Difference between revisions of "2017 USAJMO Problems/Problem 5"
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Let <math>O</math> and <math>H</math> be the circumcenter and the orthocenter of an acute triangle <math>ABC</math>. Points <math>M</math> and <math>D</math> lie on side <math>BC</math> such that <math>BM = CM</math> and <math>\angle BAD = \angle CAD</math>. Ray <math>MO</math> intersects the circumcircle of triangle <math>BHC</math> in point <math>N</math>. Prove that <math>\angle ADO = \angle HAN</math>. | Let <math>O</math> and <math>H</math> be the circumcenter and the orthocenter of an acute triangle <math>ABC</math>. Points <math>M</math> and <math>D</math> lie on side <math>BC</math> such that <math>BM = CM</math> and <math>\angle BAD = \angle CAD</math>. Ray <math>MO</math> intersects the circumcircle of triangle <math>BHC</math> in point <math>N</math>. Prove that <math>\angle ADO = \angle HAN</math>. | ||
− | ==Solution== | + | ==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("\(A\)", pA, NW); | ||
+ | dot("\(B\)", pB, W); | ||
+ | dot("\(C\)", pC, E); | ||
+ | dot("\(O\)", pO, SW, blue); | ||
+ | dot("\(H\)", pH, SW, blue); | ||
+ | dot("\(N\)", pN, NE, magenta); | ||
+ | dot("\(M\)", pM, S, red); | ||
+ | dot("\(D\)", pD, S, red); | ||
+ | dot("\(H'\)", pHprime, SW, magenta); | ||
+ | dot("\(A'\)", pAprime, SW); | ||
+ | dot("\(N'\)", pNprime, S, magenta); | ||
+ | </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>. | ||
+ | |||
+ | <b>Claim:</b> <math>\triangle AA'N' \sim \triangle ADO</math>. The proof easily follows. | ||
+ | |||
+ | <b>Proof:</b> Note that <math>\angle BAA'=\angle CAO=90^{\circ}-\angle ABC</math>. Then we have <math>\angle A'AN'=\angle BAD-\angle BAA'=\angle CAD-\angle CAO=\angle DAO</math>. So, it suffices to show that <cmath>\frac{AA'}{AN'}=\frac{AD}{AO}\rightarrow AA'\cdot AO=AN'\cdot AD.</cmath> Notice that <math>\triangle ABA' \sim \triangle AOC</math>, so that <cmath>\frac{AB}{AA'}=\frac{AO}{AC}\rightarrow AA'\cdot AO=AB\cdot AC.</cmath> Therefore, it suffices to show that <cmath>AB\cdot AC=AN'\cdot AD\rightarrow \frac{AB}{AN'}=\frac{AD}{AC}.</cmath> But it is easy to show that <math>\triangle BAN'\sim \triangle DAC</math>, implying the result. <math>\blacksquare</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | <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; | ||
+ | |||
+ | 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("$A$", 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>. | ||
Line 11: | Line 92: | ||
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. | 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 | + | 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 ND=DP\implies</math> NDP is isosceles. Angle ADN is an exterior angle, so <math>\angle ADN=\angle DNP+\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>. | 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>. | 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>. |
Latest revision as of 11:05, 7 June 2024
Contents
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 |