China Mathematical Olympiad 1989 problem4

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China Mathematical Olympiad 1989 problem4

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Source: China Mathematical Olympiad 1989 problem4

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jred

#1 h Oct 28, 2013, 9:15 AM • 2 Y

Y by Adventure10, Mango247

Given a triangle $ABC$ , points $D,E,F$ lie on sides $BC,CA,AB$ respectively. Moreover, the radii of incircles of $\triangle AEF, \triangle BFD, \triangle CDE$ are equal to $r$ . Denote by $r_0$ and $R$ the radii of incircles of $\triangle DEF$ and $\triangle ABC$ respectively. Prove that $r+r_0=R$ .

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#2 h Nov 3, 2013, 1:33 PM • 2 Y

Y by nquocthuy, Adventure10

The area and perimeter of a figure $\mathcal{F}$ will be denoted by $[\mathcal{F}]$ and $p(\mathcal{F})$ respectively.

Note that $r=\frac{2[\triangle AEF]}{p(\triangle AEF)}=\frac{2[\triangle BDF]}{p(\triangle BDF)}=\frac{2[\triangle CDE]}{p(\triangle CDE)}=\frac{2[\triangle ABC]-2[\triangle DEF]}{p(\triangle ABC)+p(\triangle DEF)}=$ $\frac{R(p(\triangle ABC))-r_0(p(\triangle DEF))}{p(\triangle ABC)+p(\triangle DEF)}=\frac{R \left(\frac{p(\triangle ABC)}{p(\triangle DEF)} \right)-r_0}{\frac{p(\triangle ABC)}{p(\triangle DEF)}+1}$ .

If incircle of $\triangle AEF$ touches $EF,AE,AF$ at $P,Q,S$ , then we have $FP=FS,EP=EQ$ and thus $EF=FS+EQ=AF+AE-AS-AQ=AF+AE-2AS$ . But since $A$ is the homothetic center of incircles of $\triangle AEF$ and $\triangle ABC$ , we get $AS=\frac{r}{R}(s-a)$ . Thus, we have $EF=AE+AF-\frac{2r}{R}(s-a)$ . Similarly, we get $DF=BF+BD-\frac{2r}{R}(s-b)$ and $DE=CD+CE-\frac{2r}{R}(s-c)$ . Adding, we have $p(\triangle DEF)=p(\triangle ABC)-\frac{2r}{R}(s-a+s-b+s-c)$ $=p(\triangle ABC) \left(1-\frac{r}{R} \right)$ $\implies \frac{p(\triangle ABC)}{p(\triangle DEF)}=\frac{R}{R-r}$ .

Thus, we get $r=\frac{\frac{R^2}{R-r}-r_0}{\frac{R}{R-r}+1}$ $=\frac{R^2-r_0R+r_0r}{2R-r} \implies 2Rr-r^2=R^2-r_0R+r_0r$ $\implies (R-r)^2=r_0(R-r)$ $\implies R-r=r_0$ $\implies r+r_0=R$ . $\blacksquare$

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komath

#3 h Aug 8, 2016, 3:21 AM • 2 Y

Y by Adventure10, Mango247

Suppose the in-circle of triangle AEF touches sides AE, AF at A'. A" respectively. In the same way B',B" and C',C" are defined and let AA'=AA"=u,
BB'=BB"=v, CC'=CC"=w. [ABC] denotes the area of triangle ABC. [ABC]=[AEF]+[BDF]+[CDE]+[DEF], that is;
R(a+b+c)=r{2u+2v+2w+2(b-u-w)+2(c-u-v)+2(a-v-w)}+r'(a-v-w+b-u-w+c-u-v)...<1>
Let I(A),I(B),I(C) be the incenters of triangles AEF ,BDF,CDE respectively.
Then by rearranging the quadrilaterals AA'I(A)A", BB'I(B)B", CC'I(C)C", we get a new triangle ABC with inradius r, similar to triangle ABC.
Hence 2(u+v+w) / (a+b+c)=r/R...<2>. From equations <1>, <2>, we get (R-r)(R-r-r')=0
Hence R=r+r'. Here r' denotes the inradius of triangle DEF.

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