143. Assess of a ratio (nice and difficult problem).

by Virgil Nicula, Oct 5, 2010, 3:12 PM

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=153668

Lemma. Let $ABC$ be a triangle and let $M\in (BC)$ be a point for which $MB=m\cdot MC$. The second tangent from the point $M$

to the incircle of $\triangle ABC$ touches it at the point $T$. Denote the intersection $X\in AT\cap BC$. Prove that $\frac{XB}{XC}=\frac{p-c}{p-b}\cdot m^{2}$.


Remark. The proposed problem and the its Pohoatza"s generalization from

http://www.artofproblemsolving.com/viewtopic.php?t=152994 prove easily (even immediately) with the above lemma.


Proof. Denote the incircle $w$ of the triangle $ABC$ and the tangent points $\left\{\begin{array}{c}D\in BC\cap w\\\ E\in AC\cap w\end{array}\right\|$. Suppose w.l.o.g. $M\in (DC)$ and denote $V\in AC\cap MT$

and observe that $\left\{\begin{array}{c}MT=MD\\\ VT=VE\end{array}\right\|$.Thus, $MD=MB-BD=\frac{ma}{1+m}-(p-b)$ $\Longleftrightarrow$ $\boxed{DM=\frac{m(p-c)-(p-b)}{1+m}}\ \ (1)$.

Apply the Brianchon's theorem to the circumscriptible quadrilateral $ABMV$ i.e. there is $S\in AM\cap BV\cap DE$.

$1\blacktriangleright$ Apply the Menelaus' theorem to the transversal $\overline{DSE}$ in $\triangle AMC$ :

$\frac{DM}{DC}\cdot\frac{EC}{EA}\cdot\frac{SA}{SM}=1$ $\Longleftrightarrow$ $\frac{m(p-c)-(p-b)}{(1+m)(p-c)}\cdot\frac{p-c}{p-a}\cdot\frac{SA}{SM}=1$ $\Longleftrightarrow$ $\boxed{\frac{SA}{SM}=\frac{(1+m)(p-a)}{m(p-c)-(p-b)}}\ \ (2)$.

$2\blacktriangleright$ Apply the Menelaus' theorem to the transversal $\overline{BSV}$ in $\triangle AMC$ :

$\frac{BM}{BC}\cdot\frac{VC}{VA}\cdot\frac{SA}{SM}=1$ $\Longleftrightarrow$ $\frac{m}{1+m}\cdot\frac{VC}{VA}\cdot\frac{(1+m)(p-a)}{m(p-c)-(p-b)}=1$ $\Longleftrightarrow$ $\boxed{\frac{VA}{m(p-a)}=\frac{VC}{m(p-c)-(p-b)}=\frac{b}{mb-(p-b)}}\ \ (3)$.

Observe that $EV=AV-AE=\frac{m(p-a)b}{mb-(p-b)}-(p-a)$ $\Longleftrightarrow$ $\boxed{EV=\frac{(p-a)(p-b)}{mb-(p-b)}}\ \ (4)$.

$3\blacktriangleright$ Apply the Menelaus' theorem to the transversal $\overline{ATX}$ in $\triangle MVC$ :

$\frac{AV}{AC}\cdot\frac{XC}{XM}\cdot\frac{TM}{TV}=1$ $\Longleftrightarrow$ $\frac{m(p-a)}{mb-(p-b)}\cdot\frac{XC}{XM}\cdot\frac{MD}{VE}=1$ $\Longleftrightarrow$ $\frac{m(p-a)}{mb-(p-b)}\cdot\frac{XC}{XM}\cdot\frac{m(p-c)-(p-b)}{1+m}\cdot\frac{mb-(p-b)}{(p-a)(p-b)}=1$

$\Longleftrightarrow$ $\boxed{\frac{XM}{m\left[m(p-c)-(p-b)\right]}=\frac{XC}{(1+m)(p-b)}=\frac{a}{(1+m)\left[m^{2}(p-c)+(p-b)\right]}}\ \ (5)$.

Therefore, $XC=\frac{a(p-b)}{m^{2}(p-c)+(p-b)}$ $\implies$ $XB=BC-XC=\frac{m^{2}(p-c)a}{m^{2}(p-c)+(p-b)}$ $\implies$ $\boxed{\frac{XB}{XC}=\frac{p-c}{p-b}\cdot m^{2}}$.
This post has been edited 8 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:04 AM

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