perplexus.info :: Just Math : Inequality in lengths of pentagon

Inequality in lengths of pentagon (Posted on 2019-04-26)

Let p denote the perimeter of a convex pentagon, and let d denote the sum of diagonals of this same pentagon.

Show that p < d < 2p

No Solution Yet Submitted by Danish Ahmed Khan

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part 2 of 2

Comment 2 of 2 |

Cute!

Nodes of pentagon at A,B,C,D,E; in sequential order

V marks the intersection of AC and BD

W marks the intersection of BD and CE

X marks the intersection of CE and DA

Y marks the intersection of DA and EC

Z marks the intersection of EB and DA

Then

AB < AV + BV

BC < BW + CW

CD < CX + DX

DE < DY + EY

EA < EZ + AZ

Summing:

p < (AV + CW) + (BW + DX) + (CX + EY) + (DY + AZ) + (EZ + BV)

< (AV + VW + CW) + (BW + WX + DX) + (CX + XY + EY)

+(DY + YZ + AZ) + (EZ + ZV + BV) = d

Posted by FrankM on 2019-04-28 21:48:26

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