Triangle Area

Geometry Level pending

Two circles are internally tangent at point A A . The larger circle has radius 3 and the smaller circle has radius 1. Find the maximum area of a triangle which has the point of tangency as a vertex and one vertex on each of the small and large circles.

2 π 2\pi 27 4 \frac{27}{4} 3 3 4 \frac{3\sqrt{3}}{4} 9 3 4 \frac{9\sqrt{3}}{4}

Brandon Chen by Brandon Chen , 30, USA

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1 solution

Ujjwal Rane
Aug 20, 2016

Max Area Triangle on Tangent Circles Max Area Triangle on Tangent Circles

Let OPT be the desired triangle. Select the coordinate frame as shown.

Suppose the base OP is given. Then to maximize the area of the triangle its height MT must be maximized. This can happen when the Tangent to the small circle at T is parallel to the base.

If θ \theta is the inclination of base OP, height will be given by h = M T = M A + A T = sin ⁡ θ + 1 h = MT = MA + AT = \sin \theta + 1

The base OP will be 2 × O B cos ⁡ θ = 6 cos ⁡ θ 2 \times OB \cos \theta = 6 \cos \theta

Area A = 1 2 O P × M T = 1 2 ( sin ⁡ θ + 1 ) ( 6 cos ⁡ θ ) A = \frac{1}{2} OP \times MT = \frac{1}{2} (\sin \theta + 1)(6 \cos \theta)

Differentiating to find the maximum: yields θ = 30 ° \theta = 30° Giving maximum area = 1 2 ( sin ⁡ θ + 1 ) ( 6 cos ⁡ θ ) = 1 2 3 2 ( 6 3 2 ) = 9 3 4 \frac{1}{2} (\sin \theta + 1)(6 \cos \theta) = \frac{1}{2} \frac{3}{2}(\frac{6\sqrt{3}}{2}) = \frac{9 \sqrt{3}}{4}

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