A Sangaku: Two Unrelated Circles
What if applet does not run? |
Chords KN and ST are perpendicular to diameter CP of a circle with center O at points Q and R. SQ intersects the circle in V. (K, S are on one side of CP, N and T on the other. Q is between P and R.) Let q be the radius of the circle inscribed into the curvilinear triangle TQV. Prove that
(1)
1/q = 1/PQ + 1/QR.
|Activities| |Contact| |Front page| |Contents| |Geometry|
Copyright © 1996-2018 Alexander Bogomolny
Solution
The solution comes from an Art of Problem Solving Forum.
For (1) to hold, the radius q must satisfy
(2)
q = PQ·QR / (PQ + QR) = PQ·QR / PR.
So assume a number q is defined by (2). Denote
(3)
OM = r - q,
where r is the radius of the given circle.
What if applet does not run? |
Start by applying the Pythagorean theorem in ΔOQM:
OM^{2} | = MQ^{2} + OQ^{2} |
= q^{2}/(sinθ)^{2} + (PQ - r)^{2} | |
= q^{2} + q^{2}/(tanθ)^{2} + (PQ - r)^{2} | |
= q^{2} + PQ^{2}·QR^{2}/(PR·tanθ)^{2} + (PQ - r)^{2} | |
= q^{2} + PQ^{2}·SR^{2}/PR^{2} + PQ^{2} - 2r·PQ + r^{2} | |
= q^{2} + PQ^{2}·(SR^{2} + PR^{2})/PR^{2} - 2r·PQ + r^{2}. |
By the Intersecting Chords Theorem,
SR^{2} = PR·(2r - PR) = 2r·PR - PR^{2}. |
Using this we continue:
OM^{2} | = MQ^{2} + OQ^{2} |
= q^{2} + PQ^{2}·(SR^{2} + PR^{2})/PR^{2} - 2r·PQ + r^{2} | |
= q^{2} + PQ^{2}·2r·/PR - 2r·PQ + r^{2} | |
= q^{2} - 2r·PQ·(PR - PQ)/PR + r^{2} | |
= q^{2} - 2r·PQ·QR/PR + r^{2} | |
= q^{2} - 2r·q + r^{2} | |
= (r - q)^{2}, |
from which OM = r - q, as expected.
It is noteworthy to observe that the radius in (1) does not depend on the radius r of the given circle, and thus not on point S on the perpendicular PC at R.
Michel Cabart has observed that this problem can be solved by a similar calculation but without first assuming the formula to be true. He proceeds as follows:
We can suppose the circle is of radius 1. Say
OQ = 1 - x,
OR = x + y - 1,
QT^{2} = y^{2} + [1 - (x + y - 1)^{2}] = 2(x + y) - 2xy - x^{2},
(sinθ)^{2} = y^{2}/[ 2(x + y) - 2xy - x^{2}].
By hypothesis OM^{2} = (1 - q)^{2} = (1 - x)^{2} + q^{2}/(sinθ)^{2}.
Move (1 - x)^{2} to left and divide all by x^{2}q^{2}.
[(1 - q)^{2} - (1 - x)^{2}]/x^{2}q^{2} | = [x - q][2 - (q + x)]/ x^{2}q^{2} (first member) |
= [2(x + y) - 2xy - x^{2}]/x^{2}y^{2} (2nd member) |
Replace X = 1/x, Y = 1/y, Q = 1/q. First member becomes
(t + Y) [t(2X - 1) + 2X(X + Y) - 2X - Y] = f(t).
Second member becomes Y[2X(X + Y) - 2X - Y] = f(0).
The trinom f(t) has all its coefficients positive, because
2X - 1 = (2 - x)/x > 0, and
2X(X + Y) - 2X - Y = QT^{2}/x^{2}y^{2}
thus its roots are negative and f(t) is growing when
A third approach is by Nathan Bowler who starts with the smaller circle:
Take coordinates such that it is the unit circle
- QV has equation xg + yh = 1.
- Q is at (1/g, 0).
- O is at (1/g, v/ug) = W/ug.
- V satisfies xg + yh = 1 and (x - 1/g)^{2} + (y - v/gu)^{2} = (1 - 1/ug)^{2}.
Further y^{2} - 2ygv/u + 2g/u - g^{2} - 1 = 0 so that
QR | = -y |
= -gv/u - sqrt(g^{2}/u^{2} - 2g/u + 1) | |
= -gv/u + 1 - g/u | |
= 1 - g(1+v)/u. |
and
PQ = 1 - 1/ug + v/ug = 1 - (1-v)/ug.
So (PQ - 1)(QR - 1) = (1-v)(1+v)/u^{2} = 1. Equivalently,
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
70793489