Look at the images and pick the first one, then prove that the blue area = the white area

here's my failed attempt:

let radius of large circle = 1, so radius of small circle = 0.5 and length of triangle = sqrt(3) and height of triangle = 1.5
area of big circle = PI, area of small circle = 0.25 PI
area of sector = PI/3
area of triangle = sqrt(3) * 1.5 /2 = 0.75 sqrt(3)
area of triangle - circle = 0.75 sqrt(3) - 0.25 PI
blue area = PI / 3 + 0.25 PI + 0.25 sqrt(3) - 0.25/3 PI = PI / 2 + 0.25 sqrt(3)

this is greater than half PI aka greater than half the area of circle, so my proof is wrong

  • 1
    $\begingroup$ (1) You can simplify by using choosing your units so that $r=1$. (2) Why are you using a sector spanning 60 degrees? Were you intending to use a segment spanning 120 degrees? $\endgroup$ – Lawrence Mar 11 '17 at 13:42
  • $\begingroup$ What was the point of including the other 3 images? And where did you get the question from? $\endgroup$ – boboquack Mar 12 '17 at 0:00

This is a visualization of the solution by JonMark Perry.


So why is the area of the big circle 4 times the area of the small circle?

Wikipedia says:

The centroid divides each median into parts in the ratio 2:1

Since it's an equilateral triangle, median and angle bisectors are the same, centroid and center of incircle are the same. So from there follows:


Meaning that the radius of the big circle is 2 times the radius of the small circle.

Inserted into the area formula: $$ 4 (\pi \cdot 1^2) = \pi \cdot 2^2 $$

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Rotate the image twice by $120^\circ$. Superimpose the images and add the sections of like colour. We have a solid blue circle with two inner blue circles and two outer white rings. Cancelling an outer ring yields three inner blue circles and one outer white ring. As the area ratio of inner circle to outer circle is 1:4, the blue area equals the area of white.


Not true. Cancel the outer blue segments with outer white ones, and there's an obvious mismatch.


The radius of the big semi-circle is $\sqrt{1-d^2}$ (by Pythagarus) and the radius of the small semi- circle is $d$, where $d$ is the distance from the diameter of the main circle to the diameter for the big semi-circle. Adding the areas of the two semi-circles gives half of the area of the whole circle.


Obvious, by either reflection or rotation.

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  • $\begingroup$ Could you provide an image for A? Finding it hard to visualise. Also, can I check this is good by you? $\endgroup$ – Beastly Gerbil Mar 11 '17 at 15:00
  • $\begingroup$ @BeastlyGerbil; i saw it last night - the colours are random, the grid is easy to draw, it's a pic of Hank III, and I don't have an image editor that can do 120deg rotations. $\endgroup$ – JMP Mar 11 '17 at 15:04
  • $\begingroup$ look what happens to the blue segment and little triangle under the rotations, they fill all the spaces $\endgroup$ – JMP Mar 11 '17 at 15:05

Another visually-oriented approach:

This is just moving congruent pieces around and recoloring:

Doubling the radius quadruples the area, but dividing the angle by four reduces the area to a fourth.

And this is just substitution:

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I'll use the diagram below while I'm solving this:
enter image description here
Let the radius of the larger circle equal $r$.

First, the area of the red section ($A_\mathrm{red}$). This can be found by using the formula for the area of a segment of a circle, $$A=\frac{r^2}{2}(\theta-\sin\theta),$$ where $\theta$ is in radians. In this case, $\theta=120^\circ = \frac{2\pi}{3} \mathrm{radians}$. So, plugging in, the area of the red sector is
$$ \begin{align*} A_\mathrm{red}&=\frac{r^2}{2} \left( \frac{2\pi}{3}-\sin\frac{2\pi}{3} \right) \\ &= \frac{r^2}{2} \left( \frac{2\pi}{3}-\frac{\sqrt3}{2} \right) \\ A_\mathrm{red}&=\left(\frac{\pi}{3}-\frac{\sqrt3}{4}\right)r^2. \end{align*} $$

Second, the area of the purple circle ($A_\mathrm{purple}$). This is rather simple; if the radius of the larger circle $OA = r$, then the radius of the purple circle $OD = \frac{r}{2}$, since $\triangle ODA$ is a 30-60-90 right triangle. Thus the area of the purple circle is $A_\mathrm{purple}=\pi \left(\frac{r}{2}\right)^2 = \frac14\pi r^2$.

Lastly, the area of the green section ($A_\mathrm{green}$). We know that $AD = \frac{\sqrt{3}}{2}r$ and $CD = \frac32 r$, thus the area of $\triangle ABC$ is $\frac{\sqrt{3}}{2}r \cdot \frac32 r = \frac{3\sqrt3}{4}r^2$. The area of the triangle minus the area of the purple circle will give us three times the area of the green segment, which we can then use to find $A_\mathrm{green}$: $$ \begin{align*} 3A_\mathrm{green} &= A_{ABC} - A_\mathrm{purple} \\ 3A_\mathrm{green} &= \frac{3\sqrt3}{4}r^2 - \frac14\pi r^2 \\ 3A_\mathrm{green} &= \frac{3\sqrt3-\pi}{4}r^2 \\ A_\mathrm{green} &= \frac{3\sqrt3-\pi}{12}r^2. \\ \end{align*} $$

We can now sum $A_\mathrm{red}$, $A_\mathrm{purple}$, and $A_\mathrm{green}$ to find the total area of the shaded region:

$$ \begin{align*} A_\mathrm{total} &= A_\mathrm{red}+A_\mathrm{purple}+A_\mathrm{green} \\ &= \left(\frac{\pi}{3}-\frac{\sqrt3}{4}\right)r^2 + \frac14\pi r^2 + \frac{3\sqrt3-\pi}{12}r^2 \\ &= \left( \frac\pi3-\frac{\sqrt{3}}{4} + \frac\pi4 + \frac{\sqrt{3}}{4} - \frac{\pi}{12} \right) r^2 \\ &= \left( \frac\pi3 + \frac\pi4 - \frac{\pi}{12} \right) r^2 \\ A_\mathrm{total} &= \frac\pi2 r^2. \\ \end{align*} $$

Therefore the total area of the shaded region is half of the area of the larger circle, and thus the area of the shaded region equals that of the unshaded region.

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Here's another mostly-visual approach. I'm not sure whether it's any better than JMP's or Zachiel's, but I like it :-).

We want to show that the red and blue areas of the LH diagram here are equal:

enter image description here

We'll transform it step by step, always adding/removing the same amount of red and blue. First, remove one red and one blue outer piece to get the middle diagram; we still want to show that the red and blue areas are equal.

Now observe that the small circle is circumscribed around the small triangle, the large circle is circumscribed around the large triangle, and the large triangle clearly has 4x the area of the small triangle. So the large circle is also 4x the area of the small one ... and the remaining blue outer piece has 4x the area of one of its smaller red counterparts. So replace the big blue segment with four small blue segments, and cancel three of them against the three small red segments we have. That leaves one blue segment, as in the right-hand diagram.

And of course in the RH diagram the areas are obviously equal because we've got one triangle of each colour and one, er, let's call it a Starfleet insignia of each colour. Done.

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An hexagon inscribed in a circle can be divided in six equilateral triangles. Inscribe a triangle inside the hexagon and its sides will intersect the side of the small six triangles exactly in half (it's the height of two adjacent equilateral triangles). Thus, the radius of the circle inscribed in this equilateral triangle is 1/2 of the outer circle, therefore its area is 1/4 of the area of the outer circle.
The outer ring has an area equal to 1-(1/4)=3/4
Let the outer area be defined as 3A (the three sectors) + 3B (the three triangle spikes) = 3 (A+B)
Since this area is 3/4 of the whole circle, A+B=1/4

There is a blue A, a blue B and a blue inner circle = 1/4 + 1/4 = 1/2 QED

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  • $\begingroup$ 'The inner circle has half the radius of the outer one' - why? I know why, but can you please explain this logic? $\endgroup$ – boboquack Mar 12 '17 at 0:00

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