Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.6 AIME Prove it Taiwan

Let $C_1$ and $C_2$ be two concentric circles, where $C_2$ is inside $C_1$ . From a point $A$ on $C_1$ , draw a tangent line $AB$ to $C_2$ , with point $B$ on $C_2$ . Let point $C$ be the other intersection of ray $AB$ with $C_1$ , and let point $D$ be the midpoint of $\overline{AB}$ . Construct a line through $A$ intersecting $C_2$ at two points $E, F$ , such that the perpendicular bisector of $DE$ and the perpendicular bisector of $CF$ meet at a point $M$ on $AB$ . Find all possible values of $AM/MC$ .

Solution

$AM/MC = 5/3$ .

因為 $AC \cdot AD = (2AB) \cdot (\frac{1}{2}AB) = AB^2 = AE \cdot AF$ , 故 $CDEF$ 四點共圓。又 $M$ 點位於 $CF$ 及 $DE$ 的中垂線上, 故 $M$ 點為 $CDEF$ 的外接圓圓心。因為 $CMD$ 在同一條直線上, 所以 $M$ 為 $CD$ 中點。故
$\frac{AM}{MC} = \frac{AD + DM}{MC} = \frac{AD + \frac{1}{2}CD}{\frac{1}{2}CD} = \frac{\frac{1}{4}AC + \frac{1}{2} \cdot \frac{3}{4}AC}{\frac{1}{2} \cdot \frac{3}{4}AC} = \frac{5}{3}.$

Since $AC \cdot AD = (2AB) \cdot (\frac{1}{2}AB) = AB^2 = AE \cdot AF$ , the four points $C, D, E, F$ are concyclic. Also, since $M$ lies on the perpendicular bisectors of $CF$ and $DE$ , $M$ is the center of the circumscribed circle of $CDEF$ . Since $C, M, D$ lie on the same line, $M$ is the midpoint of $CD$ . Therefore
$\frac{AM}{MC} = \frac{AD + DM}{MC} = \frac{AD + \frac{1}{2}CD}{\frac{1}{2}CD} = \frac{\frac{1}{4}AC + \frac{1}{2} \cdot \frac{3}{4}AC}{\frac{1}{2} \cdot \frac{3}{4}AC} = \frac{5}{3}.$

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Source: MathNet, licensed CC-BY-4.0. Statement translated into English from zh; metadata (topic, difficulty) added by this project.