Brocard, Napoleon, and Special Points

Let the position vectors of \(A,B,C\) be \(\vec a,\vec b,\vec c\), with \(|\vec a|=|\vec b|=|\vec c|=R\). Set \(\vec h=\vec a+\vec b+\vec c\). Then \((\vec h-\vec a)\cdot(\vec b-\vec c)=(\vec b+\vec c)\cdot(\vec b-\vec c)=0\), so \(AH\perp BC\). Similarly \(BH\perp CA\) and \(CH\perp AB\), hence \(H\) is the orthocenter.

The centroid has position vector \(\vec g=\frac{\vec a+\vec b+\vec c}{3}=\frac{\vec h}{3}\). Thus \(O,G,H\) are collinear, and from \(OG=\frac13 OH\), \(GH=\frac23 OH\) we get \(OG:GH=1:2\).

The homothety centered at \(H\) with ratio \(\frac12\) sends the circumcircle of \(ABC\) to the circle centered at \(N\), the midpoint of \(OH\), with radius \(\frac R2\). The points \(A,B,C\) go to the midpoints of \(AH,BH,CH\), so these three points lie on the new circle.

Let \(M_a\) be the midpoint of \(BC\). In the vector model from the previous problem, \(\vec n=\frac{\vec a+\vec b+\vec c}{2}\), \(\vec m_a=\frac{\vec b+\vec c}{2}\). Thus \(|\vec m_a-\vec n|=\frac{|\vec a|}{2}=\frac R2\). The other two side midpoints are identical. Hence all six points lie on one circle.

For an arbitrary cevian \(AX\), the sine rule in triangles \(ABX\) and \(ACX\) gives \(\frac{BX}{CX}=\frac{AB\sin\angle BAX}{AC\sin\angle XAC}\).

If \(AS\) is the isogonal image of the median \(AM\), then \(\angle BAS=\angle MAC\), \(\angle SAC=\angle BAM\). Since \(BM=CM\), the median gives \(\frac{\sin\angle BAM}{\sin\angle MAC}=\frac{AC}{AB}\). Substitution yields \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\).

Conversely, if this ratio holds, the same cevian formula shows that the isogonal line to \(AS\) divides \(BC\) equally, so it is the median. Hence \(AS\) is the symmedian.

Let \(X,Y,Z\) be the centers of the equilateral triangles on \(BC,CA,AB\). Consider the \(60^\circ\) rotation associated with the equilateral triangle on side \(AB\). It sends the direction from one constructed center to another to the next such direction.

The cyclic argument gives \(XY=YZ=ZX\), and the angle between \(XY\) and \(YZ\) is \(60^\circ\). Hence \(XYZ\) is equilateral.

Let the \(A\)-symmedian meet \(BC\) at \(A_1\). Then \(\frac{BA_1}{CA_1}=\frac{AB^2}{AC^2}=\frac{c^2}{b^2}\). Similarly, for \(B_1\in CA\), \(C_1\in AB\), we have \(\frac{CB_1}{AB_1}=\frac{a^2}{c^2}\), \(\frac{AC_1}{BC_1}=\frac{b^2}{a^2}\).

The product is \(1\). By Ceva's theorem, \(AA_1,BB_1,CC_1\) are concurrent. This point is the Lemoine point \(K\).

By the tangent-chord theorem, \(\angle PBA=\angle ACB\), \(\angle PCB=\angle CAB\), with analogous equalities for the other tangent. Let \(S=AP\cap BC\).

Applying the sine rule in triangles \(PBS\) and \(PCS\), we get \(\frac{BS}{CS}=\frac{PB\sin\angle BPS}{PC\sin\angle SPC}\). Since \(PB=PC\), and the angles at \(P\) are expressed through arcs \(AB\) and \(AC\), the sine rule in \(ABC\) gives \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\).

By the symmedian criterion, \(AS\), hence \(AP\), is the \(A\)-symmedian.

Let \(A=(u,0)\), \(B=(0,v)\), \(C=(0,0)\). Then \(a=BC=v\), \(b=CA=u\), \(c^2=AB^2=u^2+v^2\). The Lemoine point has barycentric coordinates \((a^2:b^2:c^2)\), hence its Cartesian coordinates are \(\left(\frac{uv^2}{2(u^2+v^2)},\frac{u^2v}{2(u^2+v^2)}\right)\).

The foot \(H\) of the altitude from \(C\) to \(AB\) has coordinates \(\left(\frac{uv^2}{u^2+v^2},\frac{u^2v}{u^2+v^2}\right)\). Thus the coordinates of \(K\) are half those of \(H\), so \(K\) is the midpoint of \(CH\).

Let \(L\) be the midpoint of \(B_1C_1\), and let \(AL\) meet \(BC\) at \(S\). From antiparallelism, \(\triangle AB_1C_1\sim \triangle ACB\) with opposite orientation. Hence the distance ratios from \(L\) to \(AB\) and \(AC\) are expressed through \(AB\) and \(AC\).

Using areas of triangles \(ABS\) and \(ACS\), we get \(\frac{BS}{CS}=\frac{AB^2}{AC^2}\). By the symmedian criterion, \(AS\) is the \(A\)-symmedian. Therefore \(A,L,S\) are collinear, and the symmedian passes through the midpoint of the antiparallel.

The \(60^\circ\) rotation about \(C\) sends \(B\) to \(X\), since \(BCX\) is equilateral. It also sends the line associated with \(A\) into the line associated with the construction on \(CA\). Thus \(T=AX\cap BY\) is arranged so that the directions \(TB\) and \(TC\), and then \(TC\) and \(TA\), differ by \(120^\circ\).

More explicitly, from \(CB=CX\), \(CA=CY\), and the \(60^\circ\) angles, similar triangles around \(T\) appear. They give \(\angle BTC=120^\circ\), and cyclically the other two angles. Hence \(T\) is the Fermat point.

The homothety centered at \(H\) with ratio \(\frac12\) sends \(A,B,C\) to the midpoints of \(AH,BH,CH\). Therefore the circle through these three midpoints is the image of the circumcircle of \(ABC\).

The image of the center \(O\) under this homothety is the midpoint of \(OH\). Hence this point is the center of the circle through the midpoints of \(AH,BH,CH\). Since this circle is the nine-point circle, its center is the midpoint of \(OH\).

Compare triangles \(ABC\) and \(HBC\). The midpoint of \(BC\) is a side midpoint in both triangles. The midpoints of \(BH\) and \(CH\) lie on the nine-point circles of both triangles: for \(ABC\) they are midpoints from the orthocenter to vertices, while for \(HBC\) they are side midpoints.

Three common points determine a circle, so the nine-point circles of these two triangles coincide. The same comparison applies to \(ABC\) with \(AHC\) and \(ABH\). Therefore all four nine-point circles coincide.

By the previous problem, the four triangles have the same nine-point circle. Let its center be \(N\).

In any triangle, the center of the nine-point circle lies on the Euler line: it is the midpoint between the circumcenter and the orthocenter of that triangle. Hence each of the four Euler lines passes through \(N\). Therefore they are concurrent at \(N\).

In the excentral triangle \(I_aI_bI_c\), the internal angle bisectors of the original triangle become altitudes. For instance, line \(AI\) is perpendicular to side \(I_bI_c\), so \(A\) is an altitude foot. Similarly, \(B\) and \(C\) are the other two altitude feet.

The nine-point circle of any triangle passes through its altitude feet. Hence the nine-point circle of \(I_aI_bI_c\) passes through \(A,B,C\). The unique circle through \(A,B,C\) is the circumcircle of the original triangle. Therefore the two circles coincide.

The Lemoine point has barycentric coordinates \((a^2:b^2:c^2)\), where \(a=BC\), \(b=CA\), \(c=AB\). Trilinear coordinates are proportional to distances from the sides, and the conversion is \((u:v:w)_{\text{bar}}=(a x:b y:c z)\).

Thus from \((a x:b y:c z)=(a^2:b^2:c^2)\), we get \(x:y:z=a:b:c\). Hence the distances from the Lemoine point to the sides are proportional to the corresponding side lengths.

Let the common Brocard angle be \(\varphi\). Since \(A,P,A_1\), \(B,P,B_1\), \(C,P,C_1\) are collinear, we get \(\angle CAA_1=\angle ABB_1=\angle BCC_1=\varphi\). Hence the corresponding directed arcs satisfy \(\widehat{CA_1}=\widehat{AB_1}=\widehat{BC_1}=2\varphi\).

Now compare the arcs between the new points. Passing from arc \(BC\) to arc \(A_1B_1\), both endpoints are shifted along the circle by the same directed arc \(2\varphi\). Therefore \(\widehat{A_1B_1}=\widehat{BC}\). Similarly, \(\widehat{B_1C_1}=\widehat{CA}\) and \(\widehat{C_1A_1}=\widehat{AB}\). Equal arcs give equal chords, so \(\triangle A_1B_1C_1\cong \triangle BCA\).

We have \(\operatorname{ctg}A=\frac{b^2+c^2-a^2}{4S}\), and cyclically. Adding gives \(\operatorname{ctg}A+\operatorname{ctg}B+\operatorname{ctg}C=\frac{a^2+b^2+c^2}{4S}\).

By Weitzenbock's inequality, \(a^2+b^2+c^2\ge 4\sqrt3 S\). Hence \(\operatorname{ctg}\varphi\ge \sqrt3\). Since \(\varphi\) is acute, \(\varphi\le 30^\circ\).

Let \(\vec n_a,\vec n_b,\vec n_c\) be unit outward normals to \(BC,CA,AB\). Since \(K\) is the Lemoine point, its distances to the sides are proportional to \(a:b:c\). Hence \(\overrightarrow{KA_1}=-\lambda a\vec n_a\), \(\overrightarrow{KB_1}=-\lambda b\vec n_b\), \(\overrightarrow{KC_1}=-\lambda c\vec n_c\) for some \(\lambda\).

For any triangle, \(a\vec n_a+b\vec n_b+c\vec n_c=\vec0\): the sum of outward normal vectors weighted by side lengths is zero. Therefore \(\overrightarrow{KA_1}+\overrightarrow{KB_1}+\overrightarrow{KC_1}=\vec0\). This is exactly the condition that \(K\) is the centroid of triangle \(A_1B_1C_1\).

Use barycentric coordinates. The Lemoine point is \(K=(a^2:b^2:c^2)\). The line through \(K\) parallel to \(BC\) cuts sides \(AB\) and \(AC\) at points whose coordinates are linear in \(a^2,b^2,c^2\). The same holds for the other two parallels.

Substituting the six points into the general barycentric equation of a circle shows that the same equation vanishes for all six. Synthetically, this says that the symmedian ratios make the corresponding segment pairs antiparallel, and hence they determine one circle. Therefore the six points are concyclic.

Let \(\angle ABP=\angle BCP=\angle CAP=\varphi\). Reflect \(AP,BP,CP\) in the angle bisectors of \(A,B,C\). By trigonometric Ceva, the three reflected lines are also concurrent; their point of concurrence is the isogonal conjugate \(Q\).

For example, \(\angle BAQ\) is the reflection of the angle between \(AP\) and \(AC\), hence corresponds to \(\angle CAP=\varphi\). Similarly, \(\angle ACQ\) corresponds to \(\angle BCP=\varphi\), and \(\angle CBQ\) corresponds to \(\angle ABP=\varphi\). Therefore \(\angle BAQ=\angle ACQ=\angle CBQ\), so \(Q\) is the second Brocard point.

Consider hexagon \(A B_1 C A_1 B C_1\). The sum of angles at \(A_1,B_1,C_1\) is \(2(\alpha+\beta+\gamma)=360^\circ\), so the sum of the remaining three angles is also \(360^\circ\).

Construct an auxiliary triangle congruent to one of the external isosceles triangles so that one side lies on a neighboring side of the hexagon. Then two pairs of triangles become congruent by two sides and the included angle. It follows that twice one angle of triangle \(A_1B_1C_1\) equals the corresponding angle of the hexagon, namely \(2\alpha\). Hence that angle is \(\alpha\). Cyclically, the other angles are \(\beta\) and \(\gamma\).

Construct an external equilateral triangle \(BCU\). For an arbitrary point \(X\), rotate segment \(XB\) by \(60^\circ\) so that \(B\) maps to \(U\); let \(X'\) be the image of \(X\). Then \(XB=UX'\), and \(XX'=XB\) with \(\angle BXX'=60^\circ\).

The sum \(XA+XB+XC\) becomes the length of a broken line connecting fixed points through \(X\) and \(X'\). By the triangle inequality, this broken line is at least the corresponding straight segment. Equality occurs when the broken line is straight, which is equivalent to the \(120^\circ\) angles at \(T\). Therefore the minimum is attained at the Fermat point.

Under the homothety centered at \(K\), the sides \(A'B'\), \(B'C'\), \(C'A'\) remain parallel to the sides of \(ABC\). The six side intersections form a configuration in which corresponding segments are antiparallel. Hence they lie on a Tucker circle.

As the homothety ratio varies, this circle continuously changes into a Lemoine circle. The centers of all these circles keep the same direction of variation: it is the line joining the circumcenter \(O\) to the center of the symmedian structure \(K\). In coordinate form, the equation of a Tucker circle is a linear combination of the circumcircle equation and the circle equation associated with parallels through \(K\); the centers of such circles lie on the line joining the original centers, namely \(KO\).

Build three similarly oriented figures on sides \(BC,CA,AB\). Their similarity circle has diameter \(OK\): point \(O\) corresponds to the perpendicular bisectors of the sides, while point \(K\) corresponds to the lines through the Lemoine point parallel to the sides.

For the first Brocard point, the lines \(CP,AP,BP\) are corresponding lines of these three figures. Hence their common point \(P\) lies on the similarity circle. Similarly, the second Brocard point \(Q\) comes from the opposite orientation of corresponding lines and also lies on the same circle.

Since the similarity circle has diameter \(OK\), \(P\) and \(Q\) lie on the circle with diameter \(OK\).

The vertices of the Brocard triangle are the constant points of three similar figures built on \(BC,CA,AB\). Therefore the lines through \(A,B,C\) parallel to the opposite sides of \(A_1B_1C_1\) are corresponding lines of these figures. By the similarity-circle property, their intersection \(S\) lies on the corresponding similarity circle; in this configuration, it is the circumcircle of \(ABC\).

Now \(S\) lies on the circumcircle, so it has a Simson line. The direction of this line is determined by the same corresponding lines, and the fixed direction in the similarity circle is the diameter \(OK\). Hence the Simson line of \(S\) is parallel to the Brocard diameter \(OK\).