Difference between revisions of "2017 USAJMO Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
+ | <asy> | ||
+ | import olympiad; | ||
+ | |||
+ | unitsize(100); | ||
+ | |||
+ | pair pA = dir(120); | ||
+ | pair pB = dir(225); | ||
+ | pair pC = dir(315); | ||
+ | pair pO = origin; | ||
+ | pair pH = orthocenter(pA, pB, pC); | ||
+ | pair pM = midpoint(pB--pC); | ||
+ | pair dD = bisectorpoint(pB, pA, pC); | ||
+ | pair pD = extension(pA, dD, pB, pC); | ||
+ | pair pN = intersectionpoints(pM--(3*pO-2*pM), circumcircle(pB, pH, pC))[0]; | ||
+ | pair dHprime = foot(pH, pB, pC); | ||
+ | pair pHprime = 2*dHprime-pH; | ||
+ | pair dAprime = foot(pA, pB, pC); | ||
+ | pair pAprime = 2*dAprime-pA; | ||
+ | pair dNprime = foot(pN, pB, pC); | ||
+ | pair pNprime = 2*dNprime-pN; | ||
+ | |||
+ | draw(pA--pB--pC--cycle); | ||
+ | draw(unitcircle, blue); | ||
+ | draw(pH--pHprime, magenta); | ||
+ | draw(pA--pD, red); | ||
+ | draw(circumcircle(pB, pH, pC), blue); | ||
+ | draw(pA--pH, blue); | ||
+ | draw(pA--pN, magenta); | ||
+ | draw(pA--pO, blue); | ||
+ | draw(pD--pNprime, magenta); | ||
+ | |||
+ | dot(" | ||
+ | dot(" | ||
+ | dot(" | ||
+ | dot(" | ||
+ | dot(" | ||
+ | dot(" | ||
+ | dot(" | ||
+ | dot(" | ||
+ | dot(" | ||
+ | dot(" | ||
+ | dot(" | ||
+ | </asy> | ||
+ | (original diagram by [[User:integralarefun|integralarefun]]) | ||
+ | |||
It's well known that the reflection of <math>H</math> across <math>\overline{BC}</math>, <math>H'</math>, lies on <math>(ABC)</math>. Then <math>(BHC)</math> is just the reflection of <math>(BH'C)</math> across <math>\overline{BC}</math>, which is equivalent to the reflection of <math>(ABC)</math> across <math>\overline{BC}</math>. Reflect points <math>A</math> and <math>N</math> across <math>\overline{BC}</math> to points <math>A'</math> and <math>N'</math>, respectively. Then <math>N'</math> is the midpoint of minor arc <math>\overarc{BC}</math>, so <math>A, D, N'</math> are collinear in that order. It suffices to show that <math>\angle AA'N'=\angle ADO</math>. | It's well known that the reflection of <math>H</math> across <math>\overline{BC}</math>, <math>H'</math>, lies on <math>(ABC)</math>. Then <math>(BHC)</math> is just the reflection of <math>(BH'C)</math> across <math>\overline{BC}</math>, which is equivalent to the reflection of <math>(ABC)</math> across <math>\overline{BC}</math>. Reflect points <math>A</math> and <math>N</math> across <math>\overline{BC}</math> to points <math>A'</math> and <math>N'</math>, respectively. Then <math>N'</math> is the midpoint of minor arc <math>\overarc{BC}</math>, so <math>A, D, N'</math> are collinear in that order. It suffices to show that <math>\angle AA'N'=\angle ADO</math>. | ||
Latest revision as of 11:05, 7 June 2024
Contents
[hide]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 1
(original diagram by integralarefun)
It's well known that the reflection of across , , lies on . Then is just the reflection of across , which is equivalent to the reflection of across . Reflect points and across to points and , respectively. Then is the midpoint of minor arc , so are collinear in that order. It suffices to show that .
Claim: . The proof easily follows.
Proof: Note that . Then we have . So, it suffices to show that Notice that , so that Therefore, it suffices to show that But it is easy to show that , implying the result.
Solution 2
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 |