Better proof : $F_e , A_o$ and $L_a$ are the pairwise center of similitudes of the incircle , ninepoint circle and the $A$excircle . So by d'Alembert's Theorem $F_e , A_o$ and $L_a$ are colinear .
Geometry Marathon : Season 3
Re: Geometry Marathon : Season 3
The first principle is that you must not fool yourself and you are the easiest person to fool.
Re: Geometry Marathon : Season 3
Problem 51:Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
 Anindya Biswas
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Re: Geometry Marathon : Season 3
Solution :Dustan wrote: ↑Thu Dec 10, 2020 10:01 pmProblem 51:Let $ABC$ be a triangle with $AB=AC$, and let $M$ be the midpoint of $BC$. Let $P$ be a point such that $PB<PC$ and $PA$ is parallel to $BC$. Let $X$ and $Y$ be points on the lines $PB$ and $PC$, respectively, so that $B$ lies on the segment $PX$, $C$ lies on the segment $PY$, and $\angle PXM=\angle PYM$. Prove that the quadrilateral $APXY$ is cyclic.
Let $N$ be the intersection point of circumcircle of $\triangle BXM$ and $\triangle CYM$. $\angle BXM=\angle BNM$ and $\angle CYM=\angle CNM$. So, $MN$ bisects $\angle BNC$
$\therefore \frac{BN}{CN}=\frac{BM}{CM}=1\Rightarrow BN=CN$
So, $\triangle BNC$ is an isosceles triangle. Since $M$ is the midpoint of $BC$, we get that $NM\perp BC$. Also, $N,M,A$ collinear.
Now, $\angle PYN=\angle CYN=180^{\circ}\angle CMN=90^{\circ}$
Similarly, $\angle PXN=90^{\circ}$.
So, $\angle PXN+\angle PYN=180^{\circ}$.
Therefore, $P,X,N,Y$ concyclic.
Again, since $BCPA$, we get $\angle PAN=\angle BMN=90^{\circ}=\angle PYN$.
This implies $P,A,Y,N$ concyclic.
So, $P,A,X, Y, N$ concyclic. $Q.E.D.$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann
 Anindya Biswas
 Posts: 71
 Joined: Fri Oct 02, 2020 8:51 pm
 Location: Magura, Bangladesh
 Contact:
Re: Geometry Marathon : Season 3
Problem 52 :
Let $I$ be the incenter of $\triangle ABC$. A point $P$ in the interior of $\triangle ABC$ satisfies :
$$\angle PBA+\angle PCA=\angle PBC+\angle PCB$$
Show that $AP\geq AI$ and the equality holds if and only if $P\equiv I$.
Source :
Let $I$ be the incenter of $\triangle ABC$. A point $P$ in the interior of $\triangle ABC$ satisfies :
$$\angle PBA+\angle PCA=\angle PBC+\angle PCB$$
Show that $AP\geq AI$ and the equality holds if and only if $P\equiv I$.
Source :
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
— John von Neumann
— John von Neumann

 Posts: 59
 Joined: Sat Jan 02, 2021 9:28 pm
Re: Geometry Marathon : Season 3
Proof :Anindya Biswas wrote: ↑Fri Feb 26, 2021 11:34 amProblem 52 :
Let $I$ be the incenter of $\triangle ABC$. A point $P$ in the interior of $\triangle ABC$ satisfies :
$$\angle PBA+\angle PCA=\angle PBC+\angle PCB$$
Show that $AP\geq AI$ and the equality holds if and only if $P\equiv I$.
Source :
(How to upload picture??)
Hmm..Hammer...Treat everything as nail

 Posts: 59
 Joined: Sat Jan 02, 2021 9:28 pm
Re: Geometry Marathon : Season 3
Problem 53:
In acute triangle $ABC$ $\angle B$ is greater than $\angle C$. Let $M$ is midpoint of $BC$. $D$ and $E$ are the feet of the altitude from $C$and $B$ respectively. $K$ and $L$ are midpoint of $ME$ and $MD$ respectively. If $KL$ intersect the line through $A$ parallel to $BC$ in $T$, prove that $TA=TM$.
Source:
In acute triangle $ABC$ $\angle B$ is greater than $\angle C$. Let $M$ is midpoint of $BC$. $D$ and $E$ are the feet of the altitude from $C$and $B$ respectively. $K$ and $L$ are midpoint of $ME$ and $MD$ respectively. If $KL$ intersect the line through $A$ parallel to $BC$ in $T$, prove that $TA=TM$.
Source:
Hmm..Hammer...Treat everything as nail