Problem 18: Let $ABC$ be a triangle with circumcircle $\omega$ and let $H, M$ be orthocenter and midpoint of $AB$ respectively. Let $P,Q$ be points on the arc $AB$ of $\omega$ not containing $C$ such that $\angle ACP=\angle BCQ < \angle ACQ$.Let $R,S$ be the foot of altitudes from $H$ to $CQ,CP$ respectively. Prove that the points $P,Q,R,S$ are concyclic and $M$ is the center of this circle.

Re: Geometry Marathon : Season 3

Posted: Wed Feb 01, 2017 3:07 pm

by ahmedittihad

Solution of Problem 18:
$\angle BAP=\angle CAQ$ implies that $BP=CQ$. Yielding that $BPQC$ is an isosceles trapezoid with $BC \parallel PQ$. So, $H$ lies on the altitude from $A$ to $PQ$. Let $X$ be the orthocenter of $\triangle APQ$. Also, let $E$,$F$ be the projection of $X$ onto $AQ$ and $AP$. It's known that $EFPQ$ is cyclic. It's also quite trivial that $\triangle HRS$ and $\triangle XEF$ are homothetically similar with the center of homothety being $A$. So, $\angle APQ=\angle AEF=\angle ARS$ implies $PQRS$ cyclic.
Now, as $BPQC$ is an isosceles trapezoid, we get $MP=MQ$. Let $HM$ intersect arc $BC$ not containing $A$ at $K$. It's known that $K$ is the antipode of $A$. So, $\angle AQK=90^{\circ}=\angle ARH$. Yielding, $KQRH$ a trapezoid. As, $HM=MK$ and $\angle AQK=90^{\circ}=\angle ARH$, we conclude that $M$ lies on the perpendicular bisector of RQ, yielding $MQ=MR$. So, $M$ is the center of $\odot PQRS$.

Problem 19:
Circles $W_1,W_2$ intersect at $P,K$. $XY$ is common tangent of two circles which is nearer to $P$ and $X$ is on $W_1$ and $Y$ is on $W_2$. $XP$ intersects $W_2$ for the second time in $C$ and $YP$ intersects $W_1$ in $B$. Let $A$ be intersection point of $BX$ and $CY$. Prove that if $Q$ is the second intersection point of circumcircles of $ABC$ and $AXY$
\[\angle QXA=\angle QKP\]

Re: Geometry Marathon : Season 3

Posted: Thu Feb 02, 2017 5:04 pm

by joydip

Solution of problem 19:

Let $M,N,L$ be the midpoints of $BX,XY,YC$ respectively .
$\angle AXY=\angle XPB =\angle YPC=\angle AYX$
The spiral similarity centered at $K$ ,taking $BX$ to $YC$ takes $MX$ to $LC$ .So $K \in (AML)$.
The spiral similarity centered at $Q$ ,taking $BX$ to $CY$ takes $MX$ to $LY$ .So $Q \in (AML)$.
$N$ has equal power w.r.t $W_1,W_2$ . So , $N \in KP $.
$\angle KXY=\angle KBX $ , $ \angle KYX=\angle KCY=\angle KXB$.
So $\triangle KXB \sim \triangle KYX \Rightarrow \angle KNX =\angle KMB \Rightarrow \angle MKN =\angle AXY.$

$\text{Problem 20:}$
Let $ABC$ be an arbitrary triangle,$P$ is the intersection point of the altitude from $C$ and the tangent line from $A$ to the circumcircle. The bisector of angle $A$ intersects $BC$ at $D$ . $PD$ intersects $AB$ at $K$, if $H$ is the orthocenter then prove : $HK\perp AD$

Re: Geometry Marathon : Season 3

Posted: Fri Feb 03, 2017 4:10 pm

by rah4927

$\text{Solution of Problem } 20$

This problem is readily bary-able.

Line $CP$ has the equation $x\cdot (b^2+c^2-a^2)-y\cdot (a^2-b^2+c^2)=0$ and line $AP$ has the equation $b^2z+c^2y=0$. We therefore proceed to find $K$. Redefine $K$ to be the point such that $KH$ is perpendicular to $AD$. It suffices to show that $AP,CP,KD$ are concurrent.

Extend $KH$ to intersect $AC$ at $L$. Since $AD$ is the angle bisector of angle $A$ and $AD\perp KL$, we conclude that $AK=AL=\ell$. Therefore $K=(c-\ell:\ell:0)$ and $L=(b-\ell:0:\ell)$. Using the collinearity criterion on $K,H,L$, we find that $\ell=\dfrac{bS_{AB}+cS_{CA}}{S^2}$ where $S$ is the area of triangle $ABC$. Simplifying this, we get $$\ell=\dfrac{(b^2+c^2-a^2)(b+c)}{(a+b+c)(b+c-a)}$$

Doing a reality check here, we note that the value of $\ell$ is symmetric with respect to $b$ and $c$. The equation of $KD$ computes to $x\cdot\ell c -y\cdot (c-\ell)c+z\cdot (c-\ell)b=0$. Now is a good time to compute $c-\ell$ which turns out to be surprisingly nice.

$$c-\ell=\dfrac{b(a^2-b^2+c^2)}{(a+b+c)(b+c-a)}$$

Applying the criterion for concurrence on the equations of $AP,CP,KD$, it suffices to show that

which is a minute's worth of menial labour (expand using first row, since it has a zero). We are done.

Re: Geometry Marathon : Season 3

Posted: Sat Feb 04, 2017 12:58 am

by Raiyan Jamil

rah4927 wrote:$\text{Solution of Problem } 21$

That's the solution of problem 20 actually.

Re: Geometry Marathon : Season 3

Posted: Sat Feb 04, 2017 1:02 am

by Raiyan Jamil

$\text{Problem 21:}$
Consider a circle $(O)$ and two fixed points $B,C$ on $(O)$ such that $BC$ is not the diameter of $(O)$. $A$ is an arbitrary point on $(O)$, distinct from $B,C$. Let $D,K,J$ be the midpoints of $BC,CA,AB$, respectively, $E,M,N$ be the feet of perpendiculars from $A$ to $BC$, $B$ to $DJ$, $C$ to $DK$, respectively. The two tangents at $M,N$ to the circumcircle of triangle $EMN$ meet at $T$. Prove that $T$ is a fixed point (as $A$ moves on $(O)$).

Re: Geometry Marathon : Season 3

Posted: Mon Feb 06, 2017 10:01 pm

by joydip

Solution of problem 21:

Let the circle with diameter $BC$ intersect $CO,CN,BO,BM$ again at $J,L,K,P$ respectively.Let $H$ denote the orthocenter.$U,V$ be the midpoints of $CK ,BJ$ respectively.$T'$ be the circumcenter of $UDV$. $L \in AB,P \in AC$.
$\angle LBK=\angle PBC \Rightarrow LK=PC \Rightarrow \dfrac {LK}{2}= \dfrac {PC}{2} \Rightarrow MD=NU $,
$\angle PCJ=\angle LCB \Rightarrow PJ=LB \Rightarrow \dfrac {PJ}{2}= \dfrac {LB}{2} \Rightarrow MV=ND.$
So $T'UDN \simeq T'DVM \Rightarrow T'M=T'N $, $\angle NT'M= \angle UT'D=2\angle UVD=180^\circ-\angle UDV =180^\circ-\angle COB $(as $CO\parallel DV,BO \parallel UD)$.
So $N,M,E \in (HD) \Rightarrow \angle NEM=\angle NDM=\angle BAC$. So $\angle NTM =180^\circ-2\angle BAC=\angle NT'M$. So $T=T'$ ,completing the proof.

Re: Geometry Marathon : Season 3

Posted: Mon Feb 06, 2017 10:18 pm

by joydip

Problem 22:

Prove that the circumcenter of $\triangle ABC$ and the centroid of the anti pedal $\triangle A'B'C'$ of the symmedian point of $\triangle ABC$ coincide.

Re: Geometry Marathon : Season 3

Posted: Mon Feb 13, 2017 1:36 am

by joydip

Solution of problem 22:

Lemma : Let $ \omega_1$ & $ \omega_2$ be two circles with center $O_1 $ & $O_2$ respectively. $ \omega_1 \cap \omega_2 = A ,B$.Let the tangents to $ \omega_1$ at $A , B$ meet at $E$.Let $C,D \in \omega_2$ such that $ACBD$ is harmonic .$CE \cap \omega_2 = C,F $ and $FO_2 \cap \omega_2 = F,J$ .Then $J,D,O_1$ are collinear.

Proof : Let $O_1D$ meet $(AEBO_1)$ & $ \omega_2$ at a $L ,J'$ respectively.$EL \cap AB=K$.$EBO_1A$ is harmonic . So, $-1 =(LB ,LA;LO_1,LE)=(J'B,J'A ;J'D, J'K)=(J'B,J'A;J'D ,J'C)$. So $J',C,K$ are collinear.So $KL. KE=KA .KB=KC . KJ' \Rightarrow J'CLE$ is cyclic.$ \angle J'CE=\angle J'LE =90^ \circ$.So $F,O_2,J'$ are collinear.So $J'=J$.

Solution : Let tangents to $(ABC)$ at $B,C$ meet at $P$. $S$ be the symmedian point ,$O$ be the circumcenter of $(ABC)$.Let $SP$ meet $(A'BC)$ at $M$.$N$ be a point such that $MCNB$ is harmonic.By the lemma $A',N,O$ are collinear.$(A'B,A'C;A'N,A'M)=-1.A'M \parallel B'C' \Rightarrow AO $ goes though the midpoint of $B'C'$.