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Geometry Difficulty 8.4 Shortlist Prove it Balkan Mathematical Olympiad

Given semicircle $(c)$ with diameter $AB$ and center $O$ . On the $(c)$ we take point $C$ such that the tangent at the $C$ intersects the line $AB$ at the point $E$ . The perpendicular line from $C$ to $AB$ intersects the diameter $AB$ at the point $D$ . On the $(c)$ we get the points $H$ , $Z$ such that $CD = CH = CZ$ . The line $HZ$ intersects the lines $CO$ , $CD$ , $AB$ at the points $S$ , $I$ , $K$ respectively and the parallel line from $I$ to the line $AB$ intersects the lines $CO$ , $CK$ at the points $L$ , $M$ respectively. We consider the circumcircle $(k)$ of the triangle $LMD$ , which intersects again the lines $AB$ , $CK$ at the points $P$ , $U$ respectively. Let $(e_1)$ , $(e_2)$ , $(e_3)$ be the tangents of the $(k)$ at the points $L$ , $M$ , $P$ respectively and $R = (e_1) \cap (e_2)$ , $X = (e_2) \cap (e_3)$ , $T = (e_1) \cap (e_3)$ . Prove that if $Q$ is the center of $(k)$ , then the lines $RD$ , $TU$ , $XS$ pass through the same point, which lies in the line $IQ$ .

Solution

Since $CH = CZ$ we have $OC \perp HZ$ . So from the cyclic quadrilateral $SODI$ we get
$CS \cdot CO = CI \cdot CD. \qquad (1)$

Figure 9: G9
We draw the perpendicular line $(v)$ to $HC$ at the point $H$ . Let $J$ be the intersection point of lines $(v)$ and $CO$ . Then $CJ$ is diameter of the circle $(O, OA)$ and
$CJ = 2CO. \qquad (2)$
From the right triangle $JHC$ we have
$HC^2 = CS \cdot CJ. \qquad (3)$

Therefore, from (1), (2) and (3) we get
$CS \cdot \frac{1}{2}CJ = CI \cdot CD \quad \text{or} \quad HC^2 = 2CI \cdot CD. \qquad (3)$
However $HC = CD$ and thus $CD = 2CI$ . Thus, $I$ is the midpoint of the segment $CD$ . Nevertheless, $LM \parallel OK$ , so the points $L, M$ are the midpoints of the sides $CO$ and $CK$ respectively. Therefore, the circumcircle ( $k$ ) of the triangle $LMD$ is the Euler circle of the $COK$ and thus it passes through the point $S$ .
We have $QS = QU$ and from the right triangles $OSK, OUK$ we get $PS = PU = \frac{OK}{2}$ .
Therefore, the points $P, Q$ are located on the perpendicular bisector of the segment $SU$ . Now, we conclude that $SU \parallel TX$ , because $QP \perp (e_3)$ . Similarly, we prove that $DU \parallel RT$ and $SD \parallel RX$ .
Since the triangles $SUD$ and $XTR$ are homothetic we get that the lines $RD, TU, XS$ are concurrent at the center $M$ of homothety.
The points $I$ and $Q$ are the incenters of homothetic triangles $SUD$ and $XTR$ , respectively. Thus, the line $IQ$ passes through the point $M$ . □

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