Maths Olympiad Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Saudi Arabia

Consider a circle of center OO and a chord ABAB of it (not a diameter). Take a point TT on the ray OBOB. The perpendicular at TT onto OBOB meets the chord ABAB at CC and the circle at DD and EE. Denote by SS the orthogonal projection of TT onto the chord ABAB. Prove that AS⋅BC=TE⋅TDAS \cdot BC = TE \cdot TD.

Solution

Figure 1

Denote BB′BB' the diameter corresponding to point BB and consider α\alpha the angle ABB′^\widehat{ABB'}. Writing the power of TT with respect to the circle, we get
TE⋅TD=TB⋅TB′=TB⋅(2R−TB)=2R⋅BScos⁡α−TB2=2R⋅BScos⁡α−BC⋅BS \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 RR is the radius of the circle.

We have AS⋅BC=TE⋅TDAS \cdot BC = TE \cdot TD if and only if
AS⋅BC=2R⋅BScos⁡α−BC⋅BS, AS \cdot BC = 2R \cdot \frac{BS}{\cos \alpha} - BC \cdot BS,
that is equivalent to
2R⋅BScos⁡α=AS⋅BC+BC⋅BS=BC⋅(AS+BS)=BC⋅AB. 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⋅AB2R=BScos⁡α, BC \cdot \frac{AB}{2R} = \frac{BS}{\cos \alpha},
that is BC⋅cos⁡α=BScos⁡αBC \cdot \cos \alpha = \frac{BS}{\cos \alpha}, hence BSBC=cos⁡2α\frac{BS}{BC} = \cos^2 \alpha.

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

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