# Harmonic Mean SangakuWhat Is It About? A Mathematical Droodle

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Explanation

### 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