Twin Segments in Arbelos
Here is a problem (10895) from the 2003 American Mathematical Monthly. It was proposed by August Wendijk, Jersey City, NJ. (The solution below is by Thomas Hermann, Milford, OH):
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Given a point C on the segment AB, erect semicircles on diameters AB, AC, and BC, all on the same side of AB. Let L be the line through C perpendicular to AB. Let S be the largest circle that fits into the region bounded by L and the semicircles on diameters AC and BC. Let D be the point of tangency between S and the semicircle on BC. Extend the
diameter of S through D until it hits L at E. Prove that AC and DE have the same length.
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Solution
References
- August Wendijk; Thomas Hermann, American Mathematical Monthly, Vol. 110, No. 1. (Jan., 2003), pp. 63-64.
Copyright © 1996-2009 Alexander Bogomolny
Let's introduce the following notations
- a the radius of the circle on AC as a diameter,
- G the center of the circle with diameter BC; b the radius of the circle,
- H the center of the circle with diameter AB; R the radius of the circle,
- F the center of the Archimedian twin in the problem (the largest circle that fits into the region bounded by L and the semicircles on diameters AC and BC); r - the radius of the the circle,
- M the projection of F on AB; y = FM,
- T the point of tangency of circles H(R) and F(r),
- w = DE.
We have to show that w = 2a.
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Observe that by the construction,
- CM = r,
- HM = |b - a - r|,
- GF = b + r,
- FH = a + b - r.
Triangles FGM and FHM are right-angled. The Pythagorean theorem implies:
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(b + r)² = y² + (b - r)² and
(a + b - r)² = y² + (b - a - r)²,
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from which
Triangles FGM and CEG are similar so that GF / GM = EG / CG, or
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(b + r) / (b - r) = (b + w) / b.
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From the latter
Comparing the to expressions for r yields w = 2a, as needed.
Copyright © 1996-2009 Alexander Bogomolny
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