Multiple choice

If $I$ is the centre of a circle inscribed in a triangle $ABC$, then $|\overline{BC}|\overline{IA}+|\overline{CA}|\overline{IB}+|\overline{AB}|\overline{IC}$ is

  1. $\overline{0}$
  2. $\overline{IA}+\overline{IB}+\overline{IC}$
  3. $\displaystyle \frac{\overline{IA}+\overline{IB}+\overline{IC}}{3}$
  4. $\displaystyle \frac{\overline{IA}+\overline{IB}+\overline{IC}}{2}$
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A Correct answer
Explanation

The incenter I of a triangle satisfies the vector equation a*IA + b*IB + c*IC = 0, where a, b, and c are the lengths of the sides opposite to vertices A, B, and C respectively. Thus, |BC|IA + |CA|IB + |AB|IC = 0.

AI explanation

Let the position vectors of vertices A, B, and C be a, b, and c. The incenter I has the position vector (|BC|a + |CA|b + |AB|c) divided by the perimeter P. Multiplying I by the perimeter P gives P times the vector OI equals |BC|a + |CA|b + |AB|c. Expanding the given expression |BC|IA + |CA|IB + |AB|IC gives |BC|(a - I) + |CA|(b - I) + |AB|(c - I). This simplifies to |BC|a + |CA|b + |AB|c - P(OI), which equals the zero vector. The correct answer is the zero vector.

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