Multiple choice

If $O=(0,0),A=(1,0)$ and $B=\left(\dfrac {1}{2},\dfrac {\sqrt {3}}{2}\right)$ then centre of the circle for which the lines $OA,OB$ and $AB$ are tangents is

  1. $\left(\dfrac {1}{2},\dfrac {1}{2\sqrt {3}}\right)$
  2. $\left(\dfrac {1}{2},\dfrac {1}{1\sqrt {3}}\right)$
  3. $\left(\dfrac {1}{\sqrt {3}},\dfrac {1}{\sqrt {2}}\right)$
  4. $\left(\dfrac {1}{2\sqrt {3}},\dfrac {1}{\sqrt {2}}\right)$
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A Correct answer
Explanation

The lines are OA (y=0), OB (y = sqrt(3)x), and AB (y - 0 = (sqrt(3)/2 - 0)/(1/2 - 1) * (x - 1), which is y = -sqrt(3)(x-1)). The center of the incircle of triangle OAB is needed. The vertices are (0,0), (1,0), (1/2, sqrt(3)/2). This is an equilateral triangle with side length 1. The incenter is (1/2, (sqrt(3)/2)/3) = (1/2, 1/(2*sqrt(3))).

AI explanation

Calculating the distances between the given points O(0,0), A(1,0), and B(1/2, sqrt(3)/2) shows that OA = 1, OB = 1, and AB = 1, meaning triangle OAB is equilateral. Because the required circle is tangent to all three sides, its center is the incentre of this equilateral triangle. The incentre coordinates are the average of the vertices, which calculates to (1/2, sqrt(3)/6) or (1/2, 1/(2*sqrt(3))).

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