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Given triangle $ABC$ inscribed $(O)$. Let $I$ be the incenter and $D$ be the contact point of $(I)$ with $BC$. $AD$ intersect $(O)$ at the second point $E$. Let $M$ be the midpoint of $BC$ and $N$ the midpoint of arc $BAC$. Let $EN$ intersect $(BIC)$ at $P$ ($P$ lies inside $ABC$). Prove $\angle IPM=90^o$ enter image description here

I think it's just angle chasing, but it's hard to approach. I don't know what properties the construction of $E$ made ?

RopuToran
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  • What have you tried? What do you know about the properties of inncenters and circumcenters? – daruma Aug 03 '18 at 09:10
  • The circumcenter won't work much. We should care about the cyclic quadrilateral. Also the center of $BIC$ is the midpoint of arc $BC$ not contain $A$ – RopuToran Aug 03 '18 at 09:13
  • These are just ideas. You should know that $IDB=90^{\circ}$ as $I$ is an incenter for triangle $ABC$. So, somehow if you could prove that $PID+PMD=180^{\circ}$, that means $IPMD$ is a cyclic quadrilateral and hence, you can say that $IPM=90^{\circ}$. You may want to call angle $BAC, CBA, BCA$ as $2\alpha,2\beta,2\gamma$ for convenience. $PID=PIB-BID=(180^\circ-PCM)-(\alpha+\gamma)=90^\circ+\beta-PCM$ and $PMD=180^\circ-PMC$. I agree you somehow need to use the definition of $E$ to link everything in which, unfortunately I couldn't see. – daruma Aug 03 '18 at 13:05
  • Yes. The construction of $E$ is hard to approach. I've never seen that construction before. P3 of Hong Kong TST is a little similar to the problem. See: https://artofproblemsolving.com/community/c6h1149892p5437223 – RopuToran Aug 03 '18 at 13:20
  • Where is the question from? – daruma Aug 03 '18 at 14:18
  • I simplify from Ukraine TST 2016 – RopuToran Aug 03 '18 at 15:52
  • Let us continue this discussion in chat. – RopuToran Aug 03 '18 at 16:56

2 Answers2

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Proof

The following statements will use some well-known facts, for example, the center of the circle $(BIC)$ is the midpoint of the arc $\widehat{BEC}$ and so on. For convenience, we omit the proofs for them.

Let $NE$ intersect $ID$ at the point $K$. It's easy to find $\angle AIK=\angle AEK$. Thus $A,I,E,K$ are concyclic. Hence $$BD \cdot DC=AD \cdot DE=ID \cdot DK,$$which shows that $B,I,C,K$ are also concyclic, namely, $K$ lies on the circle $(BIC)$.

Moreover, we may notice that $NC,NB$ are the tangents to the circle $(BIC)$. Therefore, $BPCK$ is a harmonic quadrilateral. Since $M$ is the midpoint of the diagonal $BC$, we may claim that $$\triangle BPM \sim \triangle KPC.$$ Therefore, $$\angle IPM=\angle IPB+\angle BPM=\angle IPB+\angle KPC=\angle IKB+\angle KBC=90^o.$$

enter image description here

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mengdie1982
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A longer way to solve:

Lemma: Given $\Delta ABC$ inscribed $(O)$ and its incenter $I$. Let $(I)$ touch $BC,CA,AB$ at $D,B',C'$. $B'C' \cap BC=T$. Let $N$ be the midpoint of arc $BAC$. Let $NT \cap (O)=${$N,Q$}. Let $A$-symmedian intersect $(O)$ at $G$. Then $(T,G,I,Q)$ concyclic.

For the proof of the lemma, please see here.

Back to the problem, let $(IBC) \cap (IDM)=${$I,P'$}. We define point $T,G,Q$ as the lemma said. (see figure below)

enter image description here

We will prove $\overline{N,P',E}$.

We have $E(AG;BC)=-1$ and $E(TD,BC)=-1$ so $\overline{T,E,G}$.

By using power of point: $TB.TC=TD.TM$, so $\overline{T,I,P'}$.

Hence, $TI.TP'=TB.TC=TQ.TN=TE.TG$.

$\Rightarrow (N,Q,I,P')$ and $(I,P',G,E)$ concyclic.

Now, $\angle IP'N + \angle IP'E= \angle TQI + \angle TGI = 180^o$, i.e. $\overline{E,P',N}$ (Q.E.D).

So $P' \equiv P$ and the problem solved.

RopuToran
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