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Figure from:On the construction of the Śrī Yantra

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Figure 10. In this figure we treat independently the problem of placing point A along the vertical axis passing through the triangle edge 1,4. We require that (A,5) and (A,6) meet the two remaining edges of the triangle at two vertically aligned points D and I. We find the foci and the directrices of the parabole p and p’. Point 28 is the intersection pn p’ projecting on Ae 1,4. two remaining edges of the triangle at two vertically aligned points D and J. We find the foci

Figure 10 In this figure we treat independently the problem of placing point A along the vertical axis passing through the triangle edge 1,4. We require that (A,5) and (A,6) meet the two remaining edges of the triangle at two vertically aligned points D and I. We find the foci and the directrices of the parabole p and p’. Point 28 is the intersection pn p’ projecting on Ae 1,4. two remaining edges of the triangle at two vertically aligned points D and J. We find the foci

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entre la Dase de /g et leS JambDes de /3 et lg — propriete (U1); Ls, lg, lg)). La méthode B, transmise par Laksmidhara, part du triangle central et l’entoure quatre fois par des rangées de 8, 10, 10 et 14 triangles comme dans le passage du Rudrayamala Tantra mentionné ci-dessus. Ainsi, les conditions de concurrence ((ii)—(iii)) sont satisfaites. Cependant, la Figure 2.b montre bien que, en suivant cette méthode, la condition (i) peut échouer (le cercle ne passe pas par les sommets latéraux de fz et f7).
entre la Dase de /g et leS JambDes de /3 et lg — propriete (U1); Ls, lg, lg)). La méthode B, transmise par Laksmidhara, part du triangle central et l’entoure quatre fois par des rangées de 8, 10, 10 et 14 triangles comme dans le passage du Rudrayamala Tantra mentionné ci-dessus. Ainsi, les conditions de concurrence ((ii)—(iii)) sont satisfaites. Cependant, la Figure 2.b montre bien que, en suivant cette méthode, la condition (i) peut échouer (le cercle ne passe pas par les sommets latéraux de fz et f7).
Figure 3. We illustrate the concurrency condition ((iii); ts, t5, t). The intersection between the legs of the (yellow) triangle t5, the base of the (yellow) triangle ts; and the legs of the (rose) triangle tg consists of exactly two points highlighted by the dashed circles.
Figure 3. We illustrate the concurrency condition ((iii); ts, t5, t). The intersection between the legs of the (yellow) triangle t5, the base of the (yellow) triangle ts; and the legs of the (rose) triangle tg consists of exactly two points highlighted by the dashed circles.
Up to rescaling and translation, the diagram depends on four parameters. In the construction presented here we fix the four points P.Q,R,S on the diameter OT = [0,1]. These are the base points of f3, tg, t7 and tg. Each choice of such parameters,”° in increasing order and within the circle, leads to a diagram provided we allow degenerate cases.
Up to rescaling and translation, the diagram depends on four parameters. In the construction presented here we fix the four points P.Q,R,S on the diameter OT = [0,1]. These are the base points of f3, tg, t7 and tg. Each choice of such parameters,”° in increasing order and within the circle, leads to a diagram provided we allow degenerate cases.
We proceed following the captions of the six diagrams appearing in Figure 4. Points P and R are the base points of the triangles t3 and t7, which can be drawn entirely by intersecting the vertica lines through P and R with the semicircle at U and V, Figure 4.a. Point S is the apex of the triangle t2, whose leg passes through point W = OV n PU, Figure 4.b. Point P is the apex of tg, whose les passes through X = SWnRV, Figure 4.b-c. Then, we get Y = PXNUT, its vertical projection E or OT, and F (resp. G), the intersection of the vertical line through Q with PX (resp. SW), Figure 4.d We get point 1 = FEn PU, point 2 = QWn PG, and point 3 (resp. point 4) by intersecting the line through the points O and 2 with the line through E and 1 (resp. the line PW); see Figure 4.e-f Thece four noints 1.2.3. and 4 allows to intradiice the Anolloniius nrgnhlem.
We proceed following the captions of the six diagrams appearing in Figure 4. Points P and R are the base points of the triangles t3 and t7, which can be drawn entirely by intersecting the vertica lines through P and R with the semicircle at U and V, Figure 4.a. Point S is the apex of the triangle t2, whose leg passes through point W = OV n PU, Figure 4.b. Point P is the apex of tg, whose les passes through X = SWnRV, Figure 4.b-c. Then, we get Y = PXNUT, its vertical projection E or OT, and F (resp. G), the intersection of the vertical line through Q with PX (resp. SW), Figure 4.d We get point 1 = FEn PU, point 2 = QWn PG, and point 3 (resp. point 4) by intersecting the line through the points O and 2 with the line through E and 1 (resp. the line PW); see Figure 4.e-f Thece four noints 1.2.3. and 4 allows to intradiice the Anolloniius nrgnhlem.
(a) Parallel lines through 5 and 6 to 1,3 and 4,3 meet the line (1,4) at 7 and 8, respectively.
(a) Parallel lines through 5 and 6 to 1,3 and 4,3 meet the line (1,4) at 7 and 8, respectively.
(b) Draw the right triangle 7,9,12, where 9 = mid(4,7). Point 12 is the vertical projection of the midpoint between 5 and the line (7, 4).
(b) Draw the right triangle 7,9,12, where 9 = mid(4,7). Point 12 is the vertical projection of the midpoint between 5 and the line (7, 4).
Figure 6. The dashed circle in the figure is the Apollonius circle externally tangent to the circle II, tangent to the line A, and passing through the point ¢. The circle is characterised by the fact that its center projects horizontally onto the vertical edge of the dashed triangle. The figure contains the entire construction, but Figure 7 details each step.
Figure 6. The dashed circle in the figure is the Apollonius circle externally tangent to the circle II, tangent to the line A, and passing through the point ¢. The circle is characterised by the fact that its center projects horizontally onto the vertical edge of the dashed triangle. The figure contains the entire construction, but Figure 7 details each step.
(a) The circle © passes through 22, 23 and ¢@ = 18. Join ¢ to 21. Set 24=On (¢,21).
(a) The circle © passes through 22, 23 and ¢@ = 18. Join ¢ to 21. Set 24=On (¢,21).
(c) Point 27 is the intersection between II and the above semicircle joining 26 and 14.
(c) Point 27 is the intersection between II and the above semicircle joining 26 and 14.
(d) The desired Apollonius circle is the dashed circle = passing through the points ¢, 24 and 27. Point 28 is its centre. (b) Set point 26, the intersection between the lines (#, 24) and (22,25).
(d) The desired Apollonius circle is the dashed circle = passing through the points ¢, 24 and 27. Point 28 is its centre. (b) Set point 26, the intersection between the lines (#, 24) and (22,25).
Figure 8. We deduce points A, B,C, D, H, I,J, K,L,M, N, and This entire part of the construction (Figures 5, 6, and 7) is about choosing the right slope of the eg of t (grey triangle).
Figure 8. We deduce points A, B,C, D, H, I,J, K,L,M, N, and This entire part of the construction (Figures 5, 6, and 7) is about choosing the right slope of the eg of t (grey triangle).
Figure 9. As the slope of the ray r varies, the center of mass D of the right trapezoid BGQC follows a line through O of slope mg/(g + mq), where m is the slope of OG, and we hav g=QGand q=0Q. This amounts to intersecting the parabola p with roots at v and I' and leading term t*/m and the parabola p’ with roots at v’ and / and leading term t*/m’. Since m < m’, there are two solutions, one before / and one within J, I’ [.
Figure 9. As the slope of the ray r varies, the center of mass D of the right trapezoid BGQC follows a line through O of slope mg/(g + mq), where m is the slope of OG, and we hav g=QGand q=0Q. This amounts to intersecting the parabola p with roots at v and I' and leading term t*/m and the parabola p’ with roots at v’ and / and leading term t*/m’. Since m < m’, there are two solutions, one before / and one within J, I’ [.