San Gaku:
Japanese Temple Geometry Problems
SOLUTIONS

San Gaku is the term given to a collection of theorems in Euclidian geometry produced by people of all social classes primarily during the Edo period in Japan. These theorems were drawn in color on wooden tablets that hung from the rooves of shrines and temples in local precincts. A collection of some of these problems and their solutions are shown here. The problems are taken from the excellent book by H. Fukagawa and D. Pedoe Japanese Temple Geometry Problems: San Gaku, The Charles Babbage Research Centre, 1989. The drawings and solutions have been worked out by Kevin Mirus.

Chapter I: Circles

Section 1.1: Two Circles

Problem 1.1.0 Solution

Show that (AB)2 = 4r1r2.

By constructing the line segments O1A, O2B, O1O2, and O2C, the right triangle O1O2C is formed with legs of length AB and r1 - r2, and hypotenuse of length r1 + r2, as shown on the illustration above. By use of the Pythagorean Theorem: (AB)2 + (r1 - r2)2 =  (r1 + r2)2 (AB)2 + r12 - 2r1r2 + r22 =  r12 + 2r1r2 + r22 (AB)2 =  4r1r2

Back to the statement of Problem 1.1.0.

Problem 1.1.1 Solution

Show that .

The solution to this problem uses the results of Problem 1.1.0and the fact that AC + CB = AB: =  =  =  =  =  =  =  =  =

Back to the statement of Problem 1.1.1.

Problem 1.1.2 Solution

Problem 1.1.3 Solution

Show that .

The results of Problem 1.1.1 can be applied to find r4: =  =  =  The results of Problem 1.1.1 can also be applied to find r5: =  =  =  From these two examples it can be seen that: =

Back to the statement of Problem 1.1.3.

Problem 1.1.4 Solution

Section 1.2: Three Circles

Problem 1.2.1 Solution

Problem 1.2.2 Solution

Problem 1.2.3 Solution

Show that .

Construct the line segments O1O2, O2O3, O3O1, O2C, and O1D as shown above. Notice that the length of O1D has been labeled y. Also label the angles q32C and q123 as shown above. The triangles formed by these constructions yield the following relationships: sinq32C =  cosq32C =  (See Problem 1.1.0) (r1 + r3)2 =  (r1 + r2)2 + (r2 + r3)2 - 2(r1 + r2)(r2 + r3) cosq123 Þ cosq123 =  These relationships can then be used to calculate d: d =  r1 + r2 + y d =  r1 + r2 + (r1 + r2)sin(q123 + q32C) d =  (r1 + r2)(1 + sin(q123)cos(q32C) + cos(q123)sin(q32C)) d =  d =  When the first and third terms of the numerator are expanded, their like terms combined, and the result then factored, this equation reduces to: d =  d =  d =  d =  d =  d =  d =  d =

Back to the statement of Problem 1.2.3.

Problem 1.2.4 Solution

Show that r12 = 4r2r3.

Construct the line segments O1O2, O2O3, O3O1, O2C, and O3D as shown above. Notice that the length of O3D has been labeled y. Also label the angles q321 and q12C as shown above. The triangles formed by these constructions yield the following relationships: sinq12C =  cosq12C =  (See Problem 1.1.0) sin(q321 + q12C) =  (r1 + r3)2 =  (r1 + r2)2 + (r2 + r3)2 - 2(r1 + r2)(r2 + r3) cosq321 Þ cosq321 =  Next write the distance between l and m: 2r1 =  r2 + y + r3 2r1 - r2 - r3 =  (r2 + r3)sin(q321 + q12C) 2r1 - r2 - r3 =  (r2 + r3)(sin(q321)cos(q12C) + cos(q321)sin(q12C)) 2r1 - r2 - r3 =  2r1 - r2 - r3 =  2r1 - r2 - r3 =  (2r1 - r2 - r3) (r1 + r2)2 - (r1 - r2)(r1(r2 - r3) + r2(r2 + r3)) =  2r13 + 2r12r2 - 4r1r2r3 =  r12 + r1r2 - 2r2r3 =  r14 + 2r13r2 + r12r22 - 4r12r2r3 - 8r1r22r3 - 4r23r3 =  0 (r1 + r2)2 (r12 - 4r2r3) =  0 r12 =  4r2r3

Back to the statement of Problem 1.2.4.

 

	© 2002 by Kevin Mirus (kmirus@madison.tec.wi.us or kjmirus@execpc.com).
	This document was last modified  .