STEP II 1995 question 7 solution - The Student Room
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STEP II 1995 question 7 solution

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Area of  \triangle AIB = \frac{1}{2} cr

Area of  \triangle BIC = \frac{1}{2} ar

Area of  \triangle CIA = \frac{1}{2} br

(Due to  \frac{1}{2} \times \text{ base } \times \text{ height } ).

Hence:

Area of  \triangle ABC = \frac{1}{2} r \left[ a + b + c \right] = sr .

 \triangle = \frac{1}{2} bc \sin{ \alpha } \implies \triangle ^{2} = \frac{1}{4} b^{2} c^{2} \sin ^{2} { \alpha } = \frac{1}{4} b^{2} c^{2} \left( 1 - \cos ^{2} { \alpha } \right)

Hence:

 \triangle ^{2} = \frac{1}{16} \left( 4b^{2} c^{2} - 4b^{2} c^{2} \cos ^{2} { \alpha } \right) = \frac{1}{16} \left( 4b^{2} c^{2} - \left[ 2bc \cos{ \alpha } \right] ^{2} \right) .

 a^{2} = b^{2} + c^{2} - 2bc \cos{ \alpha } \implies 2bc \cos{ \alpha } = b^{2} + c^{2} - a^{2} .

Hence:

 \triangle ^{2} = \frac{1}{16} \left( 4b^{2} c^{2} - \left[ b^{2} + c^{2} - a^{2} \right] ^{2} \right) = \frac{1}{16} \left( \left[ 2bc - b^{2} - c^{2} + a^{2} \right] \left[ 2bc + b^{2} + c^{2} - a^{2} \right] \right)

Hence:

 \triangle ^{2} = \frac{1}{16} \left( \left[ a^{2} - \left( b - c \right) ^{2} \right] \left[ \left( b + c \right) ^{2} - a^{2} \right] \right) .

Hence:

 \triangle ^{2} = \frac{1}{16} \left( \left[ \left( a - (b - c) \right) \left( a + (b - c) \right) \right] \left[ \left( (b + c) - a \right) \left( (b + c) + a \right) \right] \right) .

Now consider the following:

 2s = a + b + c

 2(s - a) = b + c - a

 2(s - b) = a + c - b

 2(s - c) = a + b - c

Hence:

 \triangle ^{2} = \frac{1}{16} \left( \left[ \left( 2(s - b) \right) \left( 2(s - c) \right) \right] \left[ \left( 2(s - a) \right) \left( 2s \right) \right] \right) = s(s - a)(s - b)(s - c)

Hence:

 \triangle = \sqrt{ s(s - a)(s - b)(s - c) } .

Consider that the sphere will rest in the incircle of the triangle, and hence will satisfy:

 R^{2} - r^{2} = d^{2}

(Where "d" is the distance to be calculated. [A diagram may help visualisation of this property].).

Consider also:

 \triangle = sr = \sqrt{ s(s - a)(s - b)(s - c) } \implies r = \sqrt{ \frac{ (s - a)(s - b)(s - c) }{s} } .

Hence:

 R^{2} - r^{2} = R^{2} - \frac{ (s - a)(s - b)(s - c) }{s} = d^{2} .

Hence:

 d = \sqrt{ R^{2} - \frac{ (s - a)(s - b)(s - c) }{s} } .

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