Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 7.3 National olympiad, round 2 Prove it Greece

Let $ABC$ be a scalene acute angled triangle with $AB < AC < BC$ and circumcenter $c(O, R)$ . The ex-circle ( $c_A$ ) corresponding to the vertex $A$ , has center $I$ and is tangent to the sides $BC$ , $AC$ , $AB$ at $D$ , $E$ , $Z$ , respectively. The line $AI$ intersects the circle $c(O, R)$ at $M$ and the circumcircle, say $(c_1)$ , of the triangle $AZE$ intersects the circle ( $c$ ) at $K$ . The circumcircle, say $(c_2)$ , of the triangle $OKM$ intersects the circle ( $c_1$ ) at point $N$ . Prove that the lines $AN$ and $KI$ intersect at a point of the circle ( $c$ ).

Solution

Since $AZ$ and $AE$ are tangents to the circle ( $c_A$ ), we have $IZ \perp AZ$ and $IE \perp AE$ . Hence, the circle ( $c_2$ ) contains $I$ and $AI$ is a diameter. First we will prove that $AN$ passes through $O$ , that is $KNO = KNA$ (*).

Moreover we have $KNO = KMO$ and from the isosceles triangle $OKM$ we get that: $KMO = OKM$ . Therefore: $KNO = KMO = OKM$ (1).

Figure 8

Similarly, we conclude that $KNA = KIA$ . From the right angled triangle $AKI$ we have $KIA = 90^\circ - IAK = 90^\circ - MAK$ . Since $MAK = \frac{MOK}{2}$ and from the isosceles triangle $OKM$ we have:
$OM = 90^\circ - \frac{MOK}{2}$ , finally, using (1), we find: $KNA = KIA = 90^\circ - IAK = 90^\circ - MAK = 90^\circ - \frac{MOK}{2} = OKM = KNO$ (*) Hence the points $A$ , $O$ , $N$ are collinear and let $T$ the point of intersection of $AN$ with the circle ( $c$ ).

To complete the proof, we show that the points $K$ , $T$ , $I$ are collinear, by proving that $AKI = AKT$ . In fact $AKT = 90^\circ$ , since $AT$ and $AI$ are diameters.

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