Δευτέρα, 23 Οκτωβρίου 2017

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 C(B, RB) and from Q to C(A, RA). Whenever the construction is possible, it leads to two "isosceles triangles" with one side a circular arc. The fact is that the "incircles" of the two triangles are always equal, i.e., have the same radius.
Proof
Let PT be tangent to C(B, RB), so that PT is perpendicular to BT. Let C(R, RS) be one of the two circles in question, and assume RS is also perpendicular to PT.
From the similarity of triangles PTB and PSR,
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].

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