Proving concurrency at symmedian point.

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ahmedittihad
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Proving concurrency at symmedian point.

Unread post by ahmedittihad » Wed Jan 04, 2017 9:09 pm

Let $A'$ be the midpoint of $BC$. $B'$ and $C$' are labelled similarly. Let $D$ be the feet of perpendicular of $A$ vertex on $BC$. $P$ be the midpoint of $AD$. label $Q,R$ similarly. Prove that, $ A'P, B'Q$ and $C'R $are concurrent and intersect at the symmedian point of $ ABC$.
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tanmoy
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Re: Proving concurrency at symmedian point.

Unread post by tanmoy » Wed Jan 04, 2017 10:12 pm

Let $L$ be the Lemoine Point (symmedian point) of $\triangle ABC$. We will prove that $A^{'},L,P$ collinear.(Similarly, we can prove that $B^{'},L,Q$ and $C^{'},L,R$ are collinear)

Let $\triangle XYZ$ be the tangential triangke of $\triangle ABC$ with $X$ lies opposite to $A$, $Y$ lies opposite to $B$ and $Z$ lies opposite to $C$.

Now let $AL \cap BC=T$, $AB \cap XY=S$, $A^{'}L \cap AD=P^{'}$.So, we have to prove that $P=P^{'}$.

Now it is well known that $AX,BY,CF$ concur at $L$. So,

$-1=(S,C;Y,X)=B(S,C;Y,X)$.
By intersecting these harmonic pencils with $AX$, we get that $-1=(A,T;L,X)=A^{'}(A,T;L,X)$.
By intersecting these harmonic pencils with $AD$, we get that $(A,D;P^{'},\infty)=-1$.
So, $P^{'}$ is the midpoint of $AD$ i.e. $P=P^{'}$.
Done
Last edited by tanmoy on Thu Jan 05, 2017 3:00 pm, edited 2 times in total.
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Thanic Nur Samin
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Re: Proving concurrency at symmedian point.

Unread post by Thanic Nur Samin » Wed Jan 04, 2017 10:13 pm

This problem is trivialized by barycentric coordinates.
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ahmedittihad
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Re: Proving concurrency at symmedian point.

Unread post by ahmedittihad » Sat Jan 07, 2017 5:40 pm

I'd like if you shared the analytic solution here.
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Thanic Nur Samin
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Re: Proving concurrency at symmedian point.

Unread post by Thanic Nur Samin » Tue Jan 10, 2017 2:40 pm

ahmedittihad wrote:I'd like if you shared the analytic solution here.
It's pretty straightforward to show that $P\equiv (2a^2:a^2+b^2-c^2:a^2+c^2-b^2)$, $A'\equiv (0:1:1)$ and we know that the symmedian point $S\equiv (a^2:b^2:c^2)$.

Now, we get that,

$$\begin{vmatrix}
2a^2 & a^2+b^2-c^2 & a^2+c^2-b^2 \\
a^2 & b^2 & c^2 \\
0 & 1 & 1
\end{vmatrix}$$

$$=\begin{vmatrix}
2a^2 & b^2-c^2 & c^2-b^2 \\
a^2 & b^2 & c^2 \\
0 & 1 & 1
\end{vmatrix}$$

$$=\begin{vmatrix}
a^2 & -c^2 & -b^2 \\
a^2 & b^2 & c^2 \\
0 & 1 & 1
\end{vmatrix}$$

$$=
\begin{vmatrix}
a^2 & -c^2 & -b^2 \\
0 & b^2+c^2 & b^2+c^2 \\
0 & 1 & 1
\end{vmatrix}=0$$
Hammer with tact.

Because destroying everything mindlessly isn't cool enough.

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