2002 IMO SL #G7

 The incircle $\Omega$ of the acute triangle $ABC$ is tangent to $\overline{BC}$ at a point $K$. Let $\overline{AD}$ be an altitude of triangle $ABC$, and let $M$ be the midpoint of the segment $\overline{AD}$. If $N$ is the common point of the circle $\Omega$ and the line $KM$ (distinct from $K$), then prove that the incircle $\Omega$ and the circumcircle of triangle $BCN$ are tangent to each other at the point $N$.



Let $l$ be the perpendicular bisector of $BC$ and $I_A$ be the A-excenter of $\bigtriangleup ABC$. From the midpoint of altitudes lemma, we know that $M$, $N$,$K$ and $I_A$ are collinear, also let $T = NM \cap l$. From the "Circles Inscribed in Segment" lemma, it suffices to show that $T$ lies on $(BCN)$.

Now let $Y$ be the midpoint of $KN$.

Claim: Points $B$, $Y$, $I$, $C$ and $I_A$ lie on a circle with diameter $II_A$

Since $Y$ is the midpoint, $\measuredangle I_AYI = 90^{\circ}$ and by trivial angle chasing $\measuredangle BII_A = \measuredangle I_AIC = 90^{\circ}$. Therefore clearly points $B$, $Y$, $I$, $C$ and $I_A$ lie on a circle with diameter $II_A$.


Claim: $T$ is the midpoint of $KI_A$.

Note that since $AD \perp BC$ and $BC \perp XI_A$, we have $XI_A || AD || l$. It is well-known that $K$ and $X$ are reflections across the midpoint of $BC$, (well-known, EGMO lemma 2.20), so there are two right triangles formed with hypotenuses $KT, TI_A$. For the one with hypotenuse $TI_A$, we can drop an altitude from $I_A$ to the perpendicular bisector to form the right triangle. Let the foot of that altitude be $P$. Let the third point in the right triangle with hypotenuse $KT$ be $Q$. We have $$KQ = I_AP, \angle KTQ = \angle I_ATP, \text{and} \space \angle KQT = \angle I_APT,$$so by AAS congruence we have $\triangle KQT \cong \triangle I_APT.$ The desired $KT=TI_A$ follows. 


Claim: Quadrilateral $BTCN$ is cyclic.

By Power of a Point,

$$KN \times KT = 2 \times KY \times \frac{1}{2} \times KI_A = KY \times KI_A = KB \times KC$$Therefore quadrilateral $BCTN$ is cyclic and $T$ lies on $(BCN)$, as desired.  $\square$

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