Harmonic Mean Sangaku
What Is It About?
A Mathematical Droodle
What if applet does not run? |
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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. |
What if applet does not run? |
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) = CP^{2}. |
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
- H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989, p. 48
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|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2017 Alexander Bogomolny
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