at most $d_0$ distant from the perimeter of $K$, and the perimeter of $F$ is formed by the sides of copies of $K$, surely we can cover every point at most $d_0$ far from the perimeter of $F$. But after doing so, there remains no point of distance less than $nd_0+d_0$ from $O$ inside the resulting figure. If we wish to cover some point $X$, this is therefore possible, if we choose $n>\frac{OX}{d_0}$. Lemma 2 guarantees the existence of the described sequence $L_0,L_1,\dots,L_k$, so this concludes the proof.

First, define the (initial) $\color{green} \textbf{\textit{target polygon}}$ to be $K$, and a $\color{green} \textbf{\textit{reflection axis}}$ to be the $``\text{sides}"$ we can reflect $X$ with. Additionally, let the $\color{green} \textbf{\textit{known-solved territory}}$ to be the region of points $X$; in which it is known that after a sequence of reflections among the $\color{green} \textbf{\textit{reflection axes}}$, every point within it will land inside the target polygon. Observe that the known-solved territory is obtained by reflecting $\color{green} \textbf{\textit{any target polygon}}$ and previously known-solved territory towards any $\color{green} \textbf{\textit{reflection axis}}$. Inductively, we wish to show that these sequence of reflections will bring about the whole plane $\textit{to fit inside}$ the target polygon we know. Here, instead of doing that, we create some new reflection axes, so that we can $\color{green} \textbf{\textit{transfer}}$ the identity of the target polygon to be $\textbf{any polygon}$ which belongs to the known-solved territory, which has all its sides belonging to the set of reflection axes. $\color{green} \rule{8.5cm}{2pt}$ $\color{green} \clubsuit$ $ \boxed{\textbf{Complexity Reduction: Seeing Rationals.}}$ $\color{green} \clubsuit$ $\color{green} \rule{8.5cm}{2pt}$ We Claim that (as in Lemma 1 of $\#2$) that the reflection of any two $\textit{reflection axes}$ among each other is also a reflection axis. Furthermore, we can transfer the target polygon to an $\color{green} \text{acute}$ triangle with no $72^{\circ}$ angle in its vertices. For simplicity, we call such a triangle to be $\color{green} \textbf{\textit{good}}$, and three angles of a target-triangle to be its $\color{green} \textbf{\textit{angular identity}}$. $\color{green} \rule{25cm}{0.3pt}$ $\color{green} \spadesuit$ $\boxed{\textbf{Proof.}}$ $\color{green} \spadesuit$ Let the two axes be $a$ and $r$, and we'd like to matriculate $r'$ (reflection of $r$ thru $a$) as a reflection axis. To do this, reflect across $a$, then $r$, then $a$ again; and we have reflected whatever figure w.r.t. $r'$. $\color{green} \rule{25cm}{0.3pt}$ To obtain the second assertion, we consider an initial polygon $K$ with $n$ sides. If $n \geq 4$ and $K$ is not a rectangle, there exists an obtuse angle among its $\textit{angular identity}$; say it is formed by $a$ and $b$. Using the first assertion, we get that $a'$, which must intersect $K$ in its interior, is also a reflection axis. This breaks down $K$ into two polygons with less sides, with them both able to function as target polygons. Do this over and over to create a triangle (when $K$ is a rectangle, just manually reflect $K$ across its sides rectangularly to cover the whole plane.) $\color{green} \rule{25cm}{0.3pt}$ Now, if $T$ (assuming $K$ is a triangle) is a right-angled triangle, form a new triangle with the first assertion applied to a non-right-angled angle (of its angular identity). This will form four axes $a,b,c,a'$ (where $a'$ is the reflection of $a$ to $b$); and use $a,a',c$ as the new target polygon. If $T$ is obtuse now, do the same with the angle ($C$) measuring less than $45^{\circ}$, so its angular identity will change to \[ (C,B,180-B-C) \Rightarrow (2C,B,180-2C-B); \ 2C \leq 90^{\circ} \]If $T$ has a $72^{\circ}$, we divide into two cases. If the triangle is not a $36-72-72$, let the triangle is $72-x-y$, where \[ \angle A = 72^{\circ}, \angle B = x, \angle C = y\]now reflect $b$ to $c$ and $a$ to $c$ to get $b',a'$; and reflect $a'$ $k$ times (with different axes) towards the direction of $b'$ to get $a_k'$ as an axis. We now use the triangle formed by $b,b',a_k'$ as a target polygon with angular-identity \[ 36^{\circ}, ky, 144-ky \]and $144-ky < 90$. This can be reached by setting $90-y \leq 144-ky < 90$; i.e. step-by-step increasing $ky$ until no obtuse is formed. We are almost done, except that we can be back where we started if $144-ky = 72$, or when $y \mid 72$. Now, $y > 18^{\circ}$; or else $x$ is a right or obtuse angle. Then, $y$ cannot be $36$ for the time being, so $y$ must be $24$. If so, pick the other angle, \[ x = 180^{\circ}-24^{\circ}-72^{\circ} = 84^{\circ} \]as the new $y$, and perform the same operations. Meanwhile, if the triangle is indeed a $(A,B,C) = (36,72,72)$ we can expand the triangle into four times its original size (as can be done with a $60-60-60$) by reflecting $b,c$ towards $a$, $c$ two times to each of $a$ and $b$ (to get the reflection axes), before verifying that its known-solved territory is sufficient for advancing (see $\color{green} \text{Fig 1}$). $\color{green} \rule{25cm}{0.3pt}$ Now we still focus on the (set of) two reflection axes which intersect at a point. Given a good target-polygon define an $\color{cyan} \textbf{\textit{influence circle}}$ of such a vertex to be the obtainable area only utilizing (mostly) the two reflection axes, with a bit of the third axis. $\color{cyan} \rule{6.1cm}{2pt}$ $\color{cyan} \diamondsuit$ $ \boxed{\textbf{Preliminary Establishment.}}$ $\color{cyan} \diamondsuit$ $\color{cyan} \rule{6.1cm}{2pt}$ Let the three sides of the target triangle be $a,b,c$ with $a < b < c$. Then, the influence circle of $B$ and $C$ consists of a circle with radius $a$, and the influence circle of $A$ consists of a circle with radius $b$. $\color{cyan} \rule{25cm}{0.3pt}$ First, we establish the membership of the small arcs of center $A,B,C$ with radius $b,a,a$ respectively. This can be obtained by simply reflecting $\triangle ABC$ once with the axes $a,b,c$, and noting that $\angle B,C > 45^{\circ}$ (see $\color{cyan} \text{Fig 2}$) is enough to guarantee that the distance of a vertex (say $A$) to the $\textit{reflected sides}$ (say, $b',c'$) is more than $b$, or $c$. This arc then can be reflected over and over (as in the forming of $a_k'$ in the previous Section) to form a complete circle. We are done for the first part. $\blacksquare$

We will now strengthen the influence circle, before finally considering the whole grid $\color{red} \textbf{\textit{in one direction}}$ (a.k.a. covering a strip in circles). $\color{magenta} \rule{7cm}{2pt}$ $\color{magenta} \diamondsuit$ $ \boxed{\textbf{Wide-Ranging for Looser Bounds.}}$ $\color{magenta} \diamondsuit$ $\color{magenta} \rule{7cm}{2pt}$ We claim that the influence circle of $A$ can be improved to be of radius $c$, then we establish that the small circles aren't that small either; $B$ and $C$ will have influence circles of radius $\sqrt{3}a$. $\color{magenta} \rule{25cm}{0.3pt}$ Using the same conventions, the region with radius $c$ not yet covered by $A$'s influence circle lies beyond $C$ (see $\color{magenta} \text{Fig 2'}$). Since anything in the $\text{marked area}$ has a distance less than $b$ with the vertex $A'$ (with influence circle of radius $b$), all sectors of the circle of radius $c$ is part of $A$'s influence circle. For $B$'s and $C$'s, this is a bit trickier. Consider the denominator of the angles $\angle B,C$ respectively, out of $360^{\circ}$ units. Example, if $\angle A = 75$, then its denominator is \[ \dfrac{75}{360} \ \text{units} = \dfrac{5}{24} \ \text{units} \]or $24$, in its simplest form. Note that since $\angle B,C < 90, \ne 72$, then its denominator is six or more. Next consider the reflection axes emanating from say, $B$, with respect to $a,c$ reflected over-and-over-again (until it repeats again to square one). These rays are equally spaced within one another, and there must exist rays equal to $\angle B$'s denominator. Now pick two points $X$ and $Y$ which are consecutively placed in those rays. Then, $X,Y$ is either similar to $A$ (with its influence circle) or $C$ (with its influence circle). When these circles are unified, these will create an arc with a radius of $\sqrt{3}a$ or more. This is because a $A$-like-influence-circle contains one $C$-sized, so those $C$-sized ones will create a union of a single large arc (see $\color{magenta} \text{Fig 3}$.) $\color{magenta} \rule{25cm}{0.3pt}$ Divide the whole plane into strips of width $a$ ($a$ being the shortest side of our good target triangle.) Given that we already have one vertex (along with its influence circle) inside a particular strip, we would like to prove that that strip is part of the known-solved territory. Construction-wise, let the set of $\color{red} \textbf{\textit{emanating rays}}$ of an $\color{red} \textbf{\textit{influence vertex}}$ be the rays connecting that vertex to the vertices of the other kind. When drawing an additional influence vertex (nearby) in that strip, a useful tool to consider is the influence vertex's $\color{red} \textbf{\textit{perimeter circles}}$; if that vertex is a $A$-type, draw the circle with radius $b$ as its perimeter circle, otherwise, draw the circle with radius $a$. If we're lucky, then the next influence vertex will lie on the previous' perimeter circle; otherwise, we might need to work a little harder. For now, we prove that a perimeter circle's range that lies inside the strip must be $\textbf{big enough}$. $\color{cyan} \rule{5.3cm}{2pt}$ $\color{cyan} \diamondsuit$ $ \boxed{\textbf{Sufficient-Sized Angles.}}$ $\color{cyan} \diamondsuit$ $\color{cyan} \rule{5.3cm}{2pt}$ Let there exists two triangles with size lengths $d,b,c$ and $d',b,b$ ($d \leq b \leq c ; d \leq d'$). Then, the angle $\theta$ between $b,c$ is smaller or equal than the angle $\theta'$ between $b,b$. $\color{cyan} \rule{25cm}{0.3pt}$ We know that \begin{align*} d^2 &= b^2+c^2-2bc \cos{\theta} \\ d'^2 &= b^2+b^2-2b^2\cos{\theta'} \end{align*}Since $d' \geq d$, \[ b^2-2b^2\cos{\theta'} \geq c^2-2bc\cos{\theta} \]or \[2bc\cos{\theta}-2b^2\cos{\theta'} \geq c^2-b^2 \]If $\theta > \theta'$, then $\cos{\theta} < \cos{\theta'}$. So, the LHS is at most $(2bc-2b^2)\cos{\theta'} < 2bc-2b^2$ since $c-b \geq 0$, but \[2bc-2b^2 > c^2-b^2 \Rightarrow 2bc > b^2+c^2\]yields a contradiction. $\blacksquare$ $\blacksquare$

Lastly, we deal with the algorithm described briefly earlier, and pull out a finish from putting all strips (of width $a$) into known-solved territory. $\color{red} \rule{7,5cm}{2pt}$ $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{Imperfect but Complete Covering.}}$ $\color{red} \clubsuit$ $\color{red} \rule{7.5cm}{2pt}$ We describe the algorithm as follows: let we have constructed $O_1$. Then, construct its perimeter circle, and intersect those with the strips's boundary (let at $L,M$ to the right of $O_1$.) If $O_1$ is type $B$ or $C$, then pick a $O_2$ in the direction between $L$ and $M$. This direction exists, as $\angle LO_1M \geq 60^{\circ}$ by $\color{cyan} \diamondsuit$ $ \boxed{\textbf{Sufficient-Sized Angles}}$ $\color{cyan} \diamondsuit$ by setting $b = c = d = a$. If its direction is $+60^{\circ}$ or more; or $-60^{\circ}$ or less, then pick the ones between $[-60,+60]$. If $O_1$ is type $A$, pick $\textbf{all}$ $O_{2_{(i)}}$ in those directions. They exist by setting $d = a$ in $\color{cyan} \diamondsuit$ $ \boxed{\textbf{Sufficient-Sized Angles}}$ $\color{cyan} \diamondsuit$. $\color{red} \rule{25cm}{0.3pt}$ Let the $\color{red} \textbf{\textit{strip segment}}$ of $O_1$ to be the area bounded by $O_1$'s and the rightmost $O_2$'s projections. Then, we claim that the strip segment belongs to the union of influence circles of $O_1$ and $O_2$ (and its friends, in the last case.) $\color{red} \rule{25cm}{0.3pt}$ $\color{red} \spadesuit$ $\color{red} \boxed{\textbf{Proof.}}$ $\color{red} \spadesuit$ [asy][asy]usepackage("tikz");label("\begin{tikzpicture}[scale=0.65] \definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666} \definecolor{qqwuqq}{rgb}{0,0.39215686274509803,0} \definecolor{xdxdff}{rgb}{0.49019607843137253,0.49019607843137253,1} \definecolor{ududff}{rgb}{0.30196078431372547,0.30196078431372547,1} \draw [shift={(-3.45181818181818,-0.15)},line width=0.8pt,color=qqwuqq!70!white,fill=qqwuqq!50!white] (0,0) -- (19.31609708386625:0.5454545454545454) arc (19.31609708386625:64.31609708386624:0.5454545454545454) -- cycle; \draw [line width=0.8pt] (-6.71,2.43)-- (1.0027272727272742,2.5045454545454535); \draw [line width=0.8pt] (-3.45181818181818,-0.15) circle (2.7708889663684952cm); \draw [line width=0.8pt] (-6.683219910042345,-0.34075955122860335)-- (1.0295073626849287,-0.26621409668315); \begin{small} \draw [fill=ududff] (-3.45181818181818,-0.15) circle (2.5pt); \draw[color=ududff] (-3.45181818181818,-0.15-0.5) node {$O_1$}; \draw [fill=ududff] (-3.4785982717758337,2.620759551228603) circle (2.5pt); \draw[color=ududff] (-3.4785982717758337,2.620759551228603+0.45) node {$90^{\circ}$}; \draw [fill=xdxdff] (-2.250898520945985,2.3471218696878373) circle (2.5pt); \draw[color=xdxdff] (-2.250898520945985+0.2,2.3471218696878373+0.35) node {$O_2$}; \draw [fill=xdxdff] (-0.836907938449702,0.7665533716425221) circle (2.5pt); \draw[color=xdxdff] (-0.6154545454545437+0.15,1.15) node {$O_2'$}; \end{small} \node[orange] at (-3.3,-3.8) {Fig 4.1: $\color{green} B,C \rightarrow B,C$}; \end{tikzpicture} \begin{tikzpicture}[scale=0.65] \definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666} \definecolor{xdxdff}{rgb}{0.49019607843137253,0.49019607843137253,1} \definecolor{ududff}{rgb}{0.30196078431372547,0.30196078431372547,1} \draw [line width=0.8pt] (-6.71,2.43)-- (2.784545454545456,2.448); \draw [line width=0.8pt] (-3.45181818181818,0.05) circle (2.762209552291129cm); \draw [line width=0.8pt] (-6.704633654585942,-0.3322043394914244)-- (2.789911799959516,-0.30220433949142445); \draw [line width=0.8pt] (-1.8418525690282408,-1.818595041322312) -- (-1.8418525690282408,4.329338842975201); \draw [line width=0.8pt,domain=-5.752644628099169:1.48785123966942] plot(\x,{(--7.838934982911665--2.2477023671862613*\x)/1.6055016908473307}); \draw [line width=0.8pt,domain=-7.452644628099169:2.08785123966942] plot(\x,{(-1.2061557011363853-0.3727558249369304*\x)/1.6105926551909115}); \draw [line width=0.8pt] (0.5076387863902916,-1.518595041322312) -- (0.5076387863902916,6.929338842975201); \draw [line width=0.8pt,domain=-0.37644628099169:0.63785123966942,color = orange,dashed] plot(\x,{(-0.5917829105821983-5.897508160313319*\x)/-0.6313432955562576}); \begin{small} \draw [fill=ududff] (-3.45181818181818,0.05) circle (2.5pt); \draw[color=ududff] (-3.45181818181818,0.05-0.6) node {$O_1$}; \draw [fill=xdxdff] (-1.8463164909708492,2.297702367186261) circle (2.5pt); \draw[color=xdxdff] ((-1.8463164909708492+0.2,2.297702367186261-0.5) node {$N$}; \draw [fill=uuuuuu] (-2.0653233202784573,2.4390235682580252) circle (2pt); \draw[color=uuuuuu] (-2.0653233202784573,2.4390235682580252+0.4) node {$L$}; \draw [fill=uuuuuu] (-0.7145783844671701,-0.3205670014533979) circle (2pt); \draw[color=uuuuuu] (-0.7145783844671701+0.3,-0.3205670014533979+0.35) node {$M$}; \draw [fill=uuuuuu] (-1.8412255266272683,-0.3227558249369304) circle (2pt); \draw[color=uuuuuu] (-1.8412255266272683+0.3,-0.3227558249369304+0.35) node {$P_N$}; \draw [fill=xdxdff] (0.49680184741991607,5.57806804093333) circle (2.5pt); \draw[color=xdxdff] (0.49680184741991607+0.5,5.57806804093333) node {$O_2$}; \draw [fill=ududff] (0.5093227210445281,-0.8667670920261241) circle (2.5pt); \draw[color=ududff] (0.5093227210445281+0.45,-0.8667670920261241-0.4) node {$E_{O_2}$}; \draw [fill=orange] (-0.17615072871497817,-0.7081210058081405) circle (2.5pt); \draw[color=orange] (-0.17615072871497817+0.25,-0.7081210058081405-0.4) node {$X$}; \end{small} \node[orange] at (-2.8,-3.8) {Fig 4.2: $\color{green} B,C \rightarrow A$}; \end{tikzpicture}");[/asy][/asy] In the first case, where the selected $O_2$ is also a $B,C$-type, there is almost nothing to prove since the whole strip can be covered by a circle of radius $\sqrt{a^2+a^2}$. Otherwise, let $O_2$ is an $A$-type. Define $N$ as the intersection of $O_1O_2$ with $O_1$'s perimeter circle, and $P_N$ be the drop of $N$ to the strip-edge $O_2$ is farther from. Let $O_1P_N$ intersect the perpendicular from $O_2$ at $E_{O_2}$. Observe that $|O_1P_N|$ is less than $\sqrt{2}a$, so the whole $\color{red} \text{bottom region}$ is fully covered by $O_1$'s influence circle. Since the influence circle of $O_2$ has a radius of at least $|O_1O_2|$, we just need to prove that $O_2$'s influence circle intersects the farther line before $P_N$ (instead of beyond it). For the sake of contradiction, let it intersect beyond $P_N$. By similar triangles and $|O_1P_N| < |NP_N|$, we know that $|O_2E_{O_2}|$ is less than $|O_1O_2|$. So, there must not exist a point of intersection beyond $P_N$, as it will force \[ |O_2X| \geq \max{(O_1O_2,O_2E_{O_2})} \]for some $X \in \overline{O_1E_{O_2}}$. $\color{red} \rule{25cm}{0.3pt}$ [asy][asy]usepackage("tikz");label("\begin{tikzpicture}[scale=0.6] \definecolor{xdxdff}{rgb}{0.49019607843137253,0.49019607843137253,1} \definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666} \definecolor{ududff}{rgb}{0.30196078431372547,0.30196078431372547,1} \draw [line width=0.8pt] (-6.71,2.43)-- (3.584545454545456,2.45); \draw [line width=0.8pt] (-6.704633654585942,-0.3322043394914244)-- (3.589911799959516,-0.31220433949142445); \draw [line width=0.8pt,domain=-5.959789990989293:3.006927547421655] plot(\x,{(--13.079520485686444-1.97208372087717*\x)/4.593162028254107}); \draw [line width=0.8pt,domain=-5.259789990989293:1.206927547421655,color=orange,dashed] plot(\x,{(--0.8704815082095503-4.735226302649363*\x)/4.1155894833306945}); \draw [shift={(-3.655123966942146,4.416942148760327)},line width=0.8pt,thin] plot[domain=3.9583252088283:6.277111286225671,variable=\t]({1*4.998624972924467*cos(\t r)+0*4.998624972924467*sin(\t r)},{0*4.998624972924467*cos(\t r)+1*4.998624972924467*sin(\t r)}); \draw [line width=0.8pt,thin,gray] (-7.077215313852736,0.7733803703155324)-- (-3.655123966942146,4.416942148760326); \draw [line width=0.8pt,thin,gray] (-3.655123966942146,4.416942148760326)-- (1.3434087973043822,4.386580582626733); \begin{small} \draw [fill=ududff] (-3.655123966942146,4.416942148760326) circle (2.5pt); \draw[color=ududff] (-3.525454209485203+0.2,4.745993720611245+0.05) node {$O_1$}; \draw [fill=uuuuuu] (-2.0685147821264662,-0.3231973971394191) circle (2pt); \draw[color=uuuuuu] (-1.9498645273968878,0.003927687141564329+0.1) node {$M$}; \draw [fill=uuuuuu] (0.938038061311961,2.4448584278831564) circle (2pt);\draw[color=uuuuuu] (1.0636419519370748-0.25,2.742088396790186+0.15) node {$L$}; \draw [fill=orange] (0.46046551638854893,-0.3182841538890367) circle (2.5pt); \draw[color=orange] (0.589435348590106+0.2,0.003927687141564329) node {$N$}; \draw [fill=xdxdff] (1.6405724190429178,1.0530092243295623) circle (2.5pt); \draw[color=xdxdff] (1.7673033633551574,1.3806565355682456) node {$O_{2_{(2)}}$}; \draw [fill=xdxdff] (-0.8030024328898215,0.31186548992201146) circle (2.5pt); \draw[color=xdxdff] (-0.6802145894033907-0.75,0.6464011497406823-0.05) node {$O_{2_{(1)}}$}; \end{small} \draw[green!80!white,very thin] (-2.0685147821264662,-0.3231973971394191)--(-0.8030024328898215,0.31186548992201146)--(1.6405724190429178,1.0530092243295623)--(0.938038061311961,2.4448584278831564); \node[orange] at (-2,-1.7) {Fig 4.3: $\color{green} A \rightarrow B,C$}; \end{tikzpicture}");[/asy][/asy] Now for the trickiest part of this Solution. We now no longer use denominators of $\angle A$ --- but we still use the same idea of finding a point nearly inside the strip. Also, instead of proving directly the union of a lot of influence circles to be equal to the big strip segment, we will prove this in consecutive pairs. Namely, we further Claim that \[ |MO_{2_{(1)}}|,|O_{2_{(i)}}O_{2_{(i+1)}}|, |O_{2_{(k)}}L| \leq a \]The first and last are obtained by direct cosine law (and noting that the angles between the rays must not exceed $\angle A$). Finally, the second relation is actually an equality: note that those triangles are congruent to the original $\triangle ABC$. $\color{red} \rule{25cm}{0.3pt}$ $\color{blue} \rule{8cm}{2pt}$ $\color{blue} \diamondsuit$ $\color{blue} \boxed{\textbf{Sidewise and Upwise Distances Finish.}}$ $\color{blue} \diamondsuit$ $\color{blue} \rule{8cm}{2pt}$ We Claim that in the cases of $4.1$ and $4.3$, there exists a $B,C$-type which lies inside the strip. Also, for every $B,C$-type inside the strip, we Claim that its strip segment has length $\dfrac{1}{2}a$ or more. Finally, after covering up one entire strip, we prove that we can slowly advance in two vertical directions to cover the whole plane. $\color{blue} \rule{25cm}{0.3pt}$ The second statement is the easiest to prove: if we have $O_2$ lying between $[+60,+90]$, for example, we can pick $O_2'$ out of the next lower ray (assuming the denominator of $B,C$ is six or more; so the distance between directions are $60^{\circ}$ or less.) The first statement is a bit more nuanced; it's obvious for $4.1$ and the argument in the paragraph before this; for $4.3$ we will take $O_{2_{(k)}}$ for the job. To prove this works, let $N$ be the point in the farther border so that $|O_1N| = c$. From here, we can prove that $\angle NO_1L$ is greater than $\angle A$ by simple cosine rule bash: \[ |LN| \geq a \Rightarrow \cos{\angle (NO_1L)} \leq \cos{\angle A} \Rightarrow A \leq \angle NO_1L\] Finally, consider one $B,C$-type influence vertex in the strip. Out of this, consider a ray in the direction $[+60,+120]$ for an upward lift or $[-120,-60]$ for a downward lift (and pick an $O$ there). This will guarantee a point outside the strip upon at most two iterations of this (since any interation brings the new point at least $\dfrac{\sqrt{3}}{2}a$ units upward); and we can be sure that $O$'s influence circle will pass through $O_1$ (the initial point) so we can just take a point in $O$'s perimeter circle and apply the algorithm again. We are finally done. $\blacksquare$ $\blacksquare$ $\blacksquare$

I would argue that the hardest ones to come up are $\color{green} \clubsuit$ $ \boxed{\textbf{Complexity Reduction: Seeing Rationals}}$ $\color{green} \clubsuit$ and (the details of) the ray-seeking point of $O_2$ from $O_1$ in $\color{cyan} \diamondsuit$ $ \boxed{\textbf{Sufficient-Sized Angles}}$ $\color{cyan} \diamondsuit$ and $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{Imperfect but Complete Covering}}$ $\color{red} \clubsuit$. $\color{green} \rule{25cm}{2pt}$ $\color{red} \text{Sec 1: To 60 degrees!}$ Looking back, it amazes me how the idea that prompted the green Claim also made me realize how the green part is much useless on its own. While I initially $\textbf{only}$ considered rational target-polygon angular identities (such as 120-60-120-60, for example), I wondered if there's a way to simplify the numerous copying of equilaterals and rectangles to cover the whole grid. Since the reflections of the triangles will all be $\textit{integer translations}$ of each other, I thought to create new rules on which shortcuts can be found, and there lies the first part. Unfortunately, given that all three angles are rational, there's no way to guarantee infinite rays emanating from one point (check the case $30-60-90$, for example, where there's a clear wheel of rays from vertices of type-30,60 and 90). Turning from triangles to quadrilaterals (while I haven't really formulated the $r'$ axis part), I knew that I could shift the target polygon from being $s_1,s_2,s_3,s_4$ into $s_1,s_2,s_3$ (therefore forming a triangle); but in most cases, the triangle itself has a bigger height than the quadrilateral. Playing on reflections more and more, I found that the (exact) copies of the quads seem to operate at multiples; [what if any figure doesn't have to strictly be rotated around the vertices?] I thought. $\color{green} \rule{25cm}{2pt}$ $\color{red} \text{Sec 2: Height Expanding!}$ As the easiest cases to be thrown out are obtuse angles, I begged the question: do obtuse triangles present as much trouble as they seem to pose? One problem I immediately thought of is that while the rotation around two acute vertices are easily managed, if the obtuse is $120$ degrees, there's a very limited option of possible rotation across that vertex. Another problem is the difficulty of establishing a $\color{green} \text{safe region}$ around vertices with small sides around it; therefore naturally, the influence sphere of $B,C$ (which has a $a$-side around it) must be less than the sphere of the vertex with longer sides around it. Now, instead of expanding the polygon towards all directions as in $\#2$, I found this to be ineffective; as safe-spaces in individual units won't contribute much to the ever-enlarging perimeter of the plane. Also, seeing that this problem more resembles the config of USEMO Patriotic P5 (cutting inside for individualistic expansion) rather than my initial thought of RMM 2020 P5 (outside intersection and strict border play to create a specific extension of the config), I opted to focus on one direction and found that the strip with length $a$ is more effective than one of length $h$. Following from that, a question of large angles occur: within the smallest perimeter/locus circles, how abundant are the rays? After establishing the (almost tedious and unnatural) trigonometric identities, I found that the $A$-transitions towards and from $B,C$-types are the most problematic ones, so I tried to capitalize on the existence of $N$ in $4.2$ to $4.3$ as a benchmark of size. Finally, dropping lots of perpendiculars and seeing that (almost in every case except, ironically, in $4.1$) that dropping the $72^{\circ}$ is not necessary (an influence circle of radius $\sqrt{2}a$ is enough to handle known-solved territory completion), except in the case where a $B,C$ type needs to expand sideways to another $B,C$ type. \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \]

Suppose $X$ is not in $K$. Then let $X$ be bad wrt. side $A_iA_{i+1}$, and reflect $X$ wrt. $A_iA_{i+1}$. Intuitively $X$ should get "closer" to $K$, and we can indeed verify by Pythagoras that the length $XA_j$ for all $j \neq i, i+1$ $\textbf{decreases}$ and $XA_i, XA_{i+1}$ do not change with such an operation. This leads us to think: if we keep performing such operations, can we make $X$ arbitrarily close to some point $A_j$? And if so, can we finish?

Suppose $X$ gets arbitrarily close to a vertex, WLOG it is $A_2$, then we will try to bring $X$ into $K$. Suppose $XA_2 = R$. For $R$ small enough, the sector centred at $A_2$ with radius $R$ bounded by the angle $A_1A_2A_3$ lies completely within $K$. Hence if $X$ is good wrt. both $A_1A_2, A_2A_3$, then $X$ lies in $K$. We will now only use reflections across $A_1A_2$ and $A_2A_3$ to preserve the length $XA_2$. If $X$ is good wrt. both $A_1A_2,A_2,A_3$ then by the above we are done, so suppose this is never the case. If $X$ is bad wrt. both $A_1A_2,A_2A_3$, reflect $X$ across $A_1A_2$ so $X$ is good wrt. $A_1A_2$ and bad wrt. $A_2A_3$. So in any case we may assume $X$ is good wrt. $A_1A_2$ and bad wrt. $A_2A_3$. Now reflect $X$ across $A_2A_3$, then $A_1A_2$ so $X$ is still good wrt $A_1A_2$ and bad wrt. $A_2A_3$. However, note that these two operations are equivalent to rotating $X$ by $2\angle A_1A_2A_3$ clockwise about $A_2$. Let $Y$ be a point on ray $A_1A_2$ outside the segment $A_1A_2$, then $X$ stays in the part of the plane bounded by $\angle YA_2A_3$ after such a rotation. However, this is not always possible. Note that if $2\angle A_1A_2A_3 > 180^{\circ}$ then the rotation is an anti-clockwise rotation by $a^{\circ}$ for some $a<180$. So in any case we rotate $X$ around $A_2$ by some non reflex angle. Visualising the rotation of $X$ as a process, with enough applications of the operation, $X$ has to "leave" the angle $\angle YA_2A_3 < 180^{\circ}$ at some point, and as the rotation is also $<180^{\circ}$ it will not be able to "return" to the angle$\angle YA_2A_3$, a contradiction. Thus if we can get $X$ arbitrarily close to a vertex we are done.

Suppose $X$ is bad wrt. $A_iA_{i+1} = l$. When we perform the reflection across $l$, we know the distances to the other vertices decrease, but $\textit{by how much}$? Note that we are moving $X$ perpendicular to $A_iA_{i+1}$ by twice the perpendicular distance from $X$ to $A_iA_{i+1}$ which we will suppose is $2d$, so we want to investigate $d$. Perpendicular distances get arbitrarily close to 0Suppose for infinitely many reflections the value $d$ above satisfies $d\geq c$ for some constant $c > 0$. Call such reflections $\textit{awful}$, then by pigeonhole principle WLOG there are infinitely many awful reflections across $A_1A_2$. Let $X'$ be the image of X after the reflection, then by cosine rule, $$X'A_3^2 = XA_3^2 + 4d^2 - 4d\cdot XA_3 \cos \angle X'XA_3$$Let the perpendicular distance from $A_3$ to $A_1A_2$ be $b$, then $\cos \angle X'XA_3 = \frac{d+b}{XA_3}$. So we have $$X'A_3^2 = XA_3^2 + 4d^2 - 4d(d+b) = XA_3^2 - 4db \leq XA_3^2 - 4bc$$Since $XA_3^2$ is non-increasing with every operation, and decreases by at least $4bc$ with every awful operation, it eventually becomes negative, contradiction. Thus for every $c>0$, we eventually have $d<c$ always with enough operations applied. Now suppose that for all $i$, there exist constants $a_i>0$ such that $XA_i \geq a_i$ always, otherwise $X$ gets arbitrarily close to a vertex and we are done. Then suppose $X$ is bad wrt. $A_2A_3$ infinitely many times, and let the foot of the perpendicular from $X$ to $A_2A_3$ be $D$. Suppose $D$ is not on $A_2A_3$, then we have 2 cases. $\textbf{Case 1.}$ $X$ is also bad wrt $A_1A_2$ (or $A_3A_4$). Let the foot of the perpendicular from $X$ to $A_1A_2$ be $E$, then since $XD,XE$ can be arbitrarily small, $XA_2$ must also get arbitrarily small, contradiction. $\textbf{Case 2.}$ $X$ is not bad wrt. $A_1A_2$ and $A_3A_4$. Then one of $A_1A_2A_3, A_2A_3A_4$ must be acute, WLOG it is the former. Since $D$ is not on segment $A_2A_3$, then it lies "above" the line $A_1A_2$, so $90^{\circ} < \angle XA_2A_3 < 180^{circ} - \angle A_1A_2A_3$ so $XA_2 = \frac{XD}{\sin \angle XA_2A_3}$ must also get arbitrarily small as $XD$ gets arbitrarily small, contradiction. Thus eventually $D$ must lie on segment $A_2A_3$. We can pick $c < a_2, a_3$ so that $XD < c$ eventually, then $DA_2 > \sqrt{a_2^2 - c^2}, DA_3 > \sqrt{a_3^2 - c^2}$ which are some positive constants. Then it is easy to see for any $D$ within this range on segment $A_2A_3$ and $XD$ small enough, the reflection of $X$ across $A_2A_3$ must lie in $K$, so we are done.

First, define the (initial) $\color{green} \textbf{\textit{target polygon}}$ to be $K$, and a $\color{green} \textbf{\textit{reflection axis}}$ to be the $``\text{sides}"$ we can reflect $X$ with. Additionally, let the $\color{green} \textbf{\textit{known-solved territory}}$ to be the region of points $X$; in which it is known that after a sequence of reflections among the $\color{green} \textbf{\textit{reflection axes}}$, every point within it will land inside the target polygon. Observe that the known-solved territory is obtained by reflecting $\color{green} \textbf{\textit{any target polygon}}$ and previously known-solved territory towards any $\color{green} \textbf{\textit{reflection axis}}$. Inductively, we wish to show that these sequence of reflections will bring about the whole plane $\textit{to fit inside}$ the target polygon we know. Here, instead of doing that, we create some new reflection axes, so that we can $\color{green} \textbf{\textit{transfer}}$ the identity of the target polygon to be $\textbf{any polygon}$ which belongs to the known-solved territory, which has all its sides belonging to the set of reflection axes. $\color{green} \rule{8.5cm}{2pt}$ $\color{green} \clubsuit$ $ \boxed{\textbf{Complexity Reduction: Seeing Rationals.}}$ $\color{green} \clubsuit$ $\color{green} \rule{8.5cm}{2pt}$ We Claim that (as in Lemma 1 of $\#2$) that the reflection of any two $\textit{reflection axes}$ among each other is also a reflection axis. Furthermore, we can transfer the target polygon to an $\color{green} \text{acute}$ triangle with no $72^{\circ}$ angle in its vertices. For simplicity, we call such a triangle to be $\color{green} \textbf{\textit{good}}$, and three angles of a target-triangle to be its $\color{green} \textbf{\textit{angular identity}}$. $\color{green} \rule{25cm}{0.3pt}$ $\color{green} \spadesuit$ $\boxed{\textbf{Proof.}}$ $\color{green} \spadesuit$ Let the two axes be $a$ and $r$, and we'd like to matriculate $r'$ (reflection of $r$ thru $a$) as a reflection axis. To do this, reflect across $a$, then $r$, then $a$ again; and we have reflected whatever figure w.r.t. $r'$. $\color{green} \rule{25cm}{0.3pt}$ To obtain the second assertion, we consider an initial polygon $K$ with $n$ sides. If $n \geq 4$ and $K$ is not a rectangle, there exists an obtuse angle among its $\textit{angular identity}$; say it is formed by $a$ and $b$. Using the first assertion, we get that $a'$, which must intersect $K$ in its interior, is also a reflection axis. This breaks down $K$ into two polygons with less sides, with them both able to function as target polygons. Do this over and over to create a triangle (when $K$ is a rectangle, just manually reflect $K$ across its sides rectangularly to cover the whole plane.) $\color{green} \rule{25cm}{0.3pt}$ Now, if $T$ (assuming $K$ is a triangle) is a right-angled triangle, form a new triangle with the first assertion applied to a non-right-angled angle (of its angular identity). This will form four axes $a,b,c,a'$ (where $a'$ is the reflection of $a$ to $b$); and use $a,a',c$ as the new target polygon. If $T$ is obtuse now, do the same with the angle ($C$) measuring less than $45^{\circ}$, so its angular identity will change to \[ (C,B,180-B-C) \Rightarrow (2C,B,180-2C-B); \ 2C \leq 90^{\circ} \]If $T$ has a $72^{\circ}$, we divide into two cases. If the triangle is not a $36-72-72$, let the triangle is $72-x-y$, where \[ \angle A = 72^{\circ}, \angle B = x, \angle C = y\]now reflect $b$ to $c$ and $a$ to $c$ to get $b',a'$; and reflect $a'$ $k$ times (with different axes) towards the direction of $b'$ to get $a_k'$ as an axis. We now use the triangle formed by $b,b',a_k'$ as a target polygon with angular-identity \[ 36^{\circ}, ky, 144-ky \]and $144-ky < 90$. This can be reached by setting $90-y \leq 144-ky < 90$; i.e. step-by-step increasing $ky$ until no obtuse is formed. We are almost done, except that we can be back where we started if $144-ky = 72$, or when $y \mid 72$. Now, $y > 18^{\circ}$; or else $x$ is a right or obtuse angle. Then, $y$ cannot be $36$ for the time being, so $y$ must be $24$. If so, pick the other angle, \[ x = 180^{\circ}-24^{\circ}-72^{\circ} = 84^{\circ} \]as the new $y$, and perform the same operations. Meanwhile, if the triangle is indeed a $(A,B,C) = (36,72,72)$ we can expand the triangle into four times its original size (as can be done with a $60-60-60$) by reflecting $b,c$ towards $a$, $c$ two times to each of $a$ and $b$ (to get the reflection axes), before verifying that its known-solved territory is sufficient for advancing (see $\color{green} \text{Fig 1}$). $\color{green} \rule{25cm}{0.3pt}$ Now we still focus on the (set of) two reflection axes which intersect at a point. Given a good target-polygon define an $\color{cyan} \textbf{\textit{influence circle}}$ of such a vertex to be the obtainable area only utilizing (mostly) the two reflection axes, with a bit of the third axis. $\color{cyan} \rule{6.1cm}{2pt}$ $\color{cyan} \diamondsuit$ $ \boxed{\textbf{Preliminary Establishment.}}$ $\color{cyan} \diamondsuit$ $\color{cyan} \rule{6.1cm}{2pt}$ Let the three sides of the target triangle be $a,b,c$ with $a < b < c$. Then, the influence circle of $B$ and $C$ consists of a circle with radius $a$, and the influence circle of $A$ consists of a circle with radius $b$. $\color{cyan} \rule{25cm}{0.3pt}$ First, we establish the membership of the small arcs of center $A,B,C$ with radius $b,a,a$ respectively. This can be obtained by simply reflecting $\triangle ABC$ once with the axes $a,b,c$, and noting that $\angle B,C > 45^{\circ}$ (see $\color{cyan} \text{Fig 2}$) is enough to guarantee that the distance of a vertex (say $A$) to the $\textit{reflected sides}$ (say, $b',c'$) is more than $b$, or $c$. This arc then can be reflected over and over (as in the forming of $a_k'$ in the previous Section) to form a complete circle. We are done for the first part. $\blacksquare$

We will now strengthen the influence circle, before finally considering the whole grid $\color{red} \textbf{\textit{in one direction}}$ (a.k.a. covering a strip in circles). $\color{magenta} \rule{7cm}{2pt}$ $\color{magenta} \diamondsuit$ $ \boxed{\textbf{Wide-Ranging for Looser Bounds.}}$ $\color{magenta} \diamondsuit$ $\color{magenta} \rule{7cm}{2pt}$ We claim that the influence circle of $A$ can be improved to be of radius $c$, then we establish that the small circles aren't that small either; $B$ and $C$ will have influence circles of radius $\sqrt{3}a$. $\color{magenta} \rule{25cm}{0.3pt}$ Using the same conventions, the region with radius $c$ not yet covered by $A$'s influence circle lies beyond $C$ (see $\color{magenta} \text{Fig 2'}$). Since anything in the $\text{marked area}$ has a distance less than $b$ with the vertex $A'$ (with influence circle of radius $b$), all sectors of the circle of radius $c$ is part of $A$'s influence circle. For $B$'s and $C$'s, this is a bit trickier. Consider the denominator of the angles $\angle B,C$ respectively, out of $360^{\circ}$ units. Example, if $\angle A = 75$, then its denominator is \[ \dfrac{75}{360} \ \text{units} = \dfrac{5}{24} \ \text{units} \]or $24$, in its simplest form. Note that since $\angle B,C < 90, \ne 72$, then its denominator is six or more. Next consider the reflection axes emanating from say, $B$, with respect to $a,c$ reflected over-and-over-again (until it repeats again to square one). These rays are equally spaced within one another, and there must exist rays equal to $\angle B$'s denominator. Now pick two points $X$ and $Y$ which are consecutively placed in those rays. Then, $X,Y$ is either similar to $A$ (with its influence circle) or $C$ (with its influence circle). When these circles are unified, these will create an arc with a radius of $\sqrt{3}a$ or more. This is because a $A$-like-influence-circle contains one $C$-sized, so those $C$-sized ones will create a union of a single large arc (see $\color{magenta} \text{Fig 3}$.) $\color{magenta} \rule{25cm}{0.3pt}$ Divide the whole plane into strips of width $a$ ($a$ being the shortest side of our good target triangle.) Given that we already have one vertex (along with its influence circle) inside a particular strip, we would like to prove that that strip is part of the known-solved territory. Construction-wise, let the set of $\color{red} \textbf{\textit{emanating rays}}$ of an $\color{red} \textbf{\textit{influence vertex}}$ be the rays connecting that vertex to the vertices of the other kind. When drawing an additional influence vertex (nearby) in that strip, a useful tool to consider is the influence vertex's $\color{red} \textbf{\textit{perimeter circles}}$; if that vertex is a $A$-type, draw the circle with radius $b$ as its perimeter circle, otherwise, draw the circle with radius $a$. If we're lucky, then the next influence vertex will lie on the previous' perimeter circle; otherwise, we might need to work a little harder. For now, we prove that a perimeter circle's range that lies inside the strip must be $\textbf{big enough}$. $\color{cyan} \rule{5.3cm}{2pt}$ $\color{cyan} \diamondsuit$ $ \boxed{\textbf{Sufficient-Sized Angles.}}$ $\color{cyan} \diamondsuit$ $\color{cyan} \rule{5.3cm}{2pt}$ Let there exists two triangles with size lengths $d,b,c$ and $d',b,b$ ($d \leq b \leq c ; d \leq d'$). Then, the angle $\theta$ between $b,c$ is smaller or equal than the angle $\theta'$ between $b,b$. $\color{cyan} \rule{25cm}{0.3pt}$ We know that \begin{align*} d^2 &= b^2+c^2-2bc \cos{\theta} \\ d'^2 &= b^2+b^2-2b^2\cos{\theta'} \end{align*}Since $d' \geq d$, \[ b^2-2b^2\cos{\theta'} \geq c^2-2bc\cos{\theta} \]or \[2bc\cos{\theta}-2b^2\cos{\theta'} \geq c^2-b^2 \]If $\theta > \theta'$, then $\cos{\theta} < \cos{\theta'}$. So, the LHS is at most $(2bc-2b^2)\cos{\theta'} < 2bc-2b^2$ since $c-b \geq 0$, but \[2bc-2b^2 > c^2-b^2 \Rightarrow 2bc > b^2+c^2\]yields a contradiction. $\blacksquare$ $\blacksquare$

Lastly, we deal with the algorithm described briefly earlier, and pull out a finish from putting all strips (of width $a$) into known-solved territory. $\color{red} \rule{7,5cm}{2pt}$ $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{Imperfect but Complete Covering.}}$ $\color{red} \clubsuit$ $\color{red} \rule{7.5cm}{2pt}$ We describe the algorithm as follows: let we have constructed $O_1$. Then, construct its perimeter circle, and intersect those with the strips's boundary (let at $L,M$ to the right of $O_1$.) If $O_1$ is type $B$ or $C$, then pick a $O_2$ in the direction between $L$ and $M$. This direction exists, as $\angle LO_1M \geq 60^{\circ}$ by $\color{cyan} \diamondsuit$ $ \boxed{\textbf{Sufficient-Sized Angles}}$ $\color{cyan} \diamondsuit$ by setting $b = c = d = a$. If its direction is $+60^{\circ}$ or more; or $-60^{\circ}$ or less, then pick the ones between $[-60,+60]$. If $O_1$ is type $A$, pick $\textbf{all}$ $O_{2_{(i)}}$ in those directions. They exist by setting $d = a$ in $\color{cyan} \diamondsuit$ $ \boxed{\textbf{Sufficient-Sized Angles}}$ $\color{cyan} \diamondsuit$. $\color{red} \rule{25cm}{0.3pt}$ Let the $\color{red} \textbf{\textit{strip segment}}$ of $O_1$ to be the area bounded by $O_1$'s and the rightmost $O_2$'s projections. Then, we claim that the strip segment belongs to the union of influence circles of $O_1$ and $O_2$ (and its friends, in the last case.) $\color{red} \rule{25cm}{0.3pt}$ $\color{red} \spadesuit$ $\color{red} \boxed{\textbf{Proof.}}$ $\color{red} \spadesuit$ [asy][asy]usepackage("tikz");label("\begin{tikzpicture}[scale=0.65] \definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666} \definecolor{qqwuqq}{rgb}{0,0.39215686274509803,0} \definecolor{xdxdff}{rgb}{0.49019607843137253,0.49019607843137253,1} \definecolor{ududff}{rgb}{0.30196078431372547,0.30196078431372547,1} \draw [shift={(-3.45181818181818,-0.15)},line width=0.8pt,color=qqwuqq!70!white,fill=qqwuqq!50!white] (0,0) -- (19.31609708386625:0.5454545454545454) arc (19.31609708386625:64.31609708386624:0.5454545454545454) -- cycle; \draw [line width=0.8pt] (-6.71,2.43)-- (1.0027272727272742,2.5045454545454535); \draw [line width=0.8pt] (-3.45181818181818,-0.15) circle (2.7708889663684952cm); \draw [line width=0.8pt] (-6.683219910042345,-0.34075955122860335)-- (1.0295073626849287,-0.26621409668315); \begin{small} \draw [fill=ududff] (-3.45181818181818,-0.15) circle (2.5pt); \draw[color=ududff] (-3.45181818181818,-0.15-0.5) node {$O_1$}; \draw [fill=ududff] (-3.4785982717758337,2.620759551228603) circle (2.5pt); \draw[color=ududff] (-3.4785982717758337,2.620759551228603+0.45) node {$90^{\circ}$}; \draw [fill=xdxdff] (-2.250898520945985,2.3471218696878373) circle (2.5pt); \draw[color=xdxdff] (-2.250898520945985+0.2,2.3471218696878373+0.35) node {$O_2$}; \draw [fill=xdxdff] (-0.836907938449702,0.7665533716425221) circle (2.5pt); \draw[color=xdxdff] (-0.6154545454545437+0.15,1.15) node {$O_2'$}; \end{small} \node[orange] at (-3.3,-3.8) {Fig 4.1: $\color{green} B,C \rightarrow B,C$}; \end{tikzpicture} \begin{tikzpicture}[scale=0.65] \definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666} \definecolor{xdxdff}{rgb}{0.49019607843137253,0.49019607843137253,1} \definecolor{ududff}{rgb}{0.30196078431372547,0.30196078431372547,1} \draw [line width=0.8pt] (-6.71,2.43)-- (2.784545454545456,2.448); \draw [line width=0.8pt] (-3.45181818181818,0.05) circle (2.762209552291129cm); \draw [line width=0.8pt] (-6.704633654585942,-0.3322043394914244)-- (2.789911799959516,-0.30220433949142445); \draw [line width=0.8pt] (-1.8418525690282408,-1.818595041322312) -- (-1.8418525690282408,4.329338842975201); \draw [line width=0.8pt,domain=-5.752644628099169:1.48785123966942] plot(\x,{(--7.838934982911665--2.2477023671862613*\x)/1.6055016908473307}); \draw [line width=0.8pt,domain=-7.452644628099169:2.08785123966942] plot(\x,{(-1.2061557011363853-0.3727558249369304*\x)/1.6105926551909115}); \draw [line width=0.8pt] (0.5076387863902916,-1.518595041322312) -- (0.5076387863902916,6.929338842975201); \draw [line width=0.8pt,domain=-0.37644628099169:0.63785123966942,color = orange,dashed] plot(\x,{(-0.5917829105821983-5.897508160313319*\x)/-0.6313432955562576}); \begin{small} \draw [fill=ududff] (-3.45181818181818,0.05) circle (2.5pt); \draw[color=ududff] (-3.45181818181818,0.05-0.6) node {$O_1$}; \draw [fill=xdxdff] (-1.8463164909708492,2.297702367186261) circle (2.5pt); \draw[color=xdxdff] ((-1.8463164909708492+0.2,2.297702367186261-0.5) node {$N$}; \draw [fill=uuuuuu] (-2.0653233202784573,2.4390235682580252) circle (2pt); \draw[color=uuuuuu] (-2.0653233202784573,2.4390235682580252+0.4) node {$L$}; \draw [fill=uuuuuu] (-0.7145783844671701,-0.3205670014533979) circle (2pt); \draw[color=uuuuuu] (-0.7145783844671701+0.3,-0.3205670014533979+0.35) node {$M$}; \draw [fill=uuuuuu] (-1.8412255266272683,-0.3227558249369304) circle (2pt); \draw[color=uuuuuu] (-1.8412255266272683+0.3,-0.3227558249369304+0.35) node {$P_N$}; \draw [fill=xdxdff] (0.49680184741991607,5.57806804093333) circle (2.5pt); \draw[color=xdxdff] (0.49680184741991607+0.5,5.57806804093333) node {$O_2$}; \draw [fill=ududff] (0.5093227210445281,-0.8667670920261241) circle (2.5pt); \draw[color=ududff] (0.5093227210445281+0.45,-0.8667670920261241-0.4) node {$E_{O_2}$}; \draw [fill=orange] (-0.17615072871497817,-0.7081210058081405) circle (2.5pt); \draw[color=orange] (-0.17615072871497817+0.25,-0.7081210058081405-0.4) node {$X$}; \end{small} \node[orange] at (-2.8,-3.8) {Fig 4.2: $\color{green} B,C \rightarrow A$}; \end{tikzpicture}");[/asy][/asy] In the first case, where the selected $O_2$ is also a $B,C$-type, there is almost nothing to prove since the whole strip can be covered by a circle of radius $\sqrt{a^2+a^2}$. Otherwise, let $O_2$ is an $A$-type. Define $N$ as the intersection of $O_1O_2$ with $O_1$'s perimeter circle, and $P_N$ be the drop of $N$ to the strip-edge $O_2$ is farther from. Let $O_1P_N$ intersect the perpendicular from $O_2$ at $E_{O_2}$. Observe that $|O_1P_N|$ is less than $\sqrt{2}a$, so the whole $\color{red} \text{bottom region}$ is fully covered by $O_1$'s influence circle. Since the influence circle of $O_2$ has a radius of at least $|O_1O_2|$, we just need to prove that $O_2$'s influence circle intersects the farther line before $P_N$ (instead of beyond it). For the sake of contradiction, let it intersect beyond $P_N$. By similar triangles and $|O_1P_N| < |NP_N|$, we know that $|O_2E_{O_2}|$ is less than $|O_1O_2|$. So, there must not exist a point of intersection beyond $P_N$, as it will force \[ |O_2X| \geq \max{(O_1O_2,O_2E_{O_2})} \]for some $X \in \overline{O_1E_{O_2}}$. $\color{red} \rule{25cm}{0.3pt}$ [asy][asy]usepackage("tikz");label("\begin{tikzpicture}[scale=0.6] \definecolor{xdxdff}{rgb}{0.49019607843137253,0.49019607843137253,1} \definecolor{uuuuuu}{rgb}{0.26666666666666666,0.26666666666666666,0.26666666666666666} \definecolor{ududff}{rgb}{0.30196078431372547,0.30196078431372547,1} \draw [line width=0.8pt] (-6.71,2.43)-- (3.584545454545456,2.45); \draw [line width=0.8pt] (-6.704633654585942,-0.3322043394914244)-- (3.589911799959516,-0.31220433949142445); \draw [line width=0.8pt,domain=-5.959789990989293:3.006927547421655] plot(\x,{(--13.079520485686444-1.97208372087717*\x)/4.593162028254107}); \draw [line width=0.8pt,domain=-5.259789990989293:1.206927547421655,color=orange,dashed] plot(\x,{(--0.8704815082095503-4.735226302649363*\x)/4.1155894833306945}); \draw [shift={(-3.655123966942146,4.416942148760327)},line width=0.8pt,thin] plot[domain=3.9583252088283:6.277111286225671,variable=\t]({1*4.998624972924467*cos(\t r)+0*4.998624972924467*sin(\t r)},{0*4.998624972924467*cos(\t r)+1*4.998624972924467*sin(\t r)}); \draw [line width=0.8pt,thin,gray] (-7.077215313852736,0.7733803703155324)-- (-3.655123966942146,4.416942148760326); \draw [line width=0.8pt,thin,gray] (-3.655123966942146,4.416942148760326)-- (1.3434087973043822,4.386580582626733); \begin{small} \draw [fill=ududff] (-3.655123966942146,4.416942148760326) circle (2.5pt); \draw[color=ududff] (-3.525454209485203+0.2,4.745993720611245+0.05) node {$O_1$}; \draw [fill=uuuuuu] (-2.0685147821264662,-0.3231973971394191) circle (2pt); \draw[color=uuuuuu] (-1.9498645273968878,0.003927687141564329+0.1) node {$M$}; \draw [fill=uuuuuu] (0.938038061311961,2.4448584278831564) circle (2pt);\draw[color=uuuuuu] (1.0636419519370748-0.25,2.742088396790186+0.15) node {$L$}; \draw [fill=orange] (0.46046551638854893,-0.3182841538890367) circle (2.5pt); \draw[color=orange] (0.589435348590106+0.2,0.003927687141564329) node {$N$}; \draw [fill=xdxdff] (1.6405724190429178,1.0530092243295623) circle (2.5pt); \draw[color=xdxdff] (1.7673033633551574,1.3806565355682456) node {$O_{2_{(2)}}$}; \draw [fill=xdxdff] (-0.8030024328898215,0.31186548992201146) circle (2.5pt); \draw[color=xdxdff] (-0.6802145894033907-0.75,0.6464011497406823-0.05) node {$O_{2_{(1)}}$}; \end{small} \draw[green!80!white,very thin] (-2.0685147821264662,-0.3231973971394191)--(-0.8030024328898215,0.31186548992201146)--(1.6405724190429178,1.0530092243295623)--(0.938038061311961,2.4448584278831564); \node[orange] at (-2,-1.7) {Fig 4.3: $\color{green} A \rightarrow B,C$}; \end{tikzpicture}");[/asy][/asy] Now for the trickiest part of this Solution. We now no longer use denominators of $\angle A$ --- but we still use the same idea of finding a point nearly inside the strip. Also, instead of proving directly the union of a lot of influence circles to be equal to the big strip segment, we will prove this in consecutive pairs. Namely, we further Claim that \[ |MO_{2_{(1)}}|,|O_{2_{(i)}}O_{2_{(i+1)}}|, |O_{2_{(k)}}L| \leq a \]The first and last are obtained by direct cosine law (and noting that the angles between the rays must not exceed $\angle A$). Finally, the second relation is actually an equality: note that those triangles are congruent to the original $\triangle ABC$. $\color{red} \rule{25cm}{0.3pt}$ $\color{blue} \rule{8cm}{2pt}$ $\color{blue} \diamondsuit$ $\color{blue} \boxed{\textbf{Sidewise and Upwise Distances Finish.}}$ $\color{blue} \diamondsuit$ $\color{blue} \rule{8cm}{2pt}$ We Claim that in the cases of $4.1$ and $4.3$, there exists a $B,C$-type which lies inside the strip. Also, for every $B,C$-type inside the strip, we Claim that its strip segment has length $\dfrac{1}{2}a$ or more. Finally, after covering up one entire strip, we prove that we can slowly advance in two vertical directions to cover the whole plane. $\color{blue} \rule{25cm}{0.3pt}$ The second statement is the easiest to prove: if we have $O_2$ lying between $[+60,+90]$, for example, we can pick $O_2'$ out of the next lower ray (assuming the denominator of $B,C$ is six or more; so the distance between directions are $60^{\circ}$ or less.) The first statement is a bit more nuanced; it's obvious for $4.1$ and the argument in the paragraph before this; for $4.3$ we will take $O_{2_{(k)}}$ for the job. To prove this works, let $N$ be the point in the farther border so that $|O_1N| = c$. From here, we can prove that $\angle NO_1L$ is greater than $\angle A$ by simple cosine rule bash: \[ |LN| \geq a \Rightarrow \cos{\angle (NO_1L)} \leq \cos{\angle A} \Rightarrow A \leq \angle NO_1L\] Finally, consider one $B,C$-type influence vertex in the strip. Out of this, consider a ray in the direction $[+60,+120]$ for an upward lift or $[-120,-60]$ for a downward lift (and pick an $O$ there). This will guarantee a point outside the strip upon at most two iterations of this (since any interation brings the new point at least $\dfrac{\sqrt{3}}{2}a$ units upward); and we can be sure that $O$'s influence circle will pass through $O_1$ (the initial point) so we can just take a point in $O$'s perimeter circle and apply the algorithm again. We are finally done. $\blacksquare$ $\blacksquare$ $\blacksquare$

I would argue that the hardest ones to come up are $\color{green} \clubsuit$ $ \boxed{\textbf{Complexity Reduction: Seeing Rationals}}$ $\color{green} \clubsuit$ and (the details of) the ray-seeking point of $O_2$ from $O_1$ in $\color{cyan} \diamondsuit$ $ \boxed{\textbf{Sufficient-Sized Angles}}$ $\color{cyan} \diamondsuit$ and $\color{red} \clubsuit$ $\color{red} \boxed{\textbf{Imperfect but Complete Covering}}$ $\color{red} \clubsuit$. $\color{green} \rule{25cm}{2pt}$ $\color{red} \text{Sec 1: To 60 degrees!}$ Looking back, it amazes me how the idea that prompted the green Claim also made me realize how the green part is much useless on its own. While I initially $\textbf{only}$ considered rational target-polygon angular identities (such as 120-60-120-60, for example), I wondered if there's a way to simplify the numerous copying of equilaterals and rectangles to cover the whole grid. Since the reflections of the triangles will all be $\textit{integer translations}$ of each other, I thought to create new rules on which shortcuts can be found, and there lies the first part. Unfortunately, given that all three angles are rational, there's no way to guarantee infinite rays emanating from one point (check the case $30-60-90$, for example, where there's a clear wheel of rays from vertices of type-30,60 and 90). Turning from triangles to quadrilaterals (while I haven't really formulated the $r'$ axis part), I knew that I could shift the target polygon from being $s_1,s_2,s_3,s_4$ into $s_1,s_2,s_3$ (therefore forming a triangle); but in most cases, the triangle itself has a bigger height than the quadrilateral. Playing on reflections more and more, I found that the (exact) copies of the quads seem to operate at multiples; [what if any figure doesn't have to strictly be rotated around the vertices?] I thought. $\color{green} \rule{25cm}{2pt}$ $\color{red} \text{Sec 2: Height Expanding!}$ As the easiest cases to be thrown out are obtuse angles, I begged the question: do obtuse triangles present as much trouble as they seem to pose? One problem I immediately thought of is that while the rotation around two acute vertices are easily managed, if the obtuse is $120$ degrees, there's a very limited option of possible rotation across that vertex. Another problem is the difficulty of establishing a $\color{green} \text{safe region}$ around vertices with small sides around it; therefore naturally, the influence sphere of $B,C$ (which has a $a$-side around it) must be less than the sphere of the vertex with longer sides around it. Now, instead of expanding the polygon towards all directions as in $\#2$, I found this to be ineffective; as safe-spaces in individual units won't contribute much to the ever-enlarging perimeter of the plane. Also, seeing that this problem more resembles the config of USEMO Patriotic P5 (cutting inside for individualistic expansion) rather than my initial thought of RMM 2020 P5 (outside intersection and strict border play to create a specific extension of the config), I opted to focus on one direction and found that the strip with length $a$ is more effective than one of length $h$. Following from that, a question of large angles occur: within the smallest perimeter/locus circles, how abundant are the rays? After establishing the (almost tedious and unnatural) trigonometric identities, I found that the $A$-transitions towards and from $B,C$-types are the most problematic ones, so I tried to capitalize on the existence of $N$ in $4.2$ to $4.3$ as a benchmark of size. Finally, dropping lots of perpendiculars and seeing that (almost in every case except, ironically, in $4.1$) that dropping the $72^{\circ}$ is not necessary (an influence circle of radius $\sqrt{2}a$ is enough to handle known-solved territory completion), except in the case where a $B,C$ type needs to expand sideways to another $B,C$ type. \[\begin{tabular}{p{12cm}} $\rule{4cm}{0.5pt} \hspace{1cm} \blacksquare \hspace{1cm} \rule{4cm}{0.5pt}$ \\ \end{tabular} \\ \]