Let $ABC$ be a triangle. Let $\theta$ be a fixed angle for which \[\theta<\frac12\min(\angle A,\angle B,\angle C).\]Points $S_A$ and $T_A$ lie on segment $BC$ such that $\angle BAS_A=\angle T_AAC=\theta$. Let $P_A$ and $Q_A$ be the feet from $B$ and $C$ to $\overline{AS_A}$ and $\overline{AT_A}$ respectively. Then $\ell_A$ is defined as the perpendicular bisector of $\overline{P_AQ_A}$. Define $\ell_B$ and $\ell_C$ analogously by repeating this construction two more times (using the same value of $\theta$). Prove that $\ell_A$, $\ell_B$, and $\ell_C$ are concurrent or all parallel.
Problem
Source: 2022 USA TSTST #2
Tags: geometry, USA TSTST
Inconsistent
27.06.2022 19:53
Solution due to tapir. By linearity of PoP, $XP_B^2 - XQ_C^2 = \lambda (XB^2 - XC^2)$ since $BCP_BP_C$ is an isosceles trapezoid by angles, where $\lambda$ is a constant of $\theta$. Hence by summing cyclically, $XP_A^2 + XP_B^2 + XP_C^2 = XQ_A^2+XQ_B^2+XQ_C^2$ is true for all points in the plane, so the perpendicular bisectors of $P_AQ_A, P_BQ_B, P_CQ_C$ concur or are pairwise parallel by phantom point on the intersection of any two of the three.
hukilau17
27.06.2022 21:04
Let $\triangle ABC$ be inscribed in the unit circle and positively oriented, so that $|a|=|b|=|c|=1$, and let $\alpha = e^{2i\theta}$. Then $\overleftrightarrow{AS_A}$ meets the unit circle again at $\alpha b$, and $\overleftrightarrow{AT_A}$ meets it again at $\alpha^{-1}c$. Thus using projections, we have
$$p_a = \frac{a+b+\alpha b-\alpha a}2$$$$q_a = \frac{a+c+\alpha^{-1}c-\alpha^{-1}a}2$$Now the equation of $\ell_A$ is
$$|z-p_a| = |z-q_a|$$$$(z-p_a)(\overline{z}-\overline{p_a}) = (z-q_a)(\overline{z}-\overline{q_a})$$$$\overline{z} = \frac{(p_a\overline{p_a} - q_a\overline{q_a}) - z(\overline{p_a}-\overline{q_a})}{p_a-q_a}$$Now we compute
$$p_a-q_a = \frac{b-c+\alpha b-\alpha^{-1}c-\alpha a+\alpha^{-1}a}2$$$$\overline{p_a} - \overline{q_a} = -\frac{ab-ac+\alpha ab-\alpha^{-1}ac-\alpha bc+\alpha^{-1}bc}{2abc}$$and
\begin{align*}
p_a\overline{p_a} - q_a\overline{q_a} &= \frac{c(a+b+\alpha b-\alpha a)(a+b+\alpha^{-1}a-\alpha^{-1}b) - b(a+c+\alpha^{-1}c-\alpha^{-1}a)(a+c+\alpha a-\alpha c)}{4abc} \\
&= \frac{\alpha^{-1}c(a^2-b^2) - \alpha c(a^2-b^2) + 4abc + \alpha^{-1}b(a^2-c^2) - \alpha b(a^2-c^2) - 4abc}{4abc} \\
&= \frac{(\alpha-\alpha^{-1})(b-c)(a^2-bc)}{4abc}
\end{align*}So the equation of $\ell_A$ is
$$\overline{z} = \frac{(\alpha-\alpha^{-1})(b-c)(a^2-bc) + 2z(ab-ac+\alpha ab-\alpha^{-1}ac-\alpha bc+\alpha^{-1}bc)}{2abc(b-c+\alpha b-\alpha^{-1}c-\alpha a+\alpha^{-1}a)}$$The equations of $\ell_B$ and $\ell_C$ are analogously found by cycling the variables $a,b,c$. Thus we intersect $\ell_A$ and $\ell_B$ to get
$$\frac{(\alpha-\alpha^{-1})(b-c)(a^2-bc) + 2z(ab-ac+\alpha ab-\alpha^{-1}ac-\alpha bc+\alpha^{-1}bc)}{2abc(b-c+\alpha b-\alpha^{-1}c-\alpha a+\alpha^{-1}a)}$$$$= \frac{(\alpha-\alpha^{-1})(c-a)(b^2-ca) + 2z(bc-ab+\alpha bc-\alpha^{-1}ab-\alpha ac+\alpha^{-1}ac)}{2abc(c-a+\alpha c-\alpha^{-1}a-\alpha b+\alpha^{-1}b)}$$Solving for $z$ gives
$$z = \frac{(\alpha-\alpha^{-1})\left[(b-c)(a^2-bc)(a-c+\alpha^{-1}a-\alpha c-\alpha^{-1}b+\alpha b) - (a-c)(b^2-ca)(b-c+\alpha b-\alpha^{-1}c-\alpha a+\alpha^{-1}a)\right]}{2\begin{bmatrix}
(ab-ac+\alpha ab-\alpha^{-1}ac-\alpha bc+\alpha^{-1}bc)(a-c+\alpha^{-1}a-\alpha c-\alpha^{-1}b+\alpha b) \\
- (ab-bc+\alpha^{-1}ab-\alpha bc-\alpha^{-1}ac+\alpha ac)(b-c+\alpha b-\alpha^{-1}c-\alpha a+\alpha^{-1}a)
\end{bmatrix}}$$We simplify the expression in the numerator by writing it in terms of powers of $\alpha$:
$$(a-c)(b-c)\left[(a^2-bc)-(b^2-ca)\right]$$$$+ \alpha\left[(b-c)^2(a^2-bc) + (a-b)(a-c)(b^2-ca)\right] - \alpha^{-1}\left[(a-c)^2(b^2-ca) + (b-a)(b-c)(a^2-bc)\right]$$which we rewrite as
$$-(a-b)(b-c)(c-a)(a+b+c) + \alpha(ab^3+bc^3+ca^3+a^2bc+ab^2c+abc^2-2a^2b^2-2a^2c^2-2b^2c^2)$$$$- \alpha^{-1}(a^3b+b^3c+c^3a+a^2bc+ab^2c+abc^2-2a^2b^2-2a^2c^2-2b^2c^2)$$Similarly we write the denominator as
$$\alpha^2b(a-c)(b-c) + \alpha b(a-c)^2 + \alpha a(b-c)^2 + a(c-a)(c-b) + b(a-b)(a-c) + c(b-a)(b-c)$$$$-\alpha^{-1}c(a-b)(a-c)-\alpha^{-1}a(b-a)(b-c) - \alpha^{-2}c(a-b)^2$$$$+ \alpha^2c(a-b)^2 + \alpha b(a-b)(a-c) + \alpha c(b-a)(b-c) - b(c-a)(c-b) - a(b-a)(b-c) - c(a-b)(a-c)$$$$-\alpha^{-1}b(a-c)^2 - \alpha^{-1}a(b-c)^2 - \alpha^{-2}a(a-c)(b-c)$$Combining like powers of $\alpha$ gives
$$\alpha^2(ab^2+bc^2+ca^2-3abc) + 2\alpha(a^2b+b^2c+c^2a-3abc) - 3(a-b)(b-c)(c-a)$$$$- 2\alpha^{-1}(ab^2+bc^2+ca^2-3abc) - \alpha^{-2}(a^2b+b^2c+c^2a-3abc)$$
Now the end result of this is that the numerator and denominator of $z$ are both invariant under cyclic permutations of $a,b,c$. In particular, if the denominator is nonzero, this expression gives the coordinate of a point that lies on all three of $\ell_A, \ell_B, \ell_C$, so they are concurrent at that point. If the denominator is instead zero, then intersecting any two of these three lines will give a point lying at infinity, so all three of them are parallel. $\blacksquare$
i3435
27.06.2022 23:17
Let $P_A'=(AC)\cap \overline{AP_A}$ and let $Q_A'=(AB)\cap \overline{AQ_A}$. Then angle chasing gives $P_A,P_A',Q_A,Q_A'$ cyclic, and the perpendicular bisectors of $P_AP_A'$ and $Q_AQ_A'$ are the $B,C$ midlines, so the midpoint $M$ of $BC$ is the center of $(P_AQ_AP_A'Q_A')$ and lies on the perpendicular bisector of $P_AQ_A$. Let $T_A=\overline{BP_A}\cap\overline{CQ_A}$ and define $T_B,T_C$ similarly. By Jacobi's theorem, $\overline{AT_A},\overline{BT_B},\overline{CT_C}$ concur. If $T$ is the isogonal conjugate of this intersection, since $AT_A$ is a diameter of $(AP_AQ_A)$, $\overline{AT}\perp\overline{P_AQ_A}$. Since the line between $M$ and $\frac{A+B+C-T}{2}$ is parallel to $\overline{AT}$, by symmetry the $3$ perpendicular bisectors intersect at $\frac{A+B+C-T}{2}$. Another finish, which I embarrassingly found before the above one, is that $\overline{AT_A},\overline{BT_B},\overline{CT_C}$ concurring means that hexagon $AT_CBT_ACT_B$ has an inconic. The center of this inconic lines on the line bisecting $AT_A$ and $BC$ by this, which is the perpendicular bisector of $P_AQ_A$, as desired.
cadaeibf
27.06.2022 23:43
We will show that the locus of concurrence is the Kiepert hyperbola of the medial triangle $M_aM_bM_c$. Let $U_a,V_a$ be the projections of $B,C$ on $AT_a,AS_a$ respectively. By BMO 2019/3 we know that $P_aQ_aU_aV_a$ is cyclic. Therefore, if $N_a$ is the midpoint of $P_aQ_a$, we know that $\ell_a$ is actually $N_aM_a$. Now, let us move projectively $P_a$ on $(AB)$, which consequently makes the other six points move projectively as well on the three circles which have the three sides as diameter (because of the right angles). Now, the locus of $N_a$ will be an ellipse passing through $M_a$ with a projective transformation from $P_a$ to $N_a$. This is because we can parametrize in terms of linear combinations of $\cos(2\theta)$ and $\sin(2\theta)$ both points $P_a$ and $Q_a$ on circles $(AB)$ and $(AC)$, so their midpoint will also have such a parametrisation. But a point with such parametrisation is an ellipse, which is in fact obtainable from the original circle $(AB)$ trough an affinity which maps $P_a$ to $N_a$. By taking $\theta=0$ we also see that this ellipse passes through $M_a$. Thus, we can transform projectively $M_aN_a$ into $P_a$, which maps to $P_b$, which maps to $M_bN_b$. Now, if $\theta=0$, which implies $\ell_A$ is the axis of $BC$, the three lines concur in the circumcenter $O$, which is the orthocenter of the medial triangle. Furthermore, if $\theta=\frac{\pi}{2}$, both $P_a$ and $Q_a$ tend to $A$, but since their axis always passes through $M_a$ the limit of $\ell_a$ will be $AM_a$. These three lines concur in the barycenter of $ABC$, which is also the barycenter of the medial triangle. If $\theta=\frac{\pi}{2}-B$, we'll have $P_a=V_a=H_a$ (the foot of the A-altitude), so that $\ell_a$ is also the axis of $Q_aV_a$. Since both points lie on $(AC)$, their axis passes through the center of $(AC)$, which $M_b$. Simmetrically, $\ell_c$ passes through $M_b$ as well, and of course $\ell_b$ always passes through $M_b$. This concurrence symmetrically happens for $\theta=\frac{\pi}{2}-A$ and $\frac{\pi}{2}-C$. Therefore, since by Steiner the locus of $\ell_a\cap \ell_b$ is a conic and it passes through all five points $M_a,M_b,M_c,O,G$, it must be the Kiepert hyperbola of the medial triangle. But this is also trough for $\ell_a\cap \ell_c$, so this means that $\ell_a, \ell_b,\ell_c$ all concur at a point on the Kiepert's hyperbola of the medial triangle.
MarkBcc168
28.06.2022 05:04
Great problem! It really suits my style. [asy][asy] size(10cm,0); defaultpen(fontsize(10)); pair A = (0.5,3.5); pair B = (0,0); pair C = (4,0); pair M_A = (B+C)/2; pair M_B = (A+C)/2; pair M_C = (A+B)/2; pair D = foot(A,B,C); pair O_A = (M_B+M_C)/2 + -2.5*dir(90); pair S_A = reflect(M_C,O_A)*D; pair T_A = reflect(M_B,O_A)*D; draw(A--B--C--cycle, linewidth(1)); draw(circumcircle(D,S_A,T_A),blue); draw(M_B--O_A--M_C,red); draw(M_A--M_B--M_C--cycle); draw(S_A--D--T_A,red+linewidth(0.8)); draw(arc(M_C,abs(A-M_C),degrees(A-M_C),degrees(B-M_C),CW), blue+dashed); draw(arc(M_B,abs(A-M_B),degrees(A-M_B),degrees(C-M_B),CCW), blue+dashed); draw(S_A--T_A,deepgreen); draw((2.7*M_A-1.7*O_A)--(2.7*O_A-1.7*M_A),deepgreen+linewidth(1)); dot("$A$",A,N); dot("$B$",B,SW); dot("$C$",C,SE); dot("$M_A$",M_A,3*dir(105)); dot("$M_B$",M_B,NE); dot("$M_C$",M_C,NW); dot("$D$",D,1.5*dir(270)); dot("$O_A$",O_A,dir(300)); dot("$P_A$",S_A,dir(120)); dot("$Q_A$",T_A,SE); [/asy][/asy] Let $M_A, M_B, M_C$ be the midpoints of $BC$, $CA$, $AB$, respectively. Let $D$ be the foot from $A$ to $BC$, noticing that $ABDP_A$ is concyclic with center $M_C$ and $ACDQ_A$ is concyclic with center $M_B$. Moreover, let $O_A$ be the center of $\odot(DP_AQ_A)$. Observe that $\measuredangle O_AM_CM_B = \measuredangle P_ADA = 90^\circ - \theta$. Similarly, $\angle O_AM_BM_C = 90^\circ-\theta$. Thus, $O_A$ lies on the perpendicular bisector of $M_BM_C$, which happens to be the perpendicular bisector of $DM_A$, so $D,M_A,P_A,Q_A$ are concyclic. Moreover, note that $\angle P_ADM_A = \theta = \angle M_ADQ_A$, so $M_A$ lies on $\ell_A$. Thus, $\ell_A = M_AO_A$. It then suffices to show that $M_AO_A, M_BO_B, M_CO_C$ are concurrent. However, notice that $\triangle O_AM_BM_C$, $\triangle O_BM_CM_A$, and $\triangle O_CM_AM_B$ are similar isosceles triangles (with base angle $90^\circ-\theta$), and we are done by Jacobi's theorem!
MatteD
30.06.2022 00:37
[asy][asy] import graph; size(16.cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-9.,xmax=7.,ymin=-8.,ymax=7.; pen qqffff=rgb(0.,1.,1.); pair A=(-3.8394601798198598,5.945047754642118), B=(-8.,-6.), C=(6.,-6.), M_A=(-1.,-6.), M_B=(1.0802699100900701,-0.02747612267894084), M_C=(-5.91973008990993,-0.02747612267894084), Q_A=(0.5608044489014754,-7.74789976762016), P_A=(-3.334752819088428,-5.7995170852269835), T_A=(4.974376086508739,5.945047754642119); draw(A--B,linewidth(0.8)+red); draw(B--C,linewidth(0.8)+red); draw(C--A,linewidth(0.8)+red); draw(A--T_A,linewidth(0.8)+qqffff); draw(M_A--T_A,linewidth(0.8)+qqffff); draw(M_C--M_B,linewidth(0.8)+red); draw(M_B--(1.9871880432543696,-0.027476122678940396),linewidth(0.8)+red); draw(P_A--T_A,linewidth(0.8)+linetype("4 4")+blue); draw(T_A--Q_A,linewidth(0.8)+linetype("4 4")+blue); dot(A,linewidth(2.pt)+ds); label("$A$",(-3.755726387343544,6.028781547118434),NE*lsf); dot(B,linewidth(2.pt)+ds); label("$B$",(-7.921482563040237,-5.924217328875595),NE*lsf); dot(C,linewidth(2.pt)+ds); label("$C$",(6.0829942286235195,-5.924217328875595),NE*lsf); dot(M_A,linewidth(2.pt)+ds); label("$M_A$",(-1.3693133017685544,-6.4475535318525665),NE*lsf); dot(M_B,linewidth(2.pt)+ds); label("$M_B$",(0.6193642695439372,-0.4815208179150915),NE*lsf); dot(M_C,linewidth(2.pt)+ds); label("$M_C$",(-6.602675331538269,-0.20938599236706634),NE*lsf); dot(Q_A,linewidth(2.pt)+ds); label("$Q_A$",(0.640297717663016,-7.66169352275914),NE*lsf); dot(P_A,linewidth(2.pt)+ds); label("$P_A$",(-3.8394601798198598,-5.5264818146130965),NE*lsf); dot(T_A,linewidth(2.pt)+ds); label("$T_A$",(5.057255270788655,6.028781547118434),NE*lsf); dot((1.9871880432543696,-0.027476122678940396),linewidth(2.pt)+ds); label("$T_A'$",(2.1684394303557726,-0.3559201292006184),NE*lsf); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] As everyone else, show $M_AP_A=M_AQ_A$. Now, let $T_A$ be the point on $\ell_A$ such that $AT_A$ is parallel to $BC$, and $T_A'$ the midpoint of $MT_A$. Now we can apply $\boldsymbol{Al-Kashi's Theorem}$ to triangles $\triangle AT_AP_A$ and $\triangle AT_AQ_A$ to compute $AT_A$. $$P_AT_A^2=AT_A^2+AP_A^2-2\cdot AP_A \cdot AT_A \cos{(\alpha + \gamma - \theta)}=AT_A^2+AP_A^2+2\cdot AP_A \cdot AT_A \cos{(\beta + \theta)}$$$$Q_AT_A^2=AT_A^2+AQ_A^2-2\cdot AQ_A \cdot AT_A \cos{(\gamma + \theta)}$$\begin{align*} \implies AT_A &=\frac{AQ_A^2-AP_A^2}{2(AP_A\cos(\beta+\theta)+AQ_A\cos(\gamma+\theta))}\\ &= \frac{4R^2(\sin^2 \beta - \sin^2 \gamma)\cos^2\theta}{4R(\sin \gamma \cos \theta \cos (\beta + \theta)+\sin \beta \cos \theta \cos (\gamma + \theta))}\\ &= \frac{R(\sin^2 \beta - \sin^2 \gamma)\cos\theta}{\sin \gamma \cos \beta \cos \theta - \sin \gamma \sin \beta \sin \theta + \sin \beta \cos \gamma \cos \theta - \sin \beta \sin \gamma \sin \theta}\\ &= \frac{R(\sin^2 \beta - \sin^2 \gamma)\cos\theta}{\sin \alpha \cos \theta - 2 \sin \beta \sin \gamma \sin \theta}\\ \end{align*}\begin{align*} \frac{M_BT_A'}{M_CT_A'} &=\frac{\frac{BC}{2}-AT_A}{\frac{BC}{2}+AT_A}\\ &= \frac{R \sin \alpha -\frac{R(\sin^2 \beta - \sin^2 \gamma)\cos\theta}{\sin \alpha \cos \theta - 2 \sin \beta \sin \gamma \sin \theta}}{R \sin \alpha +\frac{R(\sin^2 \beta - \sin^2 \gamma)\cos\theta}{\sin \alpha \cos \theta - 2 \sin \beta \sin \gamma \sin \theta}}\\ &= \frac{\sin \alpha (\sin \alpha \cos \theta - 2 \sin \beta \sin \gamma \sin \theta)-(\sin^2 \beta - \sin^2 \gamma)\cos\theta}{\sin \alpha (\sin \alpha \cos \theta - 2 \sin \beta \sin \gamma \sin \theta)+(\sin^2 \beta - \sin^2 \gamma)\cos\theta}\\ &=\frac{(\sin^2 \alpha - \sin^2 \beta + \sin^2 \gamma) \cos \theta -2\sin \alpha \sin \beta \sin \gamma \sin \theta}{(\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma) \cos \theta -2\sin \alpha \sin \beta \sin \gamma \sin \theta} \end{align*}Now, if $T_B'$ and $T_C'$ are defined similarly, we have that the expressions for $\frac{M_CT_B'}{M_AT_B'}$ and $\frac{M_AT_C'}{M_BT_C'}$ cancel out with $\frac{M_BT_A'}{M_CT_A'}$, so $\ell_A, \ell_B, \ell_C$ concur by $\boldsymbol{Ceva's Theorem}$. $\square$
407420
30.06.2022 01:06
MatteD wrote: Now we can apply $\boldsymbol{Al-Kashi's Theorem}$ to triangles $\triangle AT_AP_A$ and $\triangle AT_AQ_A$ to compute $AT_A$. Decirle Teorema de Al-Kashi al Teorema del Coseno es más careta que decirle Fórmula de Bhaskara a la resolvente. Translation: Writing "Al-Kashi's Theorem" is a strange way of writing "Cosine Law".
MatteD
30.06.2022 01:18
GianDR wrote: Decirle Teorema de Al-Kashi al Teorema del Coseno es más careta que decirle Fórmula de Bhaskara a la resolvente.
Darvas
30.06.2022 08:49
Use algebraic methods: Complex, or equations.
spartacle
05.07.2022 08:47
Moving points sketch because why not
Let $\theta$ be a directed angle mod $\pi$, and vary $\theta$ so that lines $AS_A$, etc., etc. all move projectively.
Let $I, J$ be the circle points. $P_A$ and $Q_A$ have degree 2 since they are the intersections of a line of degree one through $A$ with a fixed circle through $A$. Line $P_AQ_A$ has degree $\le 3$ by taking $\alpha = \pi/2$. Further, $P_AQ_A \cap IJ$ has degree $\le 1$ by considering when $AS_A = AI, AJ$. Therefore, the infinity point $\infty_A$ on $\ell_A$, which is the harmonic conjugate of $\infty_A$ wrt $IJ$, has degree $1$. We can also show $P_AI \cap Q_AJ$ has degree 1. Note that $\ell_A$ passes through the harmonic conjugate of $\infty_A$ wrt $P_AQ_A$ (i.e. the midpoint of $P_AQ_A$) and the harmonic conjugate of $\infty_A$ wrt $IJ$, so it passes through $P_AI \cap Q_AJ$. It follows that in fact $\ell_A$ has degree $\le 2$. We can now check that $\ell_A$ passes through the midpoint $M_{A}$ of $BC$ in the cases when $2\alpha \equiv \angle A$ and when $\alpha = 0$, so $\ell_A$ always passes through $M_{A}$. It follows that $\ell_A$ may be redefined as the line through $M_A$ and the harmonic conjugate of $\ell_A$ wrt $IJ$, so in fact $\ell_A$ has degree $\le 1$. Therefore, it suffices to check the problem for 4 cases of $\alpha$. We can check all cases when $\alpha$ is a multiple of $\pi/4$ (though note that for $\alpha = \pi/2$ we must argue that the midpoint of $P_BP_C$ approaches $A$, so $\ell_A$ approaches $AM_{A}$.)
Flash_Sloth
09.07.2022 17:42
My first problem to TSTST! The idea starts with a well known result that $P_A$, $Q_A$ are equal distance to the midpoint of $BC$. With that and trig-ceva can easily solve the problem.
Pathological
22.07.2022 07:55
Let $M_a, M_b, M_c$ be the midpoints of $BC, CA, AB.$ Observe that $\triangle M_a M_c P_a \cong \triangle Q_aM_bM_a$ and so $M_a$ is on the perpendicular bisector of $P_a Q_a.$
Let the perpendicular bisector of $P_aQ_a$ hit $M_bM_c$ at $X_a$, and let $\varphi$ be the homothety at $A$ with factor $2$. Let $P_a', Q_a' = \varphi(P_a), \varphi(Q_a)$ and let the perpendicular bisector of $P_a' Q_a'$ hit $BC$ at $Y_a.$ Define symmetric points analogously. Then we have $Y_a = \varphi(X_a)$ and it suffices to show $AY_a, BY_b, CY_c$ concurrent.
Recall that the function $f: \mathbb R^2 \rightarrow \mathbb R$ given by $f(P) = PP_a'^2 - PQ_a'^2$ is linear, so that we have (as $f(Y_a) = 0$):
$$\frac{BY_a}{Y_aC} = - \frac{f(B)}{f(C)} = - \frac{BP_a'^2 - BQ_a'^2}{CP_a'^2 - CQ_a'^2} = - \frac{c^2 - (a^2 + b^2 - 2 a b \cos (180 - \angle C - 2 \theta))}{a^2 + c^2 - 2 a c \cos (180 - \angle B - 2 \theta) - b^2}.$$
After obtaining symmetric expressions for $CY_b/Y_bA$ and $AY_c/Y_cB$ the product will easily cancel and we are done by Ceva.
NoctNight
22.06.2023 13:25
Let $M_a, M_b, M_c$ be the midpoints of $BC, CA, AB$ respectively and let $X, Y$ be the feet from $C$ to $AS_A$ and $B$ to $AT_A$ respectively. Notice that $$\frac{AX}{AY}=\frac{AC}{AB}=\frac{AQ_A}{AP_A}$$so by power of a point, $XYQ_AP_A$ is cyclic. But $M_a$ lies on both perpendicular bisectors of $XP_A$ and $YQ_A$ so it is $(XYP_AQ_A)$'s centre. Thus, $\ell_A$ is the line through $M_a$ perpendicular to $P_AQ_A$. Let $D=BP_A\cap CQ_A$. Define $E,F$ analogously. By trig-Ceva's and isogonal conjugation (because $\triangle M_aM_bM_c$ and $\triangle ABC$ are homothetic), it suffices to show $AD, BE, CF$ concur which follows from Jacobi's theorem because $\angle P_ABC=90^{\circ}-\theta-\angle ABC=\angle Q_CBC$ and the other angles follow similarly.
bhan2025
15.09.2023 10:09
An alternate continuation of #6 after the Claim that $(D,M_A,P_A,Q_A)$ is cyclic: Let $K_A = BP_A \cap CQ_A$ By Jacobi's theorem we can see that $AK_A$, $BK_B$, and $CK_C$ are coincident Taking the iso-conjugate of the jacobi point gives us that the lines crossing $A$, $B$, and $C$ perpendicular to $P_AQ_A$, $P_BQ_B$, and $P_CQ_C$ are coincident. The similarity of $ABC$ with $M_AM_BM_C$ gives us the wanted result.
GrantStar
22.06.2024 18:37
Let $D$ be the foot from $A$ to $BC$. Let $M_A$ and others be the midpoints. Claim: $D, M_A, P_A, Q_A$ are cyclic. Proof. Invert about $D$ passing through $A$. Since $M_A$ is sent to the point $T$ with $(TD; B’C’)=-1$. But since $AD$ bisects $\angle P_A’D Q_A’$ we finish by the right angle and angle bisectors lemma. Now it is easy to angle chase that $M_AP_A=M_AQ_A$. Let $O_A$ be the circumcenter of the circle in the claim. Note that $O_A$ lies on the perpendicular bisector of $M_BM_C$. Then intersecting $O_AM_B$ with $DP_A$ we get a right angle and similar relation should on the other side and as $\angle P_ADQ_A=2\theta $ Jacobi finishes.