Bary FTW

Attachments:
Problem 16.pdf (135kb)

Even I got the same solution

Projective cookie. Vrangr wrote: Let $AH_1$ and $BH_2$ be the altitudes of triangle $ABC$. Let the tangent to the circumcircle of $ABC$ at $A$ meet $BC$ at point $S_1$, and the tangent at $B$ meet $AC$ at point $S_2$. Let $T_1$ and $T_2$ be the midpoints of $AS_1$ and $BS_2$ respectively. Prove that $T_1T_2$, $AB$ and $H_1H_2$ concur. Solution . Define $S_3$ and $T_3$ analogously as shown in the diagram. By applying Pascal's theorem on degenerate hexagon $AABBCC,$ it follows that $S_1,S_2,S_3$ are collinear. By Gauss-Bodenmiller Line Theorem on $S_2ABS_1,$ $T_1,T_2,T_3$ are collinear. Let $K_1= BC\cap H_2H_3,$ similarly define $K_2$ and $K_3.$ Applying Desargue's theorem on perspective triangles $H_1H_2H_3$ and $ABC,$ we get $K_1,K_2,K_3$ collinear. Let $P_{\infty}$ be the point at infinity on the line $CS_3.$ It is easy to see that $\overline{K_3H_1H_2}\parallel \overline{CS_3}.$ Now, \[-1=(C,S_3;T_3,P_{\infty})\overset{K_3}{=}(C,B;K_3T_3\cap BC,H_1)\overset{K_3}{=}(C,A;K_3T_3\cap AC,H_2)\]Consider the following lemma: Lemma. Let $ABC$ be a triangle with concurrent cevians $AD,BE,CF.$ Line $EF$ meets $BC$ at $X$ (possibly at a point at infinity). Then $(X,D;B,C)$ is a harmonic bundle. By the above lemma, we know $(C,B;K_1,H_1)=-1$ and $(C,A;K_2,H_2)=-1$. Therefore, $K_3T_3\cap BC=K_1,~K_3T_3\cap AC=K_2,$ so in particular, $K_3,T_3,K_1,K_2$ are collinear, so $T_3\in\overline{K_1K_2K_3},$ similarly $T_1,T_2\in\overline{K_1K_2K_3}.$ Thus, $T_1T_2,AB$ and $H_1H_2$ concur. $\square$

Attachments:

Problem Relabeled wrote: Let $BH_2$ and $CH_3$ be the altitudes of triangle $ABC$. Let the tangent to the circumcircle of $ABC$ at $B$ meet $CA$ at point $S_2$, and the tangent at $C$ meet $AB$ at point $S_3$. Let $T_2$ and $T_3$ be the midpoints of $BS_2$ and $CS_3$ respectively. Prove that $T_2T_3$, $AB$ and $H_2H_3$ concur. Let $H_2H_3$ intersect $BC$ at $T$ and $AH_1$ be the third altitude of triangle $ABC$. We proceed with barycentric co-ordinates with sides $BC, CA, AB$ as $a, b, c$ and coordinates of $A (1, 0, 0)$, $B(0, 1, 0)$ and $C(0, 0, 1)$. The coordinate of $H_1$ is, \[H = (0 : a^2 + b^2 - c^2 : a^2 - b^2 + c^2).\]Since $(A, B; H, T)$ is harmonic, coordinate of $T$ is \[T = (0 : - a^2 - b^2 + c^2 : a^2 - b^2 + c^2)\]Since symmedian and $BS_2, CS_3$ are harmonic conjugates by Steiner's Ratio Theorem, \[S_2 = (- a^2 : 0 : c^2)\qquad S_3 = (- a^2 : b^2 : 0).\]\[T_2 = \frac{S_2 + B}{2} = \frac{(-a^2 : 0 : c^2) + (0 : -a^2 + c^2 : 0)}{2} = (-a^2 : -a^2 + c^2 : c^2)\]\[T_3 = (- a^2 : b^2 : -a^2 + b^2).\]Now, to prove $T_2, T_3, T$ to be collinear, it suffices to prove \[\begin{vmatrix}0 & - a^2 - b^2 + c^2 & a^2 - b^2 + c^2 \\ -a^2 & -a^2 + b^2 & b^2 \\ -a^2 & -a^2 + c^2 & c^2 \end{vmatrix} = 0\] However, this follows by subtracting the third row from the first and then adding the second row to the first.

This is just USA TSTST 2017 P1

Either use USA TSTST 2017 P1 (as above stated) or else: Define $H$ as the orthocenter of $\Delta ABC$, Let $H_3$ be the foot of altitude from $C$ to $AB$, Let $H_2H_3 \cap BC=X_A$, $H_1H_3 \cap AC=X_B$ and $H_1H_2 \cap AB=X_C$, Let $AS_1 \cap X_AX_B=P_1$ and $BS_2 \cap X_AX_B=P_2$ We have $AS_1||H_2H_3$ and similarly, we can have, $BS_2||H_1H_3$, Let $\mathcal{X_1}$ be the infinite point along $AS_1$, $$-1=(C,A;H_2,X_B) \stackrel{X_A}{=} (S_1,A;\mathcal{X_1},P_1)$$hence, $P_1$ is the midpoint of $AS_1$ $\Longrightarrow$ $P_1=T_1$ similarly, we can show that, $P_2=T_2$ and as a result, $T_1,T_2$ lie on the orthic axis, now note that $AB,H_1H_2$ concur on the orthic axis, hence, $T_1T_2,AB$ and $H_1H_2$ are concurrent lines

Projective...

Attachments:
Sharygin_P16_Pranjal.pdf (434kb)

Here’s a non-projective Solution. Trivial angle chasing yields that $T_2H_b$ is tangential to nine-point circle of $\Delta ABC$ and that $T_2H_b = T_2B$ So, $Pow(T_2, (H_aH_bH_c)) = Pow(T_2,(ABC))$. Thus $T_2$ lies on the Orthic axis of $\Delta ABC$. Similarly, $T_1$ lies on the Orthic axis of $\Delta ABC$. So, the required claim follows.

Problem wrote: Let $AH_1$ and $BH_2$ be the altitudes of triangle $ABC$. Let the tangent to the circumcircle of $ABC$ at $A$ meet $BC$ at point $S_1$, and the tangent at $B$ meet $AC$ at point $S_2$. Let $T_1$ and $T_2$ be the midpoints of $AS_1$ and $BS_2$ respectively. Prove that $T_1T_2$, $AB$ and $H_1H_2$ concur. Menelaus bash anyone? Let $R=H_1H_2\cap{AB}$, By menelaus theorm on $H_1,H_2,R$ we get $\dfrac{CH_1}{H_1B}\cdot \dfrac{BR}{RA}\cdot \dfrac{AH_2}{H_2C}=1$ .$BH_2=c\cdot{\cos{B}},H_2C=b\cdot{\cos{C}},AH_2=c\cdot{\cos{A}},H_2C=a\cdot{\cos{C}}$$\therefore \dfrac{BR}{RA}=\dfrac{a\cos{B}}{b\cos{A}}=\dfrac{a^2+c^2-b^2}{b^2+c^2-a^2}$ Again by menelaus theorm proving $\dfrac{KS_2}{S_2B}\cdot\dfrac{BR}{RA}\cdot\dfrac{AS_1}{S_1K}=1$ proves $R\in{S_1S_2}$ which solves the problem. Notice that $T_2BA \sim T_2CB$ and $T_1BA \sim T_1AC$,So $T_2B=\dfrac{abc}{a^2-c^2},T_1A=\dfrac{abc}{b^2-c^2}$,Also note that $KA=KB$ and $\angle{KAB}=\angle{KBA}=\angle{C} \implies \dfrac{c}{\sin{180-2C}}=\dfrac{KA}{\sin{C}}$ (from sine rule.) $\implies KA=\dfrac{c}{2\cos{C}}=\dfrac{abc}{a^2+b^2-c^2}$,$KS_2=BS_2-KB= \dfrac{abc}{2(a^2-c^2)}-\dfrac{abc}{a^2+b^2-c^2}=\dfrac{abc(b^2+c^2-a^2)}{2(a^2-c^2)(a^2+b^2-c^2)}$,$S_2B=\dfrac{abc}{2(a^2-c^2)} \implies \dfrac{KS_2}{S_2B}=\dfrac{b^2+c^2-a^2}{a^2+b^2-c^2}$similarly $\dfrac{AS_1}{S_1K}=\dfrac{a^2+b^2-c^2}{a^2+c^2-b^2}$,$\therefore \dfrac{KS_2}{S_2B}\cdot\dfrac{BR}{RA}\cdot\dfrac{AS_1}{S_1K}=\dfrac{b^2+c^2-a^2}{a^2+b^2-c^2}\cdot\dfrac{a^2+c^2-b^2}{b^2+c^2-a^2}\cdot\dfrac{a^2+b^2-c^2}{a^2+c^2-b^2}=1$ as desired.$\blacksquare$

Non-Bash solution. Nice Problem.

Attachments:
Sharygin- 16.pdf (37kb)

Let $P$ $\equiv$ $H_1H_2$ $\cap$ $AB$; $D$, $E$, $F$ be midpoint of $BC$, $CA$, $AB$, resepectively We have: $P_{T_1 / (ABC)}$ = $T_1A^2$ = $\overline{T_1E}$ . $\overline{T_1F}$ = $P_{T_1 / NPC}$, $P_{T_2 / (ABC)}$ = $T_1B^2$ = $\overline{T_1D}$ . $\overline{T_1E}$ = $P_{T_2 / NPC}$, $P_{P / (ABC)}$ = $\overline{PA}$ . $\overline{PB}$ = $\overline{PH_1}$ . $\overline{PH_2}$ = $P_{P / NPC}$ So: $T_1$, $T_2$, $P$ lie on radical axis of ($ABC$) and NPC of $\triangle$ $ABC$ or $AB$, $H_1H_2$, $T_1T_2$ concurrent at $P$

Really Nice Problem! Vrangr wrote: Let $AH_1$ and $BH_2$ be the altitudes of triangle $ABC$. Let the tangent to the circumcircle of $ABC$ at $A$ meet $BC$ at point $S_1$, and the tangent at $B$ meet $AC$ at point $S_2$. Let $T_1$ and $T_2$ be the midpoints of $AS_1$ and $BS_2$ respectively. Prove that $T_1T_2$, $AB$ and $H_1H_2$ concur. Solution:- Notations:- Let $P,Q$ be the $A,B$ Expoint of $\triangle ABC$ respectively, $H_3$ is the foot of altitude from $C$ onto $AB$ and $H$ is the orthocenter of $\triangle ABC$. Claim :-$\overline {Q-T_2-T_1-P}$ Let $QP\cap AS_1=R$. So it suffices to prove that $R\equiv T_1$. For this notice that $H_2H_3\|AS_1$. So, $$-1=(A,C;H_2,Q)\overset{P}{=}(A,S_1;\infty_{H_3H_2},R)\implies\boxed{R\equiv T_1}$$Similarly you get $\overline {P-T_2-Q}$. Hence, $\overline {P-T_1-T_2-Q}$. Let $K$ be the $C-\text{expoint}$ of $\triangle ABC$, So, $\overline{P-Q-K}$ as $\triangle ABC$ and $\triangle H_1H_2H_3$ are perspective, hence by Desargues Theorem we are done. So, $T_1T_2\cap AB\equiv K$ which is the $C-\text{expoint}$ of $\triangle ABC$ so $\overline{H_1-H_2-K}$. Hence, $T_1T_2,AB,H_1H_2$ are concurrent at the $C-\text{expoint}$ of $\triangle ABC$. $\blacksquare$.