USATST 2018/P4 : Prove that $OA\perp RA$ [Proof Verification needed]

Question

Acute triangle $ABC$ is inscribed in circle $\omega$. Let $H$ and $O$ denote its orthocenter and circumcenter, respectively. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Rays $MH$ and $NH$ meet $\omega$ at $P$ and $Q$, respectively. Lines $MN$ and $PQ$ meet at $R$. Prove that $OA\perp RA$.

I will be very grateful if someone can verify this proof . I am very new to radical axis.

Also, Please post your solutions too. We learn a lot from other's solutions too. Thanks in advance.

My Proof: Before proceeding further, I would like to state a lemma.

Lemma: Let $ABC$ be a triangle with orthocenter $H$, and suppose that $E$ and $F$ are the feet of the $B$ and $C$-altitudes. Suppose that the circumcircle of triangle $AEF$ meets the circumcircle of triangle $ABC$ again at $K$. Let $M$ be the midpoint of $BC$. Then we have $K, H,$ and $M$ are collinear.

Proof of the Lemma: Sine $HF\perp AB$ and $HE\perp AC$, we note that $H\in (AEF)$. So $\angle AKH= \angle AFH = 90^{\circ}$

Let $KH\cap(ABC)=X$. Note that since, $\angle AKH=90^{\circ}$, we have $X=$ diametrically opposite point of $A$.

But by a known lemma , we know that $H,M,X$ are collinear . So we have $K$,$ H,$M$ are collinear.

Now, using this Lemma, we claim that $MNPQ$ is cyclic

Claim: $MNPQ$ is cyclic Proof: By the above Lemma, we get $H'MHP$ and $QHN{H'}_1$ are collinear, where $H"M=HM$ and $H'$ is the antipode of $P$ wrt $(ABC)$ and $N{H'}_1=HN$ and ${H'}_1$ is the antipode of $Q$ wrt $(ABC)$.

Hence by $POP$, $\Bbb P(H, (ABC))=HH'\cdot HP=QH\cdot H{H'}_1$.

But $HM=\frac {1}{2} HH'$ and $HN=\frac {1}{2}H{H'}_1= HN \implies HM\cdot HP=QH\cdot HN$.

Hence by converse of $POP$ , we have $MNPQ$ cyclic .

Claim: $AMON$ is cyclic with diametre $AO$ .

Proof of the Claim: Just note that $AM \perp OM$ and $AN \perp ON$ .

Main Proof : Now, using the fact that pairwise radical axis of 3 circles concur,

we get that for circles $(MNPQ),(ABC),(AMON)$ ; the pairwise radical axis concur at $PQ\cap MN=R$.

But note that the radical axis of $(ABC)$ and $(AMON)$ is nothing but the line tangent to $(AMON)$ at $A$.

Since $AO$ is the diameter of $(AMON)$, hence $OA\perp RA$

Answer 1

This proof is correct. Some stylistic notes:

I think some of the structure could be improved, in that many of your claims are so small that the proof would read more easily if it were written linearly (i.e. as a sequence of statements that you show to be true, instead of a sequence of claims that you state and then prove separately). Also, your names of $H_1$ and $H_1'$ aren't especially great -- I would suggest using $C'$ and $B'$ (to indicate that they are antipodes), or maybe $H_C$ and $H_B$.