Problem
GEO-B2-M01-P107 A Tangent and a Directed Ratio
In triangle \(ABC\), \(AB=8\) and \(AC=9\). The tangent to the circumcircle at \(A\) meets line \(BC\) at point \(T\). Find the ratio \(TB:TC\) in directed lengths.
C. Hint 1. Replace the angle between the tangent and \(AB\) by angle \(ACB\).
D. Hint 2. After the sine rule, the square of the ratio \(AB:AC\) will appear.
E. Full solution. By the tangent-chord theorem, \(\angle TAB=\angle ACB\), and \(\angle TAC=\angle ABC\). These two replacements are the main step of the problem.
Apply the sine rule in triangles \(TAB\) and \(TAC\). Since the angles at \(T\) are supplementary as angles on line \(BC\), after dividing the corresponding equalities we obtain \(\frac{TB}{TC}=\frac{\sin^2\angle ACB}{\sin^2\angle ABC}\).
In triangle \(ABC\), by the sine rule, \(\frac{AB}{AC}=\frac{\sin\angle ACB}{\sin\angle ABC}\). Hence \(\frac{TB}{TC}=\frac{AB^2}{AC^2}=\frac{64}{81}\). Therefore \(TB:TC=64:81\) in directed lengths.
Method comment. the tangent replaces angles at \(T\) by angles of the original triangle, after which the sine rule works Thus the solution is not a brute-force chase of all angles in the diagram, but a deliberate choice of the right circle or tangent, after which the angles can be compared through the same chord or the same line.
If the auxiliary step is skipped, the problem looks almost arbitrary: the equal angles live in different parts of the diagram. That is why the hidden configuration is identified first, then the angle replacement is made, and only at the end the required conclusion follows.
F. Difficulty justification. This is Level 6: a medium regional-style problem. A direct angle chase quickly overloads the diagram, so one must see a hidden circle or replace an angle by a tangent argument; after that, the chain is short.
G. Check. This is not a one-step exercise: it requires 4 key ideas. First one must recognise the hidden geometric structure, then make an angle replacement or add a circle, and only after that complete the final conclusion.