136. A geometrical interpretation of a system (own).

by Virgil Nicula, Oct 2, 2010, 2:17 PM

Consider $\{m,n,p\}\subset R^*$ and the system $(*)\ \left\{\begin{array}{c}
mx+ny+pz=1\\\\
xy+yz+zx=1\end{array}\right\|$ . Prove that the system $(*)$ has an unique

solution iff $m^2+n^2+p^2+1=2(mn+np+pm)\ (**)$ and in this case its solution is $\left\|\begin{array}{c}
x=n+p-m\\\\
y=p+m-n\\\\
z=m+n-p\end{array}\right\|$ .

Consider $\triangle ABC$ with incircle $w=C(I,r)$ and the positive numbers $m$ , $n$ , $p$ which can be the lengths

of the sides of a triangle and for which $m^2+n^2+p^2+1=2(mn+np+pm)$ . Ascertain the measures

of the angles of $\triangle ABC$ if exists the relation $\frac {m}{s-a}+\frac {n}{s-b}+\frac {p}{s-c}=\frac 1r$ , where $2s=a+b+c$.


Proof. $xy+(x+y)z=1\implies z=\frac {1-xy}{x+y}\implies$ $mx+ny+p\cdot\frac {1-xy}{x+y}=1$ $\implies$ $m\cdot\underline x^2+[(m+n-p)y-1]\cdot\underline x+ny^2-y+p=0\ (1)$ .

The system $(*)$ admits an unique solution iff $\Delta (y)\equiv [(m+n-p)y-1]^2-4m\cdot (ny^2-y+p)=0$ , i.e.

$\Delta (y)=\left[\left(m^2+n^2+p^2\right)-2(mn+np+pm)\right]\cdot \underline y^2 +2(m+p-n)\cdot \underline y+(1-4mp)=0$ and

$\Delta\equiv (m+p-n)^2-(1-4mp)\left[\left(m^2+n^2+p^2\right)-2(mn+np+pm)\right]=0$ , i.e. $m^2+n^2+p^2+1=2(mn+np+pm)$

and in this case $\Delta (y)=-[y-(m+p-n)]^2$ and $\Delta (y)=0\iff$ $y=m+p-n$ . Prove easily that and $x=n+p-m$ , $z=m+n-p$ .

$\blacktriangleright$ From the well-known relations $(s-a)\tan\frac A2=(s-b)\tan\frac B2=(s-c)\tan\frac C2=r$ obtain easily $\tan\frac A2\tan\frac B2+\tan\frac B2\tan\frac C2+\tan\frac C2\tan\frac A2=0$

i.e. $x=\tan\frac A2$ , $y=\tan\frac B2$ , $x=\tan\frac C2$ verify the system $(*)$ with restriction $(**)$ . In conclusion, since this system admits an unique solution

obtain that this solution is $x=n+p-m=\tan\frac A2$ , $y=p+m-n=\tan\frac B2$ , $z=m+n-p=\tan\frac C2$ a.s.o.
This post has been edited 17 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:12 AM

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