## Geometry Marathon : Season 3

- Kazi_Zareer
**Posts:**86**Joined:**Thu Aug 20, 2015 7:11 pm**Location:**Malibagh,Dhaka-1217

### Re: Geometry Marathon : Season 3

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.

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.

We cannot solve our problems with the same thinking we used when we create them.

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

### Re: Geometry Marathon : Season 3

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 $AXY$

\[\angle QXA=\angle QKP\]

$\angle BAP=\angle CAQ$ 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 $AXY$

\[\angle QXA=\angle QKP\]

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

### Re: Geometry Marathon : Season 3

**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 $\triangle KXB \sim \triangle KYX \Rightarrow \angle KNX =\angle KMB \Rightarrow \angle MKN =\angle AXY.$

So, $\angle QKP =\angle MKN -\angle MKQ=\angle AXY-\angle QAX=\angle AYX-\angle QYX =\angle AYQ =\angle QXA$.

**The first principle is that you must not fool yourself and you are the easiest person to fool.**

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

### Re: Geometry Marathon : Season 3

**$\text{Problem 20:}$**

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 : $HK\perp AD$

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

### Re: Geometry Marathon : Season 3

$\text{Solution of Problem } 20$

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

$$\begin{vmatrix}

b^2+c^2-a^2&0&(b^2+c^2-a^2)(b+c)c\\

a^2-b^2+c^2&c^2&-cb(a^2+c^2-b^2)\\

0&b^2&b^2(a^2+c^2-b^2)

\end{vmatrix}=0$$

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

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

$$\begin{vmatrix}

b^2+c^2-a^2&0&(b^2+c^2-a^2)(b+c)c\\

a^2-b^2+c^2&c^2&-cb(a^2+c^2-b^2)\\

0&b^2&b^2(a^2+c^2-b^2)

\end{vmatrix}=0$$

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

Last edited by rah4927 on Sat Feb 04, 2017 3:59 pm, edited 4 times in total.

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

### Re: Geometry Marathon : Season 3

That's the solution of problem 20 actually.rah4927 wrote:$\text{Solution of Problem } 21$

**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

**$\text{Problem 21:}$**

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 $(O)$).

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

### Re: Geometry Marathon : Season 3

**The first principle is that you must not fool yourself and you are the easiest person to fool.**

### Re: Geometry Marathon : Season 3

**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.

**The first principle is that you must not fool yourself and you are the easiest person to fool.**

### Re: Geometry Marathon : Season 3

**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 $B'C'$.

**The first principle is that you must not fool yourself and you are the easiest person to fool.**