Hammering with barycentric coordinates
by Keith50, Feb 7, 2022, 7:21 AM
Malaysia IMO Training 1 Weekly Test 4 P1 wrote:
Let 
be a triangle with orthocenter 
and let 
be the midpoint of 
. Pick two points 
on 
respectively such that 
are collinear and 
Prove that 
be a triangle with orthocenter 
and let 
be the midpoint of 
. Pick two points 
on 
respectively such that 
are collinear and 
Prove that 
then ![\[\begin{vmatrix}
x_Y & 0 & 1-x_Y \\
\frac{1}{S_A} & \frac{1}{S_B} & \frac{1}{S_C} \\
x_X & 1-x_X & 0
\end{vmatrix} =0\]](http://latex.artofproblemsolving.com/1/d/c/1dc7f30a91fa797d36bc5c4c7f2764e540cb1b39.png)
![\[x_Y\left(-\frac{1}{S_C}(1-x_X)\right)+(1-x_Y)\left(\frac{1}{S_A}(1-x_X)-\frac{1}{S_B}x_X\right)=0\]](http://latex.artofproblemsolving.com/2/8/4/284cce4e069cec8add3b036ed7061094b8c3ff56.png)
![\[\frac{x_Y}{1-x_Y}=\frac{S_C}{S_A}-\frac{S_C}{S_B}\left(\frac{x_X}{1-x_X}\right).\]](http://latex.artofproblemsolving.com/1/c/6/1c6a2750b1b126caa7bc5d74da35f42a2494c7e4.png)
Shift 
then the displacement vector 
which can be scaled as 
and 
.
and strong EFFT, we have ![\[a^2(x_X-x_Y)+b^2(2-x_X-x_Y)+c^2(x_X+x_Y-2)=0\]](http://latex.artofproblemsolving.com/1/f/0/1f0039cd7b4fb8a3d1ccdc0632bde8c2aebf7f90.png)
![\[(a^2-b^2+c^2)x_X+2(b^2-c^2)=(a^2+b^2-c^2)x_Y\]](http://latex.artofproblemsolving.com/5/d/2/5d20427cc55ab98ad77186c5e205d7052d405233.png)
![\[x_X=\frac{c^2-b^2}{S_B}+\frac{S_C}{S_B}x_Y=\frac{S_B-S_C}{S_B}+\frac{S_C}{S_B}x_Y=1-\frac{S_C}{S_B}(1-x_Y).\]](http://latex.artofproblemsolving.com/5/5/a/55a4ec4ce40728639d306104363ebbe1f3ca1932.png)
Therefore, substituting this back to the previous equation, ![\[\frac{x_Y}{1-x_Y}=\frac{S_C}{S_A}-\frac{1-\frac{S_C}{S_B}(1-x_Y)}{1-x_Y}\]](http://latex.artofproblemsolving.com/2/9/b/29b2250ea89916bac4eafccfdfa4d8c7f7179c68.png)
![\[x_Y=\frac{S_C}{S_A}(1-x_Y)-1+\frac{S_C}{S_B}(1-x_Y)\]](http://latex.artofproblemsolving.com/0/d/1/0d16b0df637c9a1bda24c5ecc01bfee7cd7bd0d7.png)
![\[x_Y=1-\frac{2}{\frac{S_C}{S_A}+\frac{S_C}{S_B}+1}=1-\frac{2S_{AB}}{S^2}\]](http://latex.artofproblemsolving.com/7/7/4/774f62355092f78d068ecd5edeb73435eed74d46.png)
so ![\[x_X=1-\frac{S_C}{S_B}\cdot \frac{2S_{AB}}{S^2}=1-\frac{2S_{AC}}{S^2}.\]](http://latex.artofproblemsolving.com/2/2/0/2209259697a1160f03fbcc0805cb80fcefe9d141.png)
To prove that 
we just have to show that the areas ![$[YHM]$](http://latex.artofproblemsolving.com/1/1/b/11b3185020768957c876929118c74cd99a041f0d.png)
and ![$[HXM]$](http://latex.artofproblemsolving.com/d/4/b/d4b50f01ee5534b4c036bb28ca31069084f1168e.png)
are equal, so
which is obviously true, so 
(I actually attempted to use complex bashing during the test, but ended up screwing everything as the algebra are too complicated and not realising this bary bashing solution.
)This post has been edited 1 time. Last edited by Keith50, Feb 7, 2022, 7:26 AM
February 2022
May 2021