## Geometry Marathon : Season 3

- Raiyan Jamil
**Posts:**138**Joined:**Fri Mar 29, 2013 3:49 pm

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 47:}$

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 $PE^2=PY.PZ=PN.PM$.

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 $PE^2=PY.PZ=PN.PM$.

A smile is the best way to get through a tough situation, even if it's a fake smile.

- Raiyan Jamil
**Posts:**138**Joined:**Fri Mar 29, 2013 3:49 pm

### Re: Geometry Marathon : Season 3

$\text{Problem 48:}$

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.

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.

A smile is the best way to get through a tough situation, even if it's a fake smile.

### Re: Geometry Marathon : Season 3

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 .

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 more I learn, the more I realize how much I don't know.

- Albert Einstein

- Albert Einstein

### Re: Geometry Marathon : Season 3

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.

I have no problem to submit. Anybody feel free to take my turn.

The more I learn, the more I realize how much I don't know.

- Albert Einstein

- Albert Einstein

- ahmedittihad
**Posts:**147**Joined:**Mon Mar 28, 2016 6:21 pm

### Re: Geometry Marathon : Season 3

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 $AH=HN$.

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 $AH=HN$.

Frankly, my dear, I don't give a damn.

- Raiyan Jamil
**Posts:**138**Joined:**Fri Mar 29, 2013 3:49 pm

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 49:}$

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 $AH=HN$.

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.

- Raiyan Jamil
**Posts:**138**Joined:**Fri Mar 29, 2013 3:49 pm

### Re: Geometry Marathon : Season 3

$\text{Problem 50:}$

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.

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.

A smile is the best way to get through a tough situation, even if it's a fake smile.

### Re: Geometry Marathon : Season 3

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.

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 more I learn, the more I realize how much I don't know.

- Albert Einstein

- Albert Einstein

### Re: Geometry Marathon : Season 3

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.

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.

I like girls and mathematics; both are beautiful.

### Re: Geometry Marathon : Season 3

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:
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 .

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:

$$I_bM^2=MC^2=MS\cdot MP=2OM(MK+KP)=2OM(\dfrac{OM}{2} + LN)=OM(2LN + OM)$$

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 more I learn, the more I realize how much I don't know.

- Albert Einstein

- Albert Einstein