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Too Cumbersome for a Problem

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.

Note by Drex Beckman
10 months ago

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You have posted few really nice articles which i liked Aman Dubey · 8 months, 3 weeks ago

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@Aman Dubey @Aman Dubey Thanks! It does mean a lot to hear that, my friend. :) Drex Beckman · 8 months, 3 weeks 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 · 7 months ago

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@Mark C @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 · 7 months 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 · 9 months, 3 weeks ago

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@Rishabh Sood @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 · 9 months, 3 weeks ago

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@Drex Beckman That's ok, but awesome, very very awesome article Rishabh Sood · 9 months, 3 weeks ago

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@Rishabh Sood Thanks, man. Means a lot. Drex Beckman · 9 months, 3 weeks ago

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@Drex Beckman Hey in a mood of a question? Check out "this is not mathematically possible..." I hope you like it Rishabh Sood · 9 months, 3 weeks ago

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@Rishabh Sood Will do. :) Drex Beckman · 9 months, 3 weeks ago

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@Drex Beckman Isn't it awesome Rishabh Sood · 9 months, 3 weeks ago

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@Rishabh Sood Yeah, it is a cool problem. Interesting how there seems to be a more common solution, like you said. I wonder why that is. Drex Beckman · 9 months, 3 weeks ago

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@Drex Beckman 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 common Rishabh Sood · 9 months, 3 weeks ago

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@Rishabh Sood 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=1124196 Drex Beckman · 9 months, 3 weeks ago

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@Drex Beckman Terrific, by the way I lost on that one... Rishabh Sood · 9 months, 3 weeks ago

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@Rishabh Sood Was it clear enough, in your opinion? Thanks for trying it, man. Drex Beckman · 9 months, 3 weeks ago

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@Drex Beckman Yes it was clear enough, nice question Rishabh Sood · 9 months, 3 weeks ago

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@Rishabh Sood Right, I see. I guess that makes sense, since that's the same way I solved it. Cool! Drex Beckman · 9 months, 3 weeks ago

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