Difference between revisions of "2017 USAJMO Problems/Problem 5"
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==Solution== | ==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("<math>A</math>", A, dir(A)); dot("<math>B</math>", B, dir(B)); dot("<math>C</math>", C, dir(C)); dot("<math>P</math>", P, dir(P)); dot("<math>Q</math>", Q, dir(Q)); dot("<math>D</math>", D, dir(225)); dot("<math>O</math>", O, dir(315)); dot("<math>M</math>", M, dir(315)); dot("<math>N</math>", 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 <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 15:59, 31 August 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 |