arXiv:2012.05485v1 [math.MG] 10 Dec 2020 mlao 1,wihwt oecag nntto a esae sfollo as stated be can 1.1. notation in Theorem an change featured some Geometry with Classical which of [1], Journal Emelyanov the of volume 2012 The Introduction 1 Points onso xice ihtesides the with excircles of points epciey hnterdclai fcircumcircles of axis radical the Then respectively. A h ae point Nagel the 2 r age ihVrie qiitn oaPdlTriangle Pedal a to Equidistant Vertices with Triangles nthe on n oaiywt epc ocircumconics. The to respect with re circumcircles. projective polarity powerful defined and a strangely uses which two of one of generalizations, axis radical the A nti ae,w rsn ytei ouint geometric a to solution synthetic a present we paper, this In 1 B , BC 1 C , , B N 1 Let 1 i nterays the on lie etodadicne of incenter and centroid , and N A eteNglpito rageAC Let ABC. triangle a of point Nagel the be B 1 2 A nthe on 0 = uigLag vnZelich Ivan Liang, Xuming A A C A AB CA, BC, 2 CA A 0 ,B C, iue1 hoe 1.1 Theorem 1: Figure 0 = , C 0 B 1 ,C A, 1 Abstract and B ABC 0 0 = B 1 epciey e scnie i points: six consider us Let respectively. C B 2 and . 2 nthe on B ⊙ 0 ( A A = 2 1 C , B C AB utrltn sgnlconjugation isogonal relating sult 1 1 2 C C B , 1 uhthat such , 0 ) 2 = and i nterays the on lie ouinecpuae two encapsulates solution C pnpolminvolving problem open A 2 pnpolmpsdb Lev by posed problem open ⊙ C 0 0 B , ( . A 2 0 B ws and 2 C 2 ) C A assthrough passes 0 0 ,C B, etetouch the be 0 ,B A, A 1 and 0 C In this paper, we will present two generalizations to this proposition. It turns out if A0B0C0 is the pedal triangle of an arbitrary point P on the plane of ABC, then the radical axis in ques- tion would always pass through a fixed point (Theorem 2.1). Furthermore, if P lies on the line connecting the incenter and circumcenter of ABC, then the radical axis becomes fixed(Theorem 3.1). Note that A0B0C0 in the problem statement above is actually the pedal triangle of the Bevan point of ABC, which lies on the incenter-circumcenter line, so the generalization par- tially explains why the radical axis does not change. It will take additional insights from a powerful projective result (Proposition 3.3) to reveal that the radical axis coincides with the Nagel line—the line that contains the Nagel point, centroid, and incenter.

2 General Case: P is arbitrary

The purpose of this section is to prove one the main results captured by the paper, stated below.

Theorem 2.1. Let P be an arbitrary point on the plane of triangle ABC. Let AP BP CP be the pedal triangle of P on ABC. Define A1, A2, B1, B2,C1,C2 such that for some x> 0

x = AP A1 = AP A2 = BP B1 = BP B2 = CP C1 = CP C2, with A1, B1,C1 lying on rays AP C, BP A, CP B, and A2, B2,C2 lying on rays AP B, BP C,CP A. As x varies, the radical axis of ⊙A1B1C1 and ⊙A2B2C2 passes through a fixed point.

The proof will reveal the geometric nature of the triangles A1B1C1 and A2B2C2 that makes this theorem true. The approach taken here involves identifying the fixed point and proving that the it exerts equal power with respect to the triangles’ circumcircles. This was done by applying a metric lemma that relates circumradius of a reflection triangle and distances to an orthocenter. The metric conditions in the theorem translates straight-forwardly to the application of the lemma with minimal algebraic manipulations and calculations, granting the theorem a truely synthetic proof.

2.1 Preliminary Definitions and Propositions Definition 2.1. Two triangles ABC and DEF are said to be orthologic if the perpendicu- lars from A,B,C to sides EF,DF,DE are concurrent. It can be shown that this relation is symmetric, namely the perpendiculars from D,E,F to sides BC, AC, AB are also concurrent. The first point of concurrency is called the orthologic center of ABC w.r.t DEF and is denoted OABC (DEF ).

Definition 2.2. Given a triangle ABC and a point P on its circumcircle. The reflection of P over the sides of ABC lie on a line called Steiner’s line of P w.r.t ABC. In other words, the Steiner’s line is the Simson’s line of P dilated by a factor of 2. It is well known that the Steiner’s line contains the orthocenter of ABC, and is perpendicular to isogonals of P w.r.t ∠A, ∠B, ∠C.

We now introduce an important property on a local configuration in the problem.

2 Proposition 2.1. Given a triangle ABC, suppose B′,C′ are two points on AB, AC respectively such that BB′ = CC′ and they lie on opposite sides of BC. If D is the intersection of the angle bisector of ∠A with ⊙(ABC), prove that DB′ = DC′ Proof. Since AD bisects ∠BAC, we have DB = DC and by construction of B′,C′, ∠B′BD = ∠C′CD. Therefore, BB′ = CC′ implies that △DBB′ =∼ △DCC′ and thus DB′ = DC′. The following will be the central property used to compute the power of the fixed point with respect to the circumcircles of A1B1C1 and A2B2C2. Its proof is slightly complicated and will be omitted here. Proposition 2.2. [2] Let H,O be the orthocenter and circumcenter of △ABC. Suppose P, Q are isogonal conjugates w.r.t △ABC. Prove that 2 2 2 2 2 RPQ = HQ + HP + |R − OH | where R, RPQ are the circumradii of △ABC and the reflection triangle of P (or Q) w.r.t △ABC respectively. Note that |R2 − OH2| is simply the power of H w.r.t ⊙(ABC).

2.2 Proof Let I be the incenter of ABC. From P drop pedals onto AI,BI,CI and denote them by ′ ′ ′ ′ ′ ′ A , B ,C respectively. With P, A ∈ ⊙(ABP CP ), proposition 2.1 states that A B1 = A C1 and ′ ′ ′ ′ ′ ′ A B2 = A C2. Symmetrically, we can obtain B Ai = B Ci,C Ai = C Bi for i =1, 2. To rid the suspense, we claim that the fixed point that lies on the radical axis of ⊙(A1B1C1), ⊙(A2B2C2) is the orthocenter of A′B′C′, which we will denote by H′. This first lemma opens the door for A1B1C1 to relate with other triangles. ′ ′ ′ Lemma 2.2. Triangles A B C and A1B1C1 are orthologic. In addition, the two orthologic centers between the triangles are isogonal conjugates wrt A′B′C′.

Proof. Let O1 denote the circumcenter of A1B1C1. We claim that O1 = OA′B′C′ (A1B1C1). ′ ′ Indeed, A lies on the perpendicular bisector of B1C1, which O1 also lies on, so A O1 ⊥ B1C1. ′ ′ ′ ′ ′ Symmetrically, we can obtain B O1 ⊥ A1C1,C O1 ⊥ A1B1 =⇒ A B C and A1B1C1 are orthologic. ′ ′ ′ ′ ′ ′ ′ Now Let Q1 = OA1B1C1 (A B C ). Since B1Q1 ⊥ A C ,C1Q1 ⊥ A B , thus ′ ′ ′ ′ ∠(Q1B1, Q1C1)= ∠(A C , A B ) = ∠(IC′, IB′) (A′, B′,C′,P,I are concyclic) ∠A = 90 − 2 180 − ∠A = 2 ′ ′ ∠(A B1, A C1) = . 2 ′ It follows that A is the circumcenter of B1C1Q1. Thus, B1,C1 are the reflection of Q1 over ′ ′ ′ ′ ′ ′ A C , A B respectively. It follows by symmetry that A1 is the reflection of Q1 over B C . These ′ ′ ′ result can be summerized as follows: A1B1C1 is the reflection triangle of Q1 w.r.t A B C . By a well-known fact of isogonal conjugates, O1, the circumcenter of the reflection triangle of Q1, ′ ′ ′ is isogonal conjugate with Q1 w.r.t A B C ; this proves second part of the proposition.

3 Figure 2: Proof of Lemma 2.2. Initial set up for the solution.

′ ′ ′ ′ ′ ′ ′ ′ Lemma 2.3. If Q1 = OA1B1C1 (A B C ) and Q2 = OA2B2C2 (A B C ), then H Q1 = H Q2.

Proof. By taking x = 0, A1B1C1, A2B2C2 coincide with AP BP CP , so in light of Proposition ′ ′ ′ ′ ′ ′ ′ ′ ′ ′ Lemma 2.2, we define Q = OAP BP CP (A B C ). Let A H ∩ ⊙(A B C ) = HA =6 A , thus ′ ′ ′ ′ ′ H ,HA are symmetric about B C . We claim that HA ∈ P AP , which follows from angle ∠ ′ ∠ ′ ′ chasing (P A , P AP )= (P A ,PHA). In the proof above, we showed that A1, Q1(A2, Q2) are ′ ′ ′ ′ ′ ′ symmetric about B C . Therefore, a reflection over B C produces HAA1A2AP 7→ H Q1Q2Q, and the lemma follows.

Remark Notice that we have proven that H′Q is the Steiner’s line of P with respect to A′B′C′. In addition, as A1, A2 varies with x, Q1, Q2 varies on a fixed line through Q and perpendicular to H′Q. This observation will be used in the next section. We are now ready to prove the main result. ′ ′ ′ Proof. We proved in Lemma 2.2 that Q1,O1 are isogonal conjugates w.r.t A B C , and A1B1C1 ′ ′ ′ is the reflection triangle of Q1 w.r.t A B C . Applying Proposition 2.2, we have 2 ′2 ′2 O1A1 = O1H + Q1H + p where p is the power of H′ w.r.t ⊙(A′B′C′). Symmetrically, we can obtain 2 ′2 ′2 O2A2 = O2H + Q2H + p ′ ′ Since Proposition asserts that H Q1 = H Q2, it follows by subtracting the two equations that 2 ′2 2 ′2 O1A1 − O1H = O2A2 − O2H . ′ However, this equality is equivalent to H having equal power wrt ⊙(A1B1C1) and ⊙(A2B2C2). This implies H′ lies on the radical axis of the two circles and the result is proven.

4 Figure 3: Proof of Lemma 2.3

3 Special Case: When P lies on IO

This section aims to investigate the case where P lies on IO, where I,O are incenter and circumcenter of ABC. The main result is stated as follows.

Theorem 3.1. Let P be a point that lies on IO, where I and O are the incenter and cir- cumcenter of ABC respectively. Let AP BP CP be the pedal triangle of P on ABC. Define A1, A2, B1, B2,C1,C2 such that for some x> 0

x = AP A1 = AP A2 = BP B1 = BP B2 = CP C1 = CP C2, with A1, B1,C1 lying on rays AP C, BP A, CP B, and A2, B2,C2 lying on rays AP B, BP C,CP A. As x varies, the radical axis of ⊙A1B1C1 and ⊙A2B2C2 is a fixed line.

Proving this would bring us one step closer to deducing our open problem. Surprisingly, efforts to establish this theorem lead to explorations into fruitful projective results that strayed relative far from the theorem at hand. We now pick up where we left off from the previous section to analyze the current theorem. Since we proved that the radical axis contains a fixed point, the natural next step is to show that the radical axis is always parallel to a fixed line. This is equivalent to showing that the line connecting circumcenters of △A1B1C1 and △A2B2C2, O1 and O2 respectively, is always parallel to a fixed line as x varies. Keeping the same notations as the previous section, recall from Lemma 2.2 that O1,O2 are isogonal conjugates of Q1, Q2, a pair of points on a line through ′ ′ ′ ′ ′ ′ Q (defined as OAP BP CP (A B C )) and perpendicular to the Steiner’s line of P w.r.t △A B C .

5 Additionally, Q1, Q2 are equidistant from Q. Therefore, it suffices to find out what is special about the point Q and Steiner’s line of P when P lies on IO. The next proposition helps answer this.

Proposition 3.1. Suppose P is a point on IO. Define A′, B′,C′ to be the pedals from P onto AI,BI,CI respectively. Then the Steiner’s line of P w.r.t. △A′B′C′ is the Euler line of △A′B′C′.

Notice that as P moves along IO, the corresponding configurations for A′B′C′ ∪ P are homothetic through I. Therefore, it suffices to prove that this proposition is true for one point on IO. For the sake of simplicity, we will choose P ≡ Be the Bevan point of △ABC, i.e. if IAIBIC is the excenter triangle of ABC, then Be,O,I are its circumcenter, nine-point center, and orthocenter respectively. Thus we can restate Proposition 3.1 in terms of the excenter triangle.

Lemma 3.2. Given a triangle ABC with circumcenter O, nine-point center N, and orthocenter H. Define A∗, B∗,C∗ to be the pedals from O onto AH,BH,CH. Then the Steiner’s line of P w.r.t A∗B∗C∗ is the Euler line of △A∗B∗C∗.

Figure 4: Proof of Lemma 3.2

Proof. Let H∗ be the orthocenter of A∗B∗C∗. Suppose A∗H∗ ∩ ⊙(A∗B∗C∗)= D =6 A∗, then D is the reflection of H∗ over B∗C∗. Clearly N is the circumcenter of A∗B∗C∗, so let N ′ be the reflection of N over B∗C∗. By properties of the Steiner’s line, it suffices to show that O,D,N ′ are collinear.

6 Angle chasing reveals that A∗B∗C∗ ∼ ABC, thus

∠(OC∗,OD)= ∠(A∗C∗, A∗D)= ∠(AH, AC)= ∠(AB, AO)= ∠(OC∗, OA), which means O,D,A are collinear. Hence it remains to show that O, N ′, A are collinear. Let O′ be the reflection of O over BC, and define AO∩⊙(ABC)= E =6 A,EO′ ∩⊙(ABC)= F =6 E. By properties of the nine-point center, EO′ k HO. Since CE k BH,BE k CH, it follows by angle chasing with cyclic quadrilaterals that FBC ∼ OC∗B∗. Because OO′ corresponds to NN ′ between the two similar triangles, we have FBC ∪ O′ ∼ OC∗B∗ ∪ N ′. Hence, ∠(OC∗, ON ′)= ∠(FB,FO′)= ∠(AB, AO)= ∠(OC∗, OA), which implies that O, N ′, A are collinear and proves the proposition. Now that Proposition 3.1 is proven, the implication is that the Steiner’s line of P w.r.t. △A′B′C′ passes through the circumcenter of △A′B′C′. It turns out that this is the property that makes Theorem 3.1 true, as exemplified by the following generalization we found.

Proposition 3.2. Given a triangle ABC and a line l on the same plane. Let O be the circum- center of ABC, and let T be the pedal from O to l. Suppose R and S are two variable points on l satisfying TS = T R. If S′, R′ are the isogonal conjugates of S, R w.r.t ABC respectively, then S′R′ is parallel to a fixed line as S, R vary on l.

This result would finish the proof for Theorem 3.1 when applying it to triangle A′B′C′ with Q as T and the line through Q perpendicular to the Euler line (by Lemma 3.2) of A′B′C′ as l. We will devote the next subsection to proving Proposition 3.2 and give a formal proof of the theorem in the last subsection.

3.1 Consequences of a Projective Generalization Under an isogonal conjugation with respect to a fixed triangle, a line is transformed into a circumconic of the triangle and vise versa [3]. Thus we were encouraged to work on the projective plane containing conics.

Definition 3.1. With respect to a conic C, two points A, B are said to be conjugates if the polar of A passes through B or, equivalently, the polar of B passing through A.

We now present a powerful result relating isogonal conjugation and polarity.

Proposition 3.3. Given a triangle ABC and two lines l1,l2. Let ψ denote the isogonal con- jugation w.r.t △ABC. Suppose ψ(l1) = C1, ψ(l2) = C2. If X,Y are two points on l1 and ′ ′ ′ ′ ψ(X) = X , ψ(Y ) = Y , then X,Y are conjugates w.r.t C2 =⇒ X Y passes through the pole of l2 w.r.t C1

Proof. We first recall two properties of isogonal conjugation and conjugates. One, at least one of a conjugate pair is an external point, meaning two tangents can be drawn from the point to the respective conic. Two, isogonal conjugation preserves tangency, i.e. if l is tangent to C, then ψ(C) is tangent to ψ(l).

7 Figure 5: Proof of Proposition 3.3

In light of the first property, WLOG assume X is an external point w.r.t. C2; thus there exists two points V, U ∈ C2 such that XV,XU are tangents. Take the following isogonal conjugations:

′ ′ ψ(V )= V , ψ(U)= U , ψ(V U)= CX , ψ(XV )= HV , ψ(XU)= HU .

′ Since Y ∈ V U, it follows that Y ∈ CX . By the second property, HV , HU are tangent to l2 at U ′,V ′ respectively. Consider the following cross-ratio equality(or projectivity if you prefer):

′ ′ ′ ′ Y (A, B; C,V )= U (A, B; C,V ) (taken on CX ) ′ ′ = U (A, B; C, U ) (taken on HU ) ′ ′ = X (A, B; C, U ) (taken on HU )

′ ′ ′ ′ ∗ ′ ′ ′ ′ ∗ This implies X U ∩ Y V = V ∈ C1. Symmetrically, we can obtain X V ∩ Y U = U ∈ C1. ′ ′ ∗ ∗ Hence X Y intersect V U at the pole of l2 w.r.t C1.

Corollary 3.2.1. Let X,Y be conjugates conjugates w.r.t ⊙(ABC). If X′,Y ′ are isogonal conjugates of X,Y w.r.t △ABC, then X′Y ′ passes through the center of the circumconic defined by A,B,C,X′,Y ′.

Proof. This is simply a consequence of Proposition 3.3 with l2 as the line at infinity and l1 as XY , since ψ(l2)= ⊙(ABC) and the pole of l2 w.r.t C1 is the center of C1.

8 We can now prove Proposition 3.2.

Figure 6: Proposition 3.2

Proof of Proposition 3.2. Let l∞ denote the point at infinity on l. Suppose l is transformed to a circumconic C under isogonal conjugation w.r.t ABC. If T ′, L′ are the isogonal conjugates of ′ ′ ′ ′ T,l∞, then T , L ,X ,Y all lie on C. Since l∞ and T are conjugates w.r.t ⊙(ABC), Corollary 3.2.1 asserts that T ′L′ passes through the center of C, or equivalently, the pole of T ′L′, denoted by K, lies on the line of infinity. Observe that ′ ′ ′ ′ A(T , L ; S , R )= A(T,l∞; S, R)= −1. ′ ′ ′ ′ This implies S R passes through K. Hence S R is parallel to the tangent at l∞ at C, which is a fixed line.

3.2 Proof We now give a formal proof of Theorem 3.1. We will heavily refer to results already established in Section 2. Proof of Theorem 3.1. Define A′, B′,C′ to be the pedals from P onto AI,BI,CI respectively. Lemma 2.2 states that the following points exist:

′ ′ ′ ′ ′ ′ ′ ′ ′ OA1B1C1 (A B C )= Q1,OA2B2C2 (A B C )= Q2,OAP BP CP (A B C )= Q.

Furthermore, we proved in Lemma 2.3 that Q1, Q, Q2 all lie on a line, denoted by q, that is ′ ′ ′ perpendicular to the Steiner’s line of P w.r.t A B C and the points satisfy QQ1 = QQ2. By Proposition 3.1, we know the circumcenter O′ of A′B′C′ lies on the Steiner’s line. Thus, Q can ′ be viewed as the pedal from the O to q. From Lemma 2.2, we know the circumcenters O1,O2 of ′ ′ ′ △A1B1C1, △A2B2C2 are isogonal conjugates of Q1, Q2 w.r.t △A B C . Applying Proposition ′ ′ ′ 3.2 on triangle A B C with q as l and Q as T , we obtain that O1O2 is parallel to a fixed line as

9 Q1, Q2 varies l. Since the radical axis of ⊙(A1B1C1) and ⊙(A2B2C2) is perpendicular to O1O2 and passes through a fixed point, it follows that the radical axis is fixed. Proposition 3.4. Keeping the same notations as above, if q is transformed to C under isogonal ′ ′ ′ conjugation w.r.t △A B C , then we claim that P,OP ∈ C, where OP is the circumcenter of AP BP CP .

Proof. On one hand, OP is the isogonal conjugate of Q, which lies on q; so OP ∈ C. On the other hand, since P ∈ ⊙(A′B′C′), the isogonal conjugate of P is the point at infinity corresponding to the direction perpendicular to the Steiner line of P w.r.t A′B′C′. Because q is parallel to this direction, it contains the isogonal conjugate of P , which proves that P ∈ C. The following corollary follows directly from the proof of Proposition 3.2. Corollary 3.2.2. Keeping the same notations as above, the fixed radical axis is perpendicular to the tangent to OP or P on C.

4 Proof of the Open Problem

The open problem focuses on when P = Be, the Bevan point of ABC. It suffices to prove that the fixed line described in Theorem 3.1 is the Nagel line. We first prove that the fixed point characterized in section 2 lies on the Nagel line. In fact, the next proposition proves that it coincides with the Nagel point. Then, it remains to show that the Nagel line is perpendicular to the fixed line characterized in section 3. ′ ′ ′ Proposition 4.1. Let Be be the Bevan point of ABC. Define A , B ,C to be the pedals from ′ ′ ′ Be onto AI,BI,CI respectively. Then, the orthocenter of A B C coincides with the Nagel point of ABC. Similar to Proposition 3.1, Proposition 4.1 will follow from a lemma written in the perspec- tive of the excenter triangle of ABC. Lemma 4.1. Given a triangle ABC with circumcenter O, orthocenter H, and orthic triangle ∗ ∗ ∗ AH BH CH . Define A , B ,C to be the pedals from O onto AH,BH,CH respectively. If AO ∗ ∗ ∗ ∗ intersects BH CH at J, then the orthocenter H of A B C lies on AH J. Proof. Let AO∩⊙(A∗B∗C∗)= D =6 O. Recall from the proof of Lemma 3.2 that A∗D ⊥ B∗C∗; thus H∗ ∈ A∗D. Additionally, since HO is the diameter of ⊙(A∗B∗C∗), we have HD ⊥ AO, which implies D ∈ ⊙(ABH CH H). If AH ∩ ⊙(ABC)= G =6 A, then because AD ⊥ BH CH and ABC ∼ ABH CH , it follows that AJ AA = H , JD GAH implying AH J k GD. Note that it is a well known property of circumcenters that the distance of O to BC is half ∗ of AH, i.e. 2A AH = AH. Because 2GAH = HG, we get ∗ ∗ ∗ A AH AH A H = = ∗ GAH HG DH where the last equality follows from the similarity ABC ∪ H ∪ G ∼ A∗B∗C∗ ∪ H∗ ∪ D. Notice ∗ ∗ ∗ that this implies AH H k GD. Hence, AH J k AH H =⇒ H ∈ AH J.

10 Figure 7: Proof of Lemma 4.1

In the context of Proposition 4.1, note that if A0B0C0 is the pedal triangle of Be, then the Nagel point is where AA0, BB0,CC0 concur. In Lemma 4.1, O is the Bevan point of AH BH CH , and J is the pedal from O to BH CH . Thus a symmetric application of the lemma implies that ∗ H is the Nagel point of AH BH CH , and Proposition 4.1 follows by a change in notation. ′ Proposition 4.2. Let O0 to be the circumcenter of A0B0C0, and let H be the orthocenter of ′ ′ ′ ′ A B C . Then Be,H ,O0 are collinear (See Figure 8.) ′ ′ Proof. Let Be denote the isogonal conjugate of Be w.r.t ABC. According to Proposition 4.1, H ′ is the Nagel point of ABC, so by definition, ABC and A0B0C0 are perspective at H . Observe ′ ′ ′ that Be = OA0B0C0 (ABC) and Be = OABC (A0B0C0). By Sondat’s theorem, Be, Be,H are ′ collinear and the proposition follows since O0 is the midpoint of BeBe(a property of isogonal conjugates). We are finally ready to prove the open problem. ′ ′ ′ Proof of Theorem 1.1. Let A , B ,C be the pedals from Be onto AI,BI,CI respectively, and let H′ denote the orthocenter of A′B′C′. By proposition 4.1 and Theorem 3.1, H′ is the Nagel point of ABC and lies on the fixed radical axis, so it remains to show that IH′ coincides with this radical axis. ′ ′ ′ Let O0 denote the circumcenter of A0B0C0. Define C to be the circumconic of A B C that also passes through Be,O0. According to Proposition 3.4 and Corollary 3.2.2, the fixed radical axis is perpendicular to the tangent of Be or O0 at C. If b is the tangent of Be at C, then it suffices to show that IH′ ⊥ b. ′ ′ ′ Since the circumcenter of A B C is the midpoint of IBe, which coincides with the circum- center of ABC, therefore OH′ is the Euler line of A′B′C′. Suppose OH′ is transformed to

11 Figure 8: Proof of open problem conic H under isogonal conjugation w.r.t A′B′C′. According to Lemma 2.2 and the proof of Lemma 2.3, if Q0 is the isogonal conjugate of O0, then Q0 lies on the Steiner’s line of Be w.r.t ′ ′ ′ ′ ′ ′ ′ A B C . By Proposition 3.1, Q0 lies on the Euler line of A B C , which is OH , so it follows ′ ′ ′ that O0 ∈ H. On the other hand, since I, Be are antipodal on ⊙(A B C ), their Steiner’s lines are perpendicular, so the isogonal conjugate of I w.r.t A′B′C′ is the point of infinity along the ′ Steiner line of Be, which is OH . Hence, we have I ∈ H. ′ ′ ′ Now suppose b intersects ⊙(A B C ) again at L(it is possible that L ≡ Be if b is a tangent.) Consider the following cross-ratio equality,

′ ′ ′ ′ ′ ′ Be(A , B ; C , L)= Be(A , B ; C , Be) (taken on C) ′ ′ ′ = O0(A , B ; C , Be) (taken on C) ′ ′ ′ ′ = O0(A , B ; C ,H ) (by Proposition 4.2, taken on H) = I(A′, B′; C′,H′). (taken on ⊙(A′B′C′))

This implies that I,H′, L are collinear, so IH′ ⊥ b, which is what we wanted to prove. The open problem has been resolved.

References

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13