# What is the total area of the two quarter circles?

Puzzle by Catriona Agg.

The yellow circle has radius 4. What’s the total area of the two quarter circles?

• @DanielMathias Doesn't it depend if it is more puzzle or more math?
– Simd
Commented Jan 8 at 13:45
• That is subjective. I do not view this as a puzzle at all, except to realize that the length of the shared chord is irrelevant. I believe this question was asked on Math.SE, but have not found it. Commented Jan 8 at 13:56
• Nice problem with many approaches to resolve, some of them totaly match Puzzle SE.
– z100
Commented Jan 8 at 22:36
• @DanielMathias Catriona Agg's puzzles really are puzzles---they are typically amenable to brute force mathematical arguments, but are designed to be very simply solved via clever thinking. These definitely feel like puzzles to me, rather than mathematics. Commented Jan 9 at 14:55
• I think "puzzles rather than mathematics" is the wrong framing. Whether something is mathematical or not doesn't determine whether it's a puzzle or not. Some puzzles are mathematical, some not. Some mathematical questions are puzzles, some not. This one (it seems to me) is a mathematical question which is also a puzzle. Commented Jan 10 at 3:46

Note that the marked triangles in the picture above are all equivalent because of symmetry. From this we can see that angle $$BAE$$ is a right angle (to see this, rotate $$\triangle AIE$$ around point $$A$$ until it fits $$\triangle AHB$$, keeping in mind that $$AHFI$$ is a square).

Thus from Pythagorean theorem,

$$BE^2=4^2+4^2=32$$.

Total area of the quarter circles is

$$\dfrac14\pi(BF^2+FE^2)=\dfrac14\pi BE^2=8\pi$$.

Similar to the Napkin ring problem, there is a trick solution:

Given that there is nothing given with regards to the relative sizes of the two quarter discs, the answer must not depend on that. Therefore we might as well choose to enlarge the inner one until the outer one has size zero. This leaves us with a quarter disc that is inscribed in the radius 4 circle. The quarter circle has radius $$4\sqrt2$$ and hence area $$8\pi$$.

In a comment Sneftel pointed out that it is even easier to

go in the other direction, enlarging the outer quarter disc until it matches the size of the inner one. Now both quarter discs will have the same radius as the circle, i.e. $$4$$, so they have total area $$\frac{\pi\cdot4^2}2 = 8\pi$$.

• I think this definitely shows it is a puzzle, not math!
– Simd
Commented Jan 8 at 14:51
• This feels a lot like the "forbidden technique" of solving a sudoku by assuming that a unique solution exists, and then making deductions based on that assumption. There, too, you'll get a result, but you have no way of knowing whether the puzzle is broken or not.
– Bass
Commented Jan 8 at 19:55
• Just one piece is missing - a proof that the area is constant (as a function of let's say variable x).
– z100
Commented Jan 8 at 22:34
• @Bass I would call it a 'meta technique'. Wheher using that should be forbidden or rather exceptionally clever is a matter of opinion. Commented Jan 9 at 7:56
• @Bass — my exact thought! If you say the tower-power of x (that is x to the power of x to the power of x, ad infinitum is 2, then you can quickly see x² = 2 so x = √2. However, if you say the tower-power of x is 4, then x⁴ = 4, so again, x = √2. Well, is the tower-power of √2 equal to two or four? The bug in the reasoning is assuming that both equations have a solution (only the first does). Commented Jan 10 at 18:37

The area to calculate is $$(x^2+y^2)\pi/4$$ which considering the right triangle $$\Delta ACE$$ equals $$z^2\pi/4$$. As the inscribed angle $$\angle AFE$$ over the arc $$z$$ is $$\pi/4$$, the corresponding central angle $$\angle AOE$$ is $$\pi/2$$ giving rise to the right triangle $$\Delta AOE$$ whose short sides are radii. Therefore $$z^2=2r^2=32$$ and the desired area is $$8\pi$$, half that of the original circle.

• Isn't this the same solution as ACB? Commented Jan 10 at 14:28
• @justhalf Boilerplate steps (Pythagoras, obvious symmetries) are the same, but the actual interesting bit (length of z) I use inscribed angle theorem where they rely on 4 auxiliary triangles. Commented Jan 10 at 19:12