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-

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

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

Answered by mbertrand658 last updated on 16/Aug/20

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

Commented by ZiYangLee last updated on 19/Aug/20

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

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