Difference between revisions of "2017 USAJMO Problems/Problem 5"
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==Solution== | ==Solution== | ||
− | THIS SOLUTION HAS NO DIAGRAM, SOMEONE WHO HAS ASYMPTOTE SKILLS PLEASE HELP. | + | THIS SOLUTION HAS NO DIAGRAM, SOMEONE WHO HAS ASYMPTOTE SKILLS PLEASE HELP. <math>-1</math>!!! because of the diagram rule |
Suppose ray <math>OM</math> intersects the circumcircle of <math>BHC</math> at <math>N'</math>, and let the foot of the A-altitude of <math>ABC</math> be <math>E</math>. Note that <math>\angle BHE=90-\angle HBE=90-90+\angle C=\angle C</math>. Likewise, <math>\angle CHE=\angle B</math>. So, <math>\angle BHC=\angle BHE+\angle CHE=\angle B+\angle C</math>. | Suppose ray <math>OM</math> intersects the circumcircle of <math>BHC</math> at <math>N'</math>, and let the foot of the A-altitude of <math>ABC</math> be <math>E</math>. Note that <math>\angle BHE=90-\angle HBE=90-90+\angle C=\angle C</math>. Likewise, <math>\angle CHE=\angle B</math>. So, <math>\angle BHC=\angle BHE+\angle CHE=\angle B+\angle C</math>. |
Revision as of 17:23, 16 April 2018
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
THIS SOLUTION HAS NO DIAGRAM, SOMEONE WHO HAS ASYMPTOTE SKILLS PLEASE HELP. !!! because of the diagram rule
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 |