In Acute angled triangle $ABC$, let $D$ be the point where $A$ angle bisector meets $BC$. The perpendicular from $B$ to $AD$ meets the circumcircle of $ABD$ at $E$. If $O$ is the circumcentre of triangle $ABC$ then prove that $A,E$ and $O$ are collinear.

$\text{Another Solution:}$

Let $E$ lies on $AO$ so that $BE \perp AD$. We will show that $A,B,D,E$ are concyclic.

Let $P$ be the projection of $A$ on $BC$. Let $BE \cap AD=Q$, $BE \cap AP=R$. Since $H$ and $O$ are isogonal conjugates, $AD$ also bisects $\angle HAO$. As $AQ \perp RE$, $RA=AE$. Also, $R,P,D,Q$ are concyclic. Then

Problem 5:
Let $I$ be an incenter of $\triangle ABC$. Denote $D, \ S \neq A$ intersections of $AI$ with $BC, \ (ABC)$ respectively. Let $K, \ L$ be incenters of $\triangle DSB, \ \triangle DCS$. Let $P$ be a reflection of $I$ with the respect to $KL$. Prove that $BP \perp CP$.

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 11:25 pm

by rubab

Solution of problem 5:

we know,$SB=SI=SC$
since $K$ is on the angle bisector of $\angle$BSI
$KB=KI$
And since $P$ is the reflection of $I$ across $KL$
$KP=KI$
SO $K$ is the circumcentre of $\triangle$BPI
Similarly $L$ is the circumcentre of $\triangle$IPC
Now,$\angle$BPC= $\angle$BPI+$\angle$IPC= $\frac{1}{2}(\angle$BKI+$\angle$ILC)=$\frac{1}{2}$(180-2($\angle$IBK)+ 180- 2($\angle$ICL))= 1/2(180-2($\frac{1}{4}\angle$A+$\frac{1}{2}\angle$B)+ 180- 2($\frac{1}{4}\angle$A+$\frac{1}{2}\angle$C))=90

Re: Geometry Marathon : Season 3

Posted: Fri Jan 06, 2017 11:41 pm

by rubab

Problem 6

In an acute triangle $ABC$ the points $D$,$E$ and $F$ are the feet of the altitudes through A,B,C respectively. The incenters of the triangle $AEF$ and $BDF$ are $I_1$ and $I_2$ respectively; the circumcenters of the triangles $ACI_1$ and $BCI_2$ are $O_1$ and $O_2$ respectively. Prove that $I_1I_2$ and $O_1O_2$ are parallel.

Re: Geometry Marathon : Season 3

Posted: Sat Jan 07, 2017 12:50 am

by joydip

Solution of problem 6 :

Let $I_3$ be the incenter of $\triangle CDE$ . Let $CI_3 \cap I_1I_2=S$
$\triangle FBD \sim \triangle FEA \Rightarrow \triangle FBI_2 \sim \triangle FEI_1 \Rightarrow \triangle FI_2I_1 \sim \triangle FBE$.
So,$\angle BI_2I_1+\angle I_2AB=\angle BI_2F+\angle FI_2I_1+\angle I_1AB=90^\circ +\dfrac {B}{2}+90^\circ-B+\dfrac {B}{2} =180^\circ $. So $BI_2I_1A$ is cyclic. Simillerly $BI_2I_3C,CI_3I_1A$ are cyclic .
$\angle I_2I_3S +\angle I_3I_2I_1=\angle CBI_2+\angle I_1AB +\angle I_3CB=\dfrac {A}{2}+\dfrac {B}{2}+\dfrac {C}{2}= 90^\circ \Rightarrow CI_3\perp I_1I_2$,but $CI_3\perp O_1O_2 \Rightarrow I_1I_2 \parallel O_1O_2$

Re: Geometry Marathon : Season 3

Posted: Sat Jan 07, 2017 1:22 am

by rubab

problem 7:
Let $I$ be the incenter of triangle $ABC$. Prove that the nine point circles of triangle $AIB$, $BIC$ and $CIA$ are concurrent at the feuerbach point of triangle $ABC$.

Re: Geometry Marathon : Season 3

Posted: Sat Jan 07, 2017 9:57 am

by joydip

Solution to problem 7 :

Let $H$ and $O$ be the orthocenter and circumcenter of $\triangle IBC$ respectively.Let $\omega$ be the nine point circle of $\triangle IBC$,$M,N$ be the midpoints of $BC,IH $ respectively.$(I)$ touches $BC$ at $D$.Let $K$ be the reflection of $D$ W.R.T $AI$.
Then $A,I,O$ are collinear .$IN = \dfrac {IH}{2}=OM \Rightarrow INMO$ is a parallelogram .So, $IO \| MN$.Now,$KD \perp AI \Rightarrow KD \perp MN$.Let $KD$ meet $\omega$ again at $P$.As $MN$ is a diameter of $\omega$,so $MD=MP$
Let$ \psi$ be the circle with center $M$ and radios $MD$.Then the inversion W.R.T $\psi$ sends $\omega$ to $DP$ .This inversion sends the nine point circle of $\triangle ABC$ to the tangent of $(I)$ at $K$ and $(I)$ to $(I)$.So,it sends the feuerbach point($F_e$) to $K$.so,$ K\in DP \Rightarrow F_e\in \omega$,completing the proof.

Re: Geometry Marathon : Season 3

Posted: Sat Jan 07, 2017 11:13 am

by tanmoy

Problem 8:
Given a cyclic quadrilateral $ABCD$ with circumcircle $(O)$. Let $AB \cap CD=E, \ AD \cap BC=F, \ AC \cap BD=G, \ AC \cap EF=P, \ BD \cap EF=Q$. Let $M, \ N$ be midpoints of $AC, \ BD$, respectively and let $MN \cap EF=H$.

(i) Prove that $M, \ N, \ P, \ Q$ are concyclic.

(ii) Let $K$ be the center of the circumcircle of $MNPQ$. $OK \cap EF=L$, $GL$ cuts $(O)$ at two distinct points $S$ and $T$. Prove that $HS, \ HT$ are tangents from $H$ to $(O)$ at $S$ and $T$, respectively.

Re: Geometry Marathon : Season 3

Posted: Sat Jan 07, 2017 4:35 pm

by M Ahsan Al Mahir

$\text{Solution of problem 8:}$

Part (i): Lemma: If $(A, \ B; \ C, \ D)=-1$ and $M$ is the midpoint of $AB$, then $CM \times CD=CA \times CB$.

Lemma's Proof: Consider a circle with center $M$ and radius $MA=MB$. Draw a tangent $DX$ from $D$ to $\odot (M)$. So we have $XC \perp AB$. Using similar triangles, the result follows directly.

Now for our original problem, we have $(C, \ A; \ G, \ P)=(D, \ B; \ G, \ D)=-1$. From our lemma, we show that, $GN \times GQ=GB \times GD=GA \times GC=GM \times GP$. From which the result follows.

Part (ii):
As from (i), we have $MNPQ$ is cyclic. Let's denote the midpoint of $EF$ by $H$.
Let $HG \cap \odot (ABCD)=\ X, \ Y$. From the harmonic bundle $(D, \ B; \ G, \ Q)=-1$, we get that $GB \times GD = GN \times GQ = GX \times GY$. So, $\square XNYQ$ is cyclic. Similarly, $\square XMYP$ is cyclic too.
And as $MN$ meets $PQ$ at H, it implies that $X, \ Y$ lies on $\odot (MNPQ)$. So, $XY$ is the radical axis of $\odot (ABCD)$ and $\odot (MNPQ)$.

As the polar line of every point on $EF$ passes through $G$, and $XY \perp OL$, $LX, \ LY$ are tangents to $ABCD$. As $H$ lies on the polar lines of $G$ and $L$, the polar line of $H$ passes through $GL$. (Q.E.D)

Re: Geometry Marathon : Season 3

Posted: Sat Jan 07, 2017 6:29 pm

by Thanic Nur Samin

Problem 9

Let $\{P, P'\}$ and $\{Q,Q'\}$ be two pairs of isogonal conjugates of $\triangle ABC$. Let $\triangle P_AP_BP_C$ be the cevain triangle of $P$ wrt $\triangle ABC$. Define $\triangle Q_AQ_BQ_C$ similarly.

Prove that, $P_B, P_C$ and $Q'$ are collinear if and only if $Q_B,Q_C$ and $P'$ are collinear.