Return
by math_explorer, Jan 5, 2012, 9:13 AM
So I went and dug up last year olympiad practice problems. And got stuck on the very first geometry problem.
If it's any compensation, here's the second. Source discovered to be Bulgarian MO 2004.1. (I should stop being lazy and do this for all of my posts since AoPS pretty much has every problem in existence?) And yay I don't need to give a silly paraphrase.)
Well now I see my solution is pretty horrible.
First, construct the excentral triangle
of 
.
Next, through
draw the line parallel to 
(which is 
's external bisector and therefore perpendicular to 
.) Do the same thing cyclically for 
and 
, and let the triangle formed by the three drawn lines be 
.
Let
, 
, 
, 
be the respective circumcenters of 
, 
, 
, .
Now it's easily seen that, 
, 
have parallel corresponding sides and are homothetic. Since 
is the midpoint of 
and cyclic versions thereof, we can prove that is the midpoint of and . (Write out the homothecy factors and do some algebra. Annoying details or noting of some obvious shortcut I'm not seeing left to the reader.)
Meanwhile,
and are the orthocenter and nine-point center of respectively (since 
are the feet of the altitudes of ).
Thus, since the nine-point center is the midpoint of the orthocenter and the circumcenter,
iff 
iff 
bisect 
iff 
are the midpoints of iff 
is the medial triangle of iff 
and cyclic versions iff are altitudes of 
iff is the orthocenter of .
QED
Semi-relevantly: do you pronounce "cyclic" <sike-lik> or <sik-lik>?
If it's any compensation, here's the second. Source discovered to be Bulgarian MO 2004.1. (I should stop being lazy and do this for all of my posts since AoPS pretty much has every problem in existence?) And yay I don't need to give a silly paraphrase.)
Valentin Vornicu wrote:
Let 
be the incenter of triangle 
, and let 
, 
, 
be arbitrary points on the segments 
, 
, 
, respectively. The perpendicular bisectors of 
, 
, 
intersect each other at 
, 
, and 
. Prove that the circumcenter of the triangle 
coincides with the circumcenter of the triangle if and only if is the orthocenter of triangle 
.
be the incenter of triangle 
, and let 
, 
, 
be arbitrary points on the segments 
, 
, 
, respectively. The perpendicular bisectors of 
, 
, 
intersect each other at 
, 
, and 
. Prove that the circumcenter of the triangle 
coincides with the circumcenter of the triangle if and only if is the orthocenter of triangle 
.Well now I see my solution is pretty horrible.
First, construct the excentral triangle
of 
.Next, through
draw the line parallel to 
(which is 
's external bisector and therefore perpendicular to 
.) Do the same thing cyclically for 
and 
, and let the triangle formed by the three drawn lines be 
.Let
, 
, 
, 
be the respective circumcenters of 
, 
, 
, .Now it's easily seen that
, 
, 
have parallel corresponding sides and are homothetic. Since 
is the midpoint of 
and cyclic versions thereof, we can prove that is the midpoint of and . (Write out the homothecy factors and do some algebra. Annoying details or noting of some obvious shortcut I'm not seeing left to the reader.)Meanwhile,
and are the orthocenter and nine-point center of respectively (since 
are the feet of the altitudes of ).Thus, since the nine-point center is the midpoint of the orthocenter and the circumcenter,
iff 
iff 
bisect 
iff 
are the midpoints of iff 
is the medial triangle of iff 
and cyclic versions iff are altitudes of 
iff is the orthocenter of .QED
Semi-relevantly: do you pronounce "cyclic" <sike-lik> or <sik-lik>?
May 2020
January 2020