This might be big clown but wasn't this drawing two 30-30-120 triangles on each side of the n small ones and proving no overlap and computing the bound Here's what I mean

Shoot. The idea is to center a hexagon of side length $\tfrac{\sqrt{3}}{4}$ or something at the center of each unit equilateral triangle, show that if two equilateral triangles are nonoverlapping then the hexagons aren't either, and then use area. During the test I got this idea but with a circle instead, leading to an inequality that was just a bit too weak. I tried a bunch of other stuff that didn't get anywhere, and decided not to write the circle idea down. Dangit. After the test it took me 10 more minutes of thinking to get the hexagon. Nice problem though, although I would've liked it more as P6 (then I would've had time to do it)

@above

Main idea:

ok my half-sol was very sketchy but basically I proved that if u look at the set of downward facing triangles' centroids, the minimum distance between 2 of these centroids is $\frac{\sqrt{3}}{2}$ (diagram attached) so you can fit the most downward facing triangles if their centroids are arranged as centers of tangent circles of radius $\frac{\sqrt{3}}{4}$ which is the desired config it was very hand wavy so I think at most 2 points

jeteagle wrote: Main idea: hm yes i can totally comprehend that

IAmTheHazard wrote: Shoot. The idea is to center a hexagon of side length $\tfrac{\sqrt{3}}{4}$ or something at the center of each unit equilateral triangle, show that if two equilateral triangles are nonoverlapping then the hexagons aren't either, and then use area. During the test I got this idea but with a circle instead, leading to an inequality that was just a bit too weak. I tried a bunch of other stuff that didn't get anywhere, and decided not to write the circle idea down. Dangit. After the test it took me 10 more minutes of thinking to get the hexagon. Nice problem though, although I would've liked it more as P6 (then I would've had time to do it) I put down everything I did with the circles, used LoC so I think my proof was valid, but got $\frac{4\sqrt{3}}{3\pi}L^2$ instead of $\frac 23 L^2$. Do you think this could be 1-2 points? Yeah basically what 2above did but I'm pretty sure not hand-wavy

The idea is close enough to the hexagon solution that I'd say it could earn 3 points.

v4913 wrote: ok my half-sol was very sketchy but basically I proved that if u look at the set of downward facing triangles' centroids, the minimum distance between 2 of these centroids is $\frac{\sqrt{3}}{2}$ (diagram attached) so you can fit the most downward facing triangles if their centroids are arranged as centers of tangent circles of radius $\frac{\sqrt{3}}{4}$ which is the desired config it was very hand wavy so I think at most 2 points For this solution, can't you also divide the locus of the possible locations for the centroids (which is an equilateral triangle) into smaller equilateral triangles of side length $k = \frac{\sqrt{3}}{2} - \epsilon$, then argue that the number of centroids is at most the number of equilateral triangles by Pigeon Hole? oh no did I fakesolve this

bissue wrote: v4913 wrote: ok my half-sol was very sketchy but basically I proved that if u look at the set of downward facing triangles' centroids, the minimum distance between 2 of these centroids is $\frac{\sqrt{3}}{2}$ (diagram attached) so you can fit the most downward facing triangles if their centroids are arranged as centers of tangent circles of radius $\frac{\sqrt{3}}{4}$ which is the desired config it was very hand wavy so I think at most 2 points For this solution, can't you also divide the locus of the possible locations for the centroids (which is an equilateral triangle) into smaller equilateral triangles of side length $k = \frac{\sqrt{3}}{2} - \epsilon$, then argue that the number of centroids is at most the number of equilateral triangles by Pigeon Hole? oh no did I fakesolve this o yea oops thats a good idea

Given a convex body $K$, define its Minkowski symmetrization $M(K) = \{\tfrac{1}{2}(k_1 - k_2) \mid k_1,k_2 \in K\}$. It is straightforward to show that $K - K = M(K) - M(K)$ (where $A - B = \{a - b \mid a\in A, b\in B\}$), so \[(p + K) \cap (q + K) \neq \varnothing \iff q - p \in K - K = M(K) - M(K) \iff (p + M(K)) \cap (q + M(K)) \neq \varnothing.\]If $\nabla$ is an upside-down triangle, when $M(\nabla)$ is a hexagon with $\tfrac{3}{2}$ the area. By the above result, we may replace all triangles with hexagons and still have no two bodies intersect. If you do this correctly, all the hexagons must still fit inside the larger triangle, so you are done.

My idea was to show that the tiling where each vertex touches an edge of another minimizes filler area in an infinite plane: - note that each vertex must touch another edge or another vertex, otherwise i can move it there - let one vertex of one triangle touch another r from one vertex and 1-r from another. i can minimize the filler area to show r=1/2 Now for the sketchy part: showing that this is optimal in a triangle L>=5 To do this, I compared this method with the other method of using a triangular grid (i.e. the case where r=0) This is really sketchy because I said that the triangular grid minimizes wasted edge space while the r=1/2 case minimizes in-between triangle space, so at any value of L at least one of them gives the optimal value Finally, I divided the triangles in the r=1/2 case into a-type, b-type, and c-type, and found how many triangles you can fit depending on which type were in the bottom row of triangles

Post-contest solution. Let the triangles be $T_1,T_2,\ldots,T_n$. For any triangle $T_i$, consider the following region $T_i'$ created by appending another unit equilateral triangle to $T_i$: [asy][asy] size(45); import geometry; import olympiad; filldraw(dir(30)--dir(150)--dir(270)--cycle,grey,black); draw(dir(270)--(dir(270)+dir(30)-dir(150))--dir(30), black+linewidth(0.4)); label("$T_i$",origin); label("$T_i'$",dir(-30)); [/asy][/asy] Clearly, $T_i'$ lies in $\Delta$, because $T_i$ and $\Delta$ have opposite orientations. The sides of $T_i'$ are parallel to the sides of $\Delta$ and in the same orientation, so none of the sides of $T_i'$ can cross the sides of $\Delta$. Furthermore, at most one other triangle $T_j$ can intersect $T_i'$, because if there were two then either these two would intersect or one would intersect $T_i$, neither of which is possible. Now, for each $T_i'$ let $F(T_i')$ be the set of points in $T_i'$ which are not a part of $T_1,T_2,\ldots,T_n$. [asy][asy] size(60); import geometry; import olympiad; pair a, b, c, d, e, f, g, h; a = dir(30); b = dir(150); c = dir(270); d = dir(30)+1.8*dir(-40); e = dir(150)+1.8*dir(-40); f = dir(270)+1.8*dir(-40); g = extension(c, c+a-b, e, f); h = extension(d, e, a, c+a-b); filldraw(a--b--c--cycle,grey,black); filldraw(d--e--f--cycle,grey,black); draw(dir(270)--(dir(270)+dir(30)-dir(150))--dir(30), dotted+black+linewidth(0.8)); filldraw(c--g--e--h--a--cycle, mediumgrey, black); label("$T_i$",origin); label("$F(T_i')$ is shaded in a lighter grey.", (7,-0.8)); [/asy][/asy] Claim: For all $i$, $F(T_i')$ is at least $1/2$ of the area of $T_i$. Proof: Clearly, this is true if no triangle $T_k$ intersects $T_i'$. Now, assume that some triangle $T_k$ intersects $T_i'$. If the upper-left vertex of $T_k$ doesn't lie on the bottom-right side of $T_i$, then clearly the area of $F(T_i')$ can be reduced by translating $T_k$ in a direction perpendicular to this side of $T_i$. Thus, it suffices to consider the cases where $T_k$ has a vertex on $T_i$. This vertex will split this side of $T_i$ into two segments of length $t$ and $1-t$, and $F(T_i')$ will then consist of two equilateral triangles: one with side length $t$ and one with side length $1-t$. Since $t^2+(1-t)^2\ge 1/2$, the area of $F(T_i')$ is at least $1/2$ of the area of $T_i$ (equality holds iff $T_k$ has a vertex at the midpoint of the bottom-right side of $T_i$). Define $R_i=T_i\cup F(T_i')$. Since no two triangle $T_i$ and $T_k$ overlap, no two regions $F(T_i')$ and $F(T_k')$ intersect. Also, by construction, no two regions $F(T_i')$ and $T_k$ overlap. Thus, all $R_i$ are disjoint. Furthermore, the area of each $R_i$ is at least $\frac{3\sqrt{3}}{8}$. Since the total area of $\Delta$ is $\frac{\sqrt{3}L^2}{4}$, there can be no more than $\frac{\sqrt{3}L^2/4}{3\sqrt{3}/8}=\frac{2L^2}{3}$ triangles, as desired.

Ok, guess I'll share my post-contest solution since nobody seems to have posted the "center a hexagon" solution yet. It's written rather poorly. Given a unit equilateral triangle $T$ with center $O$, let $H(T)$ denote the regular hexagon with side length $\frac{1}{2}$ centered at $O$, such that the sides of the hexagon are parallel to the sides of $T$. Here is an example, with $T$ drawn and $H(T)$ shown in red: [asy][asy] pair A,B,C,O,H1,H2,H3,H4,H5,H6; size(150); A=(-4,0); B=(4,0); C=(0,4*sqrt(3)); O=(0,4*sqrt(3)/3); H1=O-(2,2*sqrt(3)); H2=O-(-2,2*sqrt(3)); H3=O-(-4,0); H6=O-(4,0); H4=O+(2,2*sqrt(3)); H5=O+(-2,2*sqrt(3)); filldraw(H1--H2--H3--H4--H5--H6--cycle,red+white,red); draw(A--B--C--cycle); dot("$O$",O,N); [/asy][/asy] I will first prove the following key lemma: Lemma: Given two nonoverlapping unit equilateral triangles $T_1=\triangle A_1B_1C_1,T_2=\triangle A_2B_2C_2$ with the same orientation, $H(T_1)$ and $H(T_2)$ do not intersect. Proof WLOG orient both triangles as shown in the above diagram, and let $T_1$ be above $T_2$ and let the points be ordered as shown: [asy][asy] size(100); pair A1,B1,C1,A2,B2,C2; A1=(4,0); B1=(-4,0); C1=(0,4*sqrt(3)); A2=A1-(2,12); B2=B1-(2,12); C2=C1-(2,12); pair O1,O2; pair H11,H12,H13,H14,H15,H16; pair H21,H22,H23,H24,H25,H26; O1=(0,4*sqrt(3)/3); O2=O1-(2,12); H11=O1-(2,2*sqrt(3)); H12=O1-(-2,2*sqrt(3)); H13=O1-(-4,0); H16=O1-(4,0); H14=O1+(2,2*sqrt(3)); H15=O1+(-2,2*sqrt(3)); H21=O2-(2,2*sqrt(3)); H22=O2-(-2,2*sqrt(3)); H23=O2-(-4,0); H26=O2-(4,0); H24=O2+(2,2*sqrt(3)); H25=O2+(-2,2*sqrt(3)); draw(H11--H12--H13--H14--H15--H16--cycle); draw(H21--H22--H23--H24--H25--H26--cycle); draw(A1--B1--C1--cycle); draw(A2--B2--C2--cycle); label("$A_1$",A1,E); label("$A_2$",A2,E); label("$B_1$",B1,W); label("$B_2$",B2,W); label("$C_1$",C1,N); label("$C_2$",C2,N); [/asy][/asy] Also let $O_1,O_2$ be the centers of $T_1,T_2$. Let $\ell_1$ denote the bottom-most side of $H(T_1)$ and let $\ell_2$ denote the top-most side of $H(T_2)$. Then, since the distance from $A_1B_2$ to $\ell_1$ is equal to the distance from $C_2$ to $\ell_2$, $H(T_1)$ and $H(T_2)$ cannot intersect: otherwise, $C_2$ lies "above" $A_1B_2$ and then $T_1,T_2$ intersect. $\blacksquare$ Now note that it is clear that given an equilateral $T$ with an opposite orientation as $\Delta$ that is completely inside $\Delta$, then $H(T)$ is also completely inside $\Delta$, since the distance from the center of $T$ to a side of $H(T)$ is less than the distance from the center of $T$ to a vertex. Also observe that the area of $H(T)$ is $\frac{3\sqrt{3}}{8}$. Now draw $n$ unit equilateral triangles $T_1,T_2,\ldots,T_n$, all contained inside $\Delta$ and with the opposite orientation as $\Delta$. Then, $H(T_1),H(T_2),\ldots,H(T_n)$ are all contained in $\Delta$. Further, since they are all disjoint, it is necessary to have: $$[H(T_1)]+[H(T_2)]+\cdots+[H(T_n)]\leq [\Delta] \implies \frac{3\sqrt{3}}{8}n \leq \frac{\sqrt{3}}{4}L^2 \implies n \leq \frac{2}{3}L^2,$$which is the desired inequality. $\blacksquare$

IAmTheHazard wrote:

This is almost exactly what I did in-contest, but I forgot to reword my Lemma and kept it as centers $\sqrt3/2$ away, so probably losing 1-2 points there. Still feels easier than 1 though.

Solution: Just call the residue for each triangle the two smaller triangles above it. See the attached file (the residues belong to T1). These are all disjoint and lie on the larger equilateral triangle, and so we are done by power mean.

This question has a lot to do with equality. I drew the "equality" case (I'm quite sure it's not actually sharp) and then figured out divvying up the rest of the areas.

I feel some solutions have two fundamental flaws: (1) Yes, you showed by one specific configuration, $\nabla$ covers 2/3 of the total area within a infinite plane. But the problem asks for inside a big triangle. A lot of space are wasted around the big triangle edges because we can not fit in a complete $\nabla$, so that you never achieve 2/3. Please see my graph of the blue area, which are completely wasted around the edge. If you take a look at the black region, it is clearly show that green color is 2/3 of the region. But again, it only achievable if it is in a infinite plane. (2) You never showed that there is no other configurations better than in (1). Above is only one configuration out of infinite amount of possibility. Maybe by changing some $\nabla$ a little bit, maybe we can fit a complete $\nabla$ along the edge? Without a proof, how you know 2/3 is the upper bound?

Hexagons are bestagons The main motivation for the disjoint hexagons solution is to consider one such equilateral triangle centered at $A$, and then consider where the other triangle centers can lie. It turns out that they have to be outside of a hexagon centered at $A$ with sidelength 1. Then at this point, you notice that if you draw a hexagon $\frac14$ the size with $\frac12$ the side length centered at each center, they have to be disjoint, YAY!

Center a hexagon of side length $\frac{1}{2}$ on the center of each equilateral triangle. Claim: If two triangles don't intersect, then their corresponding hexagons don't. Proof: Note that it is enough to show that the distance between the bottom point on the triangle and the bottom side of the hexagon is equal to the top side of the hexagon to the top side of the triangle.This is trivial. The problem now becomes how many hexagons of side length $\frac{1}{2}$ can fit into the big triangle. Bounding by using areas we get the desired result. $\blacksquare$

Note that the total area of $\triangle$ to be tiled is $\tfrac{L^2\sqrt{3}}{4}$. This includes the areas of the $n$ small triangles, which covers $\tfrac{n\sqrt{3}}{4}$ area and the remaining empty space. We will count the area of the empty space $S$, which has area $A(S) = \tfrac{\sqrt{3}}{4}(L^2 - n)$. We will lower bound the area of this empty space. For each unit equilateral triangle $T$, define its $\textit{upper compartment}$ to be the space formed by reflecting $T$ over its upper side. Note that for every such unit equilateral triangle, some portion of its upper compartment has to be empty. No matter how other unit equilateral triangles are placed, the empty part of its upper compartment has to have area at least\[A = \frac{\sqrt{3}}{4}(t^2 + (1-t)^2) \geq \frac{\sqrt{3}}{8}.\]Furthermore, the upper compartments of any two unit equilateral triangles clearly cannot intersect, and all must lie within $\triangle$ due to their opposite orientations, hence\[A(S) = \frac{\sqrt{3}}{4}(L^2 - n) \geq \frac{n\sqrt{3}}{8} \implies n \leq \frac23 L^2\]as desired. Note that the bound becomes very sharp for large $n$ if we have every unit triangle touch the one below it at the midpoint of its top side.

Let the betrothed hexagon of an equilateral triangle be the regular hexagon with side length $\frac12$ concentric with the triangle. A reason to consider these is the following claim (and the fact that the betrothed hexagon is the largest shape for which the claim is true): Claim 1: if two triangles $T_1,T_2$ have non-overlapping interiors then their betrothed hexagons $H_1,H_2$ respectively have non-overlapping interiors. Suppose $T_1,T_2$ are disjoint. Let $C_1,C_2$ be the centers of $T_1,T_2$ respectively. Draw a hexagon $H$ with side length $1$ centered at $C_1$. Then $C_2$ lies outside $H$. Consider the side of $H_2$ which $C_1C_2$ passes through. By symmetry we may assume that $C_1C_2$ passes through the top edge of $H$. Then the distance between $H_1$ and $H_2$ is equal to the distance between the bottom vertex of $T_2$ and the top edge of $T_1$. But the latter is just the minimum distance between $T_1$ and $T_2$, which is positive, so $H_1,H_2$ are disjoint also. Claim 2: if a triangle $T$ is contained inside $\Delta$, then its betrothed hexagon is contained inside $\Delta$. Let $T_a,T_b,T_c$ be the reflections of $T$ about each of its sides. Note that $T_a,T_b,T_c$ are contained in $\Delta$. Then the betrothed hexagon of $\Delta$ is a subset of $T\cup T_a\cup T_b\cup T_c$ which is in $\Delta$. So by symmetry, it's enough to show that one of $T_a,T_b,T_c$ is in $\Delta$. $\Delta$ is convex, so for any set of points $S\subseteq\Delta$ the convex hull of $S$ lies in $\Delta$ too. Thus it suffices to show that the reflection of $T$ across a side is in $S$ (call this point $T'$). Let $T$ be centered at $(a,b)$, and let the vertices of $\Delta$ be $(0,0)$, $\left(\frac L2,\frac{L\sqrt3}2\right)$, $(L,0)$. The lower vertex of $T$ is within the triangle with vertices at $(1,0)$, $\left(\frac L2,\frac{(L-2)\sqrt3}2\right)$, $(L-1,0)$. Then $T'$ lies within the triangle with vertices at $\left(1,\sqrt3\right)$, $\left(\frac L2,\frac{L\sqrt3}2\right)$, $\left(L-1,\sqrt3\right)$, so $T'\in\Delta$. So it suffices to prove the following problem instead: popcorn1 wrote: An equilateral triangle $\Delta$ of side length $L>0$ is given. Suppose that $n$ regular hexagons with side length $\frac12$ and with non-overlapping interiors are drawn inside $\Delta$, such that each hexagon has sides parallel to $\Delta$. Prove that\[n \leq \frac{2}{3} L^{2}.\] Then each hexagon has an area of $\frac{3\sqrt3}8$, so the total area of the betrothed hexagons is $\frac{3n\sqrt3}8$. This is at most the area of $\Delta$ which is $\frac{\sqrt3}4L^2$, so: $$\frac{3n\sqrt3}8\le\frac{\sqrt3}4L^2$$which gives the desired inequality: $$n\le\frac23L^2.$$ PS: See this Desmos graph.

^That's an awesome solution! I couldn't really figure it out while doing it so I'm glad that it's cleared up now.

MathGuy103 wrote: ^That's an awesome solution! I couldn't really figure it out while doing it so I'm glad that it's cleared up now. Thanks.

Sketch: Draw hexagons sticking out on top of the upside down equilateral triangles like ice-cream balls sticking out of the cones. You can basically prove if two hexagons do not touch, then their respective triangles do not touch. Then, directly dividing the area of the big triangle by the area of a hexagon yields the result.

Whenever a normal triangle is placed, place a "ghost triangle" that has the opposite orientation as the normal triangle (and thus same as the $L$ triangle) and so that it shares its bottom edge with the top edge of the normal triangle that was just placed. Note that for any normal triangle placed, its ghost triangle also lies entirely within the large triangle. For now on, define a triangle of area as $\sqrt{3}/4$ square units. Claim: A ghost triangle can only be at most half filled with other triangles. Note that each ghost triangle can only have 1 real triangle that it overlaps with. The intersection of the ghost triangle and one other normal triangle is a parallelogram, and in order to maximize the area, the paralleogram should touch the base of the ghost triangle. Then, the parallelogram is the entire ghost triangle minus 2 smaller equilateral triangles whose side length sum up to 1. By QM-AM, the sum of the squares of the two side lengths is minimized at the center, in which case the intersection has an area of 1/2 triangles. Clearly, the ghost triangles do not overlap with each other. Therefore, if we place $n$ normal triangles, there must be at least $n/2$ triangles of area that are in a ghost triangle but not in a real triangle (that are still within the large triangle), so we have $$\frac{3}{2}n \leq L^2$$$$n\leq \frac{2}{3}L^2,$$as desired.

Consider two centroids $P$ and $Q$ of equilateral triangles used in the tiling. Realize that the equilateral triangles corresponding to $P$ and $Q$ are disjoint if and only if $Q$ is outside of the hexagon of side length $1$ centered at $P$ with three of its sides parallel to the sides of $\Delta$, and analogously for $P$. Now for a given point $X$, we say that $\mathcal{T}(X)$ is the triangle with sides parallel to those of $\Delta$ and of opposite orientation with unitary length and centroid $X$, and $\mathcal{H}(X)$ is the hexagon of side length $1/2$ centered at $X$ with three of its sides parallel to the sides of $\Delta$. Taking a homothety of scale factor $1/2$ at $P$ and $Q$, respectively, $\mathcal{H}(P)$ and $\mathcal{H}(Q)$ are disjoint if and only if $\mathcal{T}(P)$ and $\mathcal{T}(Q)$ are disjoint. Realize that for all points $X$, $\mathcal{H}(X)$ has area $\tfrac{3\sqrt{3}}{8}$, and thus we have the bound \[ \frac{3\sqrt{3}}{8} \cdot n \le L^2 \cdot \frac{\sqrt{3}}{4}, \]which rewrites as \[ n \le \frac{2}{3} \cdot L^2, \]as desired.

My first formal proof for JMO written, I appreciate any feedback, thanks Claim: Any two of the equilateral triangles do not intersect iff, the two regular hexagon constructed with side length of 1/2, sides parallel to the original triangle and centers at the centers of the two triangles do not overlap. Proof: Let $T_1$, $T_2$ be two black triangles inside the triangle of side length L. Subclaim 1: $T_1$ and $T_2$ only intersect iff $T_2$ lies in the regular hexagon constructed with side length of 1 sides parallel to the original triangle and center at center of $T_1$. Reasoning: The triangles intersect only when either one of the vertices of $T_1$ is in $T_2$ or one of the vertices of $T_2$ is in $T_1$ (true for all convex polygons). When a point of $T_2$ is in $T_1$ the distance of the center of $T_2$ to the closest line must be less than $\sqrt{3}/3$. When a point of $T_1$ is in $T_2$ the distance of the center of $T_1$ to the closest line must be less than $\sqrt{3}/3$. Since the perpendicular distance from the closest side to the center of the triangle is $\sqrt{3}/6$ This makes it so that the center of the $T_1$ the distance perpendicular to the six sides formed from points and sides be $\sqrt{3}/2$ This creates a bounded area where $T_1, T_2$ intersect which is a hexagon of length of 1 sides parallel to the original triangle and center at center of $T_1$. Subclaim 2: The result from Subclaim 2 is identical to ensuring that the two regular hexagon constructed with side length of 1/2, sides parallel to the original triangle and centers at the center of $T_1$ and $T_2$ do not overlap. Reasoning: Let the centers of $T_1$ and $T_2$ be P and Q respectively Draw the line connecting P and Q The distance inside the two smaller hexagons and the large hexagon around P are the same due to scaling. The smaller hexagon around P and the line from PQ is identical to the larger hexagon around P and the line from PQ scaled down by 1/2. Whereas the the smaller hexagon around Q and the line from PQ is identical to the larger hexagon around P and the line from PQ scaled down by 1/2, transformed onto having center Q and rotated by 180 degrees. 1/2 the length + 1/2 the length is the total length which means that if these hexagons overlap then, then Q is inside the large hexagon of P. Notice that a hexagon of 1/2 length has area of $\frac{3\sqrt3}{8}$ While the main triangle has area of $\frac{L^2 \sqrt3}{4}$. Since the hexagons cannot overlap, the area of the hexagons must be $\frac{3\sqrt3n}{8}$ where n is the number of hexagons (connected to the number of black triangle). This area must obviously be less than that of the large triangle or equal to it meaning: $\frac{3\sqrt3n}{8} \leq \frac{L^2 \sqrt3}{4}$ Which simplifies to $n \leq \frac{2}{3}L^2$ and we are done. Discussed with pi271828

Let $T_1, T_2$ be two black triangles, and $C_1, C_2$ be the centers respectively. All hexagons described in this solution are regular and are oriented such that their sides are parallel to the equilateral triangle. Claim: $T_1, T_2$ intersect with one another iff $C_2$ lies within a regular hexagon of side length $1$ that is centered at $C_1$ This can be easily seen by sliding $T_2$ around $T_1$ such that $T_2$ is always touching $T_1$. It is clear the path that $C_2$ traces is a regular hexagon, and it is also clear that the side length of the hexagon it traces is $1$. Claim: $C_2$ lies within a regular hexagon of side length $1$ centered at $C_1$ iff the hexagons of side length $0.5$ centered at $C_1, C_2$ respectively intersect. Let $f(P)$ be the hexagon centered at $P$ with side length of $0.5$. Let $d(P, Q)$ be vertical distance between two points $P, Q$ where we set the bottom side of the equilateral triangle as the $x$-axis. Start with a sufficiently small hexagon centered at $C_1$ and keep expanding the size of the hexagon until $C_2$ is on the perimeter of the hexagon. Let $\chi$ be the side length of the final hexagon. WLOG that $C_2$ is on the "top" side of the hexagon. It is clear to see that if $\chi \ge 1$ then $C_2$ does not lie within a hexagon of side length $1$ centered at $C_1$ and vice versa for $\chi < 1$. Notice that the vertical distance from $C_2$ to the bottom side of $f(C_2)$ is $\frac{\sqrt{3}}{4}$ and likewise for $C_1$. Now it is clear that $d(C_1, C_2) = \frac{\chi \cdot \sqrt{3}}{2}$. Now simply comparing vertical distances the result clearly follows. Now we are simply fitting in hexagons with side length of $0.5$ within $\Delta$, so we have the following bound. $$\frac{3\sqrt{3}}{8} \cdot n \le L^2 \cdot \frac{\sqrt{3}}{4}$$ and the result immediately follows

Fakesolve? Stronger Bound!? Claim. The centers of any two such unit triangles are $\Vert\textbf{v}\Vert\ge\tfrac{\sqrt{3}}{2}$ units apart. Proof. Let $\theta$ be the angle of $\textbf{v}$ above the horizontal. In each of the intervals $\theta\in[0,\tfrac{\pi}{6}],[\tfrac{\pi}{6},\tfrac{\pi}{3}],\dots$, $\textbf{v}$ is restricted from below by a ``restricting edge" from one triangle and a ``restricting vertex" from the other. In particular, within any fixed interval, the component of $\textbf{v}$ perpendicular to the restricting edge clearly exceeds $\tfrac{\sqrt{3}}{2}$ (the height of one triangle), as claimed. Additionally it is apparent that any center is at least $\tfrac{1}{\sqrt{3}}$ units away from each edge, which restricts the region of possible centers to a central triangle $\Delta'$ of side length $L-2$. It suffices to place the maximum number $n$ of points in $\Delta'$ so that any two are at least $\tfrac{\sqrt{3}}{2}$ units apart. We shall bound $n$ from above as follows: construct a triangular grid of unit distance $\tfrac{\sqrt{3}}{2}$ parallel to $\Delta'$ with $S:=1+2+\dots+\lfloor\tfrac{L-2}{\sqrt{3}/2}\rfloor$ points so that its top point is $\tfrac{1}{2}$ units below that of $\Delta'$; and at each such point, construct a circle of diameter $1$. There can be at most one point in each such circle, so $\Delta'$ can contain at most $S$ points. Then \begin{align*} S&\le1+2+\dots+\lfloor\tfrac{L-2}{\sqrt{3}/2}\rfloor\\ &=\tfrac{1}{2}\lfloor\tfrac{L-2}{\sqrt{3}/2}\rfloor(\lfloor\tfrac{L-2}{\sqrt{3}/2}\rfloor+1)\\ &\le\tfrac{L-2}{\sqrt{3}}(\tfrac{L-2}{\sqrt{3}/2}+1)\\ &=\tfrac{2}{3}(L-2)^2+\tfrac{L-2}{\sqrt{3}}\\ &=\tfrac{2}{3}L^2-(\tfrac{8}{3}-\tfrac{1}{\sqrt{3}})L+(\tfrac{8}{3}-\tfrac{2}{\sqrt{3}}). \end{align*}The last quantity is at most $\tfrac{2}{3}L^2$ so long as $L\ge1$, but $L\le1$ is vacuously satisfied. $\square$ (In hindsight it's probably $\lceil\cdot\rceil$ instead of $\lfloor\cdot\rfloor$, but it doesn't matter -- the linear term is still negative and so the inequality holds for all nontrivial $L$.)

Assume all regular hexagons mentioned below refer to regular hexagons that is parallel to the triangle. Each triangle has a center. From now on, we will refer only to the centers. Define the "aura" of a center $P$ to be to be a regular hexagon with side length 1 centered at $P$. Claim: No center $Q \neq P$ can lie in the aura of $P$. Proof: Otherwise, triangle $Q$ will intersect triangle $P$. Define the "mana-space" of a center $P$ to be a regular hexagon with side length $\frac 12$. Claim: Two mana-spaces may not intersect. Proof: FTSoC assume there exists centers $P, Q$ whose mana-spaces intersect. Clearly, this means that $Q$ lies within the aura of $P$, contradiction. Thus, if we have $n$ equilateral triangles of side length $1$, and the area of each mana-space is $6 \cdot \frac{\frac{1}{2} ^2 \sqrt{3}}{4} $, then we must have $n \cdot 6 \cdot \frac{\frac{1}{2} ^2 \sqrt{3}}{4} \leq \frac{L^2 \cdot \sqrt{3}}{4}$, or $\boxed{n \leq \frac 23 L^2}$. Remark: In fact, considering the concept of mana-spaces, I believe the optimal construction is tiling the equilateral triangle (bar the corners) in a hexagonal grid using these mana-spaces. The number of tiles asymptotically goes towards $\frac{2}{3} L^2$ (this is also why smaller side lengths are much farther away from the actual bound).

Draw a hexagon of side length $\frac 12$ centered at the center of each triangle. Then note that if the triangles aren't overlapped then neither are the hexagons, so $$\frac{\frac{\sqrt{3}}{4}n^2}{\frac{\sqrt{3}}{8}}=\frac 23 n^2,$$so we're done$.\blacksquare$

The ``draw a hexagon" method doesn't seem like it works at first glance; I spent much too long trying much more elaborate ways of tiling the triangle. For each triangle $\mathcal T$ in the tiling, we draw a regular hexagon $\mathcal H$ such that $\mathcal T$ and $\mathcal H$ have the same center, three sides of $\mathcal H$ are parallel to the sides of $\mathcal T$, and $\mathcal H$ has side length $\frac 12$. Claim: No two hexagons $\mathcal H$ overlap. Proof: Consider any two triangles $\mathcal T_1$ and $\mathcal T_2$, and let $O_1$ and $O_2$ be their centers. There must exist a side $s$ of $\mathcal T_1$ such that $\mathcal T_2$ lies completely on the opposite side of $s$ than $\mathcal T_1$. Let $A_2$ be the vertex of $\mathcal T_2$ opposite to the side of $\mathcal T_2$ parallel to $s$. Measuring distances perpendicular to $s$, we then have \[d(O_1, O_2) = d(O_1, s) + d(s, A_2) + d(A_2, O_2) \geq d(O_1, s) + A_2O_2 = \frac{\sqrt 3}2.\]But the distance from $O_2$ to the side $a_2$ of $\mathcal H_2$ parallel to $s$ is at most $\frac{\sqrt 3}4$ and similar for $O_2$, so it follows that $\mathcal H_1$ and $\mathcal H_2$ may not overlap (as the other sides extend in different directions). $\blacksquare$ Each such hexagon has area $\frac{3\sqrt 3}8$, so it follows that there are at most $\frac{ \frac{\sqrt 3}4 L^2}{\frac{3\sqrt 3}8} = \frac 23 L^2$ such hexagons, and thus at most $\frac 23 L^2$ triangles $\mathcal T$.