An important lemma of isogonal conjugate points

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An important lemma of isogonal conjugate points

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#1 h Mar 23, 2025, 10:18 AM

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Let $P$ and $Q$ be two isogonal conjugate with respect to triangle $ABC$ . Let $S$ and $T$ be two points lying on the circle $(PBC)$ such that $PS$ and $PT$ are perpendicular and parallel to bisector of $\angle BAC$ , respectively. Prove that $QS$ and $QT$ bisect two arcs $BC$ containing $A$ and not containing $A$ , respectively, of $(ABC)$ .

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#2 h Monday at 10:16 AM • 1 Y

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Dear teacher Buratinogigle, this is my solution for this problem.
Solution

This post has been edited 1 time. Last edited by Ktoan07, Monday at 10:17 AM

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Ritwin

#3 h Monday at 7:50 PM • 1 Y

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Dear teacher Buratinogigle, I apologize for bashing this aesthetic problem.

We employ barycentric coordinates (though the setup looks like trilinear coordinates). By extraversion it suffices to show just one of the intersections, so we will only deal with the internal $A$ -bisector. (For the external $A$ -bisector, just change $a$ to $-a$ in all computations.) Set $P = (a \alpha : b \beta : c \gamma), \quad Q = (a/\alpha : b/\beta : c/\gamma),$ and abbreviate $p = a \alpha + b \beta + c \gamma$ and $q = a/\alpha + b/\beta + c/\gamma$ .
The circle $(BPC)$ is given by $u x (x+y+z) = a^2 yz + b^2 zx + c^2 xy, \quad u = b c \beta \gamma \frac{a/\alpha + b/\beta + c/\gamma} {a \alpha + b \beta + c \gamma} = b c \beta \gamma \cdot \frac qp.$ Then $S = (a \alpha - (b+c) \kappa : b (\beta+\kappa) : c (\gamma+\kappa))$ for some $\kappa$ , and $S \in (BPC)$ . Plugging this in and expanding yields $\begin{align*} \text{LHS} &= b c \beta \gamma q (a \alpha - (b+c) \kappa), \\ \text{RHS} &= a^2 yz + b^2 zx + c^2 xy \\ &= a^2 b c (\beta + \kappa) (\gamma + \kappa) + b^2 c (\gamma + \kappa) (a \alpha - (b+c) \kappa) + c^2 b (\beta + \kappa) (a \alpha - (b+c) \kappa) \\ &= a b c \alpha \beta \gamma q + b c (a^2 (\beta + \gamma) + (b+c) (a \alpha - b \gamma - c \beta)) + b c (a^2 - (b+c)^2) \kappa^2. \end{align*}$ We may ignore $\kappa = 0$ as it corresponds to $S = P$ . Thus $-(b+c) \beta \gamma q = a^2 (\beta+\gamma) + (b+c) (a\alpha - b\gamma - c\beta) + (a^2-(b+c)^2) \kappa,$ which means $\begin{align*} \kappa &= \frac{(b+c) (b\gamma + c\beta - a\alpha - \beta \gamma q) - a^2 (\beta + \gamma)}{a^2 - (b+c)^2} \\ &= \frac{a (b+c)(\alpha + \beta \gamma / \alpha) - a^2 (\beta+\gamma)} {(b+c)^2 - a^2} \\ &= \frac{(\alpha + \beta \gamma/\alpha) - \tfrac{a}{b+c} (\beta+\gamma)} {\tfrac{b+c}{a} - \tfrac{a}{b+c}}. \end{align*}$ The internal $A$ -bisector meets $(ABC)$ again at $M = (-a^2 : b(b+c) : c(b+c))$ . We want to check $\overline{QMS}$ collinear, which is equivalent to the determinant $\begin{align*} 0 &\overset{?}= \det \begin{bmatrix} Q \\ M \\ S \end{bmatrix} \\ &= \det \begin{bmatrix} a/\alpha & b/\beta & c/\gamma \\ -a^2 & b(b+c) & c(b+c) \\ a \alpha - (b+c) \kappa & b (\beta + \kappa) & c (\gamma + \kappa) \end{bmatrix} \\ &= abc \det \begin{bmatrix} 1/\alpha & 1/\beta & 1/\gamma \\ -a & b+c & b+c \\ \alpha - \tfrac{b+c}{a} \kappa & \beta+\kappa & \gamma+\kappa \end{bmatrix} \\ &= abc \det \begin{bmatrix} 1/\alpha & 1/\beta & 1/\gamma \\ -a & b+c & b+c \\ \alpha + (\tfrac{a}{b+c} - \tfrac{b+c}{a}) \kappa & \beta & \gamma \end{bmatrix} \\ &= abc \det \begin{bmatrix} 1/\alpha & 1/\beta & 1/\gamma \\ -a & b+c & b+c \\ \tfrac{a}{b+c} (\beta+\gamma) - \tfrac{\beta\gamma}{\alpha} & \beta & \gamma \end{bmatrix}. \end{align*}$ The above matrix has determinant zero because $\begin{bmatrix} \beta \gamma (b+c) \\ (\beta+\gamma) \\ (b+c) \end{bmatrix}^\intercal$ is a left eigenvector, so $M \in QS$ and we are done. $\blacksquare$

This post has been edited 1 time. Last edited by Ritwin, Monday at 7:53 PM

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