A Sangaku Follow-Up on an Archimedes' Lemma
The following Sangaku problem sports a publication trail:
X is a point on a chord L of a given circle C, and circles A and B are drawn on opposite sides of L so as to touch L at X and be tangent to C. Prove that the ratio of the radii of A and B is independent of the position of X on L.
The problem has been included in Fukagawa and Pedoe's Japanese Temple Geometry Problems, offered by Crux Mathematicorum, 1987, 194, and appeared in R. Honsberger's collection From Erdös To Kiev with two solutions from the Crux. The first two solutions below, have been adapted from Honsberger's book.
This solution is by Dan Sokolowsky, Williamsburg, VA.
Start by drawing PXQ perpendicular to L at X. Since L is a common tangent to A and B, PX and QX are diameters of A and B. If M and N are their points of tangency with C, angles at M and N subtended by PX and QX are right:
∠XMP = ∠XNS = 90o.
Extending MX and MP to meet C in R and S, we get a diameter RS (since RS subtends right angles in C.)
Now, the centers U and V of A and C are in line with the point of tangency M, and it is clear that triangles MUX and MVS are isosceles. Since they share a common base angle at M, the base angles at X and S are also equal, and the lines PXQ and RS are parallel. Thus RS is perpendicular to L.
Similarly, extending NX and NQ we'll get a diameter perpendicular to L. Therefore, R lies on NX and S on NQ.
The next step is to extend MR and NS beyond M and N to meet in T. In ΔRST, RN and SM are altitudes, so that X is the orthocenter. TX is then the third altitude and is perpendicular to RS. But L is a line through X perpendicular to RS, which leads us to conclude that T lies on L.
The final step is to consider two pairs of similar triangles
PX/RW = TX/TW = QX/SW,
PX/QX = RW/SW.
Since the right side here is independent of X, so is the left side. The conclusion follows.
I have only one remark concerning this proof. We may have started with constructing RS, a diameter perpendicular to PXQ. Then PR passes through M (and QS through N) by Lemma 1 from Archimedes' Book of Lemmas. (The situation has been explored elsewhere.)
This solution is by Sam Baethge, San Antonio, TX. It's a straightforward application of a theorem attributed to another famous Greek mathematician.
Let p and d be the horizontal and vertical components of VX. And let a, b and R be the radii of A, B, and C, respectively. By the Pythagorean theorem,
p2 + (a + d)2 = (R - a)2 and p2 + (b - d)2 = (R - b)2.
These are solved for a and b to get
a = (R2 - p2 - d2) / 2(R + d), b = (R2 - p2 - d2) / 2(R - d),
from which a/b = (R - d)/(R + d) independent of X.
This unexciting solution proved extremely useful in constructing the diagram and, ultimately, writing the applet illustration.
This simple solution that is due to Michel Cabart employs inversion which I am not sure has been available to the Japanese during the period of seclusion.
Trace the tangents D1 and D2 to circle C in R and S. Now transform the figure by the inversion with center X, and power equal to the power of X relative to C. The latter equals
Circle A is transformed into a straight line parallel to L through the image of M, which is D2. Similarly, B is mapped onto D1.
The inverse of P stays on the perpendicular to L through X, i.e., PQ, but also lies on the image of A, i.e. D2. Hence the distance from X to the image of P is R + d. The distance from P itself to X is of course 2a. Thus
- 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 49th 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
- H. Fukagawa, D. Pedoe, Japanese Temple Geometry Problems, The Charles Babbage Research Center, Winnipeg, 1989, #4.2.4
- R. Honsberger, From Erdös To Kiev, MAA, 1996.
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny