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Turkey — Team Selection Test

Geometry Difficulty 8.4 Shortlist Prove it Turkey

Points $A$ , $B$ , $C$ are given on a semicircle. The line which is tangent to the semicircle at $A$ intersects the line of the diameter at $M$ , and the line which is tangent to the semicircle at $B$ intersects the line of the diameter at $N$ . The line which passes through $A$ and is perpendicular to the diameter intersects $NC$ at $R$ , and the line which passes through $B$ and is perpendicular to the diameter intersects $MC$ at $S$ . The line $RS$ intersects the line of the diameter at $Z$ . Show that $ZC$ is tangent to the semicircle.

Solution

In the solution, we use directed lengths on the line of the diameter. Let $A'$ , $B'$ , $C'$ be the feet of the perpendiculars dropped onto the diameter from $A$ , $B$ , $C$ . We have
$\frac{ZB'}{ZA'} = \frac{|SB'|}{|RA'|} = \frac{|SB'|}{|CC'|} \cdot \frac{|CC'|}{|RA'|} = \frac{MB'}{MC'} \cdot \frac{NC'}{NA'}$
Now let $OA' = a$ , $OB' = b$ and $OC' = c$ , thus we have $OM = r^2/a$ and $ON = r^2/b$ where $r$ is the radius of the semicircle. In this notation, the above equality reads
$\frac{OZ - b}{OZ - a} = \frac{r^2/a - b}{r^2/a - c} \cdot \frac{r^2/b - c}{r^2/b - a} = \frac{r^2/c - b}{r^2/c - a}$
Now since $a \neq b$ , one has $OZ = r^2/c$ which shows that $ZC$ is tangent to the semicircle.

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