Dear Mathlinkers, for the beginning AP is the A-symmedian of ABC. Sincerely Jean-Louis

Let $ M,N,L$ be the midpoints of $ CA,AB,BC.$ $ \mathcal{M}_1 \equiv \odot(ABM),$ $ \mathcal{M}_2 \equiv \odot(ACN)$ and $\mathcal{M}_3 \equiv \odot(AMN).$ Homothety $ (A,\frac {_3}{^2})$ carries $\mathcal{M}_3$ into $\mathcal{M}_3',$ which passes through the intersections of $AC,AB$ with the parallels through $ L$ to the medians $ BM, CN.$ Now, it suffices to show thaw $ \mathcal{M}_3', \mathcal{M}_1 , \mathcal{M}_2$ are coaxal. Their barycentric equations are: $ \mathcal{M}_3' \equiv [0,\frac {_1}{^4}c^2,\frac {_1}{^4}b^2] \ , \ \mathcal{M}_1 \equiv{} [0,0,\frac {_1}{^2}b^2] \ , \ \mathcal{M}_2 \equiv{} [0,\frac {_1}{^2}c^2,0]$ Radical axis of $ \mathcal{M}_1,\mathcal{M}_2$ is the line $ b^2z - c^2y = 0$ (A-symmedian) and the radical axis of $ \mathcal{M}_3',\mathcal{M}_2$ is the line $ b^2z - c^2y = 0$ $\Longrightarrow$ $ \mathcal{M}_3', \mathcal{M}_1 , \mathcal{M}_2$ are coaxal, as desired.

uglysolutions wrote: Let $ ABC$ be a triangle, $ B_1$ the midpoint of side $ AB$ and $ C_1$ the midpoint of side $ AC$. Let $ P$ be the point of intersection ($ \neq A$) of the circumcircles of triangles $ ABC_1$ and $ AB_1C$. Let $ Q$ be the point of intersection ($ \neq A$) of the line $ AP$ and the circumcircle of triangle $ AB_1C_1$. Prove that $ \frac {AP}{AQ} = \frac {3}{2}$. Let $ CB_1\cap (AB_1C_1)=D, CP\cap (ABC_1)=L, BC\cap (ABC_1)=E, B_1C\cap BC_1=O, S$ be a point on segment $ C_1C$ such that $ \frac{C_1S}{SC}=\frac{1}{3}$ Since $ CC_1.CA=CE.CB=CD.CB_1$ we get $ B_1DEB$ is a cyclic quadrilateral. Moreover $ AB_1DC_1$ is a cyclic quadrilateral, so $ EDC_1C$ is a cyclic quadrilateral. Then $ \angle DC_1E=\angle DCE (1)$ On the other side, $ \angle BLP+\angle B_1CL=\angle BLP+\angle BAP=180^o$ then $ BL//B_1C$ Thus $ \angle LC_1E=\angle LBC=\angle BCB_1 (2)$ From (1) and (2) we have $ \angle LC_1E=\angle DC_1E$ therefore $ L,D,C_1$ are collinear. Because $ BL//OD$ then $ \frac{C_1D}{C_1L}=\frac{C_1O}{C_1B}=\frac{1}{3}=\frac{C_1S}{C_1C}$ $ \Rightarrow DS//LC$ We have $ \angle DQA=\angle DB_1A=\angle CPA \Rightarrow DQ//PC$ So Q,D,S are collinear we get $ QS//PC$ Hence $ \frac{AP}{AQ}=\frac{AC}{AS}=\frac{3}{2}$

uglysolutions wrote: Let $ ABC$ be a triangle, $ B_1$ the midpoint of side $ AB$ and $ C_1$ the midpoint of side $ AC$. Let $ P$ be the point of intersection ($ \neq A$) of the circumcircles of triangles $ ABC_1$ and $ AB_1C$. Let $ Q$ be the point of intersection ($ \neq A$) of the line $ AP$ and the circumcircle of triangle $ AB_1C_1$. Prove that $ \frac {AP}{AQ} = \frac {3}{2}$. Inversion kills it Make an inversion with radius $ r$ and center $ A$. For every point $ P$ let $ P'$ denote the inversed point. Then $ \frac{AC_1'}{AC'} = \frac{AC}{AC_1} = 2$, and likewise with $ B$, so $ B',C'$ are midpoints of sides $ A'B_1'$ and $ A'C_1'$. The circles around $ \triangle ABC_1$, $ \triangle AB_1C$, and $ \triangle AB_1C_1$ get transformed into lines passing through $ B'C_1'$, $ B_1'C$, and $ B_1'C_1'$ respectively. $ B'C_1'$ and $ B_1'C'$ intersect at $ P'$, which must be the centroid of $ \triangle AB_1'C_1'$. Then $ Q'$ must be the intersection between $ AP'$ and $ B_1'C_1'$. And it is wellknown that $ \frac{AQ'}{AP'} = \frac{3}{2}$. But $ \frac{AP}{AQ} = \frac{AQ'}{AP'} = \frac{3}{2}$. QED

Like Balkan's 2nd problem

As $ \angle B_1BP = \angle CC_1P$ and $ \angle BB_1P = \angle C_1CP$ (from the two circles) so $ \bigtriangleup B_1BP \sim \bigtriangleup CC_1P$ thus $ \frac{PB_1}{PC} = \frac{BB_1}{C_1C} = \frac{AB}{AC}$, i.e. $ AC \times B_1P = AB \times PC$ As $ \angle BAQ = \angle B_1C_1Q$ and $ \angle ABQ = \angle C_1B_1Q$ (from the two circles) so $ \bigtriangleup B_1QC_1 \sim \bigtriangleup BQA$ As $ \angle B_1C_1Q = \angle B_1AQ = \angle B_1CP$ and $ \angle B_1QC_1 = \pi - \angle B_1AC_1 = \angle BPC$ so $ \bigtriangleup B_1PC \sim \bigtriangleup B_1QC_1 \sim \bigtriangleup BQA$ thus $ \frac{AB}{B_1C} = \frac{AQ}{PC}$ i.e. $ B_1C = \frac{AB \times PC}{AQ}$ By Ptolemy, $ AP = \frac{AB_1 \times PC + AC \times B_1P}{B_1C} = \frac{AB_1 \times PC + AB \times PC}{\frac{AB \times PC}{AQ}} = \frac{3}{2}AQ$

Dear Mathlinkers, 0. Let Q the seond point of intersection of AP with the circumcircle of AB1C1 1. AP is the A-symmedian of ABC 2. the circumcircle of AB1C1 cut AQ in his midpoint like for example the circle going through A, B and tangent to AC at A... 3. Now we have to determine the nature of this last circle and we are done without any calculation... Sincerely Jean-Louis

Dear Mathlinkers, from where comes this nice problem? Sincerely jean-Louis

There're many proofs but it seems none of them tries this approach . Let $B',C'$ are centers of circumcircle $\triangle ABC_1$ , $\triangle ACB_1$ respectively $AB_2,AC_2$ are diameter of $(B');(C')$ respectively .$O$ is center of $(ABC)$ $\angle APB_2=\angle APC_2=90 \Rightarrow B_2,P,C_2$ are collinear . $\angle AB_1B'= \angle AB_1C_2=90 \Rightarrow B',B_1,C_2$ are collinear (by similar way ) $ C',C_1,B_2$ are collinear . and since that ,tt's easy to see : $B'C_2,C'B_2$ meet each other at $O$. But when we considered $\triangle AB_2C_2 $ ,we knew : $B'C_2,C'B_2$ are medians of $AB_2C_2$ HEnce $O$ is centroid of $\triangle AB_2C_2 \Rightarrow \frac{ AM}{AO}=\frac{3}{2}$(where $M$ is midpoint of $B_2C_2$ $OQ|| MP( \perp AP) \Rightarrow \frac{AM}{AO}=\frac{AQ}{AP}$ Done

brianchung11 wrote: As $ \angle B_1BP = \angle CC_1P$ and $ \angle BB_1P = \angle C_1CP$ (from the two circles) so $ \bigtriangleup B_1BP \sim \bigtriangleup CC_1P$ thus $ \frac{PB_1}{PC} = \frac{BB_1}{C_1C} = \frac{AB}{AC}$, i.e. $ AC \times B_1P = AB \times PC$ As $ \angle BAQ = \angle B_1C_1Q$ and $ \angle ABQ = \angle C_1B_1Q$ (from the two circles) so $ \bigtriangleup B_1QC_1 \sim \bigtriangleup BQA$ As $ \angle B_1C_1Q = \angle B_1AQ = \angle B_1CP$ and $ \angle B_1QC_1 = \pi - \angle B_1AC_1 = \angle BPC$ so $ \bigtriangleup B_1PC \sim \bigtriangleup B_1QC_1 \sim \bigtriangleup BQA$ thus $ \frac{AB}{B_1C} = \frac{AQ}{PC}$ i.e. $ B_1C = \frac{AB \times PC}{AQ}$ By Ptolemy, $ AP = \frac{AB_1 \times PC + AC \times B_1P}{B_1C} = \frac{AB_1 \times PC + AB \times PC}{\frac{AB \times PC}{AQ}} = \frac{3}{2}AQ$ can you explain why $\angle{ABQ}=\angle{C_1B_1Q}$ ?