150. Some (5) problems with fixed points.

by Virgil Nicula, Oct 7, 2010, 8:42 PM

PP1 ("problem of treasury"). Given are two points $A$ , $B$ and three mobile points $M$ , $N$ , $P$ so that the line $AB$ doesn't separate $M$ , $N$ ,

$P$ , the triangles $MAN$ , $MBP$ are isosceles and $AM\perp AN$ and $BM\perp BP$ . Prove that the line $MP$ pass through a fixed point.


Proof. Let $X(x)$ with affix $x\in\mathbb C$ . Thus, for $A(a)$ , $B(b)$ , $M(m)$ , $N(n)$ and $P(p)$ get $\left\{\begin{array}{c}
m=(1+i)a-in\\\\
p=(1-i)b+in\end{array}\right\|$ $\implies$ $\frac {m+p}{2}=\frac {a+b}{2}+i\cdot\frac {a-b}{2}$ , i.e. the midpoint $F$ of $[MP]$ is

fixed, more exactly $\triangle AFB$ is $F$-rightangled and isosceles because $\frac {m+p}{2}-\frac {a+b}{2}=i\cdot\left(a-\frac {a+b}{2}\right)$ $\iff$ $CF=CA=CB$ and $CF\perp AB$ , where $C$ is the midpoint of $[AB]$ .



PP2. Are given a circle $w=C(O,r)$ and a line $d$ so that $d\cap w=\emptyset$ . Denote the point $A\in d$ for which $OA\perp d$ . For a mobile point $M\in w$ the circle

with diameter $[AM]$ cut again the circle $w$ and the line $d$ in the points $X$ , $Y$ respectively. Prove that the line $XY$ pass through a fixed point (Toshio Seimiya).


Proof. Denote $F\in XY\cap OA$ and the intersection $L$ between $XM$ and the tangent line in $A$ to the circle $\rho$ with diameter $[AM]$ . Since $m\left(\widehat{AYF}\right)=$ $m\left(\widehat{AMX}\right)=$ $m\left(\widehat{LAX}\right)$ and

$m\left(\widehat{AFY}\right)=90^{\circ}-m\left(\widehat{AYF}\right)=$ $90^{\circ}-m\left(\widehat{LAX}\right)=$ $m\left(\widehat{ALX}\right)$ , obtain that $\widehat{AFY}\equiv\widehat{ALX}$ , i.e. the quadrilateral $ALXF$ is inscribed in the circle with the diameter $[AL]$ .

Therefore, $LF\perp AF$ , i.e. $LF\parallel d$ . Consider the circles $w$ , $\rho$ and the null circle $A$ (with center $A$ and the null radius). Prove easily that the point $F$ is the radical center for these three

circles. Since the circle $w$ and a point $A$ (null circle) are given obtain that the point $F$ is fixed and $AF^2=AO^2-r^2$ .



PP3. Are given a circle $w=C(I,r)$ and a line $d$ so that $d\cap w=\emptyset$ . Consider a mobile point $A\in d$ . Denote the tangent $t$ to $w$ so that $t\parallel d$

and $t$ doesn't separate $I$ , $A$ . The tangents from $A$ to $w$ cut $t$ in $B$ , $C$ . Prove that the $A$-median of $\triangle ABC$ pass through a fixed point (V.N.).


Proof. Denote the midpoint $M$ of $[BC]$ , the point $D\in t$ so that $AD\perp t$ , $F\in w\cap t$ , $L\in AM\cap IF$ and $S\in AI\cap t$ . Using the standard notations for $\triangle ABC$

obtain $MD=\frac {\left|b^2-c^2\right|}{2a}$ , $MF=\frac {|b-c|}{2}$ and $\frac {AD}{LF}=\frac {MD}{MF}=\frac {b+c}{a}=\frac {IA}{IS}$ (constant). Since $AD$ is constant obtain that $LF$ is constant, i.e. the point $L$ is fixed.



PP4. Let $ABC$ be a triangle with the circumcircle $w=C(O,R)$ . Given are two fixed points $E$ , $F$ so that $AB$ separates $E$ , $C$ and $AC$ separates $F$ , $B$ .

Consider a mobile point $M$ for which $BC$ separates $M$ , $A$ . Denote $X\in AE\cap MB$ and $Y\in AF\cap MC$ . Prove that $XY$ pass through a fixed point (V.N.).


Proof. The Pascal's theorem to the cyclical hexagon $AFBMCE\implies$ the points $\left\{\begin{array}{c}
Y\in AF\cap MC\\\\
G\in FB\cap CE\\\\
X\in BM\cap EA\end{array}\right\|$ are collinearly $\implies XY$ pass through the fixed point $G$ .


PP5. Are given a circle $w=C(O,r)$ and a line $d$ so that $d\cap w=\emptyset$ . Denote $E\in d$ for which $OE\perp d$ . For a mobile point $M\in d$ define

the tangent points $\{A,B\}\subset w$ of the tangent lines from $M$ to $w$ so that $OM$ separates $B$ , $E$ . Denote the projections $C$ , $D$ from $E$ on the

tangents $MA$ , $MB$ respectively. Prove that the line $AB$ pass through a fixed point and the line $CD$ pass through a fixed point (O.I.M. - 1995).


Proof. Denote $OE=a$ , $G\in AB$ for which $EG\perp AB$ , $N\in AB\cap OM$ , $P\in AB\cap OE$ and $F\in CD\cap OE$ . $MNPE$ is inscribed in the circle with diameter $[MP]$ .

Thus $OP\cdot OE=ON\cdot OM=$ $OB^2=r^2$ $\implies$ $OP=\frac {r^2}{a}$ , i.e. $P$ is fixed (the pole of $d$ w.r.t. $w$). Thus, $O$ , $B$ , $M$ , $E$ , $A$ belong to circle with diameter $[OM]$ . Since $E$

belongs to circumcircle of $\triangle AMB$ and $C$ , $D$ , $G$ are projections of $E$ on the sidelines of $\triangle AMB$ , from the Simson's theorem obtain that $G\in CD$ . Observe that $AECG$

is inscribed in the circle with the diameter $[AE]$ and $OAEM$ is inscribed in the circle with the diameter $[OM]$ and $EG\parallel OM$ . Therefore, $\widehat{GEF}\equiv\widehat{GEO}\equiv\widehat{MOE}\equiv$

$\widehat{MAE}\equiv\widehat{CAE}$ $\equiv\widehat{CGE}\equiv\widehat{FGE}$ $\implies$ $\widehat{GEF}\equiv\widehat{FGE}$ , i.e. in the $F$-right triangle $EFG$ have $FE=FG=FP$ . In conclusion, $F$ is the midpoint of $[EP]$ ,

where $E$ , $P$ are fixed. Thus and $F$ is fixed. More exactly, $OP=\frac {r^2}{a}$ and from $2\cdot OF=(OE+OP)$ obtain that $OF=\frac {r^2+a^2}{2a}$ .
This post has been edited 67 times. Last edited by Virgil Nicula, Dec 1, 2015, 10:38 AM

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