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):

  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

  1. August Wendijk; Thomas Hermann, American Mathematical Monthly, Vol. 110, No. 1. (Jan., 2003), pp. 63-64.

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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:

  (b + r)² = y² + (b - r)² and
(a + b - r)² = y² + (b - a - r)²,

from which

  r = ab / (a + b).

Triangles FGM and CEG are similar so that GF / GM = EG / CG, or

  (b + r) / (b - r) = (b + w) / b.

From the latter

  r = bw / (2b + w).

Comparing the to expressions for r yields w = 2a, as needed.

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