Points O,N, I lie on a line [Russia 2003]

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Points O,N, I lie on a line [Russia 2003]

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#1 h Nov 4, 2010, 1:07 PM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC, O$ is the circumcenter and $I$ the incenter. The excircle $\omega_a$ touches rays $AB,AC$ and side $BC$ at $K,M,N$ , respectively. Prove that if the midpoint $P$ of $KM$ lies on the circumcircle of $\triangle ABC$ , then points $O,N, I$ lie on a line.

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#2 h Nov 4, 2010, 5:48 PM • 1 Y

Y by Adventure10

Assume that $P$ lies on the circumcircle $(O,R)$ of $\triangle ABC.$ Hence, $P$ coincides with the midpoint of $II_a.$ From the right triangle $\triangle AMI_a$ with hypotenuse $AI_a$ and altitude $MP,$ we have

$\overline{I_aM}^2={r_a}^2=\overline{I_aP} \cdot \overline{I_aA}=p(I_a,(O))=2R \cdot r_a \Longrightarrow \ r_a=2R$

Since $OP \parallel NI_a,$ it follows that $O,N,I$ are collinear, such that $\overline{OI}=-\overline{ON}.$

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#3 h Nov 4, 2010, 10:19 PM • 2 Y

Y by Adventure10, Mango247

$(I)$ is incircle and $(I_a), (I_b), (I_c)$ are excircles of $\triangle ABC.$ Inversion in $(I_a)$ takes $A,B,C$ to midpoints of $MK, KN, NM$ and circumcircle $(O)$ of $\triangle ABC$ to 9-point circle of $\triangle NMK.$ Midpoint $P$ of $MK$ is in $(O)$ $\Longrightarrow$ $(O)$ goes to itself and it is 9-point circle of $\triangle NMK$ $\Longrightarrow$ centrally similar $\triangle NMK \sim \triangle II_cI_b$ have the same 9-point circle $(O)$ $\Longrightarrow$ they are centrally congruent with similarity center $O$ and ceofficient $-1$ $\Longrightarrow$ $\overline {ON}= -\overline{OI}.$

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#4 h Apr 25, 2020, 9:34 AM

Y by

Here is a completely synthetic solution.

Rename $N,M,K$ with $D,E,F$ respectively because personal preferences. If $AB=AC$ , the result is obvious. Thus assume that $AB\ne AC$ .

Let $M$ be the midpoint of $EF$ and $A'$ be the $A$ -antipode of $\triangle ABC$ . Obviously $A,I,M$ are colinear. Now since $ME=MF$ , we can construct point $C'$ such that $\triangle MC'F$ is directly congruent to $\triangle MCE$ . This gives
$\measuredangle MBF = \measuredangle MBA = \measuredangle MCA = \measuredangle MCE = \measuredangle MC'F \implies MBFC'\text{ concyclic}.$ Moreover $\angle BC'F = \angle BMF = 90^{\circ}-\angle AMB = 90^{\circ}-\angle C$ and analogously $\angle C'BF = 90^{\circ}-\angle B$ which means $\triangle BFC'\sim\triangle BA'C$ or
$\frac{BA'}{A'C} = \frac{BF}{FC'} = \frac{BF}{CE} = \frac{BD}{DC}$ or $A'D$ pass through midpoint $N$ of arc $BAC$ .

Now let $I_A$ be the $A$ -excenter and $J$ be the Bevan point (reflection of $\small{I}$ across $\small{O}$ ) of $\triangle ABC$ . Since $M$ is the midpoint of $II_A$ , we get $I_AJ\perp BC$ or $J\in I_AD$ . Moreover, observe that $A'J\parallel AI\parallel A'N$ thus $J\in A'D$ . As $AB\ne AC$ , this forces $D=J$ so done.

This post has been edited 1 time. Last edited by MarkBcc168, Apr 25, 2020, 9:34 AM

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