Geometry Marathon : Season 3

For discussing Olympiad level Geometry Problems
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nahin munkar
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Re: Geometry Marathon : Season 3

Unread post by nahin munkar » Tue Jan 10, 2017 1:32 am

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 $\bigodot (P)$ & $\bigodot ALB$
proof:
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 $ $\blacksquare$
Last edited by nahin munkar on Tue Jan 10, 2017 2:49 pm, edited 1 time in total.
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Thanic Nur Samin
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Re: Geometry Marathon : Season 3

Unread post by Thanic Nur Samin » Tue Jan 10, 2017 11:46 am

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 $P$.
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joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Tue Jan 10, 2017 3:06 pm

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

joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Wed Jan 11, 2017 11:13 pm

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

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Raiyan Jamil
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Re: Geometry Marathon : Season 3

Unread post by Raiyan Jamil » Wed Jan 25, 2017 6:20 pm

$\text{Problem 16:}$
$H,I,O,N$ 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.
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joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Tue Jan 31, 2017 1:00 am

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

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ahmedittihad
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Re: Geometry Marathon : Season 3

Unread post by ahmedittihad » Tue Jan 31, 2017 4:12 pm

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 $OI$.
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rah4927
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Re: Geometry Marathon : Season 3

Unread post by rah4927 » Tue Jan 31, 2017 6:16 pm

Solution to Problem 17

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

$A=(1:0:0),A_1=(0:b:c),A_2=(0:s-c:s-b)$

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 $$vb+wc=\dfrac{a^2bc}{b+c}$$

$$v(s-c)+w(s-b)=a(s-c)(s-b)$$

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.

$$\begin{vmatrix}
x&a^2S_A& a \\
y&b^2S_B& b \\
z&c^2S_C& c
\end{vmatrix}=0$$

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.

$$ac(s-b)(c+a)y-bc(s-a)(b+c)x+ab(s-c)(a-b)z=0$$

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!!
Last edited by rah4927 on Wed Feb 01, 2017 12:57 am, edited 2 times in total.

joydip
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Re: Geometry Marathon : Season 3

Unread post by joydip » Tue Jan 31, 2017 11:48 pm

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

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Raiyan Jamil
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Re: Geometry Marathon : Season 3

Unread post by Raiyan Jamil » Wed Feb 01, 2017 12:27 am

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

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