277. A nice and difficult relation with Lemoine's axis.

by Virgil Nicula, May 12, 2011, 3:45 PM

Proposed problem. The lines $AB$ and $AC$ are tangent from the point $A$ to the circle $(O)$ at $B$ and $C$ respectively and $D$ is a point

on the major/minor arc $BC$ . Denote $E\in BD\cap AC$ and $F\in DC\cap BA$ . Prove that $ EF^{2}=BF^{2}+CE^{2}-BF\cdot CE$ .

Proof. Suppose w.l.o.g. that the point $D$ is on the major arc $BC$ . Prove analogously that in the another case when $D$ is a point on the minor arc $BC$ .

Denote $AB=AC=l$ and $BF=a$ , $CE=b$ , $EF=x$ . If the the tangent $DD$ to $(O)$ through $D$ cuts $BC$ at $P$ , then $P\in EF$ are collinear on the

Lemoine's axis of $\triangle BDC$ . Apply the Menelaus' theorem to the transversal $\overline{PFE}$ and $\triangle ABC\ :\ \frac {FB}{FA}$ $\cdot\frac {EA}{EC}\cdot\frac {PC}{PB}=1\iff$ $\frac{PB}{PC}=\frac{a}{l+a} \cdot \frac{l+b}{b}$ .

Since $\frac {PB}{PC}=\left(\frac {DB}{DC}\right)^2$ and $\left\{\begin{array}{ccccc} \triangle FBD\sim \triangle FCB & \implies & \frac {BD}{CB}=\frac {FB}{FC} & \implies & DB=\frac {BC\cdot FB}{FC}\\\\ \triangle ECD\sim \triangle EBC & \implies & \frac {CD}{BC}=\frac {EC}{EB} & \implies & DC=\frac {BC\cdot EC}{EB}\end{array}\right\|$ $\implies $ $\boxed{\frac {DB}{DC}=\frac ab\cdot\frac {EB}{FC}}\ (*)$ .

Obtain that $\frac {a(l+b)}{b(l+a)}=\frac {a^2}{b^2}\cdot \left(\frac {EB}{FC}\right)^2$ $\Longrightarrow \boxed{\frac{EB^2}{FC^2}=\frac{b(l+b)}{a(l+a)}}\ (1)$ . Apply the Stewart's theorem to the cevians $EB$ , $FC$ in $\triangle AFE\ :$

$\left\{\begin{array}{c} EB^2=\frac{x^2l+a(l+b)^2-al(l+a)}{l+a}\\\\ FC^2=\frac{x^2l+b(l+a)^2-bl(l+b)}{l+b}\end{array}\right\|\implies$ $\boxed{\frac{EB^2}{FC^2}=\frac{l+b}{l+a} \cdot \frac{x^2l+a(l+b)^2-al(l+a)}{x^2l+b(l+a)^2-bl(l+b)}}\ (2)$ .

From the relations $(1)$ and $(2)$ obtain that $\frac{x^2l+a(l+b)^2-al(l+a)}{x^2l+b(l+a)^2-bl(l+b)}=$ $\frac{b}{a} \Longrightarrow \ x^2=a^2+b^2-ab$ .

Remark. I"ll show otherwise the relation $(*)$ . Denote $\left\|\begin{array}{c} m(\angle DBC)=u\\\\ m(\angle DCB)=v\\\\ m(\angle ABC)=w\end{array}\right\|$ . I will apply the theorem of Sinus in

$\triangle BDC$ , $\triangle BEC$ and $\triangle BFC\ :\ \frac {DB}{DC}=$ $\frac {\sin v}{\sin u}=\frac {\sin v}{\sin w}\cdot\frac {\sin w}{\sin u}=$ $\frac {a}{FC}\cdot \frac {EB}{b}$ $\implies$ $\frac {DB}{DC}=\frac ab\cdot\frac {EB}{FC}$ .
This post has been edited 35 times. Last edited by Virgil Nicula, Nov 22, 2015, 7:16 AM

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