Which colored area is large?

The red and blue areas are equal in the circle because each grid section with a red shape has a matching congruent grid section with a blue shape. The picture below shows how each square with a red curved shape matches each square with a congruent blue curved shape (A with A, B with B, and so on).

In addition, there are 13 red squares that correspond with 13 blue squares, and 6 red triangles that correspond with 6 blue triangles, which account for all the red and blue in the diagram.

Check out the Pizza Theorem :)

The polar equation of the circle of radius $1$ , measured from a point a distance $x$ from the centre is (use the Cosine Rule) $r \; = \; \sqrt{1 - x^2\sin^2\theta} - x\cos\theta$

and so the area of a slice of the circle for $\alpha \le \theta \le \beta$ is $\begin{aligned} A(\alpha,\beta) & = \; \tfrac12\int_\alpha^\beta r^2\,d\theta \; = \; \tfrac12\int_\alpha^\beta\big[1+ x^2\cos2\theta - 2x\cos\theta \sqrt{1 - x^2\sin^2\theta}\big]\,d\theta \\ & = \; \tfrac12(\beta-\alpha) + \tfrac14x^2(\sin2\beta-\sin2\alpha) - \int_{x\sin\alpha}^{x\sin\beta}\sqrt{1-u^2}\,du \end{aligned}$ Rather than evaluating this explicitly, we use periodicity properties of the $\sin$ function to see that $B(\alpha,\beta) \;= \; A(\alpha,\beta) + A(\alpha+\pi,\beta+\pi) \; = \; (\beta - \alpha) + \tfrac12x^2(\sin2\beta - \sin2\alpha)$ and hence $C(\alpha,\beta) \; = \; B(\alpha,\beta) + B(\alpha+\tfrac12\pi,\beta+\tfrac12\pi) \; = \; 2(\beta - \alpha)$ Thus the area of the blue region is $C(\alpha,\alpha + \tfrac14\pi) \; = \; \tfrac12\pi$ which is half the area of the circle for any value of $\alpha$ . The red and blue areas are equal.

This is not a solution but a comment - so I don't know if it will be posted. I want to say that I am SO jealous of all of you that post solutions to the problems; you ROCK. I am 66 years old and have struggled with mathematics all of my life; I so envy people like you that are good at it. I love to read your solutions to these problems - and I get to them, by my own admission, by taking wild guesses at the answers - most of which are wrong guesses (but with yes/no choices I have a 50/50 chance of being right). Kudos to all you math whizzes ! :-) $\frac{numerator}{denominator}$ (I had to add that LaTex fraction so that it would qualify as a solution and go through).

Building on David Vreken's solution, I've spent a little time in MSPaint to see if I can break it down in a way that's easier to follow.

I've taken all the blue and red pieces, separated them out, and grouped them into a couple of separate images. First I've pulled out the full squares, and the 45 degree triangles. You can verify that the number of blue and red squares are equal, and the number of blue and red triangles are equal, just by counting.

Next, I've separated out the irregularly shaped pieces into three groups.You should be able to tell the area of each blue piece is identical to the area of a similar red piece. You can use either mirror symmetry or rotational symmetry (depending on the piece) to confirm that the areas really are identical.

A nice visual solution is to first assume we're actually looking at a larger object (illustrated below) through a smaller viewer. Now, if we center this viewer on the center of our object then the answer becomes clear. Any movement of our viewer from this center point always conserves red and blue since (for example) if we move towards red, we lose red on the other side. Any circularly symmetric object will cut out the same of each color within the viewer.

I reasoned that the white corners were of equal area on all four corners therefore, the area of the colored areas must still be the same. It took me only a few seconds to come to the conclusion. I guessed and got it right. It was an intuitive response.

in the first picture,the red and the blue is equal.at the same time,the background picture is square, so when the two colour is filled in, the space should be divided half-and-half. now in the second picture, when the circle would be delimit in the square, the remain part would also be equal.

Hi! I'm not here to add a solution, I was just wondering if anyone here who knows about the divergence theorem could explain how this is true in terms of that? For example in Physics we often use Gauss' Law which says: $\oint_S E \cdot ds = Q_{enc} / \epsilon_0$

And is often illustrated like this:

And in fact it didn't need to be the case that the charge be at the very center, it could have been anywhere in the circle. I'm just wondering if the solution to this problem and the law stated above are related somehow? And if not, does anyone know why they aren't related?

Pizza theorem: https://en.wikipedia.org/wiki/Pizza_theorem

they are still equal. This is because in the top pic. blue and red have equal areas. so it will be the same as splitting them in half; red and blue. then making it a circle will mean subtracting the same from both sides. then they would be equal.

First example, Fold the square by the vertical line (5:3) then we get perfect match for overlapping areas, then for the remaining not covered area move upper blue aria (two small square) to the bottom as a result we get 8 red = 8 blue in total (with triangles). Response = Yes Second example. Logical solution >> since the answer to the first question is Yes then by induction it is yes for the second part because there is no options apart from YES or NO.

The way I like to think of it is that it "treats" every area "fairly", which means the larger area a piece has, the more area it loses due to shaping the graph into circle, regardless of the color, and vice versa.

Each line that divides the circle are at 45 degrees, thus the circle will be divided in 8 parts where blues and orange are interleaved the area for each color will be the same