The-sides-AB-BC-CA-of-a-triangleABC-have-3-4-and-5-interior-points-respectively-on-them-The-total-number-of-triangles-that-can-be-constructed-by-using-these-points-as-vertices-is-

Question Number 108407 by ZiYangLee last updated on 16/Aug/20

$$\mathrm{The}\:\mathrm{sides}\:\mathrm{AB},\:\mathrm{BC},\:\mathrm{CA}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangleABC} \\ $$$$\mathrm{have}\:\mathrm{3},\:\mathrm{4}\:\mathrm{and}\:\mathrm{5}\:\mathrm{interior}\:\mathrm{points}\:\mathrm{respectively} \\ $$$$\mathrm{on}\:\mathrm{them}.\:\mathrm{The}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{triangles} \\ $$$$\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{constructed}\:\mathrm{by}\:\mathrm{using}\:\mathrm{these} \\ $$$$\mathrm{points}\:\mathrm{as}\:\mathrm{vertices}\:\mathrm{is} \\ $$

Answered by mbertrand658 last updated on 16/Aug/20

Because all the points are bounded by a triangle, it follows that any combination of interior points forms a triangle as long as they do not all lie on one side. There are 3 × 4 × 5 = 60 unique triangles when choosing one point from each side. Choosing two points on side AB yields 3C2 × 9 = ((3!)/((3 − 2)! × 2!)) × 9 = 3 × 9 = 27 new triangles. Choosing two points on side BC yields 4C2 × 8 = ((4!)/((4 − 2)! × 2!)) × 8 = 6 × 8 = 48 new triangles. Choosing two points on side AC yields 5C2 × 7 = ((5!)/((5 − 2)! × 2!)) × 7 = 10 × 7 = 70 new triangles. Therefore, the total number of triangles that can be constructed using the given points as vertices is 60 + 27 + 48 + 70 = 205 triangles.

$$\mathrm{Because}\:\mathrm{all}\:\mathrm{the}\:\mathrm{points}\:\mathrm{are}\:\mathrm{bounded}\:\mathrm{by}\:\mathrm{a} \\ $$$$\mathrm{triangle},\:\mathrm{it}\:\mathrm{follows}\:\mathrm{that}\:\mathrm{any}\:\mathrm{combination} \\ $$$$\mathrm{of}\:\mathrm{interior}\:\mathrm{points}\:\mathrm{forms}\:\mathrm{a}\:\mathrm{triangle}\:\mathrm{as}\:\mathrm{long} \\ $$$$\mathrm{as}\:\mathrm{they}\:\mathrm{do}\:\mathrm{not}\:\mathrm{all}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}. \\ $$$$\: \\ $$$$\mathrm{There}\:\mathrm{are}\:\mathrm{3}\:×\:\mathrm{4}\:×\:\mathrm{5}\:=\:\mathrm{60}\:\mathrm{unique}\:\mathrm{triangles} \\ $$$$\mathrm{when}\:\mathrm{choosing}\:\mathrm{one}\:\mathrm{point}\:\mathrm{from}\:\mathrm{each}\:\mathrm{side}. \\ $$$$\: \\ $$$$\mathrm{Choosing}\:\mathrm{two}\:\mathrm{points}\:\mathrm{on}\:\mathrm{side}\:\mathrm{AB}\:\mathrm{yields} \\ $$$$\mathrm{3C2}\:×\:\mathrm{9}\:=\:\frac{\mathrm{3}!}{\left(\mathrm{3}\:−\:\mathrm{2}\right)!\:×\:\mathrm{2}!}\:×\:\mathrm{9}\:=\:\mathrm{3}\:×\:\mathrm{9}\:=\:\mathrm{27} \\ $$$$\mathrm{new}\:\mathrm{triangles}. \\ $$$$\: \\ $$$$\mathrm{Choosing}\:\mathrm{two}\:\mathrm{points}\:\mathrm{on}\:\mathrm{side}\:\mathrm{BC}\:\mathrm{yields} \\ $$$$\mathrm{4C2}\:×\:\mathrm{8}\:=\:\frac{\mathrm{4}!}{\left(\mathrm{4}\:−\:\mathrm{2}\right)!\:×\:\mathrm{2}!}\:×\:\mathrm{8}\:=\:\mathrm{6}\:×\:\mathrm{8}\:=\:\mathrm{48} \\ $$$$\mathrm{new}\:\mathrm{triangles}. \\ $$$$\: \\ $$$$\mathrm{Choosing}\:\mathrm{two}\:\mathrm{points}\:\mathrm{on}\:\mathrm{side}\:\mathrm{AC}\:\mathrm{yields} \\ $$$$\mathrm{5C2}\:×\:\mathrm{7}\:=\:\frac{\mathrm{5}!}{\left(\mathrm{5}\:−\:\mathrm{2}\right)!\:×\:\mathrm{2}!}\:×\:\mathrm{7}\:=\:\mathrm{10}\:×\:\mathrm{7}\:=\:\mathrm{70} \\ $$$$\mathrm{new}\:\mathrm{triangles}. \\ $$$$\: \\ $$$$\mathrm{Therefore},\:\mathrm{the}\:\mathrm{total}\:\mathrm{number}\:\mathrm{of}\:\mathrm{triangles} \\ $$$$\mathrm{that}\:\mathrm{can}\:\mathrm{be}\:\mathrm{constructed}\:\mathrm{using}\:\mathrm{the}\:\mathrm{given} \\ $$$$\mathrm{points}\:\mathrm{as}\:\mathrm{vertices}\:\mathrm{is} \\ $$$$\mathrm{60}\:+\:\mathrm{27}\:+\:\mathrm{48}\:+\:\mathrm{70}\:=\:\mathrm{205}\:\mathrm{triangles}. \\ $$

Commented by ZiYangLee last updated on 19/Aug/20

$$\mathrm{Wow}\:\mathrm{Nice}! \\ $$

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