## Geometry Marathon : Season 3

For discussing Olympiad level Geometry Problems
joydip
Posts: 48
Joined: Tue May 17, 2016 11:52 am

### Re: Geometry Marathon : Season 3

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

tanmoy
Posts: 289
Joined: Fri Oct 18, 2013 11:56 pm

### Re: Geometry Marathon : Season 3

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.
"Questions we can't answer are far better than answers we can't question"

asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm

### Re: Geometry Marathon : Season 3

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 (97.9 KiB) Viewed 2384 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.

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

### Re: Geometry Marathon : Season 3

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 $DG\perp DC$.
Frankly, my dear, I don't give a damn.

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

### Re: Geometry Marathon : Season 3

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

nahin munkar
Posts: 81
Joined: Mon Aug 17, 2015 6:51 pm
Location: banasree,dhaka

### Re: Geometry Marathon : Season 3

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_CDMis 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$Last edited by nahin munkar on Mon Jan 09, 2017 9:52 pm, edited 5 times in total. # Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss nahin munkar Posts: 81 Joined: Mon Aug 17, 2015 6:51 pm Location: banasree,dhaka ### Re: Geometry Marathon : Season 3 Now, an easy.problem Problem 12: Let$\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}$. # Mathematicians stand on each other's shoulders. ~ Carl Friedrich Gauss Raiyan Jamil Posts: 138 Joined: Fri Mar 29, 2013 3:49 pm ### Re: Geometry Marathon : Season 3 If$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'$are collinear. A smile is the best way to get through a tough situation, even if it's a fake smile. Thanic Nur Samin Posts: 176 Joined: Sun Dec 01, 2013 11:02 am ### Re: Geometry Marathon : Season 3 Solution to problem 13 We will use Cartesian coordinates. For the sake of simplicity, take$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. Hammer with tact. Because destroying everything mindlessly isn't cool enough. Thanic Nur Samin Posts: 176 Joined: Sun Dec 01, 2013 11:02 am ### Re: Geometry Marathon : Season 3 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\$.
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.