A Sangaku with an Egyptian Attachment:
What Is This About?
A Mathematical Droodle

 

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Solution

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Copyright © 1996-2018 Alexander Bogomolny

The applet purports to suggest the following sangaku [Temple Geometry, p. 3]:

  Assume points A and B are fixed on a line AB and two circles are drawn touching AB at A and B and tangent to each other. A circle Q is tangent to AB and the two circles externally. Prove that, as the two circles change, circle Q remains tangent to a fixed circle through A and B. Moreover, the radius of the latter is 5/8·AB.

 

(This is a Sangaku from the Miyagi prefecture whose tablet has disappeared long ago.)

We already met this configuration and now how to compute the radius and location of circle Q. If the radii of the circles on A and B are R1 and R2, the radius of circle Q r and the horizontal distances between the left circle and Q and Q and the right circle are x and y then

(1) x² = 4R1r,
y² = 4R2r,
(x + y)² = 4R1R2.

Let for some configuration of the circles, a circle with center O, radius R and passing through A and B also touches circle, Q as shown. This gives us two right triangles with sides

  h = OM, (x + y)/2, R and
r + h, (x - y)/2, R - r.

The Pythagorean theorem implies two identities:

  h² + (x + y)²/4 = R² and
(r + h)² + (x - y)²/4 = (R - r)².

Substituting h² from the first into the second gives

(2) 2Rr = xy - 2rh.

However, from (1), xy = 2r(x + y), which transforms (2) into

(2) 2Rr = 2r(x + y) - 2rh,

or

  R + h = x + y = 2·(x + y)/2.

It's not hard to see that the only right triangle in which twice one leg equals the sum of the hypotenuse and the other leg are equal is the famous 3-4-5 triangle, often referred to as the Egyptian triangle. Indeed, let a, b, c be the legs and the hypotenuse of a right triangle that satisfy say

(3) a + c = 2b.

Then since a² + b² = c², we have

  a² + b² = (2b - a)²,

from which

  3b = 4a,

and then (3) implies

  3c = 5a and 4c = 5b.

If we define t = a/3, then

  a = 3t,
b = 4t,
c = 5t.

This tells us that R and H and hence O do not depend on the values of x and y but only on their sum (x + y) and are invariant for a fixed pair of points A and B.

Applying inversion Michel Cabart found a shorter solution to the problem:

 

Let's transform the figure by an inversion with center A letting point B invariant. Then

  • line (AB) is globally invariant,
  • circle (A) becomes line (A') parallel to (AB),
  • circle (B) becomes circle (B') tangent to (AB) in B and to line (A'),
  • circle (Q) becomes circle (Q') tangent to (B') and to lines (A') and (AB), thus equal to (B').

The tangent (BF) to circle (Q') satisfies: tan (FBE/2) = t = 1/2. Its position is thus fixed. It is the inverse of a fixed circle in the original figure passing by A and B (red) and tangent to circle (Q). In this circle we have: AH × AC = AB² (as H is the inverse of C) or 2R x sin FBE = AB; but sin FBE = 2t/(1+t2) = 4/5. Thus R = (5/8)AB.

References

  1. H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989

    Write to:

    Charles Babbage Research Center
    P.O. Box 272, St. Norbert Postal Station
    Winnipeg, MB
    Canada R3V 1L6

Sangaku

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Copyright © 1996-2018 Alexander Bogomolny

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