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Question Number 76112 by Maclaurin Stickker last updated on 23/Dec/19
If the sides of a triangle are consecutive integers and the maximum angle is twice the minimum, determine the sides of the triangle.
$${If}\:{the}\:{sides}\:{of}\:{a}\:{triangle}\:{are}\:{consecutive} \\ $$$${integers}\:{and}\:{the}\:{maximum}\:{angle} \\ $$$${is}\:{twice}\:{the}\:{minimum},\:{determine} \\ $$$${the}\:{sides}\:{of}\:{the}\:{triangle}. \\ $$
Answered by mr W last updated on 24/Dec/19
say the sides are n−1,n,n+1 cos θ_(max) =(((n−1)^2 +n^2 −(n+1)^2 )/(2(n−1)n))=((n−4)/(2(n−1))) cos θ_(min) =(((n+1)^2 +n^2 −(n−1)^2 )/(2(n+1)n))=((n+4)/(2(n+1))) θ_(max) =2θ_(min) cos θ_(max) =2 cos^2 θ_(min) −1 ((n−4)/(2(n−1)))=2×(((n+4)^2 )/(4(n+1)^2 ))−1 ((n−4)/(n−1))=((4n+14−n^2 )/(n^2 +2n+1)) 2n^3 −7n^2 −17n+10=0 (n+2)(n−5)(2n−1)=0 ⇒n=5 ⇒sides are 4, 5 and 6.
$${say}\:{the}\:{sides}\:{are}\:{n}−\mathrm{1},{n},{n}+\mathrm{1} \\ $$$$\mathrm{cos}\:\theta_{{max}} =\frac{\left({n}−\mathrm{1}\right)^{\mathrm{2}} +{n}^{\mathrm{2}} −\left({n}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\left({n}−\mathrm{1}\right){n}}=\frac{{n}−\mathrm{4}}{\mathrm{2}\left({n}−\mathrm{1}\right)} \\ $$$$\mathrm{cos}\:\theta_{{min}} =\frac{\left({n}+\mathrm{1}\right)^{\mathrm{2}} +{n}^{\mathrm{2}} −\left({n}−\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\left({n}+\mathrm{1}\right){n}}=\frac{{n}+\mathrm{4}}{\mathrm{2}\left({n}+\mathrm{1}\right)} \\ $$$$\theta_{{max}} =\mathrm{2}\theta_{{min}} \\ $$$$\mathrm{cos}\:\theta_{{max}} =\mathrm{2}\:\mathrm{cos}^{\mathrm{2}} \:\theta_{{min}} −\mathrm{1} \\ $$$$\frac{{n}−\mathrm{4}}{\mathrm{2}\left({n}−\mathrm{1}\right)}=\mathrm{2}×\frac{\left({n}+\mathrm{4}\right)^{\mathrm{2}} }{\mathrm{4}\left({n}+\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{1} \\ $$$$\frac{{n}−\mathrm{4}}{{n}−\mathrm{1}}=\frac{\mathrm{4}{n}+\mathrm{14}−{n}^{\mathrm{2}} }{{n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}} \\ $$$$\mathrm{2}{n}^{\mathrm{3}} −\mathrm{7}{n}^{\mathrm{2}} −\mathrm{17}{n}+\mathrm{10}=\mathrm{0} \\ $$$$\left({n}+\mathrm{2}\right)\left({n}−\mathrm{5}\right)\left(\mathrm{2}{n}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{n}=\mathrm{5} \\ $$$$\Rightarrow{sides}\:{are}\:\mathrm{4},\:\mathrm{5}\:{and}\:\mathrm{6}. \\ $$
Commented by Maclaurin Stickker last updated on 24/Dec/19
I did by the same way, but I wasn′t sure. Thank you for the great answer.
$${I}\:{did}\:{by}\:{the}\:{same}\:{way},\:{but}\:{I}\:{wasn}'{t} \\ $$$${sure}.\:{Thank}\:{you}\:{for}\:{the}\:{great} \\ $$$${answer}. \\ $$

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