suppose that $O$ is the circumcenter of acute triangle $ABC$. we have circle with center $O$ that is tangent too $BC$ that named $w$ suppose that $X$ and $Y$ are the points of intersection of the tangent from $A$ to $w$ with line $BC$($X$ and $B$ are in the same side of $AO$) $T$ is the intersection of the line tangent to circumcirle of $ABC$ in $B$ and the line from $X$ parallel to $AC$. $S$ is the intersection of the line tangent to circumcirle of $ABC$ in $C$ and the line from $Y$ parallel to $AB$. prove that $ST$ is tangent $ABC$.
Problem
Source: iran tst 2014 first exam
Tags: geometry, circumcircle, geometry unsolved
nima1376
13.04.2014 16:38
easy problem let AX,AY meet O at M,N. AM=AN=BC--->BM is parallel to AC and CN parallel to AB ABC=NCA=CYS=CYA=ACS---> AYCS is cycle---> CA is tangent to O AT is tangent to O ----> line TAS is tangent to O
saturzo
25.05.2014 21:36
Very Easy! Steps: show $\triangle XAB$ isoceles; then little bit of angle chasing shows that $AXBT$ cyclic - from this we can easily deduce that $TA=TB$ ; so $TA$ touches $\odot ABC$ . Similarly it can be proved that $SA$ touches $\odot ABC$, some more angle chasing implies that $T, A, S$ collinear. So, $ST$ touches $\odot ABC$ at $A$ . $[QED]$
JuanOrtiz
25.05.2014 23:59
tangent at $A$ because of cyclic ASYC and ATBX.
sayantanchakraborty
30.06.2014 16:13
Well,it took me 10 minutes to come up with this solution: Without loss of generality let $\angle{B} \ge \angle{A}$. Let $AY \cap w=L,w \cap BC=M$ $\angle{ACS}=\angle{ABC}=B$ so $\angle{SCY}=A$.Again $SY \parallel AB \Rightarrow \angle{SYC}=180-B \Rightarrow \angle{CSY}=B-A$.Also we have $\triangle{OMC} \cong \triangle{OLA} \Rightarrow \angle{OAL}=90-A$.But $\angle{OAC}=90-B$ so $\angle{CAL}=\angle{CAY}=B-A$.It follows that points $A,C,Y,S$ are concyclic $\Rightarrow \angle{YAS}=\angle{YCS}=A$.Hence $\angle{CAS}=\angle{CAY}+\angle{YAS}=B-A+A=B=\angle{ABC}$.So $AS$ is tangent to the circumcircle of $\triangle{ABC}$.Similarly $AT$ is tangent to the same,and the result follows.
AlastorMoody
11.09.2019 21:30
Beautiful! Iran TST 2014 P1 wrote: suppose that $O$ is the circumcenter of acute triangle $ABC$. we have circle with center $O$ that is tangent too $BC$ that named $w$ suppose that $X$ and $Y$ are the points of intersection of the tangent from $A$ to $w$ with line $BC$($X$ and $B$ are in the same side of $AO$) $T$ is the intersection of the line tangent to circumcirle of $ABC$ in $B$ and the line from $X$ parallel to $AC$. $S$ is the intersection of the line tangent to circumcirle of $ABC$ in $C$ and the line from $Y$ parallel to $AB$. prove that $ST$ is tangent $ABC$. Due to Symmetry, $AX=XC$ and $AY=BY$ $\implies$ $\angle TXO=\angle OYS=90^{\circ}$ $\implies$ $\angle TBA=\angle ACB=\angle TXA$ $\implies$ $\Delta TAB$ is isosceles & Similarly, $\Delta SAC$ is Isosceles $\qquad \blacksquare$