Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Bulgaria

Problem:

Three nonintersecting circles $k_{i}(O_{i}, r_{i})$ , $i=1,2,3$ , where $r_{1}<r_{2}<r_{3}$ , are tangent to the arms of an angle. One of the arms is tangent to $k_{1}$ and $k_{3}$ at points $A$ and $B$ and the other one is tangent to $k_{2}$ at point $C$ . Let $K=AC \cap k_{1}$ , $L=AC \cap k_{2}$ , $M=BC \cap k_{2}$ and $N=BC \cap k_{3}$ . The four lines through $C$ and $P=AM \cap BK$ , $Q=AM \cap BL$ , $R=AN \cap BK$ and $S=AN \cap BL$ , meet $AB$ at the points $X, Y, Z$ and $T$ , respectively. Prove that $XZ=YT$ .

Solution

Solution:

If $E$ and $F$ are the second tangent points of $k_{1}$ and $k_{2}$ with the arms of the angle, then the equalities $AF^{2} = AL \cdot AC$ , $CE^{2} = CK \cdot CA$ and $AF = CE$ imply that $AL = CK$ . Hence $AK = CL$ and analogously $CM = BN$ .

On the other hand, Ceva's theorem gives
$\frac{AX}{XB} = \frac{AK \cdot CM}{KC \cdot MB} \text{ and } \frac{AT}{TB} = \frac{AL \cdot CN}{LC \cdot NB}$
Multiplying these equalities gives $\frac{AX}{XB} = \frac{TB}{AT}$ and therefore
$\frac{AX + XB}{XB} = \frac{TB + AT}{AT} \Longleftrightarrow AT = BX \Longleftrightarrow AX = BT$
Analogously $AZ = YB$ which implies that $XZ = YT$ .

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.