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Saudi Arabian Mathematical Competitions

Geometry Difficulty 5.6 AIME, harder Prove it Saudi Arabia

Consider a circle of center $O$ and a chord $AB$ of it (not a diameter). Take a point $T$ on the ray $OB$ . The perpendicular at $T$ onto $OB$ meets the chord $AB$ at $C$ and the circle at $D$ and $E$ . Denote by $S$ the orthogonal projection of $T$ onto the chord $AB$ . Prove that $AS \cdot BC = TE \cdot TD$ .

Solution

Denote $BB'$ the diameter corresponding to point $B$ and consider $\alpha$ the angle $\widehat{ABB'}$ . Writing the power of $T$ with respect to the circle, we get
$\begin{gathered} TE \cdot TD = TB \cdot TB' = TB \cdot (2R - TB) = 2R \cdot \frac{BS}{\cos \alpha} - TB^2 \\ = 2R \cdot \frac{BS}{\cos \alpha} - BC \cdot BS \end{gathered}$
where $R$ is the radius of the circle.

We have $AS \cdot BC = TE \cdot TD$ if and only if
$AS \cdot BC = 2R \cdot \frac{BS}{\cos \alpha} - BC \cdot BS,$
that is equivalent to
$2R \cdot \frac{BS}{\cos \alpha} = AS \cdot BC + BC \cdot BS = BC \cdot (AS + BS) = BC \cdot AB.$
It follows that the desired relation is equivalent to
$BC \cdot \frac{AB}{2R} = \frac{BS}{\cos \alpha},$
that is $BC \cdot \cos \alpha = \frac{BS}{\cos \alpha}$ , hence $\frac{BS}{BC} = \cos^2 \alpha$ .

On the other hand, it is clear that
$\cos \alpha = \frac{BS}{BT} \text{ and } \cos \alpha = \frac{BT}{BC}.$
Multiplying these relations we get $\cos^2 \alpha = BS / BC$ and we are done.

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