# Ghhm

Topics: Triangle, Pythagorean theorem, Geometry Pages: 7 (1451 words) Published: May 12, 2013
Japan Temple Geometry

Abstract

In this science project the problems, which were written at Japanese temple boards are considered. These problems are differing from the European geometry by their solutions. Translated chapters from the book of Fukagawa and Pedoe were devoted to ellipses and n-gons, different combinations of the ellipses, circumferences and quadrilaterals, spheres, spheres and ellipsoids, different combination of the spheres and n-gons. In this science project the most interesting and original solutions of geometric problems were considered, in the base of using affine transformation.

Aim:
Our aim is to show solutions of Japanese problems, which are new for Western mathematics; and to reveal these types of geometrical questions. In our opinion, solving different types of problems can help everyone to enlarge the outlook in mathematics.

CONTENTS

Chapter 1. Ellipses and triangles ……..……………….………..…...……. 3

Chapter 2. Ellipses and tetragons..……………………………….……..... 4

Chapter 3. Ellipses, circumferences and rhombuses …………………... 5

Chapter 4. Spheres …………………………………………...……………. 6

Chapter 5. Spheres and Ellipsoids ………………………………………...7

Chapter 6. Spheres, pyramids and prisms …………….…………………8

Conclusion …………………………………………………………………...9

History of sangaku.……………………………………..……………………10

Literature ………………………………………………..……………………11

Problem 1.
On ellipse O(a,b) three points (A, B and C) are chosen so that areas (S1, S2 and S3) of curvilinear triangle, are equal. Show that area of triangle ABC is [pic][pic]ab. Solution.
The Ellipse shall be converted into circle b by using affine transformations, triangle ABC becomes some triangle A'B'C'. We shall show that this triangle is equilateral. Since ratio of affine transformation is invariant, area of new curvilinear triangle, will also be equal. Si=Si(sectors)-Si(triangle), [pic]i - central angle(i =1,2,3)[pic]

Si (triangle) = [pic]R2sin[pic]i
Si (sector) = [pic][pic]i,
Si = [pic]R2 ([pic]i - sin[pic]i).
Notice, that area of segment increases with [pic]. Therefore, areas can be equal only if angles are equal, in other words [pic]. From this we conclude that triangle A'B'C' equilateral and its area is [pic]. Since S(ellipse)/S(circle)=[pic], then S(ABC)=[pic]*a/b=[pic]. Which was to be proved.

Problem 2.
Ellipse O(a,b), is inserted in rectangle ABCD, touches sides CD in point E, and AD in point F. Show that AF*ED=FD*CE.

Solution.
Convert ellipse into circle using affine transformation. Rectangle will become a rhombus around this circle. Centers of symmetries of circumferences and rhombus coincide. Really, at symmetries for the centre to circumferences line AB becomes line, tangent to circles and parallel A'B', and this is C'D'. Similarly, line A'D' becomes line C'B''. Consequently, point A' becomes point C'. Similarly other couple of points. That is to say rhombus becomes itself. By the rules of symmetry C'E'=A’E’, and tangents E'D'=D'F', A'F'=A'K'.

From that, C’E’=F’A’. Then [pic] . Since length of parallel lines is affine invariant, [pic]
From this we get required equality: AF.ED=FD .CE

Problem 3
Ai (i=1,2,3,4) - a points of the ellipse so that A1A2A3A4 - is a rhombus. Show that if Ai are changed within ellipse, inscribed circle C(r) will remain the same, and express r using a and b.

.

Solution.
The Centre C, of inserted circle complies with the centre O of ellipse. Really, at symmetries for the centre of the ellipse chord A1A2 becomes parallel equal chord, and this is chord A3A4 , since such chord only one. Therefore, rhombus after symmetries for the centre of the ellipse becomes itself.

Diagonals of rhombus...