Maximizing the area of a region involving triangles

Question

I thought of a question while making up an exercise sheet for high school students, and posted it on MathStackExchange but did not receive an answer (the original post is here), so I thought perhaps MO would be a more suitable place to attract answers.

Let $T$ be an equilateral triangle of unit area, with vertices $A_1, A_2, A_3$. Place triangles $T_1, T_2, T_3$ each of unit area such that the centroid $G_i$ of $T_i$ is equal to $A_i$ for $i = 1,2,3$. What is the maximum possible value of the area of the region $T_1 \cap T_2 \cap T_3$, and what configuration and shape of $T_1, T_2, T_3$ achieves this maximum?

Answer 1

I find a larger area than @MattF in a non-symmetric solution. A numerical approximation to that solution is:

 [[[-0.3204647107, -0.4111035999], [0.1858745389, -0.4555874153], [0.1345901718, 3.498839041]], 
  [[2.842082885, -1.172240373], [-0.31551567, -0.3800257369], [-0.2470601577, 0.2361920968]], 
  [[-2.635080319, -1.218751491], [0.2046624437, -0.3929906911], [0.1509108184, 0.2956681691]]]

This can then be turned in an exact solution by first viewing those decimals numbers as rational numbers an then rescaling each triangle to have exactly volume 1 and translating such that they exactly have the desired center. Doing all these calculations in $\mathbb{A}$ can be done, but the solutions have large minimial polynomials, too large to fit in the margin at least...

Then one can calculate the area of the intersection of these three triangles and gets $0.248595187378...$.

As given above they approximate the required conditions to at least 10 decimal digits, so even without relying on exact calculations in $\mathbb{A}$, I'm pretty confident that this larger area is not just a numerical artifact.

Here's an image of the arrangement:

I found this solution trying a out random triangles. Of course this is not the optimal solution, I just wanted to point out that it might be a good idea to look into solution not assuming $D_3$ symmetry, but perhaps rather into solution that only have the $\mathbb{Z}/2$ symmetry of the original triangle, because the solution I found looks like it is close to a local maximum where the green and the orange triangle are mirror images of each other and the purple one is different.

Update: Instead of doing random exploration, I used some gradient descent numerical optimization to look for more local optima. As suspected the solution with $\mathbb{Z}/2$-symmetry comes up. So even without imposing symmetry the optimization leads to a symmetric solution of the form $${\\((a, b), (-a, b), (0, c)),\\\\ ((d, e), (f, g), (h, i)),\\\\ ((-d,e), (-f,g), (-h,i))\\}$$ Choosing approximately $$ (a, b, c, d, e, f, g, h, i) = \\(-0.23003 , -0.571673, 3.775494, 2.701848, -0.885711, -0.229117, -0.55433, -0.19322 , 0.123974)$$ gives an area of approximately $0.2517...$. This it what it looks like:

I also didn't prescribe that the green and yellow triangles should have sides overlapping the sides of the purple triangle, but the optimization led to that too.

Looking at the numerical evindence, I would expect that the global optimum is close to the above solution. It might be possible to get exact coordinates for the local optimum.

Answer 2

I find an area of $.245463$ for the intersection when each $T_i$ has a vertex at distance $3.15835$ from the center of $T$. This seems to be maximal among the symmetric options.

The gray triangle in the diagram is $T$, the black triangles are the $T_i$ (with their specified far vertices outside the area of the diagram), and the blue nonagon is the desired intersection.

If the distance is $(1+2b)R$, where $R$ is the circumradius of $T$, we can write the area exactly as $$4\left(2b - b^2 - \frac{3}{4b^2-1} - \frac{8b^2-4b-2}{12b^2 -1}\right)$$

which is easy to maximize numerically.

For a quicker approximate answer, we can assume that the area of intersection is a circle, tangent to the two long sides of $T_1$ at $(r,\theta)$ in polar coordinates. Then we can solve for $r,R,\theta$ satisfying \begin{align} 3R^2 \sqrt{3} / 4 &= \,1, \text{ for circumradius of }T;\\ r(\sec \theta - 2)/3 &= R, \text{ for centroid of }T_1;\\ r^2(1+\sec \theta)^2 \cot \theta &= \,1, \text{ for area of }T_1 \end{align} and the area and distance quoted above are $\pi r^2\simeq .225$ and $r \sec \theta \simeq 3.17$.

Answer 3

too long for a comment: The exact numbers in @Matt's answers are all algebraic and their minimal polynomials are:

For the area: root of $$81x^9 - 5184x^8 + 60012x^7 + 1080072x^6 - 30787658x^5 + 308408248x^4 - 1618286324x^3 + 4501458360x^2 - 5533131519x + 1109800008$$ near $$0.24546347734508587544309378249...$$

Distance from center: root of $$617673396283947x^{36} - 51122005539352848x^{32} - 976537068350570496x^{28} - 6031070320116105216x^{24} + 83285306175293227008x^{20} - 71803381414983892992x^{16} + 47645844627271974912x^{12} - 12502304826882785280x^8 + 600386125323829248x^4 - 4503599627370496$$ near $$3.1583535536268193745169822798...$$ The corresponding value of $b$: root of $$2304x^9 - 2304x^8 - 1536x^7 + 1920x^6 - 1120x^5 - 512x^4 + 128x^3 + 40x^2 + 5x + 1$$ near $$1.2998723034626336283906933307...$$ None of these numbers have a radical expression.