Page 4 of 15

Re: Geometry Marathon : Season 3

Posted: Tue Jan 10, 2017 1:32 am
by nahin munkar
Thanic Nur Samin wrote: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 $AP\perp PC$.
Solution of problem 14 :

Let , the radical axis of these two mentioned circles is $PQ$. Here,$PQ \cap AB = T$. It is well-known that $AT=BT$.
& Let, $AC \cap PB = L$.

First, we draw a circle with radius $0$ centered at $P$. We denote this circle by $\bigodot (P)$ .& we draw another circle centred at $T$ with radius $AT$.
AS, $\angle ALB = 90^{\circ}$, $\triangle ALB$ lies on this circle.

$\bullet$ claim : $O_1O_2$ is the radical axis of $\bigodot (P)$ & $\bigodot ALB$
straightforward by the idea of Power of Point.
$TO_1^2-TA^2 = O_1A^2= O_1P^2$
& same, $O_2P^2=O_2B^2$
As, $O_1$ & $O_2$ have the same power wrt those circles, the desired claim follows. $$$$$$$$$$$$$$$$$\bullet$

$\bigstar$ As, $C$ lies on the radical axis $O_1O_2$ wrt $\bigodot (P)$ & $\bigodot ALB$, we get,

$ CP^2=CL \cdot CA \Longrightarrow \triangle APC \sim \triangle CPL \Longrightarrow \angle APC=90^{\circ} \Longrightarrow AP \perp PC $ $\blacksquare$

Re: Geometry Marathon : Season 3

Posted: Tue Jan 10, 2017 11:46 am
by Thanic Nur Samin
Problem 15

Let $ABC$ be a triangle with circumcircle $\omega$ and let $P$ be a variable point in its interior. The lines $PA, PB, PC$ meet $\omega$ again at $A_1, B_1, C_1$. Let $A_2$ denote the reflection of $A_1$ over $BC$, and define $B_2$ and $C_2$ similarly. Prove that the circumcircle of triangle $A_2B_2C_2$ passes through a fixed point independent of $P$.

Re: Geometry Marathon : Season 3

Posted: Tue Jan 10, 2017 3:06 pm
by joydip
Solution of problem 13 :

Let $AB \cap PA' = M $ , $AP \cap BC =N$ , $BC \cap PB' =K$ , $AA' \cap PB' = L$

$(M,A ;C',B)=(PA',PA;PC',PB)=(PA,PA';PC,PB')=A(N,A';C,K)=(P,L;B',K)$
$\Rightarrow (A'M,A'A ;A'C',A'B)=(A'P,A'L;A'B',A'k)$
So,$A',B',C'$ are colinear .

Re: Geometry Marathon : Season 3

Posted: Wed Jan 11, 2017 11:13 pm
by joydip
Solution to problem 15 :

Let $H$ be the orthocenter of $\triangle ABC$.

$\angle A_2BP= \angle A_2BC -\angle B_1BC=\angle A_1BC -\angle B_1AC=\angle A_1AC -\angle B_2AC=\angle B_2AP $
Now, $\dfrac {BA_2}{BP}=\dfrac {BA_1}{BP}=\dfrac {AB_1}{AP}=\dfrac {AB_2}{AP}$ . So, $\triangle PBA_2 \sim \triangle PAB_2 \Rightarrow \triangle PAB \sim \triangle PB_2A_2 .$
Similarly ,$\triangle PBC \sim \triangle PC_2B_2$ & $\triangle PCA \sim \triangle PA_2C_2$

Now ,$\angle BA_2C=\angle BA_1C=\angle BHC \Rightarrow BHA_2C$ is cyclic.Similarly $AHB_2C$ is cyclic.
So, $\angle A_2HB_2=\angle A_2HC + \angle CHB_2 =\angle A_2BC +\angle CAB_2=\angle A_1BC +\angle CAB_1 = \angle A_1AC +\angle CBB_1 $
$=\angle PC_2A_2 +\angle PC_2B_2 = \angle A_2C_2B_2$. So, $A_2B_2C_2H$ is cyclic ,completing the proof .

Re: Geometry Marathon : Season 3

Posted: Wed Jan 25, 2017 6:20 pm
by Raiyan Jamil
$\text{Problem 16:}$
$H,I,O,N$ are orthogonal center, incenter, circumcenter, and Nagelian point of triangle $ABC$. $I_{a},I_{b},I_{c}$ are excenters of $ABC$ corresponding vertices $A,B,C$. $S$ is point that $O$ is midpoint of $HS$. Prove that centroid of triangles $I_{a}I_{b}I_{c}$ and $SIN$ concide.

Re: Geometry Marathon : Season 3

Posted: Tue Jan 31, 2017 1:00 am
by joydip
Solution of problem 16 :

Let $G$ and $G_1$ be the centroids of $\triangle ABC$ & $\triangle I_{a}I_{b}I_{c}$ respectively .$M$ be the midpoint of $NS$ .$I, G , N$ are collinear and $\dfrac {IG}{GN}=\dfrac{HG}{GS}=\dfrac {HI}{SN}=\dfrac {1}{2}$(well known) $\Rightarrow HI \parallel NS$. As $O$ is the midpoint of $HS$ ,So $I,O,M$ are collinear . $I$ & $O$ are the orthocenter & ninepoint center of $\triangle I_{a}I_{b}I_{c}$ respectively .So, $I,O,G_1$ are collinear and $\dfrac {OG_1}{OI}=\dfrac {1}{3} \Rightarrow \dfrac{G_1M}{IG_1}= \dfrac {1}{2}$. So $G_1$ is the centroid of $\triangle SIN$.

Re: Geometry Marathon : Season 3

Posted: Tue Jan 31, 2017 4:12 pm
by ahmedittihad
Problem 17:
Let $ABC$ be a non-isosceles triangle, let $AA_1$ be its angle bisector and $A_2$ be the touching point of the incircle with side $BC$. The points $B_1$,$B_2$,$C_1$,$C_2$ are defined similarly. Let $O$ and $I$ be the circumcenter and the incenter of triangle . Prove that the radical center of the circumcircle of the triangles $AA_1A_2$,$ BB_1B_2$,$ CC_1C_2$ lies on the line $OI$.

Re: Geometry Marathon : Season 3

Posted: Tue Jan 31, 2017 6:16 pm
by rah4927
Solution to Problem 17

We will apply barycentric coordinates on $\triangle ABC$. The calculations are quite routine and requires one major insight.


Let the circumcircle of $\triangle AA_1A_2$ have the equation $$-a^2yz-b^2zx-c^2cy+(ux+vy+wz)(x+y+z)=0$$

This circle passes through $A$ making $u=0$. Plugging in the coordinates of $A_1$ and $A_2$ yields the system $$vb+wc=\dfrac{a^2bc}{b+c}$$


Solving this turns out to be surprisingly neat, contrary to what it looks like now. Alternatively, here's a trick you can use with symmetric linear equations that works sometimes in bary. The first equation suggests that both $v$ and $w$ have a $b+c$ in the denominator. The symmetric coefficients $b$ and $c$ suggest the numerators of $v$ and $w$ should have $c$ and $b$. Similarly, the symmetric coefficients in the second equation suggest the numerators also have $s-b$ and $s-c$ respectively. Putting this in the equations suggest that a factor of $a$ is missing from both, so we add that as well. The final solutions are $v=\dfrac{ac(s-b)}{b+c},w=\dfrac{ab(s-c)}{b+c}$ which indeed work.

Now finding the equation of $BB_1B_2$ simply becomes a matter of cyclically rotating $a,b,c$ in the above expressions. To find their radical axis, we simply subtract the two to get $$ac(s-b)(c+a)y-bc(s-a)(b+c)x+ab(s-c)(a-b)z=0$$

The other radical axes can be gained by cyclically rotating $a,b,c$ alongside $x,y,z$.

Let us stop here. Looks like the equations are getting out of hand. We need to show that these lines intersect on $OI$. It basically suffices to show that the equation of $OI$ is linearly dependent on the other three. Let's calculate $OI$ and hope we find something.

x&a^2S_A& a \\
y&b^2S_B& b \\
z&c^2S_C& c

which turns into $$xbc(b-c)(s-a)+yca(c-a)(s-b)+zab(a-b)(s-c)=0$$

Do the coefficients look familiar? They seem to match up with the one term from each equation of the radical axis. Here's the equation of the radical axis again.


We do the only thing we can to get that term and eliminate the rest. Add the three radical equations up. The $-bc(s-a)(b+c)$ of this equation and $+bc(s-a)(b+c)$ from another equation will cancel each other out, and so the only terms left will be the ones in our equation of $OI$. We are done!!

Re: Geometry Marathon : Season 3

Posted: Tue Jan 31, 2017 11:48 pm
by joydip
Another solution of problem 17:

Let $AA_1$ meet $(ABC)$ again $M$ and $MA_2$ meet $(ABC)$ again at $A_3$.Define $B_3,C_3$ similarly.Let $J=OI \cap A_2A_3$ . Let $\omega (M,N)$ denote the power of point $M$ w.r.t circle $N$.
An inversion with center $M$ and radius $MC$ takes $A_1 $ & $A_2$ to $A$ & $A_3$ respectively .So, $A_3\in (AA_1A_2)$.So $IA_2\parallel OM \Rightarrow \dfrac {JA_2}{JM}=\dfrac {JI}{JO}=\dfrac {r}{R}$.As ,$\dfrac {JI}{IO} =\dfrac {r}{R-r}$,so $J$ also lies in $B_2B_3 ,C_2C_3$.

$\omega (J,(AA_1A_2) )=\omega (J,(O)).\dfrac {r}{R}=\omega (J,(BB_1B_2) )=\omega (J,(CC_1C_2) )$
So $J$ is the radical center of $(AA_1A_2),(BB_1B_2),(CC_1C_2)$,completing the proof.

Re: Geometry Marathon : Season 3

Posted: Wed Feb 01, 2017 12:27 am
by Raiyan Jamil
rah4927 wrote:$A_2=(0:s-b:s-c)$
Aaaaa, sry but the coordinates of $A_2=(0:s-c:s-b)$ :mrgreen: .... ur solu is correct though...