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Re: Geometry Marathon : Season 3

Posted: Sat Jan 07, 2017 10:05 pm
by joydip
Solution of problem 9:

Assume $Q',P_B,P_C$ are collinear.
Let a circle $ \omega $, going though $B,C$ intersect $AB,AC$ at $\bar C ,\bar B$ respectively such that $A$ lies in the segment $B\bar C$ . Let $\Gamma $ be the transformation taking $X$ to $\bar X$ such that $X \cup ABC \sim \bar X \cup A\bar B\bar C$ . Let

$BQ' \cap \bar B\bar Q =G$ , $BP \cap \bar B\bar P' =H$ , $CQ' \cap \bar C\bar Q =E$ , $CP \cap \bar C\bar P' =F$
$BQ' \cap AC =L$ , $BA \cap \bar B\bar P' =J$ , $CQ' \cap AB =K$ , $AC \cap \bar C\bar P' =I$

Observe that $\angle GB\bar C=\angle G\bar B\bar C$ .So $ G \in \omega $,Similarly $(G,H,E,F) \in \omega $

$\bar C(\bar B,\bar Q_C; I,A) =C(\bar B,E;F,B)=Q'(A,K;P_C,B)=B(A,C;P_B,L)=\bar B(\bar C,C;H,G)=(\bar C,A;J,\bar Q_B)$ .So $ \bar P'(\bar B,\bar Q_C; I,A)=(\bar C,A;J,\bar Q_B)= \bar P' (J,\bar Q_B;\bar C,A)$
So,$\bar Q_C,\bar P' ,\bar Q_B$ are collinear $\Rightarrow Q_C, P' , Q_B$ are collinear.

Re: Geometry Marathon : Season 3

Posted: Sat Jan 07, 2017 10:56 pm
by tanmoy
Problem 10:
Let $I$ be the incenter of $\triangle ABC$. The incircle touches $BC$ at $D$ and $K$ is the antipode of $D$ in $(I)$.
Let $M$ be the midpoint of $AI$. Prove that $KM$ passes through the Feuerbach Point.

Re: Geometry Marathon : Season 3

Posted: Sun Jan 08, 2017 4:10 am
by asif e elahi
tanmoy wrote:Problem 10:
Let $I$ be the incenter of $\triangle ABC$. The incircle touches $BC$ at $D$ and $K$ is the antipode of $D$ in $(I)$.
Let $M$ be the midpoint of $AI$. Prove that $KM$ passes through the Feuerbach Point.
We define some new points

1. $L$ is the midpoint of $BC$
2. $N$ is the point where the $A$-excircle touches $BC$
3. $S$ is the second meeting point of the incircle and $AK$.
4. $T$ is the reflection of $K$ on $M$.
5. $P$ is the reflection of $D$ on $AI$.
6. $F$ is the point where $\vec{LP}$ meets the incircle.
7. $G=LI\cap KP$
Lemma 1: $A.K.N$ are collinear.
Proof Well-known.
Lemma 2: $F$ is the *Feurbatch point*.
Proof Well-known.
asd.png
asd.png (97.9 KiB) Viewed 2380 times
Main Proof: AS $N$ lies on $AK$, we get $A,K,S,N$ are collinear. Now $LD=LN$ and $\angle DSN=180^{\circ}-\angle DSK=90^{\circ}$. So $LS=LD$ which implies $LS$ is tangent to the incircle. Also $LI\parallel KS$ as $LI$ is the midline of $\triangle GKN$. As $DP\perp AI$ and $DP\perp KP$, we have $KP\parallel AI$ or $KG\parallel AI$. Also $AK$ and $IG$ are parallel as $LI\parallel KS$. So $AKGI$ is a parallelogram. Now $KM=TM$ and $AM=MI$, so $AKIT$ is a parallelogram. So $TI=AK=IG$. Hence $I$ is the midpoint of $TG$. Now as $TG\parallel KS$, the four lines $KT,KI,KG,KS$ form a harmonic pencil and they intersect the incircle at $F',D,P$ and $S$ respectively. So $F'DPS$ is a harmonic quadriletarel. Also $FDPS$ is a harmonic quad. So we must have $F\equiv F'$. So $KM$ passes through the Feurbatch point.

Re: Geometry Marathon : Season 3

Posted: Sun Jan 08, 2017 3:21 pm
by ahmedittihad
Problem 11:
Let $ABC$ be a triangle inscribed circle $(O)$, orthocenter $H$. $E,F$ lie on $(O)$ such that $EF\parallel BC$. $D$ is midpoint of $HE$. The line passing though $O$ and parallel to $AF$ cuts $AB$ at $G$. Prove that $DG\perp DC$.

Re: Geometry Marathon : Season 3

Posted: Sun Jan 08, 2017 4:16 pm
by Raiyan Jamil
Thanic Nur Samin wrote: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.
$\text{Another solution to Problem 9}$

We use barycentric coordinates. Taking $ABC$ as the reference triangle, we define,

$P=(x:y:z)\Rightarrow P'=(a^2/x:b^2/y:c^2/z),P_B=(x:0:z),P_C=(x:y:0),$
$Q=(p:q:r)\Rightarrow Q'=(a^2/p:b^2/q:c^2/r),Q_B=(p:0:r),Q_C=(p:q:0)$.

Now, if $P_B,P_C$ and $Q'$ are collinear, then,

$\begin{vmatrix}
x&x&a^2/p\\
0&y&b^2/q\\
z&0&c^2/r
\end{vmatrix}=0$
$\Rightarrow xypqc^2+xzprb^2+yzqra^2=0....(i)$

Again, if $Q_B,Q_C$ and $P'$ are collinear, then,

$\begin{vmatrix}
p&p&a^2/x\\
0&q&b^2/y\\
r&0&c^2/z
\end{vmatrix}=0$
$\Rightarrow xypqc^2+xzprb^2+yzqra^2=0....(ii)$

Since $(i)=(ii)$, we get $P_B,P_C$ and $Q'$ are collinear $\Longleftrightarrow$ $Q_B,Q_C$ and $P'$ are collinear. $[Proved]$

Re: Geometry Marathon : Season 3

Posted: Mon Jan 09, 2017 12:15 am
by nahin munkar
ahmedittihad wrote:Problem 11:
Let $ABC$ be a triangle inscribed circle $(O)$, orthocenter $H$. $E,F$ lie on $(O)$ such that $EF\parallel BC$. $D$ is midpoint of $HE$. The line passing though $O$ and parallel to $AF$ cuts $AB$ at $G$. Prove that $DG\perp DC$.
Solution of problem 11 :

We first denote some extra points. $H_A,H_B,H_C$ be three projections from $A,B,C$ on $BC,CA,AB$ resp. $K$ be midpoint of $AH$ , & $OG \cap AC = T$

Claim 1: $D$ lies on nine-point circle of $\triangle ABC$
Proof:
$D , K$ are midpoints of $EH,AH$ resp. Easily can be seen, By $\frac{1}{2}$ scale factor homothety of center $'O'$, $C$ sends to $D$ that lies on $\bigodot KH_AH_BH_C$, called nine point circle of $\triangle ABC$.

Claim 2:$\angle BAF = \angle CAE$
proof:
$BC\parallel EF \Rightarrow \widehat{BE}=\widehat{CF} \Longrightarrow \angle BAE =\angle CAF$.
Now, by adding $\angle BAC$ to bothsides, we get desired claim.


$\bigstar$ We extend $KD$ both sides that intersects $AB,AC$ at $L$ & $M$ resp.As, $D , K$ are midpoints of EH,AH resp. so, $DK||AE \Rightarrow LM||AE$.

Claim 3 : $LMTG$ cyclic
Proof :
\begin{align*}
\angle LGT&=\angle AGO$\\
&=180^{\circ}- \angle GAF \, ;[\text{as }OG||AF] \\
&= 180^{\circ}- \angle BAF\\
&=180^{\circ}- \angle CAE \, ;[\text{by claim }2]\\
&=180^{\circ}-\angle MAE\\
&= \angle AML\\
\Longrightarrow LMTG & \text{is cyclic.}
\end{align*}

claim 4:$CH_CDM$ is cyclic
Proof:
[N.B. Here, we use directed angles.]

\begin{align*}
\measuredangle H_CDM&=\measuredangle H_CDK\\
&=\measuredangle H_CH_AK\\
&=\measuredangle H_CCA\\
&=\measuredangle H_CCM\, ;[\text{as }H_ACAH_C]\\
\Longrightarrow CH_CDM & \text {is cyclic.}
\end{align*}

Claim 5 : $CH_CGM$ is cyclic
Proof:
Use claim 3 & Angle chasing. Left for the reader.


By combining Claim 4 & claim 5, we get $CH_CGD$ is cyclic
Thus, $\angle CH_CG=90^{\circ} \Longrightarrow \angle CDG =90^{\circ} \Longrightarrow DG \perp DC $ $\blacksquare$

Re: Geometry Marathon : Season 3

Posted: Mon Jan 09, 2017 2:59 pm
by nahin munkar
Now, an easy.problem :D

Problem 12:
Let $\triangle ABC$ be scalene, with $BC$ as the largest side. Let $D$ be the foot of the perpendicular from $A$ on side $BC$. Let points $K,L$ be chosen on the lines $AB$ and $AC$ respectively, such that $D$ is the midpoint of segment $KL$. Prove that the points $B,K,C,L$ are concyclic if and only if $\angle BAC=90^{\circ}$.

Re: Geometry Marathon : Season 3

Posted: Mon Jan 09, 2017 5:49 pm
by Raiyan Jamil
If $BKCL$ is cyclic, then $\bigtriangleup ABC$ is oppositely similar to $\bigtriangleup AKL$. Then, generalizing it becomes that, $\bigtriangleup ABC$ is a scalene triangle where the $A$-symmedian is perpendicular to $BC$. And we have to prove that $\angle BAC=90^\circ$.

Now,let us assume $\angle BAC \neq 90^\circ$meaning $B$ and $C$ are not diametrically opposite i.e. $B$,$C$ tangents do intersect; let at $T$. Then $AT \perp BC$. Again if $M$ is the midpoint of $BC$,then $TM \perp BC \Rightarrow AM \perp BC \Rightarrow \bigtriangleup ABC$ is $isosceles$; which contradicts the fact that $\bigtriangleup ABC$ is $scalene$. So, $B,C$ must be diametrically opposite i.e. $\angle BAC=90^\circ$ as desired.

$\text{Problem 13:}$

Let $P$ be a point in the plane of $\bigtriangleup ABC$, and $L$ a line passing through $P$ . Let $A',B',C'$ be the points where the reflections of lines $PA,PB,PC$ with respect to $L$ intersect lines $BC,AC,AB$ respectively. Prove that $A',B',C'$ are collinear.

Re: Geometry Marathon : Season 3

Posted: Mon Jan 09, 2017 8:02 pm
by Thanic Nur Samin
Solution to problem 13

We will use Cartesian coordinates. For the sake of simplicity, take $P$ as the origin and $L$ as $x$-axis. Also, let $A\equiv (x_1,y_1), B\equiv (x_2,y_2)$ and $C\equiv (x_3,y_3)$ Then $A_1$ would be $(x_1,-y_1)$. Define $B_1$ and $C_1$ similarly. Now, See that,

\[\displaystyle{\dfrac{\vec{BA'}}{\vec{A'C}}=\dfrac{[PBA_1]}{[PA_1X]}}=-\dfrac{x_1y_2+x_2y_1}{x_3y_1+x_1y_3}\]

Where the last equality follows from shoelace formula. Similarly, we can develop similar equalities and due to Menelaus' Theorem, we get that $A_1, B_1$ and $C_1$ are collinear.

Re: Geometry Marathon : Season 3

Posted: Mon Jan 09, 2017 8:41 pm
by Thanic Nur Samin
Problem 14

Let one of the intersection points of two circles with centres $O_1,O_2$ be $P$. A common tangent touches the circles at $A,B$ respectively. Let the perpendicular from $A$ to the line $BP$ meet $O_1O_2$ at $C$. Prove that $AP\perp PC$.