130. A triangle and a fixed point.

by Virgil Nicula, Sep 24, 2010, 1:09 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=368627
Quote:
Let $ABC$ be a triangle . Denote the midpoint $D$ of the side $[BC]$ and construct the points $E\in (BD)$ , $F\in (DC)$ so that $\frac {EB}{ED}=\frac {FD}{FC}=\frac {AB}{AC}$ . For a mobile point

$M\in AD$ define $U\in BM\cap AE$ , $V\in CM\cap AF$ . Prove that the line $UV$ pass through a fixed point $S\in BC$ for which $\frac {SE}{SF}=\frac {AB}{AC}$ and $\frac {SB}{SC}=\left(\frac {AB}{AC}\right)^2$ ,

i.e. the sideline $BC$ , the exterior $A$-angle bisector of $\triangle\ EAF$ and the tangent line in $A$ to the circumcircle of $\triangle\ ABC$ are concurrently !
Proof. Suppose w.l.o.g. that $b>c$ . Observe that $\frac {EB}{c}=\frac{ED}{b}=\frac {a}{2(b+c)}$ , $\frac {FD}{c}=\frac {FC}{b}=\frac {a}{2(b+c)}$ , i.e. $BE=DF$ , $DE=CF$ . Denote $S\in UV\cap BC$ .

Apply the Menelaus' theorem to the transversals : $\left\{\begin{array}{cc}
\overline{SUV}/AEF\ : & \frac {SE}{SF}\cdot\frac {VF}{VA}\cdot\frac {UA}{UE}=1\\\\
\overline{BUM}/AED\ : & \frac {BD}{BE}\cdot\frac {UE}{UA}\cdot\frac {MA}{MD}=1\\\\
\overline{CVM}/ADF\ : & \frac {CF}{CD}\cdot\frac {MD}{MA}\cdot\frac {VA}{FV}=1\end{array}\right\|$ $\implies$ $\frac {SE}{SF}=\frac {VA}{VF}\cdot\frac {UE}{UA}=$

$\left(\frac {CD}{CF}\cdot\frac {MA}{MD}\right)\cdot\left(\frac {BE}{BD}\cdot\frac {MD}{MA}\right)=\frac {BE}{CF}=\frac cb$ $\iff$ $\boxed {\frac {SE}{SF}=\frac cb}$ , i.e. the point $S$ is fixed and it is the intersection of the exterior $A$-angle bisector of $\triangle\ EAF$ with

the sideline $BC$ . Observe that $\frac {SE}{c}=\frac {SF}{b}=\frac {EF}{b-c}=$ $\frac {DE+DF}{b-c}=$ $\frac {1}{b-c}\cdot\left(\frac {ab}{b+c}+\frac {ac}{b+c}\right)\cdot\frac 12=$ $\frac {a}{2(b-c)}$ $\implies$ $SE=\frac {ac}{2(b-c)}$ , $SF=\frac {ab}{2(b-c)}$ .

Since $SB=SE-BE=\frac {ac}{2(b-c)}-\frac {ac}{2(b+c)}=\frac {ac^2}{b^2-c^2}$ and $SC=SF+SC=\frac {ab}{2(b-c)}+\frac {ab}{2(b+c)}=\frac {ab^2}{b^2-c^2}$ obtain $\boxed{\frac {SB}{SC}=\left(\frac cb\right)^2}$ , i.e.

the tangent line in $A$ to the circumcircle of $\triangle\ ABC$ meet the sideline $BC$ in the same point $S$ .
This post has been edited 12 times. Last edited by Virgil Nicula, Nov 23, 2015, 7:27 AM

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