60. Mediteranean M.C. 2010 Problem 3.

by Virgil Nicula, Jul 13, 2010, 3:12 AM

Quote:
Lemma. $\triangle\ ABC\ \ \wedge\ \ \left\|\begin{array}{ccc} M\in BC & , & \overline{MB}=m\cdot\overline {MC}\\\ N\in CA & , & \overline{NC}=n\cdot\overline {NA}\\\ P\in AB & , & \overline{PA}=p\cdot\overline {PB}\end{array}\right\|$ $\implies$

$\boxed {\ [MNP]=\frac {1-mnp}{(1-m)(1-n)(1-p)}\cdot [ABC]\ }$ .

Application. If $DEF$ is the triangle where $AD$ , $BE$ , $CF$ are the symmedians in $\triangle ABC$ and $s_a=AD$ , $s_b=BE$ , $s_c=CF$ , then $\frac {s_as_bs_c}{m_am_bm_c}=\frac {4\cdot [DEF]}{[ABC]}$ .
You can apply the well-known relation $s_a=\frac {2bc}{b^2+c^2}\cdot m_a\le m_a$ a.s.o.

Remark. If denote $f(m,n,p)=\frac {1-mnp}{(1-m)(1-n)(1-p)}$ , then $f\left(\frac 1m,\frac 1n,\frac 1p\right)=f(m.n.p)$ , i.e. two isotomic triangles w.r.t. $\triangle ABC$ have same area.

Quote:
Let $M\in(BC),$ $N\in(CA),P\in(AB)$ be the tangent points of the exincircles

of $\triangle ABC$ with its sides. Prove that the circumradius of $\triangle MNP$ is equal to

$\frac{1}{2r}\sqrt{2R\left(2R-h_{a}\right)\left(2R-h_{b}\right)\left(2R-h_{c}\right)}$ (see here).
Proof. Using the relation from upper lemma for $m:=-\frac {s-c}{s-b}$ , $n:=-\frac {s-a}{s-c}$ , $p:=-\frac {s-b}{s-a}$ obtain that $\boxed {\ [MNP]=\frac {r}{2R}\cdot [ABC]\ }$ , where $2s=a+b+c$ . Apply the generalized Pitagora's theorem to the side $[NP]$ in $\triangle ANP\ \ :$ $NP^2=(s-b)^2+(s-c)^2-2(s-b)(s-c)\cos A=$ $[(s-b)+(s-c)]^2-2(s-b)(s-c)(1+\cos A)=$ $a^2-4(s-b)(s-c)\cos^2\frac A2=$ $a^2-\frac {4s(s-a)(s-b)(s-c)}{bc}=$ $a^2-\frac {(2S)^2}{bc}=$ $a^2-\frac {(bc\sin A)^2}{bc}=$ $a\cdot a-bc\sin A\cdot\sin A=$ $a\cdot 2R\sin A-ah_a\sin A=$ $a(2R-h_a)\sin A$ . Therefore, $NP^2=a(2R-h_a)\sin A$ , $PM^2=b(2R-h_b)\sin B$ , $MN^2=c(2R-h_c)\sin C$ . If denote the circumradius $\rho$ of $\triangle MNP$ and $S=[ABC]$ , then $\rho=\frac {MN\cdot NP\cdot PM}{4\cdot [MNP]}$ $\implies$ $\rho^2=$ $\frac {\prod[a(2R-h_a)\sin A]}{\frac {4r^2S}{R^2}}=$ $\frac {R^2}{4r^2S^2}\cdot 4RS\cdot\prod (2R-h_a)\cdot \frac {S}{2R^2}=$ $\frac {R}{2r^2}\cdot\prod (2R-h_a)$ $\implies$ $\rho=\frac{1}{2r}\sqrt{2R\left(2R-h_{a}\right)\left(2R-h_{b}\right)\left(2R-h_{c}\right)}$ .
This post has been edited 10 times. Last edited by Virgil Nicula, Nov 23, 2015, 4:12 PM

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