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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 enter image description here 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 $PQ\cap MN=R$. enter image description here

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 $OA\perp RA$

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    $\begingroup$ There appears to be some dispute about the source of the problem - would either the OP or the editor link to the source so we can resolve this? $\endgroup$ – KReiser Aug 1 '20 at 5:47
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    $\begingroup$ The current source is correct . artofproblemsolving.com/community/c4183_2011_usa_tstst $\endgroup$ – Raheel Aug 1 '20 at 5:52
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    $\begingroup$ I think the point of the question as seen in the title is to have someone verify the rigor of his proposed proof. $\endgroup$ – Ojasw Upadhyay Aug 1 '20 at 7:25
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    $\begingroup$ @OjaswUpadhyay well, the OP can be a girl too. $\endgroup$ – Raheel Aug 1 '20 at 11:50
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    $\begingroup$ @Raheel You've linked a 2011 test but claim the title which references 2018 is correct. Can you clarify your comment? $\endgroup$ – KReiser Aug 2 '20 at 1:54
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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 $H_B$.

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  • $\begingroup$ Thank you, I will keep that in mind. $\endgroup$ – Sunaina Pati Aug 1 '20 at 10:57

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