# Pythagoras and Vecten Break Japan's Isolation

Here we have a third solution to a Sangaku problem: Five squares are arranged as in the applet. Show that the area of triangle KMN equals the area of the square BEKH. The other two can be found elsewhere. The solution has been suggested by Nathan Bowler; it does not invoke any algebraic means.

8 January 2016, Created with GeoGebra

The solution is based on two key observations:

First, as in Solution #2, we observe that there are pairs of flank triangles in Vecten configuration. Thus we see that the areas of the four triangles ABE, DEK, BCH, KHI are equal, as are the areas of triangles KMN and DIK.

B is the midpoint of XY. Thus, when square BEKH is cut into 5 pieces by vertical lines through B and K and horizontal lines through E and H, in the manner of Proof #3 of the Pythagorean theorem, points U and V being directly above B, the quadrilateral DUIV is a parallelogram (with center O.) So that triangles DOU and IOV have equal areas, from which

Area(DIK) = Area(DUVIK).

Now, taking into account the first observation, we see that

Area(DEK) = Area(ABE) = Area(BEU) and

Area(IKH) = Area(BCH) = Area(BHV),

which implies the required identity

Area(KMN) = Area(DIK) = Area(BEKH).

The last step admits a very slight modification that invokes Proof #10 of the Pythagorean theorem. Indeed, for the same reasons as above,

Area(DUVIK) = Area(AEFGHC) = Area(BEKH).

## 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

### References

- J. Konhauser, D. Velleman, S. Wagon,
*Which Way Did the Bicycle Go?*, MAA, 1996, #50

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