Geometry Marathon : Season 3

For discussing Olympiad level Geometry Problems
joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Fri Dec 29, 2017 12:45 pm

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 .
The first principle is that you must not fool yourself and you are the easiest person to fool.

Dustan
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Re: Geometry Marathon : Season 3

Unread post by Dustan » 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.

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Anindya Biswas
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Re: Geometry Marathon : Season 3

Unread post by Anindya Biswas » Fri Feb 26, 2021 1:01 am

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. $Q.E.D.$
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

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Anindya Biswas
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Re: Geometry Marathon : Season 3

Unread post by Anindya Biswas » 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
"If people do not believe that mathematics is simple, it is only because they do not realize how complicated life is."
John von Neumann

Asif Hossain
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Joined: Sat Jan 02, 2021 9:28 pm

Re: Geometry Marathon : Season 3

Unread post by Asif Hossain » Sat Feb 27, 2021 12:13 pm

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 :
Let us denote E as the excenter of $\triangle ABC$.
From,$$\angle PBA+\angle PCA=\angle PBC+\angle PCB$$
it is easy to prove $\angle IBP = \angle PCI$ implies $(BIPC)$ is cyclic.
By incenter-excenter lemma, $(IBCE)$ is cyclic. Line $AI$ is the diameter line. So, $AP\geq AI$ and the equality holds iff $P=I$
Couldn't write the proof in details due to latex inexperience :(
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Hmm..Hammer...Treat everything as nail

Asif Hossain
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Re: Geometry Marathon : Season 3

Unread post by Asif Hossain » 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
Hmm..Hammer...Treat everything as nail

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