Harmonic Mean Sangaku
What Is It About?
A Mathematical Droodle


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Explanation

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Copyright © 1996-2017 Alexander Bogomolny

Harmonic Mean Sangaku

The applet purports to suggest that in a configuration of two vertical segments AE, BD (or two ladders AD, BE inclined at the opposite walls AE, BD), the intersection P of the diagonals is at the height that depends solely on AE and BD but not the distance between the walls. In fact more is true

  1/CP = 1/AE + 1/BD.

 

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Let's denote AC = x and BC = y. Then from similar triangles ABD and ACP we have the proportion

  (x + y)/x = BD/CP = 1 + y/x,

so that

  y/x = BD/CP - 1 = (BD - CP)/CP.

In the same way from similar triangles ABE and CBP we obtain

  x/y = AE/CP - 1 = (AE - CP)/CP.

Multiplying the two yields

  (AE - CP)(BD - CP) = CP2.

Multiplying through we get after simplification

  AE·BD = CP·(AE + BD).

A division by AE·BD·CP gives the desired result:

  1/CP = 1/AE + 1/BD,

which says that CP is half of the harmonic mean of AE and BD.

References

  1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989, p. 48

Sangaku

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Copyright © 1996-2017 Alexander Bogomolny

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