I was going to make this into a problem, but I didn't know how to articulate it well enough for that purpose. Anyhow, it is kind of interesting. Looking at the diagram:

As you see, I've simply bisected the first circle, and put in another two circles half the diameter inside. Next, I use the vesica to bisect the original circle perpendicularly:

Now I draw segments from the centres of both smaller circles to the top point F:

Now, If we bisect these line segments, we get two points, H and I:

If we connect the two points, we get a line equal in distance to one-fourth of our first circle:

You will also notice that the line segments extended from bisecting the two original segments intersects at one-fourth the radius also.

We can prove this by drawing a vesica with the original circle, noting that it creates and angle of 60 degrees. Because of opposite interior angles, we can be sure that say angle ADF is also 60 degrees. Since we used perpendicular bisection, the top angle, AFD is 30 degrees. If we look at the 30, 60, 90 triangle:

The hypotenuse (for this example, that would be line segment DF) is 2x. Bisecting gives us x, which is the radius of our circle centred on point D or C. This is the measure of diameter, so half x is \(\frac{x}{2}\). Therefore, we have divided the original circle in quarters.

This is not really important, but maybe someone could make a problem out of it.

## Comments

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TopNewestYou have posted few really nice articles which i liked – Aman Dubey · 1 year, 3 months ago

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@Aman Dubey Thanks! It does mean a lot to hear that, my friend. :) – Drex Beckman · 1 year, 3 months ago

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I don't see how you get 30-60-90 since, for instance, angle ADF is arctan(2). – Mark C · 1 year, 1 month ago

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@Mark C

Hmm. You are right. That is obviously not a 30 60 90, now that you mention it. I am not sure what I was thinking, to be honest. – Drex Beckman · 1 year, 1 month ago

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I could have made an awesome question, but a s you have a ready explained so I won't – Rishabh Sood · 1 year, 4 months ago

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@Rishabh Sood Oh no, please feel free to make a question using this if you want to. I just wasn't able to articulate it well enough to make one. So if you want to, go ahead man! :) – Drex Beckman · 1 year, 4 months ago

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– Rishabh Sood · 1 year, 4 months ago

That's ok, but awesome, very very awesome articleLog in to reply

– Drex Beckman · 1 year, 4 months ago

Thanks, man. Means a lot.Log in to reply

– Rishabh Sood · 1 year, 4 months ago

Hey in a mood of a question? Check out "this is not mathematically possible..." I hope you like itLog in to reply

– Drex Beckman · 1 year, 4 months ago

Will do. :)Log in to reply

– Rishabh Sood · 1 year, 4 months ago

Isn't it awesomeLog in to reply

– Drex Beckman · 1 year, 4 months ago

Yeah, it is a cool problem. Interesting how there seems to be a more common solution, like you said. I wonder why that is.Log in to reply

– Rishabh Sood · 1 year, 4 months ago

It is because, the solution goes in a perfect numerical pattern if you notice it, and 90% people in this world will be answering on that pattern. Hence the solution is commonLog in to reply

– Drex Beckman · 1 year, 4 months ago

If you want to try a new one I just made (maybe you can check if it makes sense) https://brilliant.org/problems/is-there-even-any-area/?group=dPEuoP3ND3rj&ref_id=1124196Log in to reply

– Rishabh Sood · 1 year, 4 months ago

Terrific, by the way I lost on that one...Log in to reply

– Drex Beckman · 1 year, 4 months ago

Was it clear enough, in your opinion? Thanks for trying it, man.Log in to reply

– Rishabh Sood · 1 year, 4 months ago

Yes it was clear enough, nice questionLog in to reply

– Drex Beckman · 1 year, 4 months ago

Right, I see. I guess that makes sense, since that's the same way I solved it. Cool!Log in to reply