Colored areas in a triangle that don't share a line

Question

A friend and I came up with this puzzle and I'm looking for a proof.

Given an equilateral triangle of area 1, color parts of the triangle red, blue, and green such that

• Each color makes exactly one connected region strictly inside the triangle
• There is no line parallel to one of the sides that contains points of multiple colors

Let $X$ be the minimum area of the red, blue, and green regions. Find the maximum value of $X$ over all possible colorings.

I suspect the maximum occurs in the following arrangement of teardrop-shaped figures, which each have an area of $\frac{4}{45}$ (new bound found by Daniel Mathias). This is an awfully strange number for what seems like a nice problem, so I'm not sure if it's correct.

If you consider the triangle formed by the three points closest to the center and call $x$ the side length, $\frac{4}{45}$ can be reached when $x=\frac{4}{5\cdot 3^{3/4}}$. If $s$ is the side length of the original triangle, each region has area $x\left(\frac{s}{3}-\frac{5x\sqrt{3}}{12}\right)$. Maximizing this gives $\frac{4}{45}$.

Does anybody have an idea of a proof (or a counterexample) that this indeed gives the maximum? If it is valid, is there any intuition behind the value $\frac{4}{45}$ that makes it so special?

Also, we can look at the discrete case of this puzzle on a triangular grid with $n$ vertices on each side where we color vertices three colors. Asymptotically, this should have the same behavior as the original problem. I couldn't see a very nice pattern with small values—does anybody have a solution to this modified problem?

We tried to look for problems similar to this one; it seems like it should be well known! However, we couldn't find anything. If anybody could help us, that would be greatly appreciated.

Answer 1

Visualizations of the solutions mentioned in the comments to the question.

All solutions, including the one in the OP, ignore the region boundaries.

• $\dfrac{20}{225}=\dfrac{4}{45}=0.0\overline{8}\ldots$ solution by Daniel Mathias:

• $\dfrac{11}{121}=\dfrac{1}{11}=0.\overline{09}$ solution by Blue:

Answer 2

Consider a modified version of OP's (borderless) teardrop, here determined by points that separate the side of the triangle in the ratio $a:b:c:b:a$ for some (non-negative) $a$, $b$, $c$; for simplicity, we'll use those values as lengths, so that the side of the triangle has length $s := 2a+2b+c$.

The teardrop region itself is seen to be a parallelogram with sides $a$ and $b$, with an equilateral triangle of side $a-c$ cut from a corner. Thus, its area is $$A := ab\sin 60^\circ - \frac12(a-c)^2\sin 60^\circ = \frac14\sqrt{3}\left(2ab-(a-c)^2\right) \tag{1}$$ Ignoring the multiplied constant, we seek to maximize $2ab-(a-c)^2$ subject to $2a+2b+c$ being the constant $s$. Define

$$f(a,b) = 2ab-(a-c)^2 = 2ab - (a-(s-2a-2b))^2 \tag{2}$$

We can find the extreme value by taking appropriate partial derivatives: $$f_a = -9 a - 5 b + 3 s \qquad f_b = -5 a - 4 b + 2 s \tag{3}$$ Setting these equal to zero and solving yields the corresponding optimal solution: $$a = \frac{2}{11}s \qquad b = \frac{3}{11}s \quad\to\quad A = \frac{1}{11}\cdot\frac14\sqrt{3}s^2 =\frac1{11}\cdot(\text{area of triangle}) \tag{4}$$ Note that the vertices of this teardrop on the side of the triangle divide that side in the ratio $2:3:6$. $\square$

Note. While this argument optimizes the teardrop configuration, it does not show that three congruent teardrops comprise the overall maximal solution to the puzzle.

Answer 3

Just as a first step in proving an upper bound on the area, here is a proof I found that details $\dfrac{1}{5}$ as an upper bound, although it's clear from the proof the bound is unachievable.

For this proof I only considered two of the three sides of the triangle; that is, no two colors can lie on a line parallel to a chosen two out of the three sides. Ignoring the lines parallel to one of the sides, we proceed as follows:

First, we stretch the axes so that the triangle instead looks like an isosceles right triangle; this messes with the symmetry of the shape, but as we only consider two of those sides, let those two be the legs of the triangle so that by putting coordinate axes, no two red, blue, or green points can share a $x$ or $y$ coordinate.

Now, if we consider any square of side length $x$ with sides parallel to the axes, we first prove that the maximum possible area of the color that appears the least is $\dfrac{x^2}{9}$. To do this, let $x_r,x_b,x_g$ be the combined length of the projection of the red, green, and blue areas onto the $x$ axis, and define $y_r,y_b,y_g$ similarly. Then note $x_r+x_b+x_g \le x$ and $y_r+y_b+y_g \le x$. Thus, there exists a color $c$ such that $x_c+y_c \le \dfrac{2x}{3}$; by quadratic optimization, we can achieve $x_cy_c \le \dfrac{x^2}{9}$. Since the color is completely contained within the bounds of its projections, this proves our claim.

Now, we optimize the use of our lemma. Consider splitting a square into regions as shown below:

The black region is contained within a square of side length $x$ and thus the color that appears the least has at most area $\dfrac{x^2}{9}$. Add on the two purple areas outside to get a naive bound of $\dfrac{x^2}{9}+(s-x)^2$. We can optimize this by differentiating to get a minimum achieved when $x=\dfrac{9s}{10}$ and the area is $\dfrac{1}{10}s^2$. Since the area of the large triangle is $\dfrac{1}{2}s^2$, we get our final bound of $\dfrac{1}{5}$.

It's clear that in many aspects the bound is unachievable; the only case when equality can hold is when first of all we have equality inside the black region, which is already impossible because the part cut off from the square limits this. Then, even if equality was achieved, all colors take up an area of $\dfrac{x^2}{9}$ and then would need to split the outside area, which is also unaccounted for. However, it does come pretty close to the actual bound for only two sides. Using an optimization of the configuration below, I managed to achieve a possible area of $\dfrac{3-\sqrt{5}}{4} \approx 0.191,$ which is very close to $\dfrac{1}{5}$, and may not even be the best configuration.