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):

Solution

References

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

• 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.

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:

from which

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

From the latter

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

• Arbelos - the Shoemaker's Knife
• 7 = 2 + 5 Sangaku
• Another Pair of Twins in Arbelos
• Archimedes' Quadruplets
• Archimedes' Twin Circles and a Brother
• Book of Lemmas: Proposition 5
• Book of Lemmas: Proposition 6
• Chain of Inscribed Circles
• Concurrency in Arbelos
• Concyclic Points in Arbelos
• Ellipse in Arbelos
• Gothic Arc
• Pappus Sangaku
• Rectangle in Arbelos
• Squares in Arbelos
• The Area of Arbelos
• Twin Segments in Arbelos
• Two Arbelos, Two Chains