Difference between revisions of "2017 USAJMO Problems/Problem 5"
Mathwizard07 (talk | contribs) m (→Solution) |
Thingarfield (talk | contribs) (→Solution) |
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==Solution== | ==Solution== | ||
− | + | <asy> | |
+ | size(9cm); | ||
+ | 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; | + | 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--B--C--cycle, lightblue); |
+ | draw(A--P--Q, lightblue); | ||
+ | draw(A--N--D--O--A, lightblue); | ||
draw(A--D--N--O--cycle, red); | draw(A--D--N--O--cycle, red); | ||
− | dot(" | + | dot("$A$", A, dir(A)); |
− | + | dot("$B$", B, dir(B)); | |
− | + | dot("$C$", C, dir(C)); | |
− | + | dot("$P$", P, dir(P)); | |
− | + | dot("$Q$", Q, dir(Q)); | |
− | + | dot("$D$", D, dir(225)); | |
− | + | dot("$O$", O, dir(315)); | |
− | + | dot("$M$", M, dir(315)); | |
− | + | dot("$N$", N, dir(315)); | |
− | + | </asy> | |
− | |||
− | |||
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 21:37, 9 March 2019
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
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 |