thanhnam2902 wrote: Let circle $ \omega (I;r)$ be incircle of the $ \triangle ABC$.Let $ D$ be the point of tangency $ \omega$ with $ BC$. Let $ E, F$ be the midpoints of $ BC$ and $ AD$ respectively.Prove that the points $ I, E, F$ are collinear. Proof: Let $ K$ be the point of intersection $ ID$ with $ \omega$ and $ G$ be the point of intersection $ AK$ and $ BC$.Then it is well-known that $ GE = ED$ and $ ID = IK = r$,hence $ IE\|AG$,but at the same time $ EF\|AG$,so $ I,E,F$ are collinear. Best regards,Erken.

Hi!! I think this problem has a mass center´s solution. Is anybody know that?

Another hint: Use $ 2\overrightarrow{IE} = \overrightarrow{IB} + \overrightarrow{IC},2\overrightarrow{IF} = \overrightarrow{IA} + \overrightarrow{ID}$ And $ a\overrightarrow{IA} + b\overrightarrow{IB} + c\overrightarrow{IC} = \overrightarrow{0}$ $ \overrightarrow{ID} = \frac {p - c}{a}\overrightarrow{IB} + \frac {p - b}{a}\overrightarrow{IC}$

Hi $ Erken$, I think you not true. Can you answer me why $ GE = ED$ ? Thank you very much.

thanhnam2902 wrote: Hi $ Erken$, I think you not true. Can you answer me why $ GE = ED$ ? At first,if the fact,that was written by me,is wrong,then your problem is wrong,which is obviously not so. Here is the proof of this well-known fact: Image not found Let $ l$ be a line through $ K$,which is parallel to $ BC$,then it's tangent to $ \omega$ at $ K$.So the point $ K$,is the point of tangency $ \omega$ with $ MN$(see the image),but $ \omega$ is $ A$-exincircle of $ \triangle AMN$,hence $ G$ is the tangency point of $ A-exincirlce$ of the $ \triangle ABC$ with $ BC$,now it is obvious that $ CG = BD = s - c$,where $ s$ is a semi-perimeter,and $ c$ is a length of the side $ AB$.So from $ CG = BD$ and $ BE = EC$,we conclude that $ DE = EG$.

Oh! I see. Because your last post was very sort. Now I understand. Thank you very much. I wish you good luck.

thanhnam2902 wrote: Let $ w=(I,r)$ be the incircle of the triangle $ ABC$ . Let $ D$ be the contact point of the circle $ w$ with the side $ [BC]$ . Let $ E, F$ be the midpoints of the segments $ [BC]$ and $ [AD]$ respectively . Prove three point $ I\in EF$ . Remark. The line $ \overline {EIF}$ is the Newton's line of the degenerate convex quadrilateral $ ABDC$ circumscribed to the circle $ w$ .

Let $ P$ be a point in the plane of triangle $ ABC$ and points $ \{ P_a \} = AP \cap BC$, $ \{ P_b\} = BP \cap AC$, $ \{ P_c \} = CP \cap AB$. Let $ M_a$, $ M_b$, $ M_c$ the midpoints of math=inline]$ AP_a$[/math, math=inline]$ BP_b$[/math, math=inline]$ CP_c$[/math, respectively; and $ N_a, N_b, N_c$ the midpoints of math=inline]$ BC$[/math, math=inline]$ AC$[/math, math=inline]$ AB$[/math, respectively. Prove that $ M_aN_a$, $ M_bN_b$ and $ M_cN_c$ are concurent! The point $ \{ O \} = M_aN_a \cap M_bN_b \cap M_cN_c$ is the centre of conic touches the lines BC, CA and AB at points $ P_a, P_b, P_c$ respectively.

Thank Erken, mihai miculita, Virgil Nicula, tdl, conejita for your reply. I wish you good luck. . Good bye! See you again.

Hi tdl ! Can you post full your slove by vecto? Thank you very much.

$ 2\overrightarrow{IE} = \overrightarrow{IB} + \overrightarrow{IC},2\overrightarrow{IF} = \overrightarrow{IA} + \overrightarrow{ID}$ And well-known that: $ a\overrightarrow{IA} + b\overrightarrow{IB} + c\overrightarrow{IC} = \overrightarrow{0}\Leftrightarrow \overrightarrow{IA} =-\frac {b}{a}\overrightarrow{IB} -\frac{c}{a} \overrightarrow{IC}$ $ \overrightarrow{ID} = \frac {CD}{BC}\overrightarrow{IB} + \frac {BD }{BC}\overrightarrow{IC}=\frac {p - c}{a}\overrightarrow{IB} + \frac {p - b}{a}\overrightarrow{IC}$ Then $ 2\overrightarrow{IF} = \overrightarrow{IA} + \overrightarrow{ID}=\frac{p-b-c}{a}(\overrightarrow{IB} + \overrightarrow{IC})=2\frac{p-b-c}{a}\overrightarrow{IE}$ Then OK!

Thank tdl very much.

thanhnam2902 wrote: Let $ w = (I,r)$ be the incircle of the triangle $ ABC$ . Let $ D$ be the contact point of the circle $ w$ with the side $ [BC]$ . Let $ E, F$ be the midpoints of the segments $ [BC]$ and $ [AD]$ respectively . Prove three point $ I\in EF$ . Remark. The line $ \overline {EIF}$ is the Newton's line of the degenerate convex quadrilateral $ ABDC$ circumscribed to the circle $ w$ . Direct proof of the proposed problem. Observe easily that $ [EBD] + [EAC] =$ $ [FBD] + [FAC] =$ $ [IBD] + [IAC] = \frac S2$ . The geometrical locus of a interior point $ X$ for which $ [XBD] + [XAC] = k$ (constant) is a segment (well-known). Thus, $ I\in EF$ . ============================================================= mihai miculita wrote: Let $ P$ be a point in the plane of triangle $ ABC$ and points $ \{ P_a \} = AP \cap BC$, $ \{ P_b\} = BP \cap AC$, $ \{ P_c \} = CP \cap AB$. Let $ M_a$, $ M_b$, $ M_c$ the midpoints of math=inline]$ AP_a$[/math, math=inline]$ BP_b$[/math, math=inline]$ CP_c$[/math, respectively; and $ N_a, N_b, N_c$ the midpoints of math=inline]$ BC$[/math, math=inline]$ AC$[/math, math=inline]$ AB$[/math, respectively. Prove that $ M_aN_a$, $ M_bN_b$ and $ M_cN_c$ are concurent! The point $ \{ O \} = M_aN_a \cap M_bN_b \cap M_cN_c$ is the centre of conic touches the lines BC, CA and AB at points $ P_a, P_b, P_c$ respectively. Prove easily (with the barycentrical coordinates) that $ P(a,b,c)$ $ \Longrightarrow$ $ O(x,y,z)$ , where $ \frac {x}{a(b + c)} = \frac {y}{b(c + a)} = \frac {z}{c(a + b)}$ .

mihai miculita wrote: Let $ X$ be a point in the plane of triangle $ ABC$ . Denote the intersections $ \|\begin{array}{c} D\in AX\cap BC\\\ E\in BX\cap CA\\\ F\in CX\cap AB\end{array}$ , the middlepoints $ M$ , $ N$ , $ P$ of the sides $ [BC]$ , $ [CA]$ , $ [AB]$ and the middlepoints $ U$ , $ V$ , $ W$ of the segments $ [AD]$ , $ [BV]$ , $ [CW]$ respectively. Prove that the lines $ UM$ , $ VN$ , $ WP$ are concurently. Proof (with the barycentrical coordinates). $ \|\begin{array}{ccc} A(1,0,0) & B(0,1,0) & C(0,0,1)\\\\ M(0,1,1) & N(1,0,1) & P(1,1,0)\end{array}$ and $ \begin{array}{c} \blacktriangleright\ \ X\ (\ a\ ,\ b\ ,\ c\ )\ \ \blacktriangleleft\\\ ==================\\\ \|\begin{array}{ccc}D(0,b,c) & , & U(b+c,b,c)\\\\ E(a,0,c) & , & V(a,c+a,c)\\\\ F(a,b,0) & , & W(a,b,a+b)\end{array}\|\\\ ==================\end{array}$ . $ UM\cap VN\cap WP\ne\emptyset$ $ \Longleftrightarrow$ $ (\exists )\ Y(x,y.z)\in UM\cap VN\cap WP$ so that $ |\begin{array}{ccc} x & y & z\\\ 0 & 1 & 1\\\ b+c & b & c\end{array}|=|\begin{array}{ccc} x & y & z\\\ 1 & 0 & 1\\\ b & c+a & c\end{array}=|\begin{array}{ccc} x & y & z\\\ 1 & 1 & 0\\\ a & b & a+b\end{array}|=0\ \Longleftrightarrow$ $ \|\begin{array}{ccc} c-b & b+c & -(b+c)\\\ -(c+a) & a-c & c+a\\\ a+b & -(a+b) & b-a\end{array}\|\bullet\|\begin{array}{c} x\\\ y\\\ z\end{array}\|=\|\begin{array}{c} 0\\\ 0\\\ 0\end{array}\|$ . Observe that (in the upper system) the matrix $ \mathcal A$ of the coeficients is singularly, i.e. $ \det \mathcal A=0$ because the sum of the its lines is equally to $ 0$ . Thus this system is compatiblely and $ \frac {x}{(b+c)(c+a)+(b-c)(b+c)}=$ $ \frac {y}{(b+c)(c+a)+(b-c)(b+a)}=$ $ \frac {z}{c+a)(c+b)+(c-b)(a-c)}$ , i.e. $ \boxed {\ \frac {x}{a(b+c)}=\frac {y}{b(c+a)}=\frac {z}{c(a+b)}\ }$ . Thus the lines $ UM$ , $ VN$ , $ WP$ are concurrently in $ Y[\ a(b+c)\ ,\ b(c+a)\ ,\ c(a+b)\ ]$ .

See also http://www.mathlinks.ro/Forum/viewtopic.php?t=5811 http://www.mathlinks.ro/Forum/viewtopic.php?t=112509 http://www.mathlinks.ro/Forum/viewtopic.php?t=106337 http://www.mathlinks.ro/Forum/viewtopic.php?t=15932 for the same question with excircle instead of incircle darij

Thanks darij grinberg very much. I wish you good luck.

Here is my solution using complex numbers: Let the incircle of the triangle be the unit circle and let $P$ and $Q$ be the points of tangency to sides $AB$ and $AC$. Now, we have $a=\frac {2pq}{p+q}$, $b=\frac{2pd}{p+d}$ and $c=\frac {2qd}{q+d}$. We can easily continue our computations: $e=\frac{dq}{d+q}+\frac{dp}{d+p}$ and $f=\frac{pq}{p+q}+\frac{d}{2}$. Now, we need ony to show that $E, I, F$ are collinear, which is equivqlent to the condition $e\bar{f}=f\bar{e}$, or $(\frac{dq}{d+q}+\frac{dp}{d+p})(\frac{1}{p+q}+\frac{1}{2d})=(\frac{pq}{p+q}+\frac{d}{2})(\frac{1}{d+q}+\frac{1}{d+p})$. Multipying all by $(d+q)(d+p)(p+q)$, and expanding, we get $4dpq+2p^2q+2q^2p+2d^2p+p^2d+dpq+2d^2q+q^2d+dpq$ $=2d^2p+4dpq+2d^2q++dp^2+dpq+2p^2q+2q^2p+dpq+dq^2$, which is obvious.

Dear Mathlinkers, 1. R the point of intersection of EI and the A-altitude. Prove : AR = r Sincerely Jean-Louis

jayme wrote: Dear Mathlinkers, 1. R the point of intersection of EI and the A-altitude. Prove : AR = r Sincerely Jean-Louis My solution: Let $ D' $ be the antipode of $ D $ in $ \odot (I) $ . Since $ I $ is the complement of the Nagel point of $ \triangle ABC $ , so $ EI $ is parallel to the $ A- $ Nagel line $ AD' $ of $ \triangle ABC \Longrightarrow AD'IR $ is a parallelogram $ \Longrightarrow AR=D'I=\text{r} $ . Q.E.D