# Which area is larger, the blue area, or the white area?

In the square below, two semicircles are overlapping in a symmetrical pattern. Which is greater: the area shaded blue or the area shaded white?

# My Solution

Let the length of each side of the square be $$2r$$.
The area of the square is $$4r^2$$.

The two semi-circles have equal area.
Area of one semi-circle = $$\frac{{\pi}r^2}{2}$$.
$${\times}2 = {\pi}r^2$$

White area = $${\pi}r^2 -$$ area of the intersection of the two circles.

Let the area of the intersection of the two circles be $$t$$.
White area = $${\pi}r^2 - t$$.

The segments that make up $$t$$ are identical.
$$t$$ = area of segment $${\times}2$$.

Area of segment = Area of sector - Area of triangle.
Angle of sector = $$90^{\circ}$$ (The circles both have radius $$r$$, and the outer shape is a square.
Angle of sector $$= \frac{1}{4} * {\pi}r^2$$.
Area of triangle $$= \frac{1}{2} * r^2$$.
Area of segment $$= \frac{{\pi}r^2 - 2r^2}{4}$$. $$t = 2 {\times} \frac{{\pi}r^2 - 2r^2}{4}$$.
$$t = \frac{{\pi}r^2 - 2r^2}{2}$$.

White area $$= {\pi}r^2 - \frac{{\pi}r^2 - 2r^2}{2}$$.
White area $$= \frac{2{\pi}r^2 - {\pi}r^2 + 2r^2}{2}$$.
White area $$= \frac{{\pi}r^2 + 2r^2}{2}$$.

Blue area = $$r^2\left(4 - \frac{{\pi} + 2}{2}\right)$$.
Blue area = $$r^2\left(\frac{8 - ({\pi} + 2)}{2}\right)$$.
Blue area = $$r^2\left(\frac{6 - {\pi}}{2}\right)$$.

If White area $$-$$ Blue area $$\gt 0$$, then the White area is larger.

$$r^2\left(\frac{{\pi}+2 - (6 - {\pi}}{2}\right)$$
$$r^2\left(\frac{2{\pi} - 4}{2}\right)$$
$$r^2(\pi - 2)$$

$$\therefore$$ the white area is larger.

## What is the error in my solution?

The provided solution:

• It's a general rule of thumb, that in all puzzles which ask "which area is larger" the answer always is that both areas have the same size, even when one is visibly enormous and the other visibly tiny. – Angina Seng Jul 23 '17 at 21:11
• White area $= \pi r^2 - \color{red}{2} \times$area of the intersection of the two circles. – Donald Splutterwit Jul 23 '17 at 21:16
• Move the small blue caps to the other side of the semicircles, then the blue region makes a triangle. – Michael Burr Jul 23 '17 at 21:16
• @DonaldSplutterwit thanks for the pointer; do you want to make it into an answer, or should I do it? – Tobi Alafin Jul 23 '17 at 21:20
• What's the question? – Shuri2060 Jul 23 '17 at 21:21

If you just divide the lower left part of the blue area and move each part $90$ degree up to join them to the main blue part, then the area of the new blue shape will be equal to the white part.

• This is the solution given as "provided solution" by the OP. The OP has already seen this (neat) solution. How does this add to what is already present in the question? – James K Jul 23 '17 at 22:14
• @Seyed In the bottom of the question, it says "The provided solution". – HelloGoodbye Jul 23 '17 at 22:42
• As far as I can see, the question wasn't edited at any point – Shuri2060 Jul 23 '17 at 22:58
• @user1936752: Technically speaking, Seyed did answer the question. There is only one question mark in the post, which is for the title "which area is larger, blue or white?". It's likely that what the OP wanted is an explanation for what is the mistake in his own attempted solution, but he never actually asked that... – Meni Rosenfeld Jul 24 '17 at 9:09
• Making the neat solution available in the thread rather than just as a link to a dodgy-sounding url does add value IMO. – Especially Lime Jul 24 '17 at 10:44

There was a mistake in my earlier solution. I correct that mistake here.

Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.

The two semi-circles have equal area.
Area of one semi-circle = $\frac{{\pi}r^2}{2}$.
${\times}2 = {\pi}r^2$

White area = ${\pi}r^2 - 2 {\times}$area of the intersection of the two circles.

Let the area of the intersection of the two circles be $t$.
White area = ${\pi}r^2 - 2t$.

The segments that make up $t$ are identical.
$t$ = area of segment ${\times}2$.

Area of segment = Area of sector - Area of triangle.
Angle of sector = $90^{\circ}$ (The circles both have radius $r$, and the outer shape is a square.
Angle of sector $= \frac{1}{4} * {\pi}r^2$.
Area of triangle $= \frac{1}{2} * r^2$.
Area of segment $= \frac{{\pi}r^2 - 2r^2}{4}$. $t = 2 {\times} \frac{{\pi}r^2 - 2r^2}{4}$.
$t = \frac{{\pi}r^2 - 2r^2}{2}$.

White area $= {\pi}r^2 - 2\left(\frac{{\pi}r^2 - 2r^2}{2}\right)$.
White area $= \frac{2{\pi}r^2 - 2{\pi}r^2 + 4r^2}{2}$.
White area $= \frac{4r^2}{2}$.
White area $= 2r^2$.

Blue area = $r^2\left(4 - 2\right)$.
Blue area = $2r^2$.

The blue and white areas both have areas of $2r^2$, therefore the triangles have equal areas.

• This solution correctly states that the white area is ${\pi}r^2 - 2 \times\text{area of the intersection of the two circles}.$ There is still a question of why you previously thought that the area was ${\pi}r^2 - \text{area of the intersection of the two circles}$ and what you can do to prevent such mistakes in future exercises. – David K Jul 24 '17 at 12:11
• I'm not sure exactly why I made the mistake, but I guess I'll double check my assumptions in future questions. – Tobi Alafin Jul 24 '17 at 12:17
• Don't worry about getting assumptions wrong for us--this is a fine place to bring incorrect work to have it corrected. I suspect in this case you just did that step a little too quickly. The way I think of it, the area of the overlapping semicircles is $\pi r^2$ minus the overlap, and we subtract the overlap a second time by coloring it blue. But subtracting the same thing twice is error-prone: subtracting it once can satisfy either of the conditions we need, and we might overlook the fact that this doesn't satisfy both conditions at once. – David K Jul 24 '17 at 12:43
• Anyway, whatever the reason for your original confusion, I think this is a good question. It showed thorough and careful effort, and asked for help to understand why the effort reached a wrong conclusion. If only more questions here were like that! – David K Jul 24 '17 at 12:44

The area of the segment is “quarter of circle minus triangle”: $$\frac{1}{4}\pi r^2-\frac{1}{2}r^2=\frac{r^2}{4}(\pi-2)$$ Thus half of the white area is “quarter of circle plus triangle minus segment”: $$\frac{1}{4}\pi r^2+\frac{1}{2}r^2-\frac{r^2}{4}(\pi-2)=r^2$$ Therefore the white area is $2r^2$.