Difference between revisions of "2017 USAJMO Problems/Problem 5"
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A--D--N--O--cycle 0.1 yellow / red | A--D--N--O--cycle 0.1 yellow / red | ||
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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 17:49, 17 September 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
[asy] pair A = dir(130); pair B = dir(220); pair C = dir(320); draw(unitcircle, lightblue);
pair P = dir(-90); pair Q = dir(90); pair D = extension(A, P, B, C); pair O = origin; pair M = extension(B, C, O, P); pair N = 2*M-P;
draw(A--B--C--cycle, lightblue); draw(A--P--Q, lightblue); draw(A--N--D--O--A, lightblue);
draw(A--D--N--O--cycle, red);
dot("", A, dir(A)); dot("
", B, dir(B)); dot("
", C, dir(C)); dot("
", P, dir(P)); dot("
", Q, dir(Q)); dot("
", D, dir(225)); dot("
", O, dir(315)); dot("
", M, dir(315)); dot("
", N, dir(315));
/* TSQ Source:
A = dir 130 B = dir 220 C = dir 320 unitcircle 0.1 lightcyan / lightblue
P = dir -90 Q = dir 90 D = extension A P B C R225 O = origin R315 M = extension B C O P R315 N = 2*M-P R315
A--B--C--cycle lightblue A--P--Q lightblue A--N--D--O--A lightblue
A--D--N--O--cycle 0.1 yellow / red
[/asy]
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 |