The Squinting Eyes Theorem: What Is It?
A Mathematical Droodle
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Copyright © 1996-2018 Alexander Bogomolny
The applet suggests a statement reminiscent of the Eyeball theorem. To distinguish between the two, I'll call the current one
The Squinting Eyes Theorem
Let there be two circles C(A, RA) and C(B, RB), one with center A and radius RA, the other with center B and radius RB. Let P and Q be the farthest points of the two circles, as on the diagram below. Draw the tangents from P to
Proof
Let PT be tangent to C(B, RB), so that PT is perpendicular to BT. Let
PB/BT = PR/RS,
from where
(RA + AB)/RB = (2·RA - RS)/RS.
Solving this for RS gives
RS = 2·RA·RB/(RA + RB + AB).
In the same manner we could find the radius of the second "incircle". However, there is no need to. The formula for RS is symmetric in A and B, which means that the result would not change if we swapped A and B. Because of the symmetry, we can claim that the radius of the second circle is defined by exactly same formula, i.e. the two radii are indeed equal.
This Sangaku problem has been written on a tablet in 1842 in the Aichi prefecture [Temple Geometry, p. 82].
References
H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989
Write to:
Charles Babbage Research Center
P.O. Box 272, St. Norbert Postal Station
Winnipeg, MB
Canada R3V 1L6- H. Fukagawa, A. Rothman, Sacred Geometry: Japanese Temple Geometry, Princeton University Press, 2008, p. 102
- P. Yiu, Geometric Art Design
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- Equal Incircles Theorem
- Equilateral Triangle, Straight Line and Tangent Circles
- Equilateral Triangles and Incircles in a Square
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- Gion Shrine Problem
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- Sangakus with a Mixtilinear Circle
- Sequences of Touching Circles
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- Steiner's Sangaku
- Tangent Circles and an Isosceles Triangle
- The Squinting Eyes Theorem
- Three Incircles In a Right Triangle
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- Two Arbelos, Two Chains
- Two Circles in an Angle
- Two Sangaku with Equal Incircles
- Another Sangaku in Square
- Sangaku via Peru
- FJG Capitan's Sangaku
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Copyright © 1996-2018 Alexander Bogomolny
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