142. Inscribed/circumscribed quadrilateral w.r.t. ABC.

by Virgil Nicula, Oct 5, 2010, 11:26 AM

PP1. Let $P$ be an inner point of $\triangle ABC$ . Denote the intersections $E\in BP\cap AC$ , $F\in CP\cap AB$ . Prove that

$\boxed{AF+PE=AE+PF\iff BE+EC=BF+FC\ \iff\ PB-PC=AB-AC}$ .

Proof. $\odot$ Suppose $AEPF$ is circumscribed. Denote $X$ , $Y$ , $U$ , $V$ the tangent points of the incircle $w$ of $AEPF$ with $AB$ , $AC$ , $BP$ , $CP$ respectively.

Observe that $AX=AY$ , $PU=PV$ , $BX=BU$ , $CY=CV$ , $EY=EU$ , $FX=FV$ . Thus, $\boxed{AF+PE=AE+PF}\iff$

$\left(\underline{AX}+\underline{\underline{BX}}-BF\right)+$ $\left(BE+\underline{\underline{\underline{PU}}}-\underline{\underline{BU}}\right)=$ $\left(\underline{AY}+\underline{\underline{\underline{\underline{CY}}}}-CE\right)+$ $\left(CF+\underline{\underline{\underline{PV}}}-\underline{\underline{\underline{\underline{CV}}}}\right)$ $\iff$ $\boxed{BE+EC=BF+FC}$

$\iff$ $\left(PB+\underline{\underline{\underline{PU}}}+\underline{\underline{EU}}\right)+$ $\left(AC-\underline{AY}-\underline{\underline{EY}}\right)=$ $\left(AB-\underline{AX}-\underline{\underline{\underline{\underline{FX}}}}\right)+$ $\left(PC+\underline{\underline{\underline{PV}}}+\underline{\underline{\underline{\underline{FV}}}}\right)$ $\iff$ $\boxed{PB-PC=AB-AC}$.


PP2. Let $P$ be an inner point of $\triangle ABC$ with the normalized barycentrical coordinates $(x,y,z)$ w.r.t. the given triangle. Denote

the intersections $E\in BP\cap AC$ , $F\in CP\cap AB$ . Prove that $AEPF$ is inscribed $\iff$ $\boxed{\frac{b^2}{x+z}+\frac {c^2}{x+y}=\frac {a^2}{x}}$ .

Proof. Since $\left\{\begin{array}{c} \frac {EC}{x}=\frac {EA}{z}=\frac {b}{x+z}\ \blacktriangleleft\blacktriangleright\ \frac {FB}{x}=\frac {FA}{y}=\frac {c}{x+y}\\\\ (x+z)^2\cdot BE^2=\left(za^2+xc^2\right)(x+z)-b^2xz\\\\ PB=(x+z)\cdot BE\ ,\ \mathrm{where}\ x+y+z=1\end{array}\right\|$ obtain that the quadrilateral $AEPF$ is inscribed $\iff$

$BF\cdot BA=BP\cdot BE$ $\iff$ $\frac {xc^2}{x+y}=(x+z)\cdot BE^2$ $\iff$ $x(x+z)c^2=(x+y)\left[\left(za^2+xc^2\right)(x+z)-b^2xz\right]$ $\iff$

$xz(x+y)b^2+x(x+z)c^2=$ $z(x+y)(x+z)a^2+x(x+z)(x+y)c^2$ $\iff$ $xz(x+y)b^2+xz(x+z)c^2=$

$z(x+y)(x+z)a^2$ $\iff$ $x(x+y)b^2+x(x+z)c^2=$ $(x+y)(x+z)a^2$ $\iff$ $\frac{b^2}{x+z}+\frac {c^2}{x+y}=\frac {a^2}{x}$ .


Proposed problem. Let $ABC$ be a triangle with incenter $I$ . Denote $E\in AC\cap BI$ , $F\in AB\cap CI$ . Prove that in any

$\triangle ABC$ there is the equivalence $\frac {b^2}{a+c}+\frac {c^2}{a+b}=a\ \iff\ A=60^{\circ}\ \iff\ BA\cdot BF+CA\cdot CE=BC^2$ .

Proof. Prove easily that $\frac {b^2}{a+c}+\frac {c^2}{a+b}=a$ $\iff$ $(a+b+c)\left(b^2+c^2-bc-a^2\right)=0$ $\iff$ $a^2=b^2+c^2-bc$ $\iff$

$\cos A=\frac 12$ $\iff$ $A=60^{\circ}$ . Observe that this problem is a particular case for $P:=I(a,b,c)$ of the above problem PP2.
This post has been edited 60 times. Last edited by Virgil Nicula, Dec 1, 2015, 11:05 AM

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    by CSPAL, May 27, 2020, 4:17 PM

  • impressive :D
    awesome. 358,000 visits?????

    by OlympusHero, May 14, 2020, 8:43 PM

72 shouts
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