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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, 10 months ago

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## Comments

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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.

- 3 years, 10 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.

Staff - 3 years, 10 months ago

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