Geometry Marathon : Season 3

For discussing Olympiad level Geometry Problems
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Raiyan Jamil
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Re: Geometry Marathon : Season 3

Unread post by Raiyan Jamil » Sat Nov 25, 2017 11:56 am

$\text{Solution to Problem 47:}$

Let $AC \cap BD=X$, $AC \cap EF=Y$ and $BD \cap EF=Z$. Now $P,M,N$ are collinear by newton line. Then, $XM.XZ=XB.XD=XA.XC=XN.XY \Rightarrow MNYZ$ is cyclic and $PE^2=PY.PZ=PN.PM$.
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Raiyan Jamil
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Re: Geometry Marathon : Season 3

Unread post by Raiyan Jamil » Sat Nov 25, 2017 11:59 am

$\text{Problem 48:}$

Prove that for a scalene triangle $ABC$, one can never find two pairs of isogonal conjugates ${P,P'}$ and ${Q,Q'}$ such that $P,P',Q,Q'$ are all collinear and given that they're not self isogonal conjugates.
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joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Sun Nov 26, 2017 1:37 pm

Solution of problem 48 :

We will prove that for any line $L$ there can be at most one pair of isogonal conjugate point on that line .
Let $I$ be the incenter and $I_b$ and $I_c$ be the B and C excenters respectively . Let $$L\cap BI_c=M$$ , $$L\cap CI_b=N$$ , $$L\cap BI=X$$ ,$$L\cap CI=Y$$ .

We want points $\{P,P'\} \in L$ such that $BI$ and $CI$ bisects $\angle PBP'$ and $\angle PCP'$ respectively .Then $(P,P';Y,N)=-1$ and $(P,P';X,M)=-1$. Now invert the line $L$ wrt any point not on that line .For a point $U$ , let $\bar U$ denote its inverse . This inversion sends $L$ to a circle ,say $\odot\bar L$ .Let the tangent at $\bar Y$ and $\bar N$ to that circle meet at $E$ , tangent at $\bar X$ and $\bar M$ to that circle meet at $F$. As inversion preserves cross ratio so , $E \in \bar P\bar P'$ , similarly $F\in \bar P\bar P'$ . So $ \odot\bar L \cap EF = \{\bar P , \bar P'\}$ , So $\{P,P'\} $ , if exist ( $EF$ may not intersect $\odot\bar L$ at all ), then must be unique .
Last edited by joydip on Sun Nov 26, 2017 10:53 pm, edited 3 times in total.
The first principle is that you must not fool yourself and you are the easiest person to fool.

joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Sun Nov 26, 2017 1:40 pm

We can also prove Problem $48$ by using the fact that (we are not giving a proof for this here) , the isogonal conjugate of a line wrt a triangle is a circumconic of that triangle .So the isogonal conjugate of that line must be a circumconic .But $ABCPP'QQ'$ cannot be a circumconic, as a line can cut a conic at atmost 2 points.

I have no problem to submit. Anybody feel free to take my turn.
The first principle is that you must not fool yourself and you are the easiest person to fool.

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ahmedittihad
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Re: Geometry Marathon : Season 3

Unread post by ahmedittihad » Sun Nov 26, 2017 7:28 pm

Problem $49$

Let $ABC$ be an acute-angled triangle with $AB\not= AC$. Let $\Gamma$ be the circumcircle, $H$ the orthocentre and $O$ the centre of $\Gamma$. $M$ is the midpoint of $BC$. The line $AM$ meets $\Gamma$ again at $N$ and the circle with diameter $AM$ crosses $\Gamma$ again at $P$. Prove that the lines $AP,BC,OH$ are concurrent if and only if $AH=HN$.
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Raiyan Jamil
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Re: Geometry Marathon : Season 3

Unread post by Raiyan Jamil » Sun Nov 26, 2017 10:48 pm

$\text{Solution to Problem 49:}$

Let $BH \cap AC=E, CH \cap AB=F, AP \cap BC=X$. Now, by miquel's theorem on circle $BCEF, M,H,P$ are collinear. Also by radical axis theorem on circle $ABC, APFHE, BCEF$, we get $X,F,E$ are collinear. Now, we get that $H$ is the orthocentre of triangle $AXM$ so $XH \perp AN$. So, $X,H,O$ are collinear if and only if $OH \perp AN$ or rather $AH=HN$.
A smile is the best way to get through a tough situation, even if it's a fake smile.

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Raiyan Jamil
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Re: Geometry Marathon : Season 3

Unread post by Raiyan Jamil » Sun Nov 26, 2017 11:16 pm

$\text{Problem 50:}$

Let $\vartriangle ABC$ be a triangle, $O$ be the circumcenter of $\vartriangle ABC$, $N$ be the center of nine point circle of $\vartriangle ABC$ and $X$ be the midpoint of the line segment $ON$. Let $A',B',C'$ be the midpoints of the line segments $BC,CA,AB$, respectively. Let ${{l}_{a}},{{l}_{b}},{{l}_{c}}$ be the lines through the points $X$ and $A'$, $X$ and $B'$, $X$ and $C'$, respectively. Finally, let ${{H}_{b}}$ and ${{H}_{c}}$ be feet of perpendiculars from $B$ and $C$ on the lines $AC$ and $BA$, respectively. Prove that the perpendicular line from ${{H}_{b}}$ on ${{l}_{b}}$, the perpendicular line from ${{H}_{c}}$ on ${{l}_{c}}$ and ${{l}_{a}}$ are concurrent.
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joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Thu Nov 30, 2017 6:15 pm

Solution of Problem 50 :

Let $H$ be the orthocenter and $(N)$ be the nine point circle of $ \triangle ABC$. $NA'\cap (N)=\{A',L\}$. Let $J$ be a point on $LA'$ such that $JH_c$ is tangent to $(N)$ at $H_c$.$K$ is a point on $JH_c$ such that $KN \|A'H_c$ .Let $M$ be the midpoint of $AH_b$.Then $LM \perp AH_b$.Let $CH_c \cap LM=P$.Let $\alpha$ be the circle centered at $A$ and orthogonal to $(N)$.Let $\beta$ be the the circle with center $K$ and radius $KH_c$ .

$\angle PH_cA=\angle PMA=90^\circ$ .So $AH_cPM$ is cyclic .
$KH_c=KL\Rightarrow L\in \beta$ . Now $2\angle LPN=2\angle BAC=\angle H_cKL\Rightarrow P \in \beta$.

Both $\alpha$ and $\beta$ are orthogonal to $(N)$ . So, $pow (N, \alpha)=pow (N, \beta )$
$pow (C , \alpha )=AC^2 -AH_b\cdot AB'=AC^2 -AM\cdot AC=AC\cdot MC=H_cC\cdot PC = pow (C, \beta )$

So $CN$ is the radical axis of $\alpha$ and $\beta$ . So, $AK \perp CN \qquad (\dagger ) $
Let the dilation wrt $J$ that' takes $N \rightarrow A'$ ,take $A \rightarrow T$ . This dilation takes $K \rightarrow H_c$ .So , $AK \|TH_c $ and $TA' \| AN \qquad (\ddagger )$
Let $G$ be the centroid of $\triangle ABC$ . Then the dilation wrt $G$ that takes $C \rightarrow C'$ , takes the point $N \rightarrow X$ . So ,$CN \| l_c$ .

So , $TH_c \perp l_c $ (for $ \dagger $) , Similarly $TH_b \perp l_b$.
As , $TA' \| AN$ ( $\ddagger$ ) and $XA'\| AN \Rightarrow T \in l_a$ , as desired .

I have no problem to submit. Anybody feel free to take my turn.
The first principle is that you must not fool yourself and you are the easiest person to fool.

tanmoy
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Re: Geometry Marathon : Season 3

Unread post by tanmoy » Fri Dec 01, 2017 9:29 pm

A probelm is posted twice. So, actually this is the number 50.

Problem 50:Let the incircle touches side $BC$ of a triangle $\triangle ABC$ at point $D$. Let $H$ be the orthocenter of $\triangle ABC$ and $M$ be the midpoint of segment $AH$. Let $E$ be a point on $AD$ so that $HE \perp AD$. Let $ME \cap AI = F$.
Prove that the circle $(BHC)$ and the circle with center $F$ and radius $FE$ are tangents to each other.
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joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Thu Dec 14, 2017 5:57 pm

As i promised before , here i am going to give my solution for Problem 38 .

Solution of Problem 38 :

Lemma 1: In a triangle $\triangle ABC$ ,let $M$ be the midpoint of $BC$ .Let $D$ be the projection of $A$ on $BC$ .Let $E$ be a point such that ,$E$ and $A$ are on the same side of $BC$ , $ED \| AM$ and $BM^2=AM( 2ED +AM)$ . Then $\angle BAC =180^\circ -\dfrac{1}{2}\angle BEC$

Proof : Let $K , L$ be the reflection of $A$ wrt $BC$ and $M$ respectively . Then $BKLC$ is cyclic .Let $AL $ intersect $\odot BKLC$ again at $J$ .Then , $$ ML \cdot JM =BM\cdot MC=BC^2=AM( 2ED +AM)=ML( 2ED +AM)$$
$$\Rightarrow JM = 2ED +AM \Rightarrow JA = 2ED $$
Now , as $JA \| ED$ and $D$ is the midpoint of $AK$ , So $J ,E ,K$ are collinear and $E$ is the midpoint of $JK$ .Let $O$ be the center of $\odot BKLC$ . Then $OE \perp EK$ . Now as , $arc ~BK =arc ~LC$ .So , $JK$ is the $J$ -symmedian of $\triangle JBC$ . So tangents to $\odot BKLC$ at $B$ and $C$ meet at $S \in JK$ .Then $BEOCS
$ is cyclic .
So , $$\angle BAC=\angle BLC =180^\circ -\angle BJC =180^\circ -\dfrac{1}{2}\angle BOC =180^\circ -\dfrac{1}{2}\angle BEC$$


Now we will use this lemma to prove a beautiful result on angles related to some famous triangle centers .

A Beautiful Lemma : Let $O$ and $N$ be the circumcenter and nine point center of $\triangle ABC$ repectively . Let $I_a ,I_b$ and $I_c$ be $A$-excenter , $B$-excenter and $C$-excenter of $\triangle ABC$ repectively . Then , $\angle I_bOI_c=180^\circ -\dfrac{1}{2}\angle I_bNI_c $

Comment:
(Similerly we can also prove $\angle IOI_a=180^\circ -\dfrac{1}{2}\angle INI_a $ , but we won't need that here .)
Proof :Let the circumcircle of $\triangle ABC $ meet $I_cI_b$ again at $M$. Then $M$ is the midpoint of $I_cI_b$ . Let $L$ be the midpoint of $AM$ , then $OL \perp I_cI_b$ and $LN \perp BC$ . Now $LN \| MO$ as $MO \perp BC$ .Let $MO$ intersect $BC$ at $P$ and $\odot ABC$ again at $S$ .Let $K$ be the midpoint of $MO$ . Then $LK=\dfrac{R}{2}=NP$ , so $LKPN$ is a parallelogram .So $LN=KP$ .Now ,
$$I_bM^2=MC^2=MS\cdot MP=2OM(MK+KP)=2OM(\dfrac{OM}{2} + LN)=OM(2LN + OM)$$
Now the proof is done , by using lemma 1 .


Lemma 3: In $\triangle ABC$ , $I_aO \perp B_oC_o$

Proof : Let $T$ be the circumcenter of $\triangle I_cII_b$ , then $T,O,I_a$ are collinear as $O$ and $I_a$ are the ninepoint center and orthocenter of $\triangle I_cII_b$ respectively . $AIBI_c$ and $AICI_b$ are cyclic . So , $AC_o \cdot BC_o=IC_o \cdot I_cC_o $ and $AB_o \cdot CB_o=IB_o \cdot I_bB_o $. So $ B_oC_o$ is the radical axis of $\odot I_cII_b$ and $\odot ABC$ . So , $I_aO \perp B_oC_o$.


Lemma 4: Let the $A$-excircle , $B$-excircle and $C$-excircle of $\triangle ABC$ touch the ninepoint circle at $L_a , L_b $ and $L_c$ repectively . Then $\triangle L_aL_bL_c \sim \triangle A_oB_oC_o$

Proof : By using lemma 2 we get ,
$$\angle L_aL_bL_c =\dfrac{1}{2} \angle L_aNL_c =\dfrac{1}{2} \angle I_aNI_c =180^\circ - \angle I_bOI_c $$

Using lemma 3 we get , $$\angle A_oB_oC_o = 180^\circ - \angle I_bOI_c $$
So , $$\angle A_oB_oC_o = \angle L_aL_bL_c $$
Similarly , we can show this for other angles .So , $\triangle L_aL_bL_c \sim \triangle A_oB_oC_o$


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

Proof: Let $M_a$ be the midpoint of $BC$ .$H_a$ be the projection of $A$ on $BC$ .The incircle and $A$ -excircle touches $BC$ at $X,Y$ respectively .Consider the inversion with center $M_a$ and radius $M_aY$.For a point $W$ , let $\bar W$ denote its inverse .$(H_a,A_o;X,Y)=-1$. So $H_a$ is the inverse of $A_o$ . Now , $\bar F_e\bar L_a$ is the other internal common tangent of the incircle and $A$ -excircle. $\bar L_aA_o \cdot A_o\bar F_e=XA_o \cdot A_oY=H_aA_o \cdot A_oM_a $ . So , $M_a\bar F_eH_a\bar L_a$ is cyclic . So , $F_e , A_o$ and $L_a$ are colinear .


Back To The Main Problem : $\measuredangle B_oF_eC_o =\measuredangle L_bF_eL_c=\measuredangle L_bL_aL_c= \measuredangle B_oA_oC_o$ . So $A_o , B_o , C_o , F_e$ are concyclic .
The first principle is that you must not fool yourself and you are the easiest person to fool.

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