**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 $\triangle ABC$.

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

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.

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.

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

$\text{Solution to Problem 39: }$

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.

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

$\text{Problem 40:}$

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 $ABC$

$\text{Solution to Problem 40:}$

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.

$\text{Problem 41}$

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:**181**Joined:**Mon Mar 28, 2016 6:21 pm

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:**181**Joined:**Mon Mar 28, 2016 6:21 pm

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

$a)$ $P$ 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.