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Question Number 516 by 112358 last updated on 25/Jan/15
Find all triangles with consecutive  integer sides and having an angle twice  another angle.
$${Find}\:{all}\:{triangles}\:{with}\:{consecutive} \\ $$$${integer}\:{sides}\:{and}\:{having}\:{an}\:{angle}\:{twice} \\ $$$${another}\:{angle}. \\ $$
Commented by prakash jain last updated on 22/Jan/15
a=n  b=n+1  c=n+2  cos A=(((n+5))/(2(n+2)))  cos B=((n^2 +2n+3)/(2n(n+2)))  cos C=(((nāˆ’3))/(2n))  B=2A, C=2B, C=2A  All equation give no integer solution  except n=1, which does not form  a triangle.  So the given question does not have a   solution.
$${a}={n} \\ $$$${b}={n}+\mathrm{1} \\ $$$${c}={n}+\mathrm{2} \\ $$$$\mathrm{cos}\:{A}=\frac{\left({n}+\mathrm{5}\right)}{\mathrm{2}\left({n}+\mathrm{2}\right)} \\ $$$$\mathrm{cos}\:{B}=\frac{{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{3}}{\mathrm{2}{n}\left({n}+\mathrm{2}\right)} \\ $$$$\mathrm{cos}\:{C}=\frac{\left({n}āˆ’\mathrm{3}\right)}{\mathrm{2}{n}} \\ $$$${B}=\mathrm{2}{A},\:{C}=\mathrm{2}{B},\:{C}=\mathrm{2}{A} \\ $$$$\mathrm{All}\:\mathrm{equation}\:\mathrm{give}\:\mathrm{no}\:\mathrm{integer}\:\mathrm{solution} \\ $$$$\mathrm{except}\:{n}=\mathrm{1},\:\mathrm{which}\:\mathrm{does}\:\mathrm{not}\:\mathrm{form} \\ $$$$\mathrm{a}\:\mathrm{triangle}. \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{given}\:\mathrm{question}\:\mathrm{does}\:\mathrm{not}\:\mathrm{have}\:\mathrm{a}\: \\ $$$$\mathrm{solution}. \\ $$

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