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Olympiad problem-solving technique

The Incircle Configuration

Tangent lengths turn sides into s − a.

The two tangents from a point to a circle are equal, so the incircle's touch points cut the sides into three tangent lengths sas-a, sbs-b, scs-c — which sum to ss and multiply with it into Heron's formula. Translating side data into tangent-length data, and back, is the opening move of most incircle problems.

Two more handles complete the kit. The incentre angle: BIC=90+A2\angle BIC = 90^\circ + \frac{A}{2}, always obtuse, blind to how BB and CC split the remainder. And the arc-midpoint lemma: the midpoint MM of arc BCBC satisfies MB=MI=MCMB = MI = MC, so BB, II, CC lie on a circle centred at MM — converting incentre statements into circumcircle arcs.

The excircles obey the same equal-tangent bookkeeping with ss in place of sas-a, which is why the four quantities ss, sas-a, sbs-b, scs-c keep appearing together.

Where it appears on Lemma: Level 6 (the incircle configuration).

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