STEP II 1995 question 7 solution

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STEP II 1995 question 7 solution

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TSR Wiki > Study Help > Exams and Qualifications > A Levels > STEP > STEP II 1995 question 7 solution

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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