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. | ||
− | {{ | + | 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>. |
+ | <math>BHCN'</math> is cyclic, so <math>\angle BN'C=180-\angle BHC=180-\angle B-\angle C=\angle A</math>. Also, <math>\angle BAC=\angle A</math>. These two angles are on different circles and have the same measure, but they point to the same arc! Hence, the two circles must be congruent. (This is also a well-known result) | ||
+ | |||
+ | We know, since <math>M</math> is the midpoint of <math>BC</math>, that <math>OM</math> is perpendicular to <math>BC</math>. <math>AH</math> is also perpendicular to <math>BC</math>, so the two lines are parallel. <math>AN</math> is a transversal, so <math>\angle HAN=\angle ANO</math>. We wish to prove that <math>\angle ANO=\angle ADO</math>, which is equivalent to <math>AOND</math> being cyclic. | ||
+ | |||
+ | Now, assume that ray <math>OM</math> intersects the circumcircle of <math>ABC</math> at a point <math>P</math>. Point <math>P</math> must be the midpoint of <math>\stackrel{\frown}{BC}</math>. Also, since <math>AD</math> is an angle bisector, it must also hit the circle at the point <math>P</math>. The two circles are congruent, which implies <math>MN=MP\implies OD=DP\implies</math> ODP is isosceles. Angle ADN is an exterior angle, so <math>\angle ADN=\angle DOP+\angle DPO=2\angle DPO</math>. | ||
+ | Assume WLOG that <math>\angle B>\angle C</math>. So, <math>\angle DPO=\angle APO=\frac{\angle B+\angle C}{2}-\angle C=\frac{\angle B-\angle C}{2}</math>. | ||
+ | In addition, <math>\angle AON=\angle AOP=\angle AOB+\angle BOP=2\angle C+\angle A</math>. Combining these two equations, <math>\angle AON+\angle ADN=\angle B-\angle C+2\angle C+\angle A=\angle A+\angle B+\angle C=180</math>. | ||
+ | |||
+ | Opposite angles sum to <math>180</math>, so quadrilateral <math>AOND</math> is cyclic, and the condition is proved. | ||
+ | |||
+ | -william122 | ||
==See also== | ==See also== | ||
{{USAJMO newbox|year=2017|num-b=4|num-a=6}} | {{USAJMO newbox|year=2017|num-b=4|num-a=6}} |
Revision as of 08:56, 22 April 2017
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.
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 arc! 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 ODP 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 |