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Difference between revisions of "2017 USAJMO Problems/Problem 5"

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==Solution==
 
==Solution==
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[asy] pair A = dir(130); pair B = dir(220); pair C = dir(320); draw(unitcircle, lightblue);
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<asy>
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size(9cm);
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pair A = dir(130);  
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pair B = dir(220);  
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pair C = dir(320);  
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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;
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pair P = dir(-90);  
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pair Q = dir(90);
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pair D = extension(A, P, B, C);  
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pair O = origin;  
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pair M = extension(B, C, O, P);  
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pair N = 2*M-P;
  
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draw(A--B--C--cycle, lightblue); draw(A--P--Q, lightblue); draw(A--N--D--O--A, lightblue);
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draw(A--B--C--cycle, lightblue);  
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draw(A--P--Q, lightblue);  
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draw(A--N--D--O--A, lightblue);
  
 
draw(A--D--N--O--cycle, red);
 
draw(A--D--N--O--cycle, red);
  
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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));
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dot("$A$", A, dir(A));  
−
 
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dot("$B$", B, dir(B));  
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/* TSQ Source:
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dot("$C$", C, dir(C));  
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dot("$P$", P, dir(P));  
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A = dir 130 B = dir 220 C = dir 320 unitcircle 0.1 lightcyan / lightblue
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dot("$Q$", Q, dir(Q));  
−
 
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dot("$D$", D, dir(225));  
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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
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dot("$O$", O, dir(315));  
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dot("$M$", M, dir(315));  
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A--B--C--cycle lightblue A--P--Q lightblue A--N--D--O--A lightblue
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dot("$N$", N, dir(315));
−
 
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</asy>
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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 22:37, 9 March 2019

Problem

Let $O$ and $H$ be the circumcenter and the orthocenter of an acute triangle $ABC$. Points $M$ and $D$ lie on side $BC$ such that $BM = CM$ and $\angle BAD = \angle CAD$. Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$. Prove that $\angle ADO = \angle HAN$.

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; 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 $OM$ intersects the circumcircle of $BHC$ at $N'$, and let the foot of the A-altitude of $ABC$ be $E$. Note that $\angle BHE=90-\angle HBE=90-90+\angle C=\angle C$. Likewise, $\angle CHE=\angle B$. So, $\angle BHC=\angle BHE+\angle CHE=\angle B+\angle C$. $BHCN'$ is cyclic, so $\angle BN'C=180-\angle BHC=180-\angle B-\angle C=\angle A$. Also, $\angle BAC=\angle A$. These two angles are on different circles and have the same measure, but they point to the same line $BC$! Hence, the two circles must be congruent. (This is also a well-known result)

We know, since $M$ is the midpoint of $BC$, that $OM$ is perpendicular to $BC$. $AH$ is also perpendicular to $BC$, so the two lines are parallel. $AN$ is a transversal, so $\angle HAN=\angle ANO$. We wish to prove that $\angle ANO=\angle ADO$, which is equivalent to $AOND$ being cyclic.

Now, assume that ray $OM$ intersects the circumcircle of $ABC$ at a point $P$. Point $P$ must be the midpoint of $\stackrel{\frown}{BC}$. Also, since $AD$ is an angle bisector, it must also hit the circle at the point $P$. The two circles are congruent, which implies $MN=MP\implies ND=DP\implies$ NDP is isosceles. Angle ADN is an exterior angle, so $\angle ADN=\angle DNP+\angle DPO=2\angle DPO$. Assume WLOG that $\angle B>\angle C$. So, $\angle DPO=\angle APO=\frac{\angle B+\angle C}{2}-\angle C=\frac{\angle B-\angle C}{2}$. In addition, $\angle AON=\angle AOP=\angle AOB+\angle BOP=2\angle C+\angle A$. Combining these two equations, $\angle AON+\angle ADN=\angle B-\angle C+2\angle C+\angle A=\angle A+\angle B+\angle C=180$.

Opposite angles sum to $180$, so quadrilateral $AOND$ 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
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