China National-2013-1

[Tahmid Hasan]

Two circles $K_1$ and $K_2$ of different radii intersect at two points $A$ and $B$, let $C$ and $D$ be two points on $K_1$ and $K_2$, respectively, such that $A$ is the midpoint of the segment $CD$. The extension of $DB$ meets $K_1$ at another point $E$, the extension of $CB$ meets $K_2$ at another point $F$. Let $l_1$ and $l_2$ be the perpendicular bisectors of $CD$ and $EF$, respectively.
(1) Show that $l_1$ and $l_2$ have a unique common point (denoted by $P$).
(2) Prove that the lengths of $CA$, $AP$ and $PE$ are the side lengths of a right triangle.

[Tahmid Hasan]

My solution:
(1). $\angle CAE=\angle AEB+\angle ADB=\angle ACB+\angle AFB=\angle DAF$. So $AP$ bisects $\angle EAF \Rightarrow P \in \odot AEF$.
$\angle EPE=180^{\circ}-\angle EAF=180^{\circ}-(180^{\circ}-\angle EAC-\angle DAF)=\angle EBC+\angle DBF=360^{\circ}-2\angle EBF$.
Hence $P$ is the circumcentre of $\triangle BEF$.
(2). $2CA^2=CA.2CA=CA(CA+DA)=CA.CD=CB.CF=CP^2-PE^2=CA^2+AP^2-PE^2$
$\Rightarrow AP^2=CA^2+PE^2$.

[sourav das]

Solution: Let $K,L$ be midpoints of $BE$ and $BF$. Let $P'$ be the center of circle $BEF$. Now $CA.CD=CB.CF=CL^2-LF^2=CP'^2-P'F^2$ and $DA.DC=DE.DB=DK^2-BK^2=DP'^2-BP'^2$ using power of point. But as $CA=AD$ and $P'F=P'B$; so $CP'=P'D$ and so $P=P'$. Also, since $2CA^2=CA.CD=CL^2-LF^2=CP'^2-P'F^2=AP^2+CA^2-PE^2$ and so, $AP^2=CA^2+PE^2$ (Proved)
@Tahmid; if you include your figure that will be very helpful. Some of your arguments are confusing according to my figure.

[Tahmid Hasan]

@Tahmid; if you include your figure that will be very helpful. Some of your arguments are confusing according to my figure.

Actually most of the time I draw my figures by hand, that's why I don't attach figures to my solution.
But since you asked, I generated one

[Phlembac Adib Hasan]

I am a bit confused about the statement. @Tahmid vai & Sourav vai, what did you mean/realize by 'unique' for the first part? Does it mean $P$ does not depend on the choice of $CD$? I didn't found the answer and read your solutions. But still I am confused. Both of you showed $P$ is the circumcenter of $BEF$. My question is how does it solve the problem?

[Tahmid Hasan]

I am a bit confused about the statement. @Tahmid vai & Sourav vai, what did you mean/realize by 'unique' for the first part? Does it mean $P$ does not depend on the choice of $CD$? I didn't found the answer and read your solutions. But still I am confused. Both of you showed $P$ is the circumcenter of $BEF$. My question is how does it solve the problem?

What you are referring to is called constant point in mathematical literature. What the question asked was to prove that $P$ is unique i.e. well-defined.
In my solution I defined $P$ two ways- (1) Midpoint of arc $EF$ in $\odot AEF$ (2) The circumcentre of $\triangle BEF$. You have to admit, that is quite well-defined.

[*Mahi*]

I think what Adib wanted to state in his post that "well defined" is not actually that much "well defined", and for this it may vary from person to person. I agree with him on that point.
Also when a question asks to prove something well defined, what should we do? Define the point as well as I can and hope for the best? It is not certainly a good thing to ask for in a problem solving contest.

[zadid xcalibured]

Well,We can ensure the well definition.Let $\triangle{AEF}$ be fixed.As $AC^2=AE.AF$.So the length of $AC$ is fixed.And as $C,B,F$ and $D,B,E$ are collinear,this property forces $AC$ to be perpendicular to the angle bisector of $\angle{A}$.
By restating the problem another way we can define the point $P$ well.