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Particular Pentagon Point shared by Matt Enlow

I think this problem has more than one solution.Please refer to attached file.

Note by Pratyay Bhattacharya
3 years, 2 months ago

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I'll call the middle point M. Because of the Law of Sines, we must have \(\frac{\sin \angle EAM}{EM}=\frac{\sin \angle EMA}{EA}\), and also \(\frac{\sin \angle EDM}{EM}=\frac{\sin \angle EMD}{ED}\).

Rearranging and combining these, and using the fact that \(EA=ED\), we must have \(\frac{\sin \angle EMA}{\sin \angle EAM}=\frac{\sin \angle EMD}{\sin \angle EDM}\). This equation holds in the diagram on the left, but not on the right. Matt Enlow · 3 years, 2 months ago

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The point defined in the problem is uniquely determined. You cannot claim that \( \angle EDM \) is any arbitrary angle.

Note: I've cropped your image, so that the pentagons which you drew are now much larger. Calvin Lin Staff · 3 years, 2 months ago

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