5
$\begingroup$

Let $O$ be the circumcenter of $\triangle ABC$ with $\angle A=60^{\circ}$, $P$ be an arbitary point on the circumcircle of $\triangle BOC$, and $D,E,F$ be the circumcenters of $\triangle BPC,\triangle CPA, \triangle APB$ respectively. Prove $AD,BE,CF$ are concurrent.

Some intermediate results:

  • $AD$ bisects $\angle BAC$,and $OD \perp BC$;
  • $O,P$ are isogonal conjugate points of $\triangle DEF$.

enter image description here

$\endgroup$

2 Answers 2

1
$\begingroup$

Let us denote by $A',B',C$ respectively the intersections of $AD$, $BE$, $CF$ with the sides $BC$, $CA$, $AB$.

Mathematics stackexchange, 3561134, a concurrence of three lines

We have also denoted by $a_1,a_2;b_1,b_2,c_1,c_2$ the (lengths of the) angles delimited by lines $AA'D$, $BB'E$, $CC'F$ from $\hat A, \hat B,\hat C$. Then we have the relations: $$ \begin{aligned} a_1+a_2 &=\hat A =60^\circ\ ,\\ b_1+c_2 &=180^\circ-\widehat{BPC}=180^\circ-\widehat{BOC}=180^\circ-2\hat A % =180^\circ-120^\circ\\ & =60^\circ\ ,\\ b_2+c_1 &= 180^\circ-a_1-a_2-b_1-c_2 =60^\circ\ . \end{aligned} $$ Then working in the circle $(F)=(ABP)$ we have $$ \widehat{AFB} =\overset\frown{APB} =\overset\frown{AP} +\overset\frown{PB} =2b_2+2a_1\ . $$ So $\widehat{FAB}=90^\circ-\frac 12\widehat{AFB}=90^\circ-a_1-b_2$, giving $\widehat{FAC}=90^\circ+a_2-b_2$.

Similarly, $\widehat{FBC}=(90^\circ-a_1-b_2)+(b_1+b_2)=90^\circ+b_1-a_1$.

Let us find a formula for the proportion $C'A:C'B$ using (only) the above data. $$ \begin{aligned} \frac{C'A}{C'B} &= \frac {\operatorname{Area}AFC} {\operatorname{Area}BFC} = \frac {AF\cdot AC\cdot \sin \widehat{FAC}} {BF\cdot BC\cdot \sin \widehat{FBC}} = \frac {AC\cdot \sin (90^\circ+a_2-b_2)} {BC\cdot \sin (90^\circ+b_1-a_1)} \ . \\[3mm] &\qquad\text{ Similarly:} \\[3mm] \frac{B'A}{B'C} &= \frac {\operatorname{Area}AEB} {\operatorname{Area}CEB} = \frac {AE\cdot AB\cdot \sin \widehat{EAB}} {CE\cdot CB\cdot \sin \widehat{ECB}} =\frac {AB\cdot \sin (90^\circ+a_1-c_1)} {CB\cdot \sin (90^\circ+c_2-a_2)} \ . \end{aligned} $$ As already observed in the OP, the point $D$ is on the angle bisector of $\hat A$, which gives the third needed proportion $D'A:D'B$. Putting all together, and with omission of signs, $$ \begin{aligned} \frac{B'C}{B'A}\cdot \frac{C'A}{C'B}\cdot \frac{A'B}{A'C} &= \frac {BC\cdot \sin (90^\circ+c_2-a_2)} {AB\cdot \sin (90^\circ+a_1-c_1)} \cdot \frac {AC\cdot \sin (90^\circ+a_2-b_2)} {BC\cdot \sin (90^\circ+b_1-a_1)} \cdot \frac {AB} {AC} \\ &= \frac {\sin (90^\circ+c_2-a_2)} {\sin (90^\circ+a_1-c_1)} \cdot \frac {\sin (90^\circ+a_2-b_2)} {\sin (90^\circ+b_1-a_1)} \\ &=1\ , \end{aligned} $$ because we can express all angles in terms of $a_1,b_1,c_1$, and then $$ \begin{aligned} c_2-a_2=(60^\circ-b_1)-(60^\circ-a_1)=a_1-b_1\ ,\\ a_2-b_2=(60^\circ-a_1)-(60^\circ-c_1)=c_1-a_1\ , \end{aligned} $$ (and since the two values $\sin(90\pm ?)$ coincide,) we obtain the cancellations of sine factors.

Now use the reciprocal of the theorem of Ceva to obtain $AA'$, $BB'$, $CC'$ concurrent.

$\square$

$\endgroup$
0
$\begingroup$

Another way to prove this is to use projective geometry maps.

Observe that $E$ and $F$ are on fixed lines (they lie on perpendicular segment bisector for $AC$ and $AB$ respectively, name them $e$ and $f$). So when moving $P$ on circle $BOC$ we see that angle $\angle FDE$ is constant ($=60^{\circ}$), so the transformation $DE\mapsto DF$ is a projective map (induced by rotation around $D$ for $60^{\circ}$) of pencil of lines through $D$ to it self. This one induce new projective map $E\mapsto F$ from $e$ to $f$. But now we have new projective map from a pencil of lines through $B$ to a pencil of lines through $C$: $$\pi: BE\longmapsto CF$$ and let $X$ be their intersection point. Since $BC$ goes to it self so $\pi$ is perspective so $X$ describes some line. So we need only two position of $P$ to show that $X$ is on $AD$ and then you arte done.

Now this you could easly check if you put say $P=I$ and $P=O$ and you are done.

$\endgroup$
8
  • $\begingroup$ I think, for the problem, what we need to do is only show that $BA,BE,BC,BD$ and $CA,CF,CB,CD$ are perspective. $\endgroup$ Commented Feb 28, 2020 at 14:26
  • $\begingroup$ And, since $BC$ and $CB$ are corresponding to itself, so we only need to show the two pencils above are projective. $\endgroup$ Commented Feb 28, 2020 at 14:32
  • $\begingroup$ Yes that is correct, I see now, if you put $P=I$ is also obvious. $\endgroup$
    – nonuser
    Commented Feb 28, 2020 at 14:44
  • $\begingroup$ But I fail to comprehend well what you said, say, "But for this we need only three differnt positions for $E$ ". $\endgroup$ Commented Feb 28, 2020 at 14:58
  • $\begingroup$ Yes, you are right, if $BC$ goes to it self then $X$ describes some line, so we need only two position of $P$ to show that $X$ is on $AD$ and then you arte done. $\endgroup$
    – nonuser
    Commented Feb 28, 2020 at 15:00

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .