Comptes Rendus Mathématique

. The ´Sr¯ı Yantra (or ´Sr¯ı Cakra ) is a sacred diagram of Tantric Hinduism. Its study stimulated a vast e ﬀ ort of specialists from di ﬀ erent ﬁelds. In mathematics, its construction sets an elementary and nontrivial problem. In this note, we work out a straightedge and compass method for constructing concurrent models of ´Sr¯ıYantras . The question is equivalent to the circle-line-point problem of Apollonious. Résumé. Le ´sr¯ıyantra (ou ´sr¯ıcakra ) est un diagramme sacré dans les traditions hindoues tantriques. Il a fait l’objet de nombreuses études dans di ﬀ érentes disciplines. En mathématiques, sa construction pose un problème élémentaire et non trivial. Dans cette note, on fournit une méthode de construction à la règle et au compas. La question est équivalente à celle d’un problème d’Apollonius qui consiste à trouver un cercle tangent à un cercle donné, à une droite donnée et passant par un point donné.


La construction
On présente la construction dans les Figures 5, 6

Introduction
Yantras are sacred objects in Hinduism.Remarkably, many of them have very interesting mathematical properties.The Śrī Yantra is one of the most popular yantras; it is used in yoga meditation.It can be described from the exterior to the interior as follows.The outermost motive is the square of defence or bh ūpura.A triple circle circumscribes the core of the diagram.Then, sixteen petals of lotus surround an eight-petalled lotus and a polygonal diagram containing a central point, the bindu (see Figure 1).This article deals with the polygon, which is indeed the union of nine maximal triangles t 1 , . . ., t 9 possessing a common vertical axis of symmetry and ordered according to the position of their base from the top downwards.In Figures 2 and 8, for the edges of t 1 , . . ., t 9 , we use the following sequence of colours: gray (t 1 ), violet (t 2 ), blue (t 3 ), green (t 4 ), yellow (t 5 ), rose (t 6 ), orange (t 7 ), red (t 8 ), and brown (t 9 ).With these notations t i points downward if i ≤ 5 and points upward otherwise.
The following concurrency properties are required to be satisfied (see Figures 1 and 3).Throughout, base point denotes the midpoint of the base of an isosceles triangle.
(i) The triangles t 3 and t 7 share the same circumscribed circle.
(ii) The apex of the triangle t is the base point of the triangle t for (t , t ) = (t We will refer to the above conditions as (i), ((ii); t i , t j ), and ((iii); t i , t j , t k ).Figures 1 and 3 are based on a handmade realisation of the diagram that enables us to both check the concurrency conditions and appreciate the slight imperfections, as for instance ((iii); t 3 , t 7 , t 8 ).
In the GeoGebra [10] programme (available here [https://www.geogebra.org/m/zdvxtdvv]and based on the present paper) we can move the triangles without breaking the concurrency conditions.This allows us to appreciate conditions (i), (ii) et (iii) and carry out experimentation (see also Appendix A for more information and GeoGebra programmes illustrating the Śrī Yantra and its construction).

Figure 3.
We illustrate the concurrency condition ((iii); t 5 , t 5 , t 6 ).The intersection between the legs of the (yellow) triangle t 5 , the base of the (yellow) triangle t 5 and the legs of the (rose) triangle t 6 consists of exactly two points highlighted by the dashed circles.

The origins
The origins of the Śrī Yantra are unknown.We have to trace them back among the rituals that have been associated to it (see Padoux [17]).
Michaël [14] describes one of these rituals based on the text Saundarya Laharī ("The waves of beauty") attributed to a disciple of the 8 th century philosopher Ādi Śa ṅkara.
In [6], R. C. Gupta considers a large variety of yantras of great mathematical interest (some of them, such as the chautisa yantra, have been already studied from a purely mathematical point of view, see G. Bhowmik [1] and A. Navas [16]).Gupta's presentation of the properties of the Śrī Yantra enables us to reconstruct some of the crucial moments of its definition throughout history.We mostly follow Gupta's presentation, but also refer to Huet [11], Mookerjee and Khanna [15], Rao [19], and Zimmer [26].
Gupta [6, p. 180] quotes the Rudrāyamala Tantra and in particular a verse referring to a yantra consisting of a bindu, a central triangle, and then "enclosures formed by 8, 10, 10, and 14 triangles," and finally "three circles, and three bh ūpura."This does indeed describe the Śrī Yantra, see Figure 2.b in which the central triangle is t 1 ∩ t 5 and the "enclosures" are highlighted by two different shades.Traditionally these circular rows of triangles are described as cakras (wheels). 11  11 The Śrī Yantra is described in terms of cakras for instance in Mookerjee and Khanna [15] and Zimmer [26], (see Gupta also mentions three three-dimensional variants, k ūrma (turtle), padma (lotus) and meru (the fabulous mountain), of the planar Śrī Yantra considered here, usually referred to as bh ū (the Earth).In the k ūrma version the segments are replaced by cross sections of a sphere (see [6, p. 181] and references to Gaurīyāmala Tantra).
Kulaichev refers to the work of de Casparis [3, p. 34 and p. 41] reporting mentions of the Śrī Yantra in 7 th century inscriptions dating back to the time of the kingdom of Śrīvijaya in south Sumatra.
Kulaichev [13, p. 279] also alludes to a hymn from Atharva Veda (c. 12 th century BC) dedicated to a Śrī Yantra-like figure consisting of nine triangles.He also mentions a representation dated to the 17 th century to be found in the religious institution of Śr .ṅgārī Mat .ha established by Ādi Śa ṅkara, [13].
Finally, we recall, as Huet does in [11, p. 622], that although it is hinted in several sources that this symbol is very old we do not know of any published representation before the 17 th century.Therefore, the problem of determining the date of creation remains open.The interest that is brought to this question is well illustrated by the following quote of Mookerjee and Khanna [15]: "the Śrī Yantra, in its formal content, is a visual masterpiece of abstraction, and must have been created through revelation rather than by human ingenuity and craft."

The history of the construction of the Śrī Yantra
Drawing a Śrī Yantra is an elementary, but nontrivial, mathematical problem.In 2007, Gupta lists the various known methods [6].

The traditional methods
Two traditional methods are reported by two commentators of Saundarya Laharī: Laks .mīdharaet Kaivalyāśrama, see Figure 2 and [22].
Method A, is due to Kaivalyāśrama: it is applied in Figure 2.a. 12 The vertices of the triangles are placed within a frame consisting of the vertical diameter subdivided in 48 equal units and the circle (Gupta mentions also a 42-unit version).As Figure 2.a shows, the unit is used to specify the distance from the circle along the horizontal parallel lines.Figure 2.a shows that condition ((iii); t 3 , t 8 , t 9 ) can fail (and indeed there is no reason for any concurrency of (iii) to hold).
Method B is due to Laks .mīdhara: it is applied in Figure 2.b. 13 It starts from the central triangle and surrounds it four times by sequences of 8, 10, 10, and 14 triangles as in the above quote from Rudrāyamala Tantra.The concurrency conditions ((ii)-(iii)) are satisfied by construction.However Figure 2.b shows that condition (i) can fail (the circle does not pass through the side vertices of t 3 and t 7 ).
These methods are not precise, as illustrated by Figures 2.a and 2.b, but they have the merit of yielding an archetype: a precisely identified set of geometric figures.As a simple example, we can compare the collection of Śrī Yantras to the set of isosceles triangles up to rescaling and congruence.This set is, of course, defined by a much more elementary symmetry property, but also contains several nonequivalent forms (acute triangles, obtuse triangles . . . ) even after also Jung [12]).The nine cakras (wheels) surrounding the bindu are the bh ūpura, the triplet of circles, the 16 petals, the 8 petals, the enclosure of 14 triangles, than that of 10, again 10, and 8 triangles, and finally the central triangle.Gupta points out that the term cakra is also attributed sometimes to the nine triangles t i . 12The caption reads: we draw the picture by placing the vertices within a fixed grid.The concurrency conditions, such as those between the blue, red and maroon triangles ((iii); t 3 , t 8 , t 9 ), fail. 13The caption reads: we can impose all concurrency conditions ((ii)-(iii)) to the triangles.In general, condition (i) requiring that the exterior circle passes through the vertices of t 3 and t 7 fails.
C. R. Mathématique -2021, 359, n 4, 377-397 reducing by rescaling and congruence.The properties of the Śrī Yantra are not as elementary, but these two methods clarify which minimal set of conditions of contact and intersection should be imposed today in order to consider a Śrī Yantra admissible, or-as we will write here-concurrent.These properties were stated above as ((i)-(iii)).Since 1970 other authors have formalised the problem in different but equivalent terms.We now review the modern approaches.

Straightedge and compass construction
Although imprecise, the traditional methods can be performed with straightedge and compass.During the seventies, Nicolas J. Bolton and D. Nicol G. Macleod [2, p. 68-69] and Fonseca [5, p. 35-36] rewrote independently the list of vertex coordinates of Method A. The first paper also contains an optimisation of Method B: the table and the figure [2, tab.2, Fig. 2] attributed to A. West produces a collection of diagrams satisfying all the concurrency conditions ((ii)-(iii)).The method "has no errors in the intersections" of the triangles [2, p. 7], but does not take into account the external circle condition (i).A simple examination allows us to notice that the forms obtained in this way (up to rescaling and congruence) depend on six parameters; imposing condition (i) would reduce to a four-parameter space of solutions. 14 Nicolas J. Bolton and D. Nicol G. Macleod [2] and Fonseca [5] propose a straightedge and compass construction based on the golden ratio. 15These approaches fail to meet concurrency conditions in a way that is not visible to the eye.This motivated the hypothesis still considered today of a possible analogue, perhaps perfect, construction believed to have originated in ancient Egypt (numerous attempts of this kind can be found on the internet).

The germ of a variety of Śrī Yantras
In 1990, G. Huet provided a formal definition of the mathematical problem posed by the Śrī Yantra.He listed a sequence of logical and geometrical links between the vertices [11, p. 611].This is equivalent to our concurrency conditions ((i)-(iii)).Indeed, in order to start his sequence one has to fix five parameters. 16All conditions ((i)-(iii)) except ((ii); 2,6) can be satisfied as long as each of the five parameters are chosen in a sufficiently small interval. 17He finally translates ((ii); 2,6) into an equation relating the initial parameters.The computer programme issued from this argument expresses the coordinates of the vertices by Newton approximation and draws all possible Śrī Yantras. 18Note that the set of drawings of Śrī Yantras produced by five parameters subject to an equation has four degrees of freedom. 14Instead of fixing all the vertices we just set the position (along the vertical diameter) of the bases of t 1 , t 2 , t 4 , t 6 , t 8 and t 9 as well as their lengths for triangles t 4 and t 6 .In order to complete the solution, to these eight variables, we would have to impose two independent conditions equivalent to condition (i).This happens because, in order to coincide, the circumscribed circles should be concentric and congruent.We observe that, up to rescaling (one less parameter) and translation (again one less parameter), we obtain a set of solutions depending on 8 − 2 − 1 − 1 = 4 parameters in total. 15The first construction is approached by replacing at a first stage the circumscribed circle by a square.The second construction is inscribed in a circle and appears as a refinement of the first. 16Huet's parameters are the distances of the three base points of t 3 , t 6 and t 7 from the apex of t 3 on the diameter and the distance from the diameter of the concurrency points identified above as ((iii); t 1 , t 3 , t 8 ) and ((iii); t 3 , t 8 , t 9 ).In his text, these are referred to as Y Q , Y P , Y J , X F , X A . 17 Huet fixes the midpoint of this neighbourhood at Y Q = 0.668; Y P = 0.463; Y J = 0.398; X F = 0.126; X A = 0.187 with respect to the length-1 diameter of the circle. 18One of these automated drawings is the cover of a volume in the honour of the computer scientist Maurice Nivat.This is where Huet's work from the eighties [11]   19 He shows that it should be possible to obtain all these conditions simultaneously (p.284) by iterated approximation.He discusses whether the solution is unique (p.285), estimates the complexity, and argues that the answer is out of reach.This problem is not considered here, but the complexity estimated by Kulaichev is reduced.This has been the subject of recent work: there is a short addendum by Kulaichev listing further methods of approximation suggested by readers (the English translation of this text is easily available on the internet under the title "Addition to Sri Yantra and its mathematical properties").
Hoping to gain more flexibility and overcome the computational difficulties of the twodimensional problem, Kulaichev also studies the problem of constructing three-dimensional models (namely, he considers the k ūrma version mentioned above).This line of attack has been systematically investigated by C. S. Rao [20].We point out that the hypothesis of a threedimensional origin would have been reinforced by the statement that the Śrī Yantra could not be constructed on a plane by compass and straightedge but only through projection to the plane of a three-dimensional construction.Instead, the constructibility shown here implies that the coordinates of the diagram are contained in the the field of constructible numbers, [25].This may illuminate the attempt by [13, Section Analysis], [18,21,23,24] to write explicit equations for it.

Our construction and the problem of Apollonius
To the best of my knowledge, the natural question of whether the Śrī Yantra is constructible by straightedge and compass has remained open until now.Here, we treat the problem under the same hypotheses as Huet [11], i.e. without imposing any extra conditions.We construct with straightedge and compass the family of all Śrī Yantras satisfying the minimal concurrency requirements ((i)-(iii)).

Apollonius
The construction of the Śrī Yantra turns out to reduce to the solution of the so-called circle-linepoint problem posed by Apollonius from Perga in the 3 rd century BC.In its general form, the problem consists in finding the circles tangent to three given plane circles.The circle-line-point problem is analogous, but imposes the passage through a given point, and tangency to a given straight line and to a given circle.Apollonius solved these problems, but his work was lost as documented by Pappus from Alexandria in the 4 th century.The solutions were worked out again by Viète in 1600.
Below, we see that the problem arises in the construction of the Śrī Yantra starting from the base points of t 3 , t 6 , t 7 and t 9 along the vertical diameter (see Figure 4).Through an elementarybut not straightforward-argument detailed in Figures 5 and 6, these parameters identify a line ∆ and-within one of the ∆-bounded half-planes-a circle Π and a point φ exterior to the circle. 19The extra conditions are the following: (a) triangles t 3 and t 7 are congruent, (b) the circumcircle of t 1 coincides with the external cercle E (circumscribed to t 3 and t 7 ), (c) the circumcircle of t 9 also coincides with E , (d) the incircle to the innermost triangle (i.e.t 1 ∩ t 5 ) is concentric to E .In some works using this approach, condition (a) is replaced with a condition imposing that t 1 should be equilateral.Let us point out that the congruence t 3 ≡ t 7 is also considered in [2] but not in [5].It is immediately satisfied by our construction if we modify the initial steps in such a way that d (O, P ) = d (R, T ).This is a good spot to mention that it might be interesting to study the interplay between conditions (a-d) and an extra condition ((iii), t 4 , t 5 , t 6 ).
C. R. Mathématique -2021, 359, n 4, 377-397 In this configuration, there are exactly four circles Ξ satisfying the required contact with φ, ∆, and Π.The exact drawing of the Śrī Yantra is based on one of them (we explain this in Section 2.2.3).We conclude by reviewing the original Śrī Yantra problem in the light of the Apollonius problem.
In the eighties, the enormous developments in the enumerative geometry of algebraic curves brought us at least two beautiful geometric pictures representing the Apollonius problem.Eisenbud and Harris [4, Section 2.3] describe the space parametrising all the solutions of the Apollonius problem as φ, ∆, and Π vary.This family of solutions gives rise to a multiple-sheeted cover of a space parametrising conics, see [4,Section 2.3].Harris [7] provides another interpretation of the same space in termes of moduli spaces of algebraic spin curves, i.e. curves C equipped with a theta characteristic L satisfying This space, with its two descriptions, completes the variety of Śrī Yantras described by Huet.Previously, we could only see a family of Śrī Yantras parametrised by a small enough neighbourhood lifting to such a multi-sheeted cover.Now all points of this cover admit a geometric interpretation extending the family of Śrī Yantras.

The construction
In this section, we present the construction following Figures 4, 5, 6, 7, and 11.

An anticlockwise turn
From now on, and in all the figures except the final Figure 11, we consider only the right-hand half of the diagram.We place it horizontally after a right angle rotation in anticlockwise direction.

The setup
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 t 3 , t 6 , t 7 and t 9 .Each choice of such parameters, 20 in increasing order and within the circle, leads to a diagram provided we allow degenerate cases.   Choosing the base point of t 3 , t 6 , t 7 and t 9 is equivalent to choosing the base points of t 3 , t 6 and t 7 and the concurrency point ((iii); t 3 , t 8 , t 9 ) as in Huet.Furthermore Huet's choice is equivalent to drawing the bases t 3 , t 6 , t 7 and t 9 at position 0.332; 0.537; 0.602; 0.835 on the segment OT = [0, 1].In the figures below we prefer to use 0.324; 0.517; 0.692; 0.866 for sake of clarity in the final drawings of the construction.Here we label points by numbers; (n, m) is the line through n and m, n, m is the segment joining n and m, mid(n, m) is its middle point, and n, m, . . ., k is the polygon whose vertices are n, m, . . ., k.

From the Śrī Yantra to Apollonius
We state a circle-line-point (CLP) Apollonius problem with respect to a point φ, a line ∆, and a circle Π determined in Figures 5 and 6.
Consider the triangle of vertices 1, 3, 4 in Figure 5 and the points 5 = Z and 6 = Q.We define points 7, . . ., 19 allowing us to determine the relevant circle Π, line ∆ and point φ.The construction is two-fold: on the one hand it starts from point 5 and its crucial steps are points 7, 10, 12, 14 and 15.On the other hand it starts from point 6 and its crucial steps are points 8, P, 16, 18 and 19.
The first part is as follows: the line through 1 and 4 intersects at 7 the line through 5 = Z directed by 1, 3; see Figure 5     Indeed, Figure 7 is an application of the classical sequence of nested solutions of Apollonius problems (see [8,9]): in order to solve the Apollonius problem we begin by reducing the Circle-Line-Point problem to a Circle-Point-Point problem, and then to the elementary Point-Point-Point problem, whose solution is the (unique) circle passing through three given points.

From Apollonius to the Śrī Yantra
We conclude the construction in Figure 8 by applying the concurrency constraints ((i)-(iii)).As mentioned above the setup (1-14) determines the points O, P , Q, R, S, T , U , V , W , X , Y , Z , E , F , and G.We project horizontally the center of Apollonius circle Ξ and we obtain A. We deduce B , C , D, H , I , J , K , L, M , N , and, finally, point by a sequence of simple straightedge operations.Point B is where AQ meets OV .It projects vertically to C on OT .We set D = CG ∩ AQ.The line Z A meets the horizontal axis at H .We set I = Z A ∩ E F , L = BC ∩ SW , and K = BQ ∩ h, where h is the vertical line through H . Point M is where the vertical line through J meets CG.We set N = H A ∩ EY and, finally, = P Y ∩ s, where s is the vertical line through S.

Justification of the construction
On the plane, the abscissa and the ordinate are taken with respect to P .By π, we denote the vertical projection (x, y) → (x, 0).

Justification of the setup
In Figure 4.a-e, we apply concurrency conditions ((i)-(iii)) directly.We provide an example: at  The second point is not drawn in the picture; it leads to a second circle externally tangent to Π, whose center does not project within the edge 1, 4 as required.There are two more solutions to this CLP Apollonius problem; they are given by exchanging "rightmost" and "leftmost" in the above instructions.These solutions should be ignored, because the corresponding circles contain Π.

The key part of the construction
This entire part of the construction (Figures 5, 6, and 7) is about choosing the right slope of the leg of t 1 (grey triangle).
This amounts to identifying a suitable ray r stemming from Q. Notice that the choice of r determines D and I .Point D is defined as the center of mass of the polygon of vertices G, B = r ∩ OV , C = π(B ), and Q; in other words D is the intersection r ∩ Gπ(r ∩ OV ) between the legs of t 1 (grey) and t 6 (pink).Point I is the intersection of the legs of t 4 (green) and t 8 (red); hence I = Z (PU ∩ r ) ∩ E F , see Figure 8.By conditions ((iii); t 1 , t 4 , t 6 ) and ((iii); t 4 , t 4 , t 8 ), D and I lie on the base of t 4 (green), which is perpendicular to the diameter; therefore, points I and D should have the same abscissae.
We simplify the definition of D as follows.Indeed, when the slope of r varies, D moves along a ray stemming from O as Figure 9 shows.In Figure 4.f, we set point 2 = W Q ∩ PG; then, D can be described as D = O2 ∩ r.

The triangle 1, 3, 4
We refer to the triangle 1, 3, 4, which appears in Figure 5.a.The choice of the ray r is equivalent to the choice of the point A on the line PU so that r = Q A. Therefore, the problem can be restated as follows.We should choose A on the vertical axis PU in such a way that Z A and Q A intersect the sides 1, 3 and 4, 3 of the triangle 1, 3, 4 of Figure 5.a at two points (I and D) with the same abscissae x 1 and x 2 .The reader can refer to Figures 10 or 8, which contain the solution.
Let us denote the ordinates of the points 8, 4, 1, 7 by v, l , l , v .Let m and m be the abscissae of Z = 5 and Q = 6 (note that m > m for our initial choices).Let t be the ordinate of A along PU .In this way the abscissae of I = Z A ∩ (13) and D = Q A ∩ (43) can be written 21  This amounts to intersecting the parabola p with roots at v and l and leading term t 2 /m and the parabola p with roots at v and l and leading term t 2 /m .Since m < m , there are two solutions, one before l and one within ]l , l [. , where m is the slope of OG, and we have g = QG and q = OQ.

Two parabolae
The directrix and the focus of p (resp.p ) are ∆ and φ (resp.∆ and φ ).Indeed, since t 2 /m and t 2 /m are the leading terms of the parabolae, the distance d (φ, ∆) and d (φ , ∆ ) equal m/2 and m /2.By the conditions p (0, v) and p (0, v ), each focus determines the corresponding directrix.Indeed, by p (0, v), the line ∆ is the vertical axis through the leftmost point of the circle centred in (0, v) and passing through φ.The same holds for ∆ .The axis of each parabola is determined by the roots.It remains to place φ on the axis so that d (φ, ∆) = m/2.The geometric construction of Figure 10 identifies a unique choice.This happens because the polygon of vertices 8, 16, 18, 19 is a parallelogram and the distance of 18 = φ from the vertical line ∆ through 19 is the same as the distance of 8 from the vertical line through 16.By construction, the latter is m/2.So we have d (φ, ∆) = m/2.The argument for p is identical.
We now show that point 28 in the construction is indeed p ∩ p .A point κ of the parabola p is, by definition, the center of a circle tangent to the line ∆, passing through φ, and having ray ρ = d (κ, φ) = d (κ, ∆).If κ also lies on p , its distance from φ equals the distance from the directrix ∆ , which is a vertical line lying on the left of ∆.Hence d (κ, ∆ ) = (x 19 − x 15 ) + d (κ, ∆) = (x 19 − x 15 ) + ρ.In this way the circle Ξ of centre κ and radius ρ intersects the circle Π of centre φ and radius x 19 − x 15 at a single point.This amounts to saying that Ξ is the Apollonius circle tangent to ∆, externally tangent to Π, and passing through φ.By the external tangency condition, there are exactly two such circles (see [9]).By the argument above we know that only one Apollonius circle is centred at a point whose ordinate fit within ]l 2 , l 1 [.Apollonius CLP problem is solved in Figure 7 by following [9,Section 7.38
et 7 servent à poser et à résoudre le problème d'Apollonius.Dans cette digression, au lieu des lettres, on emploie des nombres de 1 à 28 pour dénoter les points : (n, m) est la droite qui passe par n et m, n, m est le segment de n à m, n, m, . . ., k est le polygone de sommets n, m, . . ., k. Cette digression a le seul but (crucial) de définir la position du point 28 qui permet de reprendre la construction en plaçant le point A. D'Apollonius au śrīyantra On conclut la construction sur la Figure 8 en revenant aux points dénotés par des lettres.On dispose des points O, P , Q, R, S, T , U , V , W , X , Y , Z , E , F , G. On résume brièvement les étapes finales en renvoyant le lecteur à la Section 2.2 pour des instructions détaillées.On projette horizontalement le centre (point numéro 28) du cercle Ξ, solution au problème d'Apollonius.On obtient ainsi le point A sur PU .On déduit B en prolongeant le segment AQ jusqu'à OV .Puis on projette B verticalement sur OT et on obtient C .On pose D = CG ∩ AQ.La droite Z A rencontre l'axe OT au point H . On pose I = Z A ∩ E F .On pose L = BC ∩ SW et K = BQ ∩ h, où h est la droite verticale qui passe par H . On définit M comme le point où la droite verticale passant par J rencontre C D. On pose N = H A ∩ EY et enfin = P Y ∩ s, où s est la droite verticale qui passe par S.

( a )
The vertical lines through P and R meet the halfcircle at U and V .Join V to O and U to T .(b) Join S to W = PU ∩ V O. Set X = SW ∩ RV .

Figure 4 .
Figure 4.The first steps -Part 1 C. R. Mathématique -2021, 359, n 4, 377-397 (c) The ray from P through X intersects U T at Y .The point Y projects vertically to E in OT .(d) The vertical line passing through Q intersects P Y and SW at F and G. (e) The ray originating at E and containing F intersects U P at point 1. (f ) Draw a ray from O to the intersection 2 of the diagonals of PW GQ.It meets the line (E , 1) at 3.

Figure 5 .
Figure 5. Setting up the Apollonius problem

Figure 6 .
Figure 6.The dashed circle in the figure is the Apollonius circle externally tangent to the circle Π, tangent to the line ∆, and passing through the point φ.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 7 .
Figure 7.We solve the Apollonius CLP problem.Point 23 is the horizontal projection of φ on ∆.The circle Θ passes through point φ, point 23, and the leftmost point of Π.We draw a line through φ and the rightmost point of Π, which we label by 21.Such a line meets Θ at point 24 (and, of course, at φ). Point 26 is the intersection between the line through φ and 24 and the line through the two points of Π∩Θ.The circle of diameter 26, φ intersects Π at two points.The first point is 27 in Figure (c) and the desired Apollonius circle is the dashed circle through φ, 24, and 27 appearing in Figure (d).The second point is not drawn in the picture; it leads to a second circle externally tangent to Π, whose center does not project within the edge 1, 4 as required.There are two more solutions to this CLP Apollonius problem; they are given by exchanging "rightmost" and "leftmost" in the above instructions.These solutions should be ignored, because the corresponding circles contain Π.
in terms of tas x = m(t − l )/(t − v) and x = m (t − l )/(t − v ).The position of A is the solution of m t − l t − v = m t − l t − v .

Figure 9 .
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 have g = QG and q = OQ.

Figure 10 .
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 parabolae p and p .Point 28 is the intersection p ∩ p projecting on A ∈ 1, 4.