Solution of a Sangaku ``Tangency" Problem via Geometric Algebra
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This document presents a solution to a tangency problem from a 19th-century Japanese sangaku using geometric algebra. The author derives necessary equations and proves that specified relationships between circles and lines hold under the conditions of the problem. Two solutions are identified, with one confirming the required tangency and the other deemed extraneous.
Solution of a Sangaku ``Tangency" Problem via Geometric Algebra
- 1. Solution of a Sangaku “Tangency” Problem via Geometric Algebra December 11, 2018 James Smith nitac14b@yahoo.com https://mx.linkedin.com/in/james-smith-1b195047 Abstract Because the shortage of worked-out examples at introductory levels is an obstacle to widespread adoption of Geometric Algebra (GA), we use GA to solve one of the beautiful sangaku problems from 19th-Century Japan. Among the GA operations that prove useful is the rotation of vectors via the unit bivector i. “The center of the red circle and the base of the isosceles triangle lie along the same diameter of the green circle. The blue circle is tangent to the other three figures. Prove that the line connecting its center to the point of contact between the red circle and the triangle is perpendicular to the above-mentioned diameter.” 1
- 2. Figure 1: The center of the red circle and the base of the isosceles triangle lie along the same diameter of the green circle. The blue circle is tangent to the other three figures. Prove that the line connecting its center to the point of contact between the red circle and the triangle is perpendicular to the above-mentioned diameter. 1 Problem Statement In Fig. 1, the center of the red circle and the base of the isosceles triangle lie along the same diameter of the green circle. The blue circle is tangent to the other three figures. Prove that the line connecting its center to the point of contact between the red circle and the triangle is perpendicular to the above-mentioned diameter. 2 Formulation of the Problem in Geometric-Algebra Terms Fig. 2 defines the vectors that we will use. Note the notation used to distinguish between points and vectors: for example, c1 is the vector from the origin to the point c1. Also, c2 1 denotes c1 2 . In GA terms, we are to prove that c3 · ˆb = 0. Other formulations are possible; for example, that c3 ˆb = ˆbc3. 3 Solution Strategy We will derive an equation that that is satisfied by two circles. For one of them, c3 · ˆb = 0.
- 3. Figure 2: The vectors and frame of reference that we will use in our solution. 3
- 4. 4 Observations Two key observations are that r3 = c3 · ˆa (4.1) and that, in turn, ˆa = ˆpi = − r2 1 − r2 2 2r1 (r1 − r2) ˆb + r1 − r2 2r1 (r1 − r2) ˆbi . (4.2) We also see that by expressing the distance between c1 and c3 as r1 − r3 and c3 − c1 , we obtain (c3 − c1) 2 = (r1 − r3) 2 , which after simplification becomes c2 3 + 2 (2r2 − r1) c3 · ˆb + 4r2 (r2 − r1) = r2 3 − r3r1 . (4.3) Similarly, because c3 − c2 = r2 + r3, c2 3 + 2r2c3 · ˆb = r2 3 + 2r3r2 . (4.4) 5 Derivation of the Equation that We Seek We begin by subtracting Eq. (4.4) from Eq. (4.3), then solving for r3: r3 = r1 − r2 r1 + r2 2r2 + c3 · ˆb . (5.1) Substituting that expression for r3 in Eq. (4.4), then simplifying, c2 3 − r1 − r2 r1 + r2 2 c3 · ˆb 2 − 4r1r2 r1 − r2 (r1 + r2) 2 c3 · ˆb = 8r1r2 2 (r1 − r2) (r1 + r2) 2 . Now, we write c2 3 as c3 · ˆb 2 + c3 · ˆbi 2 , obtaining c3 · ˆbi 2 + 4r1r2 (r1 + r2) 2 c3 · ˆb 2 − 4r1r2 r1 − r2 (r1 + r2) 2 c3 · ˆb = 8r1r2 2 (r1 − r2) (r1 + r2) 2 . (5.2) 4
- 5. An expression for c3 · ˆbi 2 in terms of c3 · ˆb by equating the expressions for r3 given by Eqs. (4.1) and (5.1), c3 · ˆa = r1 − r2 r1 + r2 2r2 + c3 · ˆb , then expressing ˆa via Eq. (4.2): c3 · − r2 1 − r2 2 2r1 (r1 − r2) ˆb + r1 − r2 2r1 (r1 − r2) ˆbi = r1 − r2 r1 + r2 2r2 + c3 · ˆb . (5.3) Thus, c3 · ˆbi 2 = 2r1 (r1 − r2) r1 + r2 2 c3 · ˆb 2 + 4r2 2r1 (r1 − r2) (r1 + r2) 2 + 2r1 r1 + r2 c3 · ˆb + 8r1r2 (R1 − R2) (r1 + r2) 2 . (5.4) Substituting that expression for c3 · ˆbi 2 in Eq. (5.2), 0 = 2r1 (r1 − r2) r1 + r2 + 2r1 r1 + r2 2 + 4r1r2 (r1 + r2) 2 c3 · ˆb 2 + 4r2 3r1 (r1 − r2) (r1 + r2) 2 + 2r1 r1 + r2 c3 · ˆb. (5.5) The two roots are 1. c3 · ˆb = 0, with c3 · ˆbi = 2r2 2r1 (r1 − r2) r1 + r2 and r3 = 2r2 (r1 − r2) r1 + r2 ; and 2. c3 · ˆb = −4r2 (r1 − r2) r1 + 2r1 (r1 + r2) 2 (r1 − r2) 2r1 (r1 + r2) + 3r2 1 + r2 2 , c3 · ˆbi = − 2r2 (r1 + r2) 2r1 (r1 − r2) + 4r1r2 r2 1 − r2 2 2 (r1 − r2) 2r1 (r1 + r2) + 3r2 1 + r2 2 , and r3 = 2r1r2 (r1 − r2) 2 (r1 − r2) 2r1 (r1 + r2) + 3r2 1 + r2 2 . The first solution is in red in Fig. 3; the second is in magenta. The first demonstrates that which was to be proved, but the second is extraneous: the magenta circle is tangent to the extension of the side of the isosceles triangle, but not to the triangle itself. 5
- 6. Figure 3: The two solutions to Eq. (5.5). For our purposes, the magenta circle is extraneous: it is tangent to the extension of the side of the isosceles triangle, but not to the triangle itself. 6