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How-many-triangles-can-be-formed-from-non-adjacent-vertices-of-a-regular-polygon-that-it-angle-is-177-

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Question Number 180773 by Acem last updated on 17/Nov/22
How many triangles can be formed from non−adjacent vertices of a regular polygon that it angle is 177^( °) ?
$${How}\:{many}\:{triangles}\:{can}\:{be}\:{formed}\:{from} \\ $$$$\:{non}−{adjacent}\:{vertices}\:{of}\:{a}\:{regular}\:{polygon} \\ $$$$\:{that}\:{it}\:{angle}\:{is}\:\mathrm{177}^{\:°} \:? \\ $$
Answered by mr W last updated on 17/Nov/22
n=((360)/(180−177))=120 vertices we can form (1/3)×n×(C_2 ^(n−3) −(n−4)) =((n(n−4)(n−5))/6) =266 800 triangles
$${n}=\frac{\mathrm{360}}{\mathrm{180}−\mathrm{177}}=\mathrm{120}\:{vertices} \\ $$$${we}\:{can}\:{form} \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}×{n}×\left({C}_{\mathrm{2}} ^{{n}−\mathrm{3}} −\left({n}−\mathrm{4}\right)\right) \\ $$$$=\frac{{n}\left({n}−\mathrm{4}\right)\left({n}−\mathrm{5}\right)}{\mathrm{6}} \\ $$$$=\mathrm{266}\:\mathrm{800}\:{triangles} \\ $$
Commented by Acem last updated on 17/Nov/22
Morning Sir!, the number is rather close to the correct answer, if we apply the formula above to a hexagon or otherwise we wouldn′t get the real answer.
$${Morning}\:{Sir}!,\:{the}\:{number}\:{is}\:{rather}\:{close}\:{to} \\ $$$$\:{the}\:{correct}\:{answer},\:{if}\:{we}\:{apply}\:{the}\:{formula}\:{above} \\ $$$$\:{to}\:{a}\:{hexagon}\:{or}\:{otherwise}\:{we}\:{wouldn}'{t}\:{get}\:{the} \\ $$$$\:{real}\:{answer}. \\ $$
Commented by mr W last updated on 17/Nov/22
i see my mistake now. i′ll fix it. the vertrices of triangle should not be adjacent vertrices of the polygon. i didn′t follow this correctly.
$${i}\:{see}\:{my}\:{mistake}\:{now}.\:{i}'{ll}\:{fix}\:{it}. \\ $$$${the}\:{vertrices}\:{of}\:{triangle}\:{should}\:{not} \\ $$$${be}\:{adjacent}\:{vertrices}\:{of}\:{the}\:{polygon}. \\ $$$${i}\:{didn}'{t}\:{follow}\:{this}\:{correctly}. \\ $$
Commented by Acem last updated on 17/Nov/22
Yes Sir, you will do it! thank you Note: The number ∈ ]265 349, 266 801[
$${Yes}\:{Sir},\:{you}\:{will}\:{do}\:{it}!\:{thank}\:{you} \\ $$$$ \\ $$$$\left.\:{Note}:\:{The}\:{number}\:\in\:\right]\mathrm{265}\:\mathrm{349},\:\mathrm{266}\:\mathrm{801}\left[\right. \\ $$
Commented by mr W last updated on 17/Nov/22
i got 266800. please check my fixed solution.
$${i}\:{got}\:\mathrm{266800}.\:{please}\:{check}\:{my}\:{fixed}\: \\ $$$${solution}. \\ $$
Commented by Acem last updated on 17/Nov/22
Shakehands! Whatever Regular Polygon { ((C_3 ^( n) − n(n−3))),(( ((n(n−4)(n−5))/6))) :} ∗ If we want to count only the triangles that are formed from adjacent vertices, apply n(n−3)
$${Shakehands}! \\ $$$${Whatever}\:{Regular}\:{Polygon\begin{cases}{{C}_{\mathrm{3}} ^{\:{n}} \:−\:{n}\left({n}−\mathrm{3}\right)}\\{\:\frac{{n}\left({n}−\mathrm{4}\right)\left({n}−\mathrm{5}\right)}{\mathrm{6}}}\end{cases}} \\ $$$$ \\ $$$$\ast\:{If}\:{we}\:{want}\:{to}\:{count}\:{only}\:{the}\:{triangles}\:{that}\:{are} \\ $$$$\:{formed}\:{from}\:{adjacent}\:{vertices},\:{apply}\:\boldsymbol{{n}}\left(\boldsymbol{{n}}−\mathrm{3}\right) \\ $$$$ \\ $$

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