It's known that $N$ is the reflection of $P$ across $M$. The main claim is that $ADNO$ is cyclic. To see this let $P$ and $Q$ be the arc midpoints of $\widehat{BC}$, so that $ADMQ$ is cyclic. Then $PN \cdot PO = PM \cdot PQ = PD \cdot PA$ as advertised. [asy][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("$A$", 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)); /* 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][/asy] To finish, note that $\measuredangle HAN = \measuredangle ONA = \measuredangle ODA$. Remark: The orthocenter $H$ is superficial and can be deleted basically immediately. One can reverse-engineer the fact that $ADNO$ is cyclic from the truth of the problem statement.

Wow! Circumcenter/triangle geo for #3 and #5

This was quite simple with complex numbers as well. Indeed, reflect everything across the midpoint of $BC$. You can easily show that $\angle ADO = \angle M_aA'H'$.

Let $F$ be the midpoint of arc $BC$ not containing $A$ on the circumcircle of $\bigtriangleup ABC$. We know that $F$ lies on $AD$. It is a well-known lemma that the circumcircle of $\bigtriangleup BHC$ is the reflection of the circumcircle of $\bigtriangleup ABC$ in $\overline{BC}$. It follows that $N$ is the reflection of $F$ across $M$. From $AH \perp BC$ and $ON \perp BC$, we have $AH\parallel ON$. We deduce that $\angle HAN=180^\circ-\angle ANO$. Our condition is equivalent to proving that $ANOD$ is cyclic. We do so by the converse of power of a point, specifically that of point $F$. We compute a few lenghs. Let $R$ be the circumradius of $\bigtriangleup ABC$ and let $a,b,c$ be the measure of sides $BC, AC, AB$ respectively. Clearly, $FO=R$. Using the Law of Sines in $\bigtriangleup ABF$, we have $BF=2R\sin{\frac{\angle{A}}{2}}$. We have $FN=2FM=2\times BF \times \sin{\frac{\angle{A}}{2}}=4R\sin^2{\frac{\angle{A}}{2}}$. Thus $FO\times FN = 4R^2\sin^2{\frac{\angle{A}}{2}}$. By Ptolemy, we have $$a\times AF =BF\times(b+c)\iff$$$$AF=\frac{b+c}a\times BF=\frac{b+c}{a}\times 2R\sin{\frac{\angle{A}}{2}}$$. By the angle-bisector theorem, $BD=\frac{ac}{b+c}$. By the Law of Sines in $\bigtriangleup BDF$, we have $$DF=\sin{\frac{\angle{A}}{2}}\times BD \times \sin{\angle BFA}=\frac{a}{b+c}\times \sin{\frac{\angle{A}}{2}} \times\frac{c}{\sin{\angle{C}}}=\frac{a}{b+c}\times 2R\times\sin{\frac{\angle{A}}{2}}$$. Thus, $FD\times FA=4R\sin^2{\frac{\angle{A}}{2}}=FO\times FN$, which proves that $ANOD$ is a cyclic quadrilateral and the proposition follows. edit: slightly better finish: Notice that $FD\times FA=FB^2$.

Darn, proved $ADNO$ is cyclic, but didn't have time to wrap it up

v_Enhance wrote: 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 $\angle ADO = \angle HAN$. Did you propose this one?

Consider the mapping $f$ consisting of an inversion centered at $A$ with radius $\sqrt{bc}$, followed by a reflection about the angle bisector of $\angle BAC$. Then $O$ gets sent to the reflection of $A$ over $BC$, call it $A'$. Let $N'$ be the midpoint of arc $BC$ not containing $A$. Then $N'$ is the reflection of $N$ over $BC$, since $(BHC)$ is the reflection of $(ABC)$ over $BC$. Then $ANN'A'$ is an isosceles trapezoid with $\angle AA'N'=\angle A'AN$. We also have $f(D)=N'$, so we conclude that \[ \angle ADO = \angle AA'N' = \angle A'AN = \angle HAN, \]as desired.

skipiano wrote: Darn, proved $ADNO$ is cyclic, but didn't have time to wrap it up Same also doesn't MO hit the circumcircle of BHC at two points...? the problem should've specified

mathguy5041 wrote: skipiano wrote: Darn, proved $ADNO$ is cyclic, but didn't have time to wrap it up Same also doesn't MO hit the circumcircle of BHC at two points...? the problem should've specified That's why it said ray MO. How much points for proving $ADNO$ cyclic?

It's a ray. Reflect over BC and use the spiral similarity lemma at A, done.

Solution credited to wu2481632 and amplreneo: Let $F$ be the midpoint of arc $BC$ not containing $A$ and let $A'$ be the reflection of $A$ across $BC$. Notice that there is a spiral similarity centred at $A$ that maps $\bigtriangleup AOF$ to $\bigtriangleup ADA'$. This implies, by the spiral similarity comes in pairs lemma, that there is another one centred at $A$ that maps $\bigtriangleup AA'F$ to $\bigtriangleup ADO$, thus $\angle ADO=\angle AA'F$, but notice that $AA'FN$ is an isosceles trapezoid, from which it follows that $\angle AA'F=\angle A'AN=\angle HAN$, proving the result.

skipiano wrote: mathguy5041 wrote: skipiano wrote: Darn, proved $ADNO$ is cyclic, but didn't have time to wrap it up Same also doesn't MO hit the circumcircle of BHC at two points...? the problem should've specified That's why it said ray MO. How much points for proving $ADNO$ cyclic? i should learn how to read

skipiano wrote: mathguy5041 wrote: skipiano wrote: Darn, proved $ADNO$ is cyclic, but didn't have time to wrap it up Same also doesn't MO hit the circumcircle of BHC at two points...? the problem should've specified That's why it said ray MO. How much points for proving $ADNO$ cyclic? ????

Probably 2 at most.

Can you actually say its well known that $N$ is reflection of $P$ over $M?$ Also @venhance in your proof shows that directed angles $HAN$ and $ADO$ are equal but don't you have to show that the angles are actually equal not just equal mod 180. Like I was going to do that to deal with the config issue but I thought proving the directed angles were equal could still mean they are like supplements technically.

Homothety WRT M and inversion WRT A...

Uh just prove that (BHC) is the reflection of (ABC) over BC and it works lol

wu2481632 wrote: Probably 2 at most. Doesn't it finish the problem though, like the rest is really easy

It "finishes" the problem ok maybe 5 darn

wow i have changed

We claim that $ADNO$ is cyclic. Note that $(BHC)$ is $(ABC)$ but reflected from $BC$. Thus, we have that $\angle{ADN} = 2\angle{HAD}$ because $A'$, $D$, and $N$ are collinear. Note that \[\angle{ADN} = 2\angle{HAD} = \angle{A}+2\angle{B}-180\]We also have that \[\angle{AON} = 2\angle{C}+\angle{A}\]From this, we get that \[\angle{AON}+\angle{ADN} = 2(\angle{A}+\angle{B}+\angle{C})-180 = 180\]which proves that $ADNO$ is cyclic. Thus, \[\angle{ADO} = \angle{ANO} = \angle{HAN}\]because $OM\parallel AH$.

Since I bashed #3 using Complex Numbers Coordinates, why not do it again for #5? Set $\triangle ABC$ to be inscribed in the unit circle and pretend it is a excellent idea to bash a question with 2 circles It works. Full proof here: https://infinityintegral.substack.com/p/usajmo-2017-contest-review

Note that $MO$ is the perpendicular bisector of $BC$. Since $\angle BHC=180^{\circ}-\angle A$, $\angle BNM=90^{\circ}-\frac12\angle A$, so $\angle NBC=\frac12\angle A$. Also, since $AH$ and $MO$ are both perpendicular to $BC$, they are parallel. Claim. $\angle BND=\angle C$. Proof. Reflect $N$ over $BC$ to $N'$. Then $N'$ is the arc midpoint of arc $BC$ not containing $A$, so by fact 5, $A$, $D$, and $N$ are collinear. Therefore, \[\angle BND=\angle BN'D=\angle BN'A=\angle BCA\,\blacksquare\] Therefore, \[\angle NDC=\angle NBC+\angle BND=\frac12\angle A+\angle C.\]But \[\angle DAO=\angle BAO-\angle BAD=90^{\circ}-\angle C-\frac12\angle A=90^{\circ}-\angle NDC=\angle DNO,\]so quadrilateral $ADON$ is cyclic. Therefore, \[\angle ADO=180^{\circ}-\angle ANO=\angle HAN\,\blacksquare\]

Nice problem. How did I not see $ADON$ cyclic? Either way, fun force overlay solution. Let $AH$ intersect $(BHC)$ a second time at point $X.$ (Alternatively, since $(BHC)$ and $(ABC)$ are reflections of each other across $BC$ by reflecting the orthocenter, $X$ is the reflection of $A$ across $BC.$) We claim that $\triangle XAN \sim \triangle ADO.$ There are two parts to doing so: Part 1: Proving $\angle DAO = \angle AXN.$ Since $AH$ and $AO$ are isogonal and $XD = DA,$ we get $\angle DAO = \angle HAD = \angle XAD = \angle AXD = \angle AXN.$ Part 2: Proving $\frac{AD}{AO} = \frac{XA}{XN}.$ Cross-multiplying, we get $AD \cdot XN = AO \cdot XA.$ If we let $Z$ be the midpoint of arc $BC$ on $(ABC)$ not containing $A$, then $XN = AZ$ by reflections, so it suffices to show that $AO \cdot AX = AD \cdot AZ.$ However, this immediately follows by an inversion centered at $A$ with radius $\sqrt{AB \cdot AC},$ followed by a reflection across $AD.$ Therefore, by SAS similarity, we have that $\triangle XAN \sim \triangle ADO.$ Finally, $\angle ADO = \angle XAN = \angle HAN,$ as desired.

Observe ANOD cyclic Then just angle chase

https://www.youtube.com/watch?v=YTQV48V44Sw $\newline$ Ado. Let $X$ be the midpoint of the minor arc $BC$, and let $Y$ be the midpoint of the major arc $BC$. Then it follows that $N$ is the reflection of $X$ over $M$. Clearly, $X$ lies on $AD$, because of arc midpoints. We will prove that $ADMY$ is cyclic. $\newline$ Because $XY$ is a diameter, we have $\measuredangle YMD = \measuredangle DAY = 90^{\circ}$. $\newline$ By Power of a Point, we have \[XD \cdot XA = XM \cdot XY = XN \cdot XO\]so $ADON$ is cyclic. Then $\measuredangle ADO = \measuredangle ONA = \measuredangle ANY$. Then because $AH \parallel OM$, we have $\angle ANY = \angle HAN$ so we are done.

you can also just define the arc midpoint $P$ then \[\triangle PDN\sim \triangle POA\]so $ADNO$ cyclic, so \[\measuredangle HAN=\measuredangle ONA=\measuredangle ODA\]

Let $T$ be the midpoint of arc $BC$, without $A$. It is well-known that $N$ is the reflection of $T$ about $M$. Claim: $ADNO$ is cyclic Proof: Letting $P$ be the antipode of $T$, we get: \[\angle ADN=180=\angle NDT=2\angle ATO=\angle AOP=180-\angle AON\]$\square$ We finish with angle-chasing: \[\angle HAN=90-\angle C-\angle A/2+\angle DAN\]\[=90-\angle AOT/2+\angle DOT\]\[=\angle ATO+\angle DOT\]\[=\angle ADO,\]as desired $\blacksquare$

$\sqrt{bc}$-invert takes $\angle ADO$ to the reflection over $BC$ of $\angle HAN.$

Let $N'$ be the midpoint of arc $BC$, so that $N$ and $N'$ are reflections over $\overline{BC}$. So, $\triangle NDN'$ is isosceles and obviously $\triangle AON'$ is isosceles too. Since these triangles share an angle, it follows that $AOND$ is cylic, from which it follows that \[ \angle ADO = \angle ANO = \angle HAN.\]

First, note that $HA\parallel MN$, so $\angle HAN = 180^\circ - \angle ANM = 180^\circ - \angle ANO$. Hence, showing $\angle HAN = \angle ADO$ is equivalent to show $\angle ANO + \angle ADO = 180^\circ$, i.e. that $ANOD$ is cyclic. To do this, use complex numbers with $(ABC)$ as the unit circle. We can pick $x,y,z$ so that $a = x^2, b = y^2, c =z^2$. Then the midpoint of arc $BC$ is $-yz$. Since $(BHC)$ is the reflection of $(ABC)$ over $BC$, it follows that $n$ is the reflection of $-yz$ over $BC$. However, this is just the reflection of $-yz$ over the midpoint of $BC$, so we have \[ n = 2\left(\frac{y^2 + z^2}{2}\right) - (-yz) = y^2 + z^2 + yz.\]We have that $d$ is the intersection of the line through $x^2, -yz$ and the line through $y^2, z^2$. From complex intersection formula, this is just \[ d = \frac{y^2z^2(x^2 -yz) + x^2yz(y^2 + z^2)}{y^2z^2 + x^2yz} = \frac{yz(x^2 - yz) + x^2y^2 + x^2z^2}{yz + x^2}.\]Finally, we verify $ANOD$ cyclic iff \[ \mathbb{R}\ni \frac{o - a}{o-n}\div \frac{d-a}{d-n} = \frac{x^2}{y^2 + yz +z^2}\cdot \frac{yz(y+z)^2}{(x^2-y^2)(x^2-z^2)}.\]It's easy to check that this is self-conjugating, so $ANOD$ is cyclic and we are done.

$ $

Define $AD \cap (ABC) = E$. Claim: $ADON$ is cyclic. Proof: Notice that $N$ is the reflection of $E$ over $BC$, a result that directly follows from reflecting the orthocenter. We prove that $ADON$ is cyclic by showing that $\angle DAO = \angle DNO = \angle AEN$. The first and third terms being equal follows from $OA = OE$ by radii in $(ABC)$, and the second and third terms being equal follows from $D$ being on $BC$, which we established to be the perpendicular bisector of $NE$. Thus the proof of our claim is complete. $\square$ To finish, define the antipode of $E$ in $(ABC)$ to be $X$, and from our cyclicity condition we have $$\angle ADO = \angle ANX = \angle HAN$$, where the last equality is from $AH$ being parallel to $NO$ (both are perpendicular to $BC$ trivially). Thus we are done.