Proving concurrency at symmedian point.
- ahmedittihad
- Posts:181
- Joined:Mon Mar 28, 2016 6:21 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$.
Frankly, my dear, I don't give a damn.
Re: Proving concurrency at symmedian point.
Last edited by tanmoy on Thu Jan 05, 2017 3:00 pm, edited 2 times in total.
"Questions we can't answer are far better than answers we can't question"
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: Proving concurrency at symmedian point.
This problem is trivialized by barycentric coordinates.
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.
- ahmedittihad
- Posts:181
- Joined:Mon Mar 28, 2016 6:21 pm
Re: Proving concurrency at symmedian point.
I'd like if you shared the analytic solution here.
Frankly, my dear, I don't give a damn.
- Thanic Nur Samin
- Posts:176
- Joined:Sun Dec 01, 2013 11:02 am
Re: Proving concurrency at symmedian point.
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 $S\equiv (a^2:b^2:c^2)$.ahmedittihad wrote:I'd like if you shared the analytic solution here.
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.
Because destroying everything mindlessly isn't cool enough.