Easy but Thought-Provoking

Geometry Level 4

In the diagram, point $A$ outside a circle with center $O$ .

Draw tangents $AB, AC$ to the circle, and line $ADE$ intersects $(O)$ at $D, E$ such that, $BE$ and $AC$ are parallel.

$BD$ intersects $AC$ at $I$ .

Let $S$ be the point on $AO$ such that $BSIA$ (the yellow quadrilateral) is concylic.

Find $\dfrac{SA}{SO}$ .

The answer is 3.

1 solution

X X
Jun 26, 2020

I will use cross ratios and inversion.

$BCDE$ is a harmonic quadrilateral, so $-1=(B,C;D,E)=_B(A,C;I,P_\infty)$ , hence $I$ is the midpoint of $AC$ .

Notice that $ACOB$ is a cyclic quadrilateral, so invert about $A$ with radius $AB$ .

$O^*$ is the midpoint of $BC$ , $I^*$ is the reflection of $A$ about $C$ , $S^*$ is the intersection of $AO$ and $BI^*$ . Let the midpoint of $AO$ be $M$ , so $M^*$ is the reflection of $A$ about $O^*$ .

Since inversion preserves cross ratio, $(O,M;S,P_\infty)=(O^*,M^*;S^*,A)=_{I^*}(O^*,P_\infty;B,C)=-1$ , hence $S$ is the midpoint of $OM$ .

Easy but Thought-Provoking

The answer is 3.

1 solution

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