# Proportions in Square

A 1835 Sangaku from the Miyagi prefecture {Fukagawa & Pedoe, 3.1.5] reached us with no solution. Two solutions have been posted by J. Marshall Unger from Ohio State University in his write-up of sangaku problems, (Problem #23).

Given square ABCD, with M, N the midpoints of AB, CD, inscribe a circle in kite QRSM and in triangle APQ.

Prove that the radius of the smaller circle is half that of the larger circle.

I find both solutions enticingly elegant and sufficiently simple to deserve inclusion in my collection. The problem also links to a separate problem of dividing a segment into equal parts which served as an extra inducement for putting up the current page.

### 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

|Contact| |Front page| |Contents| |Geometry| |Up| |Store|

Copyright © 1996-2017 Alexander Bogomolny

Given square ABCD, with M, N the midpoints of AB, CD, inscribe a circle in kite QRSM and in triangle APQ.

Prove that the radius of the smaller circle is half that of the larger circle.

### Solution 1

Observe that, for any inscriptible shape, the inradius could be found as the ratio of the area and the semiperimeter. We shall juxtapose ΔAQP and the kite MQRS.

Point P is the intersection of the diagonals in rectangle ACNM; hence P is exactly as far from AM as R which is the center of the square ABCD. It follows that triangles AMP and AMR have the same area and so are triangles AQP and MQR obtained from the former by removing the common part - ΔAMQ. We conclude that the area of the kite MQRS is twice that of ΔAQP.

Triangles MQR and CQA are similar and, since

2·MR = AC, MQ + QR + RS + MS = 2(MQ + QR) = CQ + AQ.

But, by symmetry, CQ = AT = AP + PT, while clearly PT = PQ. Therefore,

CQ + AQ = AT + AQ = AP + PQ + AQ.

A combination of the two identities shows that ΔAQP and the kite MQRS have equal perimeters and, hence, semiperimeters, thus proving the claim.

### Solution 2

Note that the incircle of the kite is also the incircle of ΔDQM. Further, AN||DM, implying similarity of triangles DQM and AQP. Since DM is twice AP, the triangles are similar with the coefficient of 2, so that their inradii are in the same proportion.

### Remark

This is a well known fact (that arises in the problem of division of a segment into equal parts) that

2·AQ = QR. Thus also2·MQ = CQ. Now, sinceCP = MP, 2·PQ = QM. This remark could be employed in both of the solutions above.|Contact| |Front page| |Contents| |Geometry| |Up| |Store|

Copyright © 1996-2017 Alexander Bogomolny

62344048 |