#8) ELMO Coordinate Bash
by TheDarkPrince, Feb 11, 2018, 6:03 PM
Problem: Let
be a triangle with orthocenter
and let
be the midpoint of
Suppose that
and
are distinct points on the circle with diameter
different from
such that
lies on line
Prove that the orthocenter of
lies on the circumcircle of 
Proposed by Michael Ren
Solution












Proposed by Michael Ren
Solution
Let
be the orthocenter of
. As
lie on the circle with diameter
,
. So,
and
. As
is the orthocenter of
,
and
.
. So,
is a parallelogram.
Now we apply coordinate bash.
Let
.
So
.
Coordinates of
: Slope of
. As
, slope of
.
Equation of
or
.
As
is parallel to the
,
has the same x-coordinate as that of
.
So,
.
Equation of circle with diameter
: Let
be the center of circle with diameter
. As
is the midpoint of
,
. Let
.
So, equation of circle with diameter
is:
.
Coordinates of
: As
is the origin, let the equation of line
be
. So, let
Simplification of coordinates of
As
lie on the circle with diameter
,

.
Subtracting these equations and using the identity
,

As
we get after simplifying,
Coordinates of
: As
is a parallelogram,
.
Substituting the value for
we get,
Equation of circumcircle of
: Let
be the center of circle
. As
is on the y-axis and
, the x-coordinate of
is
. Let the y-coordinate of
be
. Let
be the midpoint of
. So,
.
As
, the product of the slopes of the two lines is
.
So,
.
Solving for
we get
. So,
.
.
So equation of
is
Putting the coordinates of
in the equation of the circle:
We need to show:
We cancel the denominators.
Simplifying and taking similar terms together we get,
So,
lies on
.
Proved













Now we apply coordinate bash.
Let

So

Coordinates of




Equation of

or

As




So,

Equation of circle with diameter







So, equation of circle with diameter


Coordinates of










Subtracting these equations and using the identity


As





Substituting the value for














As


So,

Solving for




So equation of



We need to show:





Proved
This post has been edited 2 times. Last edited by pro_4_ever, Mar 1, 2018, 10:46 AM