Rectangle in Arbelos: What is this about?
A Mathematical Droodle

Explanation

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Let point C lie on a segment AB. Draw (similarly oriented) semicircles with diameters AB, AC, and CB. The three semicircles enclose a shape, known as Arbelos - the shoemaker's knife - that has a long list of wonderful and unexpected properties. The applet attempts to suggest two of those.

Let D lie on the outer semicircle such that CD⊥AB. Let E be the intersection of AD with the left of the smaller semicircles, F the intersection of DB with the right one. Then

1. CEDF is a rectangle,
2. EF is the common tangent to the two small semicircles.

Proof

Angles AEC, ADB, CFB are inscribed in respective semicircles and subtend their diameters. The three angles are therefore right. This implies property #1.

Let O_a be the center of the left semicircle. Triangle AO_aE is isosceles so that

Angle ADC is complementary to both ∠DCE (in the right ΔCDE and ∠CAD (in the right ΔACD). Thus we also obtain

Also, since CEDF is a rectangle,

from where

The last identity implies that angles AEC (which is right) and O_aEF are equal. The latter is therefore right.

Similarly, we can show that angle O_bFE is also right. Thus EF is indeed the common tangent to the two semicircles.

Remark

As was pointed out by Nathan Bowler, there is a simpler proof of the result. The part of the picture that consists of the circle AEC and the rectangle is symmetric with respect to the perpendicular bisector of the chord EC. In particular, the diagonal EF is the symmetric image of the diagonal CD. Since the latter is tangent to the circle at C, the former is tangent to the circle at E.

Another possibility is to consider the center O of the rectangle. Then OE = OC. And, since the tangents to a circle from a point are equal, and two different circles may intersect in at most 2 points, the fact that OC is a tangent implies that OE is also a tangent.

References

1. S. L. Greitzer, ARBELOS: Special Geometry Issue, V. 6, 1988, MAA

• 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
• A Newly Born Pair of Siblings to Archimedes' Twins

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|