https://doi.org/10.53656/math2025-1-3-apo

Резюме. The definition and some of the properties of the Apollonian circle in the plane find their analogies in the Euclidean three-dimensional space. Thus, we manage to introduce a new concept in solid geometry that we call an “Apollonian sphere”. It appears that the Apollonian sphere not only possesses classical properties similar to the Apollonian circle such as orthogonality and coaxiality, but also analogies of its lesser-known connection with the Lemoine point and the circumcenter. We also discover two notable properties of stereographic projection that we prove with an Apollonian sphere. They include collinearity of the projection point with the Lemoine points of the projection and the projected triangles or with the centers of their Apollonian circles. Moreover, we connect the newly introduced concepts and the rich configurations they generate with Olympiad geometry.

Ключови думи: Apollonian sphere; stereographic projection; harmonic tetrahedron; Lemoine point; center of Apollonian circle; Olympiad geometry

1. Introduction

The first goal of the paper is to look for and investigate the properties of the 3D space version of the Apollonian circle that we name an “Apollonian sphere”. The second goal of the article is to search for applications of the Apollonian sphere in 3D space, which brings us to the discovery of two notable properties of stereographic projection. The third goal of the paper is to find applications of the Apollonian sphere and the newly discovered properties of stereographic projection in Olympiad geometry.

In Section 2, we define the Apollonian sphere, and we look at its properties. In Section 3, we prove two new properties of stereographic projection via an Apollonian sphere, constituting of collinearity of notable points (Lemoine points/centers of Apollonian circles) from the projection and the projected triangles with the projection point. Finally, in Section 4, we examine the applications of the new concepts and configurations in Olympiad geometry. We formulate an Olympiad problem based on the discovered properties of stereographic projection, and we provide an original solution to another one via an Apollonian sphere \({ }^{1}\).

2. Apollonian Sphere

The familiar properties of the Apollonian circle in the plane such as orthogonality and coaxiality, (Andreescu et al. 2016, pp. 275 – 279), find their analogies in 3D space. We call the resulting figure an “Apollonian sphere”. In stereometry, however, we investigate more in-depth connections with notable points, lines, and planes, which we discuss in Section 2.

Theorem 2.1. Let \(\triangle A B C\) be a scalene triangle. Then the locus of points \(X\), satisfying \(\tfrac{X A}{X B}=\tfrac{C A}{C B}\), is a sphere \(\Sigma_{C}\), which we will call the Apollonian sphere for the triangle corresponding to the vertex \(C\).

Proof. 1) Existence. Let \(C L_{1}\) and \(C L_{2}\) be the interior and the exterior angle bisector of \(\angle A C B\) in \(\triangle A B C\) (see fig. 1). Let \(M\) be the midpoint of \(L_{1} L_{2}\), and let us construct the sphere \(\Sigma_{C}(M, r)\) where

\[ r=M C=M L_{1}=M L_{2} \] Let point \(X \in \Sigma_{C}\), and let us construct the plane passing through point \(X\) and the line \(A B\). We note that

\[ \tfrac{B L_{1}}{L_{1} A}: \tfrac{B L_{2}}{L_{2} A}=\tfrac{B C}{C A}: \tfrac{B C}{C A}=1 \Rightarrow\left(B, L_{1}, A, L_{2}\right)=1 . \]

But \(\angle L_{2} X L_{1}=90^{\circ}\left(L_{1} L_{2}\right.\)– diameter in \(\left.\Sigma_{C}\right)\). From the properties of the harmonic division, (Boev 2010, p. 80), it follows that \(X L_{1}\) is an angle bisector of \(\angle A X B \Rightarrow \tfrac{X A}{X B}=\tfrac{A L_{1}}{L_{1} B}=\tfrac{A C}{B C}\).

2) Uniqueness. Let point \(Y\) satisfy \(\tfrac{Y A}{Y B}=\tfrac{A C}{B C}\). Point \(Y\) and the line \(A B\) form a plane. But \(\tfrac{A C}{B C}=\tfrac{A L_{1}}{B L_{1}} \Rightarrow \tfrac{Y A}{Y B}=\tfrac{A L_{1}}{B L_{1}} \Rightarrow Y L_{1}\) is an angle bisector of \(\angle A Y B\), and \(\left(B, L_{1}, A, L_{2}\right)=1\). From the properties of the harmonic division, (Boev 2010, p. 80), it follows that \(\angle L_{1} Y L_{2}=90^{\circ} \Rightarrow Y \in \Sigma_{C}\left(M, r=M L_{1}\right)\). □

We will prove an analogue in 3D space of the coaxiality of the Apollonian circle.

Theorem 2.2. The three Apollonian spheres for a given scalene \(\triangle A B C\) intersect at a circle.

Proof. Let us construct the spheres \(\Sigma_{C}\) and \(\Sigma_{B}\) (see fig. 2). They are intersecting since the corresponding Apollonian circles are intersecting, and they are great circles in the Apollonian spheres. Let \(\Sigma_{C} \cap \Sigma_{B}=\sigma\), and let point \(X \in \sigma \xrightarrow{2.1 .}\) \[ \tfrac{X A}{X B}=\tfrac{C A}{C B} \quad \cap \quad \tfrac{X C}{X A}=\tfrac{B C}{B A} . \]

If we multiply the two equations, we get that \(\tfrac{X C}{X B}=\tfrac{A C}{A B}\). By Theorem 2.1., it follows that point \(X \in \Sigma_{A}\). The same applies for each point on \(\sigma\).

Note. As the centers of the three Apollonian spheres coincide with the centers of the three Apollonian circles, we know from planimetry that the centers lie on a line (the Lemoine axis).

The next statement is the stereometric equivalent of the orthogonality of the Apollonian circle.

Theorem 2.3. The Apollonian spheres for a given scalene \(\triangle A B C\) are orthogonal to every sphere passing through points \(A, B\), and \(C\).

Proof. Firstly, we will introduce the following fundamental Lemma:

Lemma: Let point \(M\) lie on a circle \(k\) that lies on a sphere \(\Sigma\). Let \(t\) be the tangent line to \(k\) at \(M\), and let \(\pi_{M}\) be the tangent plane to \(\Sigma\) at \(M\). Then the line \(t\) lies on the plane \(\pi_{M}\).

Back to our main problem: let us take an arbitrary sphere \(\Sigma\) passing through points \(A, B\), and \(C\) (see fig. 3). We will prove that \(\Sigma_{C} \perp \Sigma\) (analogously for the other two Apollonian spheres). It is clear that the spheres \(\Sigma_{C}\) and \(\Sigma\) are intersecting. Let \(\Sigma_{C} \cap \Sigma=\omega\), and let us arbitrarily take point \(X \in \omega\). As \(X \in \Sigma_{C}\), by Theorem 2.1., \(\tfrac{X A}{X B}=\tfrac{C A}{C B}(*)\).

Let us define the points as in Theorem 2.1. Point \(X\) and the line \(A B\) determine a plane ⇒

\[ \tfrac{X A}{X B} \stackrel{(*)}{=} \tfrac{C A}{C B}=\tfrac{A L_{1}}{L_{1} B} \Rightarrow X L_{1} \text { is an angle bisector of } \angle A X B . \]

Let us denote \(\angle A B X\) by \(\varphi\), and \(\angle A X B\) by \(2 \psi\). It is clear that we have the equalities \(\angle A X L_{1}=\angle B X L_{1}=\psi\). Therefore,

\[ \angle M X L_{1}=\angle M X A+\angle A X L_{1}=\angle M X A+\psi . \] But as an exterior angle for \(\triangle X B L_{1}\),

\[ \angle M L_{1} X=\angle L_{1} X B+\angle L_{1} B X=\psi+\varphi . \]

It remains to be considered that since \(M X=M L_{1}\), then we have the equality \(\angle M X L_{1}=\angle M L_{1} X \Rightarrow\)

\[ \angle M X A+\psi=\psi+\varphi \Leftrightarrow \angle M X A=\varphi . \]

Therefore, \(M X\) is the tangent line for the circle \(k[A B X]\), and \(k[A B X] \in \Sigma\). Applying the Lemma, it follows that the tangent plane \(\pi_{X}\) to \(\Sigma\) at point \(X\) passes through the center of \(\Sigma_{C}\), point \(M\). Similarly, the result applies for every point \(Y \in \omega\) : the tangent plane \(\pi_{Y}\) to \(\Sigma\) at point \(Y\) passes through the center of \(\Sigma_{C}\), point \(M \Rightarrow \Sigma_{C} \perp \Sigma\).

The next theorem is a 3D equivalent of the connection of the Apollonian circle with the notable points of the triangle. In this stereometric configuration, we prove the existence of new notable elements, and we examine their

Theorem 2.4. Let \(\triangle A B C\) be a scalene triangle with circumcenter \(O\), Lemoine point \(L\), and Apollonian spheres \(\Sigma_{A}(M), \Sigma_{B}(N)\), and \(\Sigma_{C}(P)\). Let the sphere \(\Sigma\) pass through points \(A, B\), and \(C\). Let \(\Sigma\) intersect \(\Sigma_{A}, \Sigma_{B}\), ΣB, and \(\Sigma_{C}\) at the circles \(k^{\prime}, k^{\prime \prime}\), and \(k^{\prime \prime \prime}\), lying on the planes \(\alpha, \beta\), and \(\gamma\), respectively. Let us denote the line through the points \(M, N\), and \(P\) by \(s\). Then the following facts apply:

a) \(\alpha \cap \beta \cap \gamma=\) line \(q\); b) \(L \in q\); c) \(s \perp q\);

d) The orthogonal projection of line \(q\) onto the plane through \(\triangle A B C\) passes through \(O\).

Proof.

a) The orthogonality in Theorem 2.3. leads to the polar reciprocation,

(Prasolov 2010b, pp. 51-53), between the centers of the Apollonian spheres and their radical planes with \(\Sigma\) with respect to \(\Sigma\) (see fig. 4). Therefore, \(\pi_{\Sigma}(M)=\alpha, \pi_{\Sigma}(N)=\beta\), and \(\pi_{\Sigma}(P)=\gamma\). But points \(M, N\), and \(P\) are collinear. Thus, \(\alpha \cap \beta \cap \gamma=\) line \(q\).

Corollary. The lines \(s\) and \(q\) are polar (Smith 1893, p. 40).

b) Without loss of generality, let us take the circles \(k^{\prime}\) and \(k^{\prime \prime}\) (fig. 5).

Circles \(k^{\prime}\) and \(k\) (the circumcircle of \(\triangle A B C\) ) lie on a sphere (\(\Sigma\) ) and have a common point (\(A\) ), and as it is clear that they do not coincide, then they have a second common point, point \(A_{1}\). But \(A_{1} \in k^{\prime} \in \Sigma_{A}\), and \(A_{1}\) lies on the plane through \(\triangle A B C\). Therefore, \(A_{1}\) lies on the Apollonian circle corresponding to the vertex \(A\). Therefore, from planimetry, (Andreescu et al. 2016, p. 277), we know that \(A A_{1}\) is a symmedian in \(\triangle A B C\).

Circles \(k^{\prime \prime}\) and \(k\) lie on a sphere (\(\Sigma\) ) and have a common point (\(B\) ), and as it is clear that they do not coincide, then they have a second common point, point \(B_{1}\). B But \(B_{1} \in k^{\prime \prime} \in \Sigma_{B}\), and \(B_{1}\) lies on the plane through \(\triangle A B C\). Therefore, \(B_{1}\) lies on the Apollonian circle corresponding to the vertex \(B\). Therefore, from planimetry, we know that \(B B_{1}\) is a symmedian in \(\triangle A B C\).

Therefore, \(A A_{1} \cap B B_{1}=L\).

It remains to be considered that as \(L \in \alpha\), and \(L \in \beta\), then \(L \in q\) (the intersection of the two planes).

Note. Another approach to proving the geometric fact is via properties of polar reciprocation. From planimetry, (Andreescu et al. 2016, p. 279), we know that \(L\) is the pole of the line \(s\), and from a), we know that the lines \(s\) and \(q\) are polar, meaning that \(L \in q\).

c) Let \(k^{\prime} \cap k^{\prime \prime}=\{K, T\}\) (fig. 6). Then we know that \(K \in\) \((\alpha \cap \beta)\), and \(T \in(\alpha \cap \beta)\), from where it follows that \(K T \equiv q\). We note that \(q \in \gamma\), that is, \(K T \in \gamma\). But \(\{K, T\} \in\) \(\Sigma \Rightarrow\{K, T\} \in(\gamma \cap \Sigma) \equiv k^{\prime \prime \prime}\). Also \(\{K, T\} \in k^{\prime}, k^{\prime \prime}, k^{\prime \prime \prime} \in\) \(\Sigma_{A}, \Sigma_{B}, \Sigma_{C} \Rightarrow\{K, T\} \in \Sigma_{A} \cap\) \(\Sigma_{B} \cap \Sigma_{C}\), which by Theorem 2.2., is a circle, which we will denote by \(\Omega\), and the plane passing through it - by \(\mu\). Therefore, \(K T \in \mu . K T \equiv q \Rightarrow q \in \mu\). But \(\mu\) is the radical plane of the three Apollonian spheres. Therefore, the line through their centers \((s)\) is) is perpendicular to \(\mu\). Thus, it is perpendicular to every line in it, and \(q\) is a line in it. Therefore, \(s \perp q\).

Note. The statement could also be proved via the property of the polar reciprocation in 3D space that two polar lines with respect to a given sphere are perpendicular, as we know from a) that the lines \(s\) and \(q\) are polar.

d) Let us denote the orthogonal projection of the line \(q\) onto the plane passing through \(\triangle A B C, \delta_{A B C}\), δABC, by \(t\) (fig. 7). From planimetry, we know that the Lemoine axis is perpendicular to the Brocard axis: \(s \perp O L\). We notice that \(s \perp \mu\), and \(s \in \delta_{A B C}\), from where it follows that \(\delta_{A B C} \perp \mu\). From the orthogonality of the two planes, we deduce that \(t \in \mu\) since \(t\) is their intersection. From \(s \perp \mu\), it follows that \(s \perp t\). As \(L \in \delta_{A B C}\) and \(L \in q\), it is clear that \(L \in t\). But there is a single line in \(\delta_{A B C}\), perpendicular to \(s\) and passing through \(L\), and \(O L\) is such. Therefore, \(O L \equiv t \Leftrightarrow O \in t\).

Note. The statement that \(s \perp t\) could also be derived by the threeperpendiculars theorem since \(s \perp q\).

3. Two Notable Properties of Stereographic Projection

Alongside the analogies from the plane in 3D space, the Apollonian sphere finds additional applications in proving new notable properties of stereographic projection. Stereographic projection is a projection of a sphere upon a plane such that the center of the projection lies on the sphere, and the plane is perpendicular to the diameter of the sphere through the given point. For stereographic projection, it is well known that the center of projection, the centroid of the projection triangle, and the centroid of the projected triangle are not collinear, as well as for their orthocenters and circumcenters, but it appears that the centers of the Apollonian circles and the Lemoine point fulfill this property.

Theorem 3.1. Let under stereographic projection with a projection point \(O\), a reference sphere \(\Sigma\), and a projection plane \(\mu, \triangle A B C\) map to \(\triangle A^{\prime} B^{\prime} C^{\prime}\). Then point \(O\), the center of an Apollonian circle for \(\triangle A B C\), and the center of the corresponding Apollonian circle for \(\triangle A^{\prime} B^{\prime} C^{\prime}\) are collinear.

Proof. Firstly, we will introduce the following Lemma (Prasolov 2010, p. 313):

Lemma: Let under inversion with a point of inversion \(O\), the circle \(k\) map to the circle \(k^{\prime}\). Let the sphere \(\Sigma\) pass through point \(O\) and the circle \(k\). Then point \(O\), the pole of the plane passing through the circle \(k\) with respect to the sphere \(\Sigma\), and the center of \(k^{\prime}\) are collinear.

Back to our main problem: as by condition, \(\Sigma\) is a reference sphere, and \(\mu\) – a projection plane, then, by definition of stereographic projection, it follows that the plane \(\mu\) is perpendicular to the diameter in \(\Sigma\) through \(O\). Therefore, there exists a single sphere \(i(O, r)\) with a center point \(O\) such that the plane \(\mu\) is the radical plane of the spheres \(\Sigma\) and \(i\). Therefore, under inversion \(p\) with an inversion sphere \(i(O, r), \Sigma \xrightarrow{\mathrm{p}} \mu\) (and so \(A \xrightarrow{\mathrm{p}} A^{\prime}, B \xrightarrow{\mathrm{p}} B^{\prime}\), and \(C \xrightarrow{\mathrm{p}} C^{\prime}\) ).

Let point \(M\) be the center of an Apollonian circle for \(\triangle A B C\), and let \(O M \cap \mu=S\) (see Fig. 8). We will prove that point \(S\) is a center of an Apollonian circle for \(\triangle A^{\prime} B^{\prime} C^{\prime}\).

Let us denote an Apollonian circle for \(\triangle A B C\) by \(k_{C}\), and the corresponding Apollonian circle for \(\triangle A^{\prime} B^{\prime} C^{\prime}\) by \(\Gamma\). Let us also construct the Apollonian sphere \(\Sigma_{C}\) for \(\triangle A B C\). As we have taken \(k_{C}\) with a ratio \(\tfrac{C A}{C B}\), then, \(\Sigma_{C}\) and \(\Gamma\) will refer to the ratios \(\tfrac{C A}{C B}\) and \(\tfrac{C^{\prime} A^{\prime}}{C^{\prime} B^{\prime}}\), respectively. We note that point \(M\) is also the center of \(\Sigma_{C}\) (the Apollonian circles are great circles in the Apollonian spheres). Let \(\Sigma \cap \Sigma_{C}=k_{0}\).

Firstly, we will prove that \(k_{0} \xrightarrow{\mathrm{p}} \Gamma(*)\).

Let us take an arbitrary point \(X\) from \(k_{0}\), and let \(X \xrightarrow{\mathrm{p}} X^{\prime}\) (see fig. 9). It suffices to show that \(X^{\prime} \in \Gamma\).

\(\begin{aligned} & X \in k_{0} \in \Sigma \Rightarrow X^{\prime} \in \mu \equiv \Sigma^{\prime} \Rightarrow O X \cap \mu=X^{\prime} \\ & X \in k_{0} \in \Sigma_{C} \xrightarrow{2.1 .} \\ & \tfrac{X A}{X B}=\tfrac{C A}{C B}~~~~ {(1)} \end{aligned}\)

Without loss of generality, let \(r=1\).

The lines \(A^{\prime} O\) and \(X^{\prime} O\) form a plane. We have that \(O A \cdot O A^{\prime}=r^{2}=1=\)

\(O X \cdot O X^{\prime} \Rightarrow \tfrac{O A}{O X}=\tfrac{O X^{\prime}}{O A^{\prime}}\) and \(\angle A O X=\angle X^{\prime} O A^{\prime} \Rightarrow \triangle O A X \sim \triangle O X^{\prime} A^{\prime} \Rightarrow\)

\(\tfrac{A X}{X^{\prime} A^{\prime}}=\tfrac{O A}{O X^{\prime}} \Leftrightarrow X^{\prime} A^{\prime}=\tfrac{A X \cdot O X^{\prime}}{O A} ~~~~ {(2)}\)

The lines \(B^{\prime} O\) and \(X^{\prime} O\) form a plane. We have that \(O B \cdot O B^{\prime}=r^{2}=1=\)

\(O X \cdot O X^{\prime} \Rightarrow \tfrac{O B}{O X}=\tfrac{O X^{\prime}}{O B^{\prime}}\) and \(\angle B O X=\angle X^{\prime} O B^{\prime} \Rightarrow \triangle O B X \sim \triangle O X^{\prime} B^{\prime} \Rightarrow\)

\(\tfrac{B X}{X^{\prime} B^{\prime}}=\tfrac{O B}{O X^{\prime}} \Leftrightarrow X^{\prime} B^{\prime}=\tfrac{B X \cdot O X^{\prime}}{O B}~~~~ {(3)}\) By (2) and (3) ⇒

\[ \tfrac{X^{\prime} A^{\prime}}{X^{\prime} B^{\prime}}=\tfrac{A X}{B X} \cdot \tfrac{O B}{O A} \stackrel{(1)}{=} \tfrac{C A}{C B} \cdot \tfrac{O B}{O A} . \] But point \(C^{\prime} \in \Gamma\), and by applying the metric property of inversion, (Johnson 1960, p. 48), it follows that:

\[ \begin{aligned} \tfrac{C^{\prime} A^{\prime}}{C^{\prime} B^{\prime}} & =\tfrac{\tfrac{C A}{O C \cdot O A}}{\tfrac{C B}{O C \cdot O B}}=\tfrac{C A}{C B} \cdot \tfrac{O B}{O A} \\ & \Rightarrow \tfrac{X^{\prime} A^{\prime}}{X^{\prime} B^{\prime}}=\tfrac{C^{\prime} A^{\prime}}{C^{\prime} B^{\prime}} \end{aligned} \] It remains to be considered that \(X^{\prime} \in \mu\). Therefore, from planimetry, we know that point \(X^{\prime}\) lies on the Apollonian circle in the plane \(\mu\) with a ratio \(\tfrac{C^{\prime} A^{\prime}}{C^{\prime} B^{\prime}} \equiv \Gamma . \Rightarrow X^{\prime} \in \Gamma\) that we wanted to prove.

Secondly, we will prove that point \(M\) is the pole of the plane passing through the circle \(k_{0}\) with respect to \(\Sigma\). (∗∗)

By Theorem 2.3., we know that \(\Sigma_{C} \perp \Sigma\), and by construction, \(k_{0}=\Sigma_{C} \cap \Sigma\). This, combined with the fact that point \(M\) is the center of \(\Sigma_{C}\), proves the statement.

By \((*)\) and \((* *) \xrightarrow{\text { Lemma }}\) The points \(O, M\), and the center of \(\Gamma\) are collinear. Therefore, the line \(O M\) intersects the plane through the circle \(\Gamma\) at its center. But \(\Gamma\) and \(\triangle A^{\prime} B^{\prime} C^{\prime}\) lie on a plane (\(\mu\) ). Consequently, \(O M\) intersects \(\mu\) at the center of the Apollonian circle for \(\triangle A^{\prime} B^{\prime} C^{\prime}\), and by construction, \(O M \cap \mu=S\) ⇒ point \(S\) is the center of \(\Gamma\) that we wanted to show.

Note. Point \(M\) does not project onto point \(S\).

Let us prove the following notable fact as well, which resembles and is partly based on the previous one.

Theorem 3.2. Let under stereographic projection with a projection point \(O\), a reference sphere \(\Sigma\), and a projection plane \(\mu, \triangle A B C\) map to \(\triangle A^{\prime} B^{\prime} C^{\prime}\). Then point \(O\), the Lemoine point for \(\triangle A B C\), and the Lemoine point for \(\triangle A^{\prime} B^{\prime} C^{\prime}\) are collinear.

Proof. Firstly, we will introduce and prove the following Lemma.

Lemma: Let \(p\) be an inversion with an inversion circle \(i(O, R)\), and let \(k\) be a circle not passing through \(O\), on which lie the points \(A, B, C, D, E\), and \(F\) such that \(A D \cap B E \cap C F=Q\). Let under \(p\), the points \(A, B, C, D, E\), and \(F\) map to the points \(A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}, E^{\prime}\), and \(F^{\prime}\). Then the lines \(A^{\prime} D^{\prime}, B^{\prime} E^{\prime}\), \(C^{\prime} F^{\prime}\), and \(O Q\) concur at one point.

Proof. Let us denote the inverse point of \(Q\) by \(Q^{\prime}\) (see fig. 10).

By condition, the circle \(k\) does not pass through the center of inversion, point \(O\). Therefore, it will be mapped again to a circle. And so the points lying on \(k\) will be sent to points lying on \(k^{\prime}\)– the image of \(k\). Thus, the points \(A^{\prime}, B^{\prime}, C^{\prime}, D^{\prime}, E^{\prime}\), and \(F^{\prime}\) lie on a circle, \(k^{\prime}\).

We know that under inversion \(p\), the line \(A D\) will be sent to a circle, passing through \(O\). Therefore, the points \(A^{\prime}, Q^{\prime}, D^{\prime}\), and \(O\) lie on a circle, which we will denote by \(k_{1}\). Analogously, the points \(B^{\prime}, Q^{\prime}, E^{\prime}\), and \(O\) lie on the circle \(k_{2}\), and the points \(C^{\prime}, Q^{\prime}, F^{\prime}\), and \(O\) lie on the circle \(k_{3}\).

Let us consider the circles \(k^{\prime}, k_{1}\), and \(k_{2}\). As the three are intersecting at three pairs of points, then their common chords are intersecting at one point (Prasolov 2010a, p. 269). Therefore, \(Q^{\prime} O \cap B^{\prime} E^{\prime} \cap A^{\prime} D^{\prime}=S\).

Analogously, we consider the circles \(k^{\prime}, k_{2}\), and \(k_{3}\), which also intersect at three pairs of points, meaning that their common chords are intersecting at one point. Thus, \(Q^{\prime} O \cap C^{\prime} F^{\prime} \cap B^{\prime} E^{\prime}=S^{\prime}\).

But \(S^{\prime}=Q^{\prime} O \cap B^{\prime} E^{\prime}=S \Rightarrow S^{\prime} \equiv S \Rightarrow\) As \(O Q \equiv O Q^{\prime}, O Q \cap A^{\prime} D^{\prime} \cap\) \(B^{\prime} E^{\prime} \cap C^{\prime} F^{\prime}=S\), which we wanted to show.

Back to our main proof: as by condition, \(\Sigma\) is a reference sphere, and \(\mu\)a projection plane, then, by definition of stereographic projection, it follows that the plane \(\mu\) is perpendicular to the diameter in \(\Sigma\) through \(O\). Therefore, there exists a single sphere \(i(O, r)\) with a center point \(O\) such that the plane \(\mu\) is the radical plane of the spheres \(\Sigma\) and \(i\). Therefore, under inversion \(p\) with an inversion sphere \(i(O, r), \Sigma \xrightarrow{\mathrm{p}} \mu\) (and so \(A \xrightarrow{\mathrm{p}} A^{\prime}, B \xrightarrow{\mathrm{p}} B^{\prime}\), and \(C \xrightarrow{\mathrm{p}} C^{\prime}\) ).

Let point \(L\) be the Lemoine point for \(\triangle A B C\), and let \(O L \cap \mu=T\) (see fig. 11). We will prove that point \(T\) is the Lemoine point for \(\triangle A^{\prime} B^{\prime} C^{\prime}\).

Let us define \(k_{C}, \Gamma, \Sigma_{C}\), Γ, ΣC, and \(k_{0}\) as in Theorem 3.1. Let us denote the circumcircles of \(\triangle A B C\) and \(\triangle A^{\prime} B^{\prime} C^{\prime}\) by \(k_{A B C}\) and \(k_{A^{\prime} B^{\prime} C^{\prime}}\), respectively. Since a circle not passing through the center of inversion is sent to a circle, and the p points \(A, B\), and \(C\) are sent to the points \(A^{\prime}, B^{\prime}\), and \(C^{\prime}\), then, \(k_{A B C} \xrightarrow{\mathrm{p}} k_{A^{\prime} B^{\prime} C^{\prime}}\).

Let \(k_{A B C} \cap k_{C}=\{C, X\}\). P oint \(X \in k_{C} \Rightarrow X \in \Sigma_{C}\). But point \(X \in k_{A B C} \Rightarrow X \in \Sigma\). Therefore, point \(X \in\left(\Sigma_{C} \cap \Sigma\right)=k_{0}\). Hence, as \(k_{A B C} \xrightarrow{\mathrm{p}} k_{A^{\prime} B^{\prime} C^{\prime}}\), and \(X \in k_{A B C}\), then \(X^{\prime}\) (the inverse point of \(X\) ) \(\in k_{A^{\prime} B^{\prime} C^{\prime}}\).

However, since by Theorem 3.1., \(k_{0} \xrightarrow{\mathrm{p}} \Gamma\), and \(X \in k_{0}\), then \(X^{\prime} \in \Gamma\). Consequently, \(X^{\prime} \in\left(k_{A^{\prime} B^{\prime} C^{\prime}} \cap \Gamma\right)\), and so \(k_{A^{\prime} B^{\prime} C^{\prime}} \cap \Gamma=\left\{C^{\prime}, X^{\prime}\right\}\).

It remains to be considered that from planimetry, we know that \(C X\) and \(C^{\prime} X^{\prime}\) are symmedians in \(\triangle A B C\) and \(\triangle A^{\prime} B^{\prime} C^{\prime}\), respectively. Analogously, we define the points \(Y\) and \(Z\), from which it follows that \(B Y\) and \(A Z\) are symmedians in \(\triangle A B C\), and \(B^{\prime} Y^{\prime}\) and \(A^{\prime} Z^{\prime}\)– in \(\triangle A^{\prime} B^{\prime} C^{\prime}\). Furthermore, this means that \(C X \cap B Y \cap A Z=L\), and \(C^{\prime} X^{\prime} \cap B^{\prime} Y^{\prime} \cap A^{\prime} Z^{\prime}=T^{\prime}\), which are Lemoine points for \(\triangle A B C\) и \(\triangle A^{\prime} B^{\prime} C^{\prime}\), respectively, and we also know that the points \(A, B, C, X, Y, Z\) lying on a circle are mapped to the points \(A^{\prime}, B^{\prime}, C^{\prime}, X^{\prime}, Y^{\prime}, Z^{\prime}\), and the lines formed by the opposite points concur at one point (\(L\) and \(T^{\prime}\), respectively) \(\xrightarrow{\text { Lemma }} T^{\prime} \in O L\). But \(T^{\prime} \in \mu\), and the line \(O L\) intersects the plane \(\mu\) at a single point, which by construction is point \(T\).

Therefore, T ′ ≡ T , which we wanted to prove.

Note. Point \(L\) does not project onto point \(T\).

4. Applications in Olympiad Geometry

The Apollonian sphere and its key results seen in Section 3 connect with Olympiad geometry. Firstly, we discover a configuration of Olympiad-level complexity by finding the planimetric equivalent of the aforementioned notable facts around stereographic projection. Secondly, we provide an original solution to another Olympiad problem using the Apollonian sphere.

We could reformulate the statements in Theorems 3.1. and 3.2. by the following way: “Let us take in 3D space the intersecting circles \(k_{1}, k_{2}\), and \(k_{3}\) such that \(k_{1} \cap k_{2}=\{A, D\}, k_{2} \cap k_{3}=\{B, E\}\), and \(k_{1} \cap k_{3}=\{C, F\}\). Let \(A D \cap B E \cap C F=O\). Then point \(O\), the center of an Apollonian circle/Lemoine point for \(\triangle A B C\), and the center of the corresponding Apollonian circle/Lemoine point for \(\triangle D E F\) are collinear.” Point \(O\) lies on the radical line of the three circles, and so it will have the same power with respect to them, \(S\). Therefore, we can take an inversion with an inversion sphere with a center point \(O\) and a radius \(\sqrt{S}\), at which \(\triangle A B C\) will map to \(\triangle D E F\). It is clear that there exists a stereographic projection with a projection point \(O\), a reference sphere passing through the points \(A, B, C\), C, and \(O\), and a projection plane passing through the points \(D, E\), and \(F\). As a result, the restatement is valid.

As a consequence, for the particular case when the three circles lie on a plane, we will get the following problem:

Problem 4.1. Let us take in the plane the intersecting circles \(k_{1}, k_{2}\), and \(k_{3}\) such that \(k_{1} \cap k_{2}=\{A, D\}, k_{2} \cap k_{3}=\{B, E\}\), and \(k_{1} \cap k_{3}=\{C, F\}\). Let \(A D \cap B E \cap C F=O\). Prove that point \(O\), the Lemoine point for \(\triangle A B C\), and the Lemoine point for \(\triangle D E F\) are collinear (fig. 12).

Using Apollonian sphere, we give an authentic solution to the following Olympiad problem.

Problem 4.2. (30-th Bulgarian Mathematical Olympiad 1981, Fourth Round, Second Day, Sixth Problem) Planes \(\alpha, \beta, \gamma, \delta\) are tangent to the circumsphere of a tetrahedron \(A B C D\) at points \(A, B, C, D\), respectively. Line \(p\) is the intersection of \(\alpha\) and \(\beta\), and line \(q\) is the intersection of \(\gamma\) and \(\delta\). Prove that if lines \(p\) and \(C D\) meet, then lines \(q\) and \(A B\) lie on a plane.

Proof. Let \(\Sigma\) denote the circumsphere of \(A B C D\). Let us take \(\triangle B C D\) with interior and exterior angle bisectors of \(\angle C B D B L_{1}\) and \(B L_{2}\), respectively (see fig. 13). Let point \(M\) be the midpoint of \(L_{1} L_{2}\). It is clear that we have \(\Sigma_{B}\left(M, r=M B=M L_{1}=M L_{2}\right)\). Since by condition, the points \(B, C, D \in \Sigma\) and by construction, \(\Sigma_{B}\) is an Apollonian sphere for \(\triangle B C D\), then by Theorem 2.3., it follows that \(\Sigma_{B} \perp \Sigma\). Therefore, \(M \in \beta\). B But point \(M \in C D\), and \(C D \cap \beta\) at a single point ⇒ as by condition, \(C D \cap \beta \cap \alpha\), then \(M \in \alpha \Rightarrow A \in \Sigma_{B} \xrightarrow{\text { 2.1. }} \tfrac{A C}{A D}=\tfrac{B C}{B D} \Leftrightarrow \tfrac{A C}{B C}=\tfrac{A D}{B D}=\zeta\).

If \(\zeta=1, \gamma \| A B\), and \(\delta \| A B\), meaning that \(q \| A B\), and so \(q\) and \(A B\) lie on a plane.

If \(\zeta \neq 1\), by Theorem 2.1., point \(C\) lies on the Apollonian sphere for \(\triangle B A D\) with a ratio \(\tfrac{D A}{D B}, \Sigma_{D}\). Let us take \(\triangle B A D\) with interior and exterior angle bisectors of \(\angle B D A D J_{1}\) and \(D J_{2}\), respectively. Let point \(N\) be the midpoint of \(J_{1} J_{2}\). It is clear that \(\Sigma_{D}\left(N, r=N D=N J_{1}=N J_{2}\right)\). By Theorem 2.3., \(\Sigma_{D} \perp \Sigma \Rightarrow N \in \delta\), and \(N \in \gamma \Rightarrow N \in(\delta \cap \gamma) \equiv q \Rightarrow N \in q\). But by construction, \(N \in A B \Rightarrow A B \cap q=N \Rightarrow A B\) and \(q\) lie on a plane.

Note: This tetrahedron is the 3D equivalent of the harmonic quadrilateral, (Boev 2010, pp. 76 – 87), due to its distinctive connection between tangency and metric relation, and so we can call it a harmonic tetrahedron.

5. Conclusion

As we see, the Apollonian sphere is the equivalent of the Apollonian circle in 3D space due to the similarity of their definitions and some of their properties, namely orthogonality and coaxiality. Additionally, we find a stereometric parallel of the connection between the Apollonian circle and the notable points of the triangle. We examine the intersecting planes of the Apollonian spheres with a sphere through the given triangle, which helps us establish the connection between the Apollonian sphere, the Lemoine point, and the circumcenter of the triangle. The planes intersect at a line, \(q\), that we prove is polar and thus, perpendicular to the central axis of the spheres, and its orthogonal projection coincides with the Brocard axis.

The Apollonian sphere holds significant applicability in solid geometry. Using it, we prove the two newly discovered notable properties of stereographic projection illustrated in Section 3. We prove collinearity between the two centers of corresponding Apollonian circles or two Lemoine points with the projection center. We note that under stereographic projection the two Lemoine points and the two centers of corresponding Apollonian circles are not image and preimage. That is why we call their properties notable. In the proof of the given stereometric construction, we use the Apollonian sphere both to prove that its common circle with the reference sphere projects onto the second Apollonian circle (in Theorem 3.1 and 3.2) and to prove that the center of the first Apollonian circle is the pole of the plane through the common circle between the Apollonian sphere and the reference sphere with respect to the reference sphere (in Theorem 3.1).

We establish the connection of these new geometric constructions with Olympiad geometry, as seen in Section 4. First, we reformulate the two theorems from Section 3 such that we include the particular case when the configuration is planimetric, which we separate as an Olympiad problem. Second, we apply the Apollonian sphere to another Olympiad problem, which gives us an authentic proof of it.

Consequently, in this paper, we managed to: define the concept Apollonian sphere (Theorem 2.1), discover and prove its properties (Theorems 2.2, 2.3, and 2.4), find the geometric configurations around the two notable properties of stereographic projection, (Theorems 3.1 and 3.2), prove them via an Apollonian sphere, formulate an Olympiad problem derived from those properties (Problem 4.1), and an original solution to another Olympiad problem via an Apollonian sphere (Problem 4.2).

In conclusion, we define a new concept, and after revealing its properties, we show its significance in geometry through the revelation of new configurations. We also formulate an Olympiad problem and solve another one using the newly discovered concepts. The aforementioned configurations would also be crucial both for the development of Olympiad geometry and for stimulating others to develop further on the topic.

Acknowledgments

I want to express my gratitude to Ch. Assist. Prof. Stoyan Boev (American University in Bulgaria) for his help in navigating the development of the various stages in the paper and for his advice on improving and refining it. I want to thank him for the given knowledge in the field of synthetic geometry without which the development of such a project would be impossible.

I want to thank my long-time mathematics teacher, Ms. Rumyana Karadjova, who has always ignited the flame for knowledge in mathematics. She is the person who has never stopped stimulating my interest in mathematics all these years. I thank her for her dedication, patience, knowledge, and expertise in this field. I further thank her for her contributions to this paper in terms of providing me with useful materials and literature, connected to inversion and polar reciprocation in solid geometry.

I thank my mathematics teacher, Dr. Katerina Velcheva, who encourages me to deepen my knowledge in the field. I am grateful for her support and for what I have learned from her.

I would also like to thank Mr. Stanislav Chobanov for his assistance in developing the paper through his valuable advice, comments, and recommendations.

I also thank HSSIMI and SchI-BAS for their support and encouragement to engage in research.

NOTES

1. All statements in this paper refer to the Euclidean three-dimensional space unless otherwise stated.

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