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

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<b>Claim:</b> <math>\triangle AA'N' \sim \triangle ADO</math>. The proof easily follows.
 
<b>Claim:</b> <math>\triangle AA'N' \sim \triangle ADO</math>. The proof easily follows.
  
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<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>
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<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==
 
==Solution 2==

Revision as of 17:31, 21 September 2021

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 1

It's well known that the reflection of $H$ across $\overline{BC}$, $H'$, lies on $(ABC)$. Then $(BHC)$ is just the reflection of $(BH'C)$ across $\overline{BC}$, which is equivalent to the reflection of $(ABC)$ across $\overline{BC}$. Reflect points $A$ and $N$ across $\overline{BC}$ to points $A'$ and $N'$, respectively. Then $N'$ is the midpoint of minor arc $\overarc{BC}$, so $A, D, N'$ are collinear in that order. It suffices to show that $\angle AA'N'=\angle ADO$.

Claim: $\triangle AA'N' \sim \triangle ADO$. The proof easily follows.

Proof: Note that $\angle BAA'=\angle CAO=90^{\circ}-\angle ABC$. Then we have $\angle A'AN'=\angle BAD-\angle BAA'=\angle CAD-\angle CAO=\angle DAO$. So, it suffices to show that \[\frac{AA'}{AN'}=\frac{AD}{AO}\rightarrow AA'\cdot AO=AN'\cdot AD.\] Notice that $\triangle ABA' \sim \triangle AOC$, so that \[\frac{AB}{AA'}=\frac{AO}{AC}\rightarrow AA'\cdot AO=AB\cdot AC.\] Therefore, it suffices to show that \[AB\cdot AC=AN'\cdot AD\rightarrow \frac{AB}{AN'}=\frac{AD}{AC}.\] But it is easy to show that $\triangle BAN'\sim \triangle DAC$, implying the result. $\blacksquare$

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 $OM$ 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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