老封(Ye):If $D$ is the midpoint of $BC$, $F$ is the midpoint of $AB$, than $T{T_B} \cap {T_A}{T_C}$ is on the $DF$.

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Not very good solutuion : Lemma 1 : given quadrangle ABCD and line l , let l intersect AB , BC , CA , BD , AD , CD at points X_1 , X_2 , X_3 , X_4 , X_5 , X_6 , then for every point P circles (PX_1X_6) , (PX_2X_5) , (PX_3X_4) intersects at another point P' and for points P_1 , P_2 , P_3 circles (P_1X_1X_6) , (P_2X_2X_5) , (P_3X_3X_4) have Radical center R and R is intersection of l and PP' Lemma 2 : let circle w tangent to sides AB , BC and CA at points C_1 , A_1 , B_1 , C' A' B' are midpoints of sides . w is tangent to (C'A'B') at point F , then lenes C_1F , A'B' , A_1B_1 intersects at one point Descovered by F.Ivlev Solution : Let E is midpoint of CA see picture let A_1C_1 and A_2C_2 intersect DF at points X and X' let C_3A_4 and A_3C_4 intersect DF at points Y and Y' Use Lemma 2 So B_2T_B goes thru X , TB_3 goes thru X' , T_CB_4 goes thru Y' and B_1D goes thru Y Use Reim's Theorem , so TX'XT_B and Y'YT_AT_C are cyclic Let A_4C_3 intersect A_1C_1 and A_2C_2 at points P and Q Let C_4A_3 intersect A_1C_1 and A_2C_2 at points P' and Q' Well known that AQ'QQCPP' are cyclic and CA is diameter and Q'QPP' is rectangle , so Q'P and QP' are intersected at point E And easy to see that D is on QP' and F is on Q'P Use Lemma 1 for quadrangle Q'QPP' and line DF , so Radical center of (TX'XT_B) , (Y'YT_AT_C) and Nine point circle is on line DF , let it is point R So TT_b and T_AT_C intersects at point R on DF . Done

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ca you prove the lemmas please? is the official solution on the internet?

Theorem: $F$ is the Feuerbach point of $\triangle ABC$ and $F_A,F_B,F_C$ are the three additional Feuerbach points against $A,B,C.$ Then $FF_A$ and $F_BF_C$ intersect at the A-Schroeter point $U_A$ of $\triangle ABC$ (intersection of the A-sidelines of the medial and orthic triangle). Since $F,F_A$ and the foot $V_A$ of the A-internal bisector are collinear due to Monge & d'Alembert theorem, then it suffices to show that $F \in U_AV_A.$ Using barycentric coordinates with respect to $\triangle ABC,$ the equations of the A-sidelines of the medial and orthic triangle of ABC are then $y+z-x=0 \ , \ S_By+S_Cz-S_Ax=0 \Longrightarrow U_A \equiv (c^2-b^2:c^2-a^2:a^2-b^2)$ We verify that $U_A,$ $V_A \equiv (0:b:c)$ and $F \equiv X_{11} \equiv [(b+c-a)(b-c)^2]$ are collinear $\left[\begin{array}{ccc} 0 & b & c\\ c^2-b^2 & c^2-a^2 & a^2-b^2\\ (b+c-a)(b-c)^2 & (c+a-b)(c-a)^2 & (a+b-c)(a-b)^2\end{array}\right]=0$ The reasoning for $F_B$ and $F_C$ is analogous. Hence, we conclude that $U_A \equiv FF_A \cap F_BF_C.$

My solution to Luis' Theorem: Lemma 2 : let circle w tangent to sides AB , BC and CA at points C_1 , A_1 , B_1 , C' A' B' are midpoints of sides . w is tangent to (C'A'B') at point F , then lenes C_1F , A'B' , A_1B_1 intersects at one point Descovered by F.Ivlev Solution : Let E is midpoint of CA see picture let A_1C_1 and A_2C_2 intersect DF at points X and X' let C_3A_4 and A_3C_4 intersect DF at points Y and Y' Use Lemma 2 So B_2T_B goes thru X , TB_3 goes thru X' , T_CB_4 goes thru Y' and B_1D goes thru Y Use Reim's Theorem , so TX'XT_B and Y'YT_AT_C are cyclic Let A_4C_3 intersect A_1C_1 and A_2C_2 at points P and Q Let C_4A_3 intersect A_1C_1 and A_2C_2 at points P' and Q' Let H_C and H_A are foots of C and A altitudes of triangle ABC Let Line H_CH_A intersects segments C_4A_3 and A_4C_3 at points U and W Easy to see that YY'UW is Isosceles trapezoid ,so YY'UW is cyclic Well known that B_1B_4T_CT_A is cyclic , so after using Reim's Theorem (B_4B_1 || Y'Y) , so YY'T_CT_A is cyclic Let C_5A_5 is second external tangent to (C_4B_4A_3) and (B_1A_4C_3) Let prove that U is on T_CC_5 and W is on T_AA_5 Let CB and AB intersects C_5A_5 at points A* and C* Let H_C*C* is Altitude of triangle BA*C* Let lines CH_C and H_C*C* intersects at point F Let S is foot of perpendicular from F on A*C* Easy to see that EH_C = AE = EC , so angle H_CCE = EH_CC = SC*F = SH_CF (H_CFSC* is cyclic) , so H_C is on ES Let M is midpoint of FI_A Easy to see that M is on Perpendicular bissectors of segments A_4HC* and H_CC_3 and SA_5 , so M is center of (A_4H_CH_C*C_3) and centers of (H_CSA_5) and (A_4H_CH_C*C_3) are on Perpendicular bissector of segment SA_5 Let H_C' is second intersection point of (A_4H_CH_C*C_3) and (H_CSA_5) H_CH_C' is perpendicular to center line of this circles , so H_CH_C' is || to A*C* Not hard to prove that H_CH_A is || to A*C* , so H_A is on H_CH_C' EH_C = EH_A , so tangent to 9 point circle is || to H_CH_A || A*C* , so center of Homothety of circles (EH_CH_AT_A) and (B_1A_4A_5) is on EA_5 , so T_A is on EA_5 Angle AH_CE = EAH_C = BA*C* , so BH_CSA* is cyclic Angle N_CT_AE + N_CDE = DH_CB = C*BA* = H_CSA* , so T_AH_CSA_5 is cyclic So lines H_CH_AH_C' , T_AA_5 , A_4C_3 are Concurrent as Radical lines of circles (B_1A_4A_5) , (A_4H_CH_C*C_3) and (T_AH_CSA_5) Easy to see that they concur at point W Not hard to prove that T_AT_CC_5A_5 is cyclic , use Reim's Theorem , so T_AT_CUW is cyclic , so lines Y'Y , T_CT_A and UW will be Concurrent at Radical center of (YY'UW) , (YY'T_CT_A) , (T_AT_CUW) Problem done

Here's a less geometric solution... Let $D,E,F$ be the intersections of the internal angle bisectors of $\triangle{ABC}$ with the sides. Then by d'Alembert's theorem, $TDT_A,TET_B,TFT_C$ are lines (with points trivially in that order). Thus it suffices to show that segment $TE$ intersects segment $DF$, i.e. $T\in\triangle{BFD}$ or $[TBF]+[TBD]<[BFD]$. It's well-known that \begin{align*} [TBC] : [TCA] : [TAB] &= a(1-\cos(B-C)) : b(1-\cos(C-A)) : c(1-\cos(A-B)) \\ &= a\sin^2(\beta-\gamma) : b\sin^2(\gamma-\alpha) : c\sin^2(\alpha-\beta) \\ &= (b+c-a)(b-c)^2 : (c+a-b)(c-a)^2 : (a+b-c)(a-b)^2, \end{align*}so the inequality is equivalent to \begin{align*} \frac{a}{a+b}\cdot\frac{c}{b+c} &> \frac{1}{\sum{(b+c-a)(b-c)^2}}\cdot\left(\frac{a}{a+b}(a+b-c)(a-b)^2 + \frac{c}{b+c}(b+c-a)(b-c)^2\right) \\ ac\sum{(b+c-a)(b-c)^2} &> a(b+c)(a+b-c)(a-b)^2 + c(a+b)(b+c-a)(b-c)^2 \\ ac(c+a-b)(c-a)^2 &> ab(a+b-c)(a-b)^2 + cb(b+c-a)(b-c)^2 \\ ac(c+a-b)((a-b)+(b-c))^2 &> ab(a+b-c)(a-b)^2 + cb(b+c-a)(b-c)^2 \\ 2ac(c+a-b)(a-b)(b-c) &> [ab(a+b-c)-ac(c+a-b)](a-b)^2 + [cb(b+c-a)-ac(c+a-b)](b-c)^2 \\ 2ac(c+a-b)(a-b)(b-c) &> a(a+b+c)(b-c)(a-b)^2 + c(a+b+c)(b-a)(b-c)^2 \\ 2ac(c+a-b) &> a(a+b+c)(a-b) + c(a+b+c)(c-b) \\ 2ac(c+a-b) &> (a+b+c)(a^2-ab+c^2-cb+2ac)-2ac(a+b+c) \\ 4ca(c+a) &> (a+b+c)(c+a)(c+a-b) \\ 4ca &> (c+a)^2-b^2 \\ (a+b-c)(b+c-a) &> 0, \end{align*}where we have used the fact that $a>b>c$.

My Solution: Lemma 1: The nine-point center $N$ lies in the region of the plane determined by rays $IA$ and $IC$ Proof: It suffices to prove that $N$ is in the half-plane determined by line $IA$ containing $C$ and that $N$ is in the half-plane determined by line $IC$ containing $A$. To prove the first part, we let $A$ be the origin and $B$ be on the positive x-axis. We need to prove that $\overrightarrow{AI}\times\overrightarrow{AN}>0$. Note that ray $AI$ can be expressed in the form $(k(1+\cos A), k\sin A)$, where $k>0$, and we get that the coordinates of $N$ are $(R(\sin C+2\sin B\cos A)/2,R(\cos C+2\cos B\cos A)/2)$, which by product-to-sum is equal to $(R\sin C+R \sin(B-A)/2, R\cos(B-A)/2)$, so \begin{eqnarray*} \overrightarrow{AI}\times\overrightarrow{AN}&=&kR/2[(1+\cos A)(\cos(B-A))-(\sin A)(2\sin C+\sin(B-A))]\\ &=&kR/2[\cos(B-A)+\cos A\cos(B-A)-2\sin A\sin C-\sin A\sin(B-A)]\\ &=&kR/2[\cos(B-A)+\cos B-(\cos(A-C)-\cos(A+C))]\\ &=&kR/2(\cos(A-B)-\cos(A-C))>0 \end{eqnarray*} Thus $N$ lies in the half-plane determined by $IA$ containing $C$. Similarly, $N$ lying in the half-plane determined by $IC$ containing $A$ is equivalent to $\cos(C-B)-\cos(C-A)>0$, so we are done. (end lemma) Lemma 2: $T_B$ is not in the portion of the plane determined by rays $NT_A$ and $NT_B$. Proof: If $M$ is the midpoint of $AH$ and N is the midpoint of $BC$, it is not hard to see that $T_B$ and $T_AT_C$ are on different sides of $MN$, which is a diameter of the nine-point circle. Thus if we can prove that $T$ is in the region of the plane subtended by rays $NT_A$ and $NT_C$, then we are done, as then $T$ and $T_B$ lie on opposite sides of $T_AT_C$, so since they are concyclic, $TT_B$ must intersect $T_AT_C$. Since $N$ is in the region of the plane determined by $IA$ and $IC$, $I$ must be in the region of the plane determined by rays $NI_A$ and $NI_C$, where $I_A$ and $I_C$ are the A- and C- excenters, respectively. However, rays $NI_A$ and $NT_A$ point in the same direction and $NI_C$ and $NT_C$ point in the same direction. Thus $I$ lies in the region of the plane determined by rays $NT_A$ and $NT_C$. Since the incircle is smaller than the nine-point circle, $T$ must be on ray $NI$, so $T$ must be in the region of the plane determined by $NT_A$ and $NT_C$. yay

Let $N$ be the nine-point center, and let $I, I_A, I_B, I_C$ be the incenter and the three excenters. Let $D,E,F$ be the midpoints of the sides $BC, CA, AB$, respectively. Since $T$, $T_A$, $T_B$, and $T_C$ are concyclic, it suffices to show that $T$ and $T_B$ are on opposite sides of line $T_CT_A$. Note that $T$ is the intersection of ray $\overrightarrow{NI}$ with the nine point circle and that $T_A$ is the intersection of ray $\overrightarrow{NI_A}$ with the nine-point circle (and similarly for the other excenters). It thus suffices to show that $I$ lies inside non-reflex angle $I_CNI_A$, or that $N$ lies inside non-reflex angle $AIC$. To show that $N$ and $B$ lie on opposite sides of $CI$, let $P$ be the point of intersection of the angle bisector of $\angle ACB$ and the perpendicular bisector of $EF$, the latter of which $N$ must lie on. Drop perpendiculars from $P$ to $AC$ and $BC$, and let the feet of these perpendiculars be $P_B$ and $P_A$, respectively. It is easy to see that $P_A$ lies on the perpendicular bisector of $EF$. Thus $P$ and $C$ must lie on opposite sides of $D$ on $BC$, since $AB < AC$. Since $CP_A$ = $CP_B$ and $CD < CE$, $P_B$ and $C$ must lie on opposite sides of $E$ on $AC$. Now note that $CP_A = CP_B$, and that $CD > CE$. Thus $P_AF < P_BE$. Noting also that $PP_A = PP_B$, we see that $PF < PE$, so this point $P$ is too close to $FD$ to be the nine-point center. Also note that this result implies that $\angle PDE > \angle PE$, and any other point on the perpendicular bisector of $EF$ on the "wrong side" of $CI$ will yield a greater $\angle PDE$ and a smaller $\angle PED$. Thus it is also impossible for any other point on the "wrong side" of $CI$ on the perpendicular bisector of $EF$ to be $N$. Thus $N$ must lie on the claimed side of $CI$. To show that $N$ is on the claimed side of $AI$, we can repeat this process with the angle bisector of $\angle BAC$ and the perpendicular bisector of $DE$.

Solution for Lemma 2 Today i found this Post : http://nguyenvanlinh.files.wordpress.com/2010/11/fontene-theorem-and-some-corollaries.pdf By Linh Nguyen Van First problem is Generalisation of this Lemma My solution to Fontene theorem 1 : Let lines C_1B_1 and C_2B_2 intersects at point Q Let (A_2B_2C_2) intersect A_2Q at points A_2 and X Let A_2P intersect C_1B_1 at point A_3 Let (A_3C_2B_2) intersect C_1B_1 at points A_3A_4 Easy to see that QX*A_2Q = QA_3*QA_4 = QC_2*QB_2 , so A_2XA_3A_4 is cyclic , so angle A_2XA_4 = 90 Let line QP" intersects (PC_2A) at points Y and P" Use Reim's Theorem , so C_2C_1QY are cyclic Use Miquel Theorem , so Y is on (B_1C_1A) Let make Simmetry of point Y wrt ine C_1B_1 Angle QYA_4 = 90 (QA_3*QA_4 = QB_2*QC_2 = QY*QP"A_4A_3YP" , so is cyclic) , so X -> Y and (B_1C_1A) -> (A_1C_1B_1) , so X is on 9point circle too . done

Luis González wrote: Theorem: $F$ is the Feuerbach point of $\triangle ABC$ and $F_A,F_B,F_C$ are the three additional Feuerbach points against $A,B,C.$ Then $FF_A$ and $F_BF_C$ intersect at the A-Schroeter point $U_A$ of $\triangle ABC$ (intersection of the A-sidelines of the medial and orthic triangle). My solution to Luis's Theorem : Lemma : Let $ (O) $ be the circle tangent to $ (O_1),(O_2) $ at $ A,B $ . Let two common tangent (both external or internal) cut $ (O) $ at $ C,D $ and $ E,F $ . Then $ AB, CE, DF $ are concurrent . Proof of the lemma : From 3 circles with common tangency point we get there exist a circle $ (T) $ tangent to $ CE,DF $ and tangent to $ (O) $ at $ A $ . Similarly, there exist a circle $ (S) $ tangent to $ CE,DF $ and tangent to $ (O) $ at $ B $ By D'Alembert theorem we get $ AB, CE, DF $ are concurrent at the similarity center of $ (T), (S) $ . Back to the main proof : Let $ I, I_A, I_B, I_C $ be the incenter and the three excenters and $ N $ be the nine-point center . Apply the lemma to $ (N), (I), (I_A) $ we get $ FF_A $ pass through A-Schroeter point . Apply the lemma to $ (N), (I_B), (I_C) $ we get $ F_BF_C $ pass through A-Schroeter point . Q.E.D

I think we just have to show that $ N $ is in $ \triangle{IAC} $ which is very easy.