## Geometry Marathon : Season 3

### Re: Geometry Marathon : Season 3

Problem 38 :

In $\triangle ABC$ let the angle bisector of $\angle BAC$ meet $BC$ at $A_o$. Define $B_o,C_o$ similarly.Prove that

the circumcircle of $\triangle A_oB_oC_o$ goes though the Feuerbach point of $\triangle ABC$.

In $\triangle ABC$ let the angle bisector of $\angle BAC$ meet $BC$ at $A_o$. Define $B_o,C_o$ similarly.Prove that

the circumcircle of $\triangle A_oB_oC_o$ goes though the Feuerbach point of $\triangle ABC$.

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

- Albert Einstein

- Albert Einstein

- Thanic Nur Samin
**Posts:**176**Joined:**Sun Dec 01, 2013 11:02 am

### Re: Geometry Marathon : Season 3

Here is an alternate solution to problem 36

We apply cartesian coordinates.

Note that the problem is equivalent to proving $F',D,E$ collinear in the diagram.

Let $U\equiv (a,1), V\equiv (-a,1), T\equiv (at,t), S\equiv (-as,s)$.

Now, let $E \equiv (p,q)$. We get, $\dfrac{q-t}{p-at}=-a$ and $\dfrac{q-s}{p+as}=a$. From these, we get $E \equiv (\dfrac{(a^2+1)(t-s)}{2a},\dfrac{(a^2+1)(t+s)}{2})$.

Again, $D$ lies on the $y$-axis. From $DU\perp OU$, we get that $D\equiv (a^2+1,0)$.

Now for the *most difficult* part, the coordinates of $F$. The slope of $ST$ is $\dfrac{t-s}{a(t+s)}$. So the slope of $OF$ is $\dfrac{a(t+s)}{s-t}$. So for some real $m$, $F\equiv (m(s-t),ma(s+t))$.

Now, $FS\perp OT$, so $\dfrac{ma(s+t)-s}{m(s-t)+as}=-a$. So we get $m=\dfrac{1-a^2}{2a}$. So $F\equiv (\dfrac{(1-a^2)(s-t)}{2a},\dfrac{(1-a^2)(s+t)}{2})$. So $F'\equiv (\dfrac{(1-a^2)(s-t)}{2a},2-\dfrac{(1-a^2)(s+t)}{2})$.

Now, apply the shoelace formula to $F',D,E$. The det is $0$, so they are collinear.

Time taken: 30 minutes while simultaneously taking english notes.

We apply cartesian coordinates.

Note that the problem is equivalent to proving $F',D,E$ collinear in the diagram.

Let $U\equiv (a,1), V\equiv (-a,1), T\equiv (at,t), S\equiv (-as,s)$.

Now, let $E \equiv (p,q)$. We get, $\dfrac{q-t}{p-at}=-a$ and $\dfrac{q-s}{p+as}=a$. From these, we get $E \equiv (\dfrac{(a^2+1)(t-s)}{2a},\dfrac{(a^2+1)(t+s)}{2})$.

Again, $D$ lies on the $y$-axis. From $DU\perp OU$, we get that $D\equiv (a^2+1,0)$.

Now for the *most difficult* part, the coordinates of $F$. The slope of $ST$ is $\dfrac{t-s}{a(t+s)}$. So the slope of $OF$ is $\dfrac{a(t+s)}{s-t}$. So for some real $m$, $F\equiv (m(s-t),ma(s+t))$.

Now, $FS\perp OT$, so $\dfrac{ma(s+t)-s}{m(s-t)+as}=-a$. So we get $m=\dfrac{1-a^2}{2a}$. So $F\equiv (\dfrac{(1-a^2)(s-t)}{2a},\dfrac{(1-a^2)(s+t)}{2})$. So $F'\equiv (\dfrac{(1-a^2)(s-t)}{2a},2-\dfrac{(1-a^2)(s+t)}{2})$.

Now, apply the shoelace formula to $F',D,E$. The det is $0$, so they are collinear.

Time taken: 30 minutes while simultaneously taking english notes.

Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

Because destroying everything mindlessly isn't cool enough.

### Re: Geometry Marathon : Season 3

As problem 38 remained unsolved for many days , i will give another problem. (I will try to find a solution for this later when I get enough time . )

Problem 39 :Let $ABC$ be a triangle with circumcenter $O$.Let $BC$ be the smallest side. $(BOC)$ cuts $CA, AB$ at $E$ and $F$. $BE$ cuts $CF$ at $M$. Denote $J$ is the $M$-excenter of $\triangle BMC, I$ is the incenter of $\triangle ABC$ and $L$ is the intersection of two tangents at $B, C$ of $(O)$. Prove that $J, I, L$ are collinear.

Problem 39 :Let $ABC$ be a triangle with circumcenter $O$.Let $BC$ be the smallest side. $(BOC)$ cuts $CA, AB$ at $E$ and $F$. $BE$ cuts $CF$ at $M$. Denote $J$ is the $M$-excenter of $\triangle BMC, I$ is the incenter of $\triangle ABC$ and $L$ is the intersection of two tangents at $B, C$ of $(O)$. Prove that $J, I, L$ are collinear.

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

- Albert Einstein

- Albert Einstein

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

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 39: }$

We do angle chase and get $\angle LCJ=\angle IBC$ and $\angle LBJ=\angle ICB$. And the we use $\text{trig ceva}$ w.r.t. $\bigtriangleup BJC$ and the points $I$ and $L$ and we get that $\frac{sin \angle CJL}{sin \angle BJL}=\frac {sin \angle CJI} {sin \angle BJI} \Rightarrow J,L,I$ are collinear.

We do angle chase and get $\angle LCJ=\angle IBC$ and $\angle LBJ=\angle ICB$. And the we use $\text{trig ceva}$ w.r.t. $\bigtriangleup BJC$ and the points $I$ and $L$ and we get that $\frac{sin \angle CJL}{sin \angle BJL}=\frac {sin \angle CJI} {sin \angle BJI} \Rightarrow J,L,I$ are collinear.

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

$\text{Problem 40:}$

Let the Nagel point of triangle $ABC$ be $N$. We draw lines from $B$ and $C$ to $N$ so that these lines intersect sides $AC$ and $AB$ in $D$ and $E$ respectively. $M$ and $T$ are midpoints of segments $BE$ and $CD$ respectively. $P$ is the second intersection point of circumcircles of triangles $BEN$ and $CDN$. $l_1$ and $l_2$ are perpendicular lines to $PM$ and $PT$ in points $M$ and $T$ respectively. Prove that lines $l_1$ and $l_2$ intersect on the circumcircle of triangle $ABC$

Let the Nagel point of triangle $ABC$ be $N$. We draw lines from $B$ and $C$ to $N$ so that these lines intersect sides $AC$ and $AB$ in $D$ and $E$ respectively. $M$ and $T$ are midpoints of segments $BE$ and $CD$ respectively. $P$ is the second intersection point of circumcircles of triangles $BEN$ and $CDN$. $l_1$ and $l_2$ are perpendicular lines to $PM$ and $PT$ in points $M$ and $T$ respectively. Prove that lines $l_1$ and $l_2$ intersect on the circumcircle of triangle $ABC$

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

### Re: Geometry Marathon : Season 3

$\text{Solution to Problem 40:}$

Since $N$ is the Nagel point, we can easily prove that $BE=CD$ i.e. $BM=CT$. $P$ is the center of spiral similarity which sends $BE$ to $CD$ thus it also sends $M$ to $T$. Since $BE=CD$, we have the dilation factor at $1$ thus $PM=PT$. Now since a spiral similarity centered at $P$ sends $BM$ to $DT$, $AMPT$ is cyclic ($A$ is the intersection of $BM$ and $DT$).

Let $Q$ be the intersection point of the circumcircles of $ABC$ and $AMT$ (Other than $A$). Then $Q$ is the center of spiral similarity which sends $MB$ to $TC$. But then we have $QM=QT$ since $MB=TC$. This means $PQ$ is a diameter of the circumcircle of $AMN$. So $Q$ is the intersection of $l_1$ and $l_2$ and it lies on the circumcircle of $ABC$.

Geo is for whimps.

Since $N$ is the Nagel point, we can easily prove that $BE=CD$ i.e. $BM=CT$. $P$ is the center of spiral similarity which sends $BE$ to $CD$ thus it also sends $M$ to $T$. Since $BE=CD$, we have the dilation factor at $1$ thus $PM=PT$. Now since a spiral similarity centered at $P$ sends $BM$ to $DT$, $AMPT$ is cyclic ($A$ is the intersection of $BM$ and $DT$).

Let $Q$ be the intersection point of the circumcircles of $ABC$ and $AMT$ (Other than $A$). Then $Q$ is the center of spiral similarity which sends $MB$ to $TC$. But then we have $QM=QT$ since $MB=TC$. This means $PQ$ is a diameter of the circumcircle of $AMN$. So $Q$ is the intersection of $l_1$ and $l_2$ and it lies on the circumcircle of $ABC$.

Geo is for whimps.

### Re: Geometry Marathon : Season 3

$\text{Problem 41}$

Let $ABC$ be a triangle, and $I$ the incenter, $M$ midpoint of $ BC $, $ D $ the touch point of incircle and $ BC $. Prove that perpendiculars from $M, D, A $ to $AI, IM, BC $ respectively are concurrent.

Let $ABC$ be a triangle, and $I$ the incenter, $M$ midpoint of $ BC $, $ D $ the touch point of incircle and $ BC $. Prove that perpendiculars from $M, D, A $ to $AI, IM, BC $ respectively are concurrent.

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

### Re: Geometry Marathon : Season 3

Solution to Problem 41

Claim: Let $IM$ intersect the perpendicular from $A$ to $BC$ at $G$. $GD \parallel AI$.

Proof: As $AG \parallel ID$, it would suffice to prove that $AG=ID$. Hence yielding $AGDI$ a parallelogram.

We prove $AG=ID$ by barycentric coordinates.

Plugging in points $M=(0,1/2,1/2)$ and $I=( (a/(a+b+c), b/(a+B+C), c/(a+b+c) )$ into the line equation, we get that every point $P=(x, y, z)$ on line $IM$ satisfies the equation $((c-b)x/a)+y-z=0$.

Let $G=(l, m, n)$.

As $G$ lies on the perpendicular from $A$ to $BC$, Using EFFT on $\vec{AG}=(1-l,-m,-n)$ and $\vec{BC}=(0,1,-1)$, we get that $a^2(m-n)+b^2(l-1)+c^2(1-l)=0$

or, $a^2/(b^2-c^2)=(1-l)/(m-n)$

or, $m/n=(a^2+b^2-c^2)/(a^2-b^2+c^2)$.

Inputting $m=(a^2+b^2-c^2)$ and $n=(a^2-b^2+c^2)$ into the line equation of $IM$, we get

$(c-b)l/a+a^2+b^2-c^2+a^2-b^2+c^2=0$

or,$l=2a(b+c)$.

but, the coordinates are non normalized here. So, the normalized $l= 2a(b+_c)/2a(a+b+c)=(1-a)/(a+b+c)$.

So, we have proved our claim.

Proof of the main problem:

Let $GD$ intersect the perpendicular from $M$ to $AI$ at $Z$. We get $\angle GZM= 90^{\circ}$. We are done as the desired concurrency is the orthocenter of $\triangle GZM$.

Claim: Let $IM$ intersect the perpendicular from $A$ to $BC$ at $G$. $GD \parallel AI$.

Proof: As $AG \parallel ID$, it would suffice to prove that $AG=ID$. Hence yielding $AGDI$ a parallelogram.

We prove $AG=ID$ by barycentric coordinates.

Plugging in points $M=(0,1/2,1/2)$ and $I=( (a/(a+b+c), b/(a+B+C), c/(a+b+c) )$ into the line equation, we get that every point $P=(x, y, z)$ on line $IM$ satisfies the equation $((c-b)x/a)+y-z=0$.

Let $G=(l, m, n)$.

As $G$ lies on the perpendicular from $A$ to $BC$, Using EFFT on $\vec{AG}=(1-l,-m,-n)$ and $\vec{BC}=(0,1,-1)$, we get that $a^2(m-n)+b^2(l-1)+c^2(1-l)=0$

or, $a^2/(b^2-c^2)=(1-l)/(m-n)$

or, $m/n=(a^2+b^2-c^2)/(a^2-b^2+c^2)$.

Inputting $m=(a^2+b^2-c^2)$ and $n=(a^2-b^2+c^2)$ into the line equation of $IM$, we get

$(c-b)l/a+a^2+b^2-c^2+a^2-b^2+c^2=0$

or,$l=2a(b+c)$.

but, the coordinates are non normalized here. So, the normalized $l= 2a(b+_c)/2a(a+b+c)=(1-a)/(a+b+c)$.

So, we have proved our claim.

Proof of the main problem:

Let $GD$ intersect the perpendicular from $M$ to $AI$ at $Z$. We get $\angle GZM= 90^{\circ}$. We are done as the desired concurrency is the orthocenter of $\triangle GZM$.

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

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

### Re: Geometry Marathon : Season 3

Problem 42

Let $ABC$ be a triangle with altitudes $AD,BE,CF$ with $D,E,F$ are on sides $BC,CA,AB$, resp. Let $O$ be the circumcenter of triangle $ABC$. $P,Q$ are respectively on $DE,DF$ such that $FP\perp AC$ and $EQ\perp AB$.

a) Prove that the midpoint $K$ of $AO$ is circumcenter of triangle $DPQ$.

b) The perpendicular line through $O$ to $PQ$ intersects $CA,AB$ at $U,V$, reps. Euler line of $ABC$ intersects $CA,AB$ at $S,T$, reps. Prove that $TU$ and $SV$ intersect at $X$ on perpendicular bisector of $BC$.

Let $ABC$ be a triangle with altitudes $AD,BE,CF$ with $D,E,F$ are on sides $BC,CA,AB$, resp. Let $O$ be the circumcenter of triangle $ABC$. $P,Q$ are respectively on $DE,DF$ such that $FP\perp AC$ and $EQ\perp AB$.

a) Prove that the midpoint $K$ of $AO$ is circumcenter of triangle $DPQ$.

b) The perpendicular line through $O$ to $PQ$ intersects $CA,AB$ at $U,V$, reps. Euler line of $ABC$ intersects $CA,AB$ at $S,T$, reps. Prove that $TU$ and $SV$ intersect at $X$ on perpendicular bisector of $BC$.

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

Solution to Problem 42:

$a)$ $P$ is just the reflection of $F$ under $AC$ and $Q$ is the reflection of $E$ under $AB$. $AFDCP$ and $AEDBQ$ cyclic. The rest is trivial by taking homothety with centre $A$ and ratio 2.

$b)$ Let $H_1$ be the reflection of $H$ under $A$. We can find that the perpendicular from $O$ to $PQ$ is actually $OH_1$. Now, let $M$ be the midpoint of $BC$. So, $OH,OH_1,OA,OM$ is a harmonic pencil. Let $TU$ intersect $OM$ at $X$ and $AO$ at $Y$. We get $(T,U;X,Y)=-1$. Let $AX,AO$ intersect $BC$ at $Z,W$ respectively. We get, by pencil $A$, $(B,C;Z,W)=-1$. The rest is trivial.

$a)$ $P$ is just the reflection of $F$ under $AC$ and $Q$ is the reflection of $E$ under $AB$. $AFDCP$ and $AEDBQ$ cyclic. The rest is trivial by taking homothety with centre $A$ and ratio 2.

$b)$ Let $H_1$ be the reflection of $H$ under $A$. We can find that the perpendicular from $O$ to $PQ$ is actually $OH_1$. Now, let $M$ be the midpoint of $BC$. So, $OH,OH_1,OA,OM$ is a harmonic pencil. Let $TU$ intersect $OM$ at $X$ and $AO$ at $Y$. We get $(T,U;X,Y)=-1$. Let $AX,AO$ intersect $BC$ at $Z,W$ respectively. We get, by pencil $A$, $(B,C;Z,W)=-1$. The rest is trivial.

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