Triangle $ ABC$ is acute. It has two bisectors $ AP,BQ$ and two altitudes $ AD,BE.$ $ O,I$ are the circumcenter and incenter of triangle $ ABC$ respectively. Show that $ P,O,Q$ are collinear if and only if $ D,I,E$ are collinear. Approach by TTsphn: We will prove that i) $ P,O,Q$ are collinear if and only if : $ \cos{C} = \cos{A} + \cos {B}$ $ \vec{OP} = \frac {1}{AB + BC}.(BC.\vec{OA} + AB.\vec {OC}$ $ \vec{OQ} = \frac {1}{AB + AC}(AC.\vec{OB} + AB.\vec{OC})$ $ O,P,Q$ are collinear if and only if : $ \vec{OP} \bigwedge \vec{OQ} = 0$ $ \Leftrightarrow (BC.\vec{OA} + AB.\vec{OC})\bigwedge (AC.\vec{OB} + AB.\vec{OC}) = 0$ $ \Leftrightarrow BC.AC.sin(\vec{OA},\vec{OB}) + AB.AC.sin(\vec{OC,OB}) + BC.AB.\sin(\vec{OA},\vec{OC}) = 0$ $ \Leftrightarrow \cos {C} = \cos {A} + \cos{B}$ ii) $ D,I,E$ are collinear if and only if $ \cos{C} = \cos{A} + \cos{B}$ This is siminar to above problem. So problem claim.

$ \mbox{In barycentric coordinate, we: } I(a;b;c); P(0;b;c); Q(a;0;c) \\ H(\tan{A};\tan{B};\tan{C}); D(0;\tan{B};\tan{C});E(\tan{A};0;\tan{C}) \mbox{ and } O(\sin{2A};\sin{2B};\sin{2C}). \\ I\in{DE}\Leftrightarrow \left|\begin{array}{ccc} a & b & c \\ 0 & {\tan{B}} & {\tan{C}} \\ {\tan{A}} & 0 & {\tan{C}} \end{array}\right| = 0\Leftrightarrow \left|\begin{array}{ccc} a & b & c \\ 0 & {\frac {b}{2R.\cos{B}}} & {\frac {c}{2R.\cos{C}}} \\ {\frac {a}{2R.\cos{A}}} & 0 & {\frac {c}{2R.\cos{C}}} \end{array}\right| = 0\Leftrightarrow \\ \Leftrightarrow \left|\begin{array}{ccc} 1 & 1 & 1 \\ 0 & {\frac {1}{\cos{B}}} & {\frac {1}{\cos{C}}} \\ {\frac {1}{\cos{A}}} & 0 & {\frac {1}{\cos{C}}} \end{array}\right| = 0\Leftrightarrow \cos{A} + \cos{B} = \cos{C}; \\ \mbox{and} \\ O\in{PQ}\Leftrightarrow \left|\begin{array}{ccc} 0 & b & c \\ a & 0 & c \\ {\sin{2A}} & {\sin{2B}} & {\sin{2C}} \end{array}\right| = 0\Leftrightarrow \\ \Leftrightarrow \left|\begin{array}{ccc} 0 & b & c \\ a & 0 & c \\ {\frac {a}{R}.\cos{A}} & {\frac {b}{R}.\cos{B}} & {\frac {c}{R}.\cos{C}} \end{array}\right| = 0\Leftrightarrow \left|\begin{array}{ccc} 0 & 1 & 1 \\ 1 & 0 & 1 \\ {\cos{A}} & {\cos{B}} & {\cos{C}} \end{array}\right| = 0\Leftrightarrow \\ \Leftrightarrow \cos{A} + \cos{B} = \cos{C}. \\ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - \\ \Rightarrow [I\in{DE}\Leftrightarrow O\in{PQ}].$

Lemma: Given triangle $ ABC, I$ lies inside and $ I(\alpha, \beta, \gamma). BI\cap AC=\{B'\}, CI\cap AB=\{C'\}$. P is an arbitrary point on $ B'C'. d_a, d_b, d_c$ be the distances from P to three sides $ BC, CA, AB$. We have $ \frac{a}{\alpha}.d_a=\frac{b}{\beta}d_b+\frac{c}{\gamma}.d_c$ Proof: $ B'(\alpha, 0, \gamma)$ and $ C'(\alpha, \beta, 0) \Rightarrow B'C': \beta.\gamma.x+\alpha.\gamma. y+\alpha.\beta.z=0$ $ \Leftrightarrow \frac{x}{\alpha}=\frac{y}{\beta}+\frac{z}{\gamma}$ $ P\in B'C'$ therefore $ \frac{S_{BPC}}{\alpha}=\frac{S_{APC}}{\beta}+\frac{S_{APB}}{\gamma}$ or $ \frac{a}{\alpha}.d_a=\frac{b}{\beta}d_b+\frac{c}{\gamma}.d_c$ Return to our problem: Let $ d_a, d_b, d_c$ be the distances from O to three sides $ BC, CA, AB$. Since $ I(a,b,c)$ we get $ O$ lies on $ PQ$ iff $ d_a=d_b+d_c$ iff $ \frac{a}{tan A}=\frac{b}{tan B}+\frac{c}{tan C}$ iff I lies on $ DE$ (because $ H(tan A, tan B, tan C)$ and the distances from I to three sides are equal)

isnt there anyother solutions?

WJ.JamshiD wrote: isnt there anyother solutions? See the topic Collinear for the generalization and solution

Dear Mathlinkers, you can see http://jl.ayme.pagesperso-orange.fr/Docs/Deux%20couples%20de%20points%20isogonaux.pdf p. 8 Sincerely Jean-Louis

Thanks