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Topic: japanese temple geometry 3 (Read 1523 times) |
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klbarrus
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Posts: 29
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japanese temple geometry 3
« on: Jul 27th, 2002, 11:13pm » |
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1) r1, radius of the large inscribed circle
drop a line segment from center of c1, use right triangles:
(a-r1)^2 = r1^2 + (a/2)^2
which solves to r1 = 3a/8
2) r2, radius of the small inscribed circle
drop a line segment from center of c2, use right triangles
(a-r2)^2 = r2^2 + x^2 (x = length of dropped segment)
or x^2 = a^2 - 2ar2
call y = length of bottom segment (a = y + r2)
(a + r2)^2 = x^2 + y^2
sub in:
a^2 + 2ar2 + r2^2 = a^2 - 2ar2 + a^2 - 2ar2 + r2^2
a^2 = 6ar2
or r2 = a/6
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