A Geometer's Fallacy

Let the equilateral triangles have side length 1.

In triangle $A$ , the circle radius can be found by finding the length of a side of the triangle in terms of r. $r=\frac{1}{2+2\sqrt{3}}$ . And so the area $3 \pi r^{2} = \frac{6-3\sqrt{3}}{8}\pi \approx 0.10048 \pi$

In triangle $B$ the center of the large circle is the centroid so $r_{1}=\frac{\sqrt{3}}{6}$ and the smaller circles are inscribed in a triangle one-third as large so $r_{2}=\frac{\sqrt{3}}{18}$ . And so the area $(r_{1}^{2}+2r_{2}^{2})\pi = \frac{11}{108}\pi \approx 0.10185 \pi$

In triangle $C$ the circle radius can be found in a manner similar to the first. $r=\frac{\sqrt{3}}{12}$ . [In retrospect this is easier to see as half of $r_{1}$ from triangle $B$ ] And so the area $4\pi r^{2} = \frac{1}{12}\pi \approx 0.0833333 \pi$

So the greatest shaded area is in figure $\boxed{B}$ , but just barely.

Note: $A$ and $B$ are so close that setting an approximate equality gives the close approximation $\sqrt{3} \approx \frac{140}{81}$

The question is equivalent to finding out which packing of circles fill the largest proportion of the triangle.

Packing C can be broken up into 4 equilateral triangles with an inscribed circle in each.

Packing B is one inscribed circle plus some extra little circles. These little extra circles make packing B is better than C.

Packing A can be broken up into nine equilateral triangles. Some of these are completely red. For each fully coloured in triangle there are 3 extra circle segments.

Packing B can be broken up into 4 equilateral triangles. Some of these are completely red. For each fully coloured in triangle there are 3 extra circle segments. Again, packing B has extra little circles making it better than packing A.

Okay, this is not a super precise proof but good enough.

Edit: the description of Packing A is not right. From the Jeremy's answer, it is clear a more precise comparison is needed before the answer is obvious.

Ordered by most to the least red area: B, A, C. The B and A is the tricky part, because they are seemingly the same in the red area, where B seems to mislead you that it has a little bit less area. The way you differentiate between is by comparing corners of blue area fused with the tangent blue area if needed to come close to equal, we find out that if we fuse A's upper corner and the two tangent blue sides we will get area approximately equal to that of B's upper corner without the tangents. if we compare the lower left and right corners of A with the lower fused corners with tangent sides of B we will see that there will be an extra centrepiece of blue area and a little bit of leftover of the blue area of A, that's how only by reasoning and sight we can conclude that the correct answer is B, since it has the least amount of blue area, that it has the highest amount of red area.

The way I solved it was finding which triangle has the least blue. "c" has lots of blue so it's red must have the least red. "a" has three blue triangles at the top that looks to be equal to "b"s first triangle, ”b"s six other triangles looks to be equal to "a"s bottom left and right triangles, so now "a" still has a triangle in the middle and bottom, so "b"has a bigger red area.