A Sangaku with an Egyptian Attachment:
What Is This About?
A Mathematical Droodle
What if applet does not run? |
|Activities| |Contact| |Front page| |Contents| |Geometry|
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 R_{1} and R_{2}, 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 = 4R_{1}r, y² = 4R_{2}r, (x + y)² = 4R_{1}R_{2}. |
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
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:
References
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
- Sangaku: Reflections on the Phenomenon
- Critique of My View and a Response
- 1 + 27 = 12 + 16 Sangaku
- 3-4-5 Triangle by a Kid
- 7 = 2 + 5 Sangaku
- A 49^{th} Degree Challenge
- A Geometric Mean Sangaku
- A Hard but Important Sangaku
- A Restored Sangaku Problem
- A Sangaku: Two Unrelated Circles
- A Sangaku by a Teen
- A Sangaku Follow-Up on an Archimedes' Lemma
- A Sangaku with an Egyptian Attachment
- A Sangaku with Many Circles and Some
- A Sushi Morsel
- An Old Japanese Theorem
- Archimedes Twins in the Edo Period
- Arithmetic Mean Sangaku
- Bottema Shatters Japan's Seclusion
- Chain of Circles on a Chord
- Circles and Semicircles in Rectangle
- Circles in a Circular Segment
- Circles Lined on the Legs of a Right Triangle
- Equal Incircles Theorem
- Equilateral Triangle, Straight Line and Tangent Circles
- Equilateral Triangles and Incircles in a Square
- Five Incircles in a Square
- Four Hinged Squares
- Four Incircles in Equilateral Triangle
- Gion Shrine Problem
- Harmonic Mean Sangaku
- Heron's Problem
- In the Wasan Spirit
- Incenters in Cyclic Quadrilateral
- Japanese Art and Mathematics
- Malfatti's Problem
- Maximal Properties of the Pythagorean Relation
- Neuberg Sangaku
- Out of Pentagon Sangaku
- Peacock Tail Sangaku
- Pentagon Proportions Sangaku
- Proportions in Square
- Pythagoras and Vecten Break Japan's Isolation
- Radius of a Circle by Paper Folding
- Review of Sacred Mathematics
- Sangaku à la V. Thebault
- Sangaku and The Egyptian Triangle
- Sangaku in a Square
- Sangaku Iterations, Is it Wasan?
- Sangaku with 8 Circles
- Sangaku with Angle between a Tangent and a Chord
- Sangaku with Quadratic Optimization
- Sangaku with Three Mixtilinear Circles
- Sangaku with Versines
- Sangakus with a Mixtilinear Circle
- Sequences of Touching Circles
- Square and Circle in a Gothic Cupola
- Steiner's Sangaku
- Tangent Circles and an Isosceles Triangle
- The Squinting Eyes Theorem
- Three Incircles In a Right Triangle
- Three Squares and Two Ellipses
- Three Tangent Circles Sangaku
- Triangles, Squares and Areas from Temple Geometry
- Two Arbelos, Two Chains
- Two Circles in an Angle
- Two Sangaku with Equal Incircles
- Another Sangaku in Square
- Sangaku via Peru
- FJG Capitan's Sangaku
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
63794042 |