Isogonals all over the place
by math_explorer, Apr 17, 2011, 9:38 AM
In acute triangle 
, the circumcenter and orthocenter are 
and 
respectively. The reflections of about 
are 
respectively. 
intersect at resp. 
. Prove: 
concur
(actually this holds for any pair of isogonal conjugates heh)
Anyway. Ceva + area comparison means we want to prove
![\[ \frac{[BHO_1][CHO_2][AHO_3]}{[CHO_1][AHO_2][BHO_3]} = 1 \]](//latex.artofproblemsolving.com/1/8/8/188c4d690033968829e375b48d3fcf75b4d2a7f8.png)
Angle chase to discover that
bisects 
, and since 
, is the perpendicular bisector of 
. (Or if you're feeling courageous, just cite "property of isogonals".) That is to say, ![$[AHO_3]/[AHO_2] = 1$](//latex.artofproblemsolving.com/8/d/a/8da679f6218c7d2daaf05e655d6e114f2a5d3fd5.png)
. The other terms are symmetric, so we're done.
must do harder problems
, the circumcenter and orthocenter are 
and 
respectively. The reflections of about 
are 
respectively. 
intersect at resp. 
. Prove: 
concur(actually this holds for any pair of isogonal conjugates heh)
Anyway. Ceva + area comparison means we want to prove
![\[ \frac{[BHO_1][CHO_2][AHO_3]}{[CHO_1][AHO_2][BHO_3]} = 1 \]](http://latex.artofproblemsolving.com/1/8/8/188c4d690033968829e375b48d3fcf75b4d2a7f8.png)
Angle chase to discover that
bisects 
, and since 
, is the perpendicular bisector of 
. (Or if you're feeling courageous, just cite "property of isogonals".) That is to say, ![$[AHO_3]/[AHO_2] = 1$](http://latex.artofproblemsolving.com/8/d/a/8da679f6218c7d2daaf05e655d6e114f2a5d3fd5.png)
. The other terms are symmetric, so we're done.must do harder problems
May 2020
January 2020