Particular Pentagon Point shared by Matt Enlow

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

#Geometry

Note by Pratyay Bhattacharya
7 years, 2 months ago

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Comments

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

Rearranging and combining these, and using the fact that EA=EDEA=ED, we must have sin⁡∠EMAsin⁡∠EAM=sin⁡∠EMDsin⁡∠EDM\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 - 7 years, 2 months ago

The point defined in the problem is uniquely determined. You cannot claim that ∠EDM \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 - 7 years, 2 months ago
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