Yesterday I went to a museum and waited in line for a very long time for my turn to spend 5 minutes staring at a piece of art that completely mystified me. I guess it’s good art if I’m still thinking about it a day later, even if the thoughts I’m having mostly revolve around my own resignation that I will never really understand art. But it wasn’t a complete loss! There was a circular pattern on the floor that I spent a lot of time staring at while in line, and it ended up inspiring this challenge question.
In the figure above, two congruent circles are tangent at point D. Points D, E, and F are the midpoints of AB, AC, and BC, respectively. If AB = 12, what is the area of the shaded region?
UPDATE: Congratulations to John, who got it first. Solution below the cut…
When we’re asked to solve for the areas of weirdly shaped shaded regions, we’re almost always going to find the area of a larger thing that we know how to calculate, and then subtract small things we know how to calculate until we’re left with the weird shaded bit:
The first thing we should do is mark this bad boy up. We know AB = 12, and D is the midpoint of AB and also the endpoint of two radii. We also know E and F are endpoints of two radii, and midpoints of AC and BC, respectively.
At this point, we actually know a great deal. First, we know the radius of each circle is 6. That means each circle has an area of π(6)2 = 36π. We’ll come back to this in a minute.
It should also be obvious that ABC is an equilateral triangle. This is awesome, because equilateral triangles are easily broken into 30º-60º-90º triangles, which is what we’ll do to find the triangle’s area.
So triangle ABC has a base of 12 and a height of 6√3.
Now that we have that, all we need to do is subtract the areas of the circle sectors (in green below) that aren’t included in the shaded region.
Areas of sectors are easy to calculate. All we do is figure out what fraction of the whole circle the sector covers by using the central angle. In this case, the angles are 60º, so we’re dealing with 60/360 = 1/6 of each circle.
We need to subtract two sectors from the area of triangle ABC to find our shaded region:
And there you have it! Cool, right?