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noway

#1 h Dec 5, 2019, 4:32 AM • 2 Y

Y by Adventure10, Mango247

$\triangle ABC$ inscribe $(O)$ . $M,N,P$ is the midpoint of $BC,CA,AB$ . $S_a$ is the intersection of the tangent of $(O)$ at $B$ and $C$ . The tangent of $(O)$ at $A$ intersects $NP$ at $R_a$ . Simillarity, we have $S_b,S_c,R_b,R_c$ .
Prove that $(OS_aR_a), (OS_bR_b), (OS_cR_c)$ have second common point (not $O$ ).

https://scontent.fdad1-1.fna.fbcdn.net/v/t1.15752-9/78417712_450594615642658_6055837001834299392_n.png?_nc_cat=100&_nc_ohc=cNSRtfWdFN4AQkduZDt7O1IXrF7agAP1jNlhtrM7dkj6hraY2lwUpOteg&_nc_ht=scontent.fdad1-1.fna&oh=711383ec4e0f0c76299ab1c96572e1f5&oe=5E42B99C

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LKira

#2 h Dec 5, 2019, 8:52 AM • 1 Y

Y by Adventure10

My geogebra say no, it not true, may you check again your problem ?

This post has been edited 1 time. Last edited by LKira, Dec 5, 2019, 8:52 AM

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noway

#3 h Dec 5, 2019, 1:29 PM • 1 Y

Y by Adventure10

LKira wrote:

My geogebra say no, it not true, may you check again your problem ?

I have checked carefully, it's true ^^

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LKira

#4 h Dec 5, 2019, 2:26 PM • 1 Y

Y by Adventure10

noway wrote:

LKira wrote:

My geogebra say no, it not true, may you check again your problem ?

I have checked carefully, it's true ^^

Then see VMO 2018 P7

This post has been edited 1 time. Last edited by LKira, Dec 5, 2019, 2:26 PM

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#5 h Dec 6, 2019, 10:49 AM • 2 Y

Y by Adventure10, Mango247

Your second common point of the three circles has the first baricentric coordinate

a^2 (2 a^16-6 a^14 (b^2+c^2)+a^12 (-4 b^4+38 b^2 c^2-4 c^4)+2 a^10 (9 b^6-20 b^4 c^2-20 b^2 c^4+9 c^6)+a^8 (-67 b^6 c^2+174 b^4 c^4-67 b^2 c^6)+4 a^4 (b^2-c^2)^2 (b^8+9 b^6 c^2-32 b^4 c^4+9 b^2 c^6+c^8)-(b^4-c^4)^2 (2 b^8-9 b^6 c^2+10 b^4 c^4-9 b^2 c^6+2 c^8)+2 a^2 (b^2-c^2)^2 (3 b^10-18 b^8 c^2+11 b^6 c^4+11 b^4 c^6-18 b^2 c^8+3 c^10)-2 a^6 (9 b^10-51 b^8 c^2+46 b^6 c^4+46 b^4 c^6-51 b^2 c^8+9 c^10))

This triangle center is not currently in https://faculty.evansville.edu/ck6/encyclopedia/ETC.html

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#6 h Dec 15, 2019, 2:36 AM • 2 Y

Y by Adventure10, Mango247

Is this correct?

We begin with the main claim
$\textbf{Claim}$ : Given a $\triangle ABC$ and point $P$ on its euler line. Let line $AP$ , $BP$ , $CP$ intersect the $\odot ABC$ again at $A'$ , $B'$ , and $C'$ respectively. Define $D,E,F$ to be the midpoint of $BC$ , $AC$ and $AB$ respectively and define also $X$ , $Y$ , $Z$ to be the midpont of $AA'$ , $BB'$ and $CC'$ respectively. Then we have that $DX$ , $EY$ and $FZ$ are concurrent.

$\textbf{Proof}$ :
Animate point $P$ on the euler line, therefore we will get that the degree of point $A',B',C'$ are all 2. Hence the degree of $X,Y,Z$ are also 2. Since point $D,E,F$ are all fixed we must have that the degree of line $DX,EY,FZ$ are all 2. So if we want to prove that $DX,EY,FZ$ are concurrent we only need to check that for $2+2+2+1=7$ possibilities of point $P$ , $DX,EY,FZ$ are always concurrent. It's easy to check that for $P$ is the circumcenter, centroid the two intersections of euler line with the circumcircle of $ABC$ , intersection of euler line with $BC$ , $CA$ , $AB$ , the three lines $DX,EY,FZ$ always concurrent. Therefore, $\textbf{QED}$

We go back to the original problem
Define $X_{a},X_{b},X_{c}$ be the point on $(O)$ different from $A,B,C$ such that $X_{a}R_{a}$ , $X_{b}R_{b}$ and $X_{c}R_{c}$ are all tangent to $(O)$ . Now define $Y_{a},Y_{b},Y_{c}$ to be the midpoint of $AX_{a}$ , $BX_{b}$ and $CY_{c}$ respectively.
Invert the figure WRT to $(O)$ , then the original problem is equivalent to proving that $MY_{a}$ , $NY_{b}$ and $PY_{c}$ are concur at one point.
However, it's a classic result from IMO 2011 G4 that $AX_{a},BX_{b}$ and $CX_{c}$ are concurrent at point $W$ , the isogonal conjugate of isotomic conjugate of orthocenter $H$ of $\triangle ABC$ . Moreover it's well known that this point $W$ is on the euler line. Therefore by the first claim, the problem is done.

This post has been edited 5 times. Last edited by Sugiyem, Dec 23, 2019, 10:11 PM

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