BdMO Online Forum

joydip wrote: ↑

Thu Dec 14, 2017 5:57 pm

Lemma 5: Let $F_e$ be the feuerbach point of $\triangle ABC$. Then $F_e , A_o$ and $L_a$ are colinear .

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 .

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.

Dustan wrote: ↑

Thu Dec 10, 2020 10:01 pm

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.

Solution :
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 $BC||PA$, we get $\angle PAN=\angle BMN=90^{\circ}=\angle PYN$.
This implies $P,A,Y,N$ concyclic.
So, $P,A,X, Y, N$ concyclic. $Q.E.D.$

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 :

Anindya Biswas wrote: ↑

Fri Feb 26, 2021 11:34 am

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 :

IMO 2006 P1

Proof : Couldn't write the proof in details due to latex inexperience
(How to upload picture??)

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:

Asif Hossain wrote: ↑

Sat Feb 27, 2021 12:24 pm

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:

Iran TST 2010 No.1

Hint(At least in my solution):

Asif Hossain wrote: ↑

Sat Feb 27, 2021 12:24 pm

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:

Iran TST 2010 No.1

Thanks to mahanmath from AoPS for the solution.
Let $\omega$ be the circumcircle of $ADE$. By Three tangents lemma, $AT, CD, BE$, are tangents to $\omega$. Since $K,L$ are midpoints of $ME,MD$ respectfully, $KLT$ is the radical axis of $\omega$ and point circle $M$. So, $Pow_{\omega}(T)=Pow_{M}(T)\Rightarrow TA=TM$

Since nobody posting any problem here is a easy one:
Let $ABC$ be a triangle. The incircle of $ABC$ is to $AB$ and $AC$ at $D$ and $E$ respectively. Let $O$ denote the circumcenter of $BCI$ Prove $\angle ODB = \angle OEC$
Source

Through Incenter and excenter lemma A,I,O are collinear.
$ADO\cong AEO$
we get $\angle ADO=\angle AEO$
so,$\angle ODB=\angle OEC$