2017 USAJMO Problems/Problem 5
Contents
Problem
Let and
be the circumcenter and the orthocenter of an acute triangle
. Points
and
lie on side
such that
and
. Ray
intersects the circumcircle of triangle
in point
. Prove that
.
Solution 1
It's well known that the reflection of across
,
, lies on
. Then
is just the reflection of
across
, which is equivalent to the reflection of
across
. Reflect points
and
across
to points
and
, respectively. Then
is the midpoint of minor arc
, so
are collinear in that order. It suffices to show that
.
Claim:
. The proof easily follows.
Proof: Note that
. Then we have
. So, it suffices to show that
Notice that
, so that
Therefore, it suffices to show that
But it is easy to show that
, implying the result.
Solution 2
Suppose ray intersects the circumcircle of
at
, and let the foot of the A-altitude of
be
. Note that
. Likewise,
. So,
.
is cyclic, so
. Also,
. These two angles are on different circles and have the same measure, but they point to the same line
! Hence, the two circles must be congruent. (This is also a well-known result)
We know, since is the midpoint of
, that
is perpendicular to
.
is also perpendicular to
, so the two lines are parallel.
is a transversal, so
. We wish to prove that
, which is equivalent to
being cyclic.
Now, assume that ray intersects the circumcircle of
at a point
. Point
must be the midpoint of
. Also, since
is an angle bisector, it must also hit the circle at the point
. The two circles are congruent, which implies
NDP is isosceles. Angle ADN is an exterior angle, so
.
Assume WLOG that
. So,
.
In addition,
. Combining these two equations,
.
Opposite angles sum to , so quadrilateral
is cyclic, and the condition is proved.
-william122
See also
2017 USAJMO (Problems • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAJMO Problems and Solutions |