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Question Number 34760 by Tinkutara last updated on 10/May/18

Answered by Rasheed.Sindhi last updated on 11/May/18

Answer in more technical style •At least two sides are equal and each side is not greater than 4. •Sum of any of two sides>Third side. So the triangles with integer sides are (1,1,c) c≤4 ∧ c<1+1=2⇒c<2 c=1 (1 triangle) (2,2,c) c≤4 ∧ c<2+2=4⇒c<4 c=1,2,3 (3 triangles) (3,3,c) c≤4 ∧ c<3+3=6⇒c≤4 c=1,2,3,4 (4 triangles) (4,4,c) c≤4 ∧ c<4+4=8⇒c≤4 c=1,2,3,4 (4 triangles) Number of required triangles =1+3+4+4=12 triangles.

$$\mathrm{Answer}\:\mathrm{in}\:\mathrm{more}\:\mathrm{technical}\:\mathrm{style} \\ $$$$\bullet\mathrm{At}\:\mathrm{least}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{and}\:\mathrm{each} \\ $$$$\mathrm{side}\:\mathrm{is}\:\mathrm{not}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{4}. \\ $$$$\bullet\mathrm{Sum}\:\mathrm{of}\:\mathrm{any}\:\mathrm{of}\:\mathrm{two}\:\mathrm{sides}>\mathrm{Third}\:\mathrm{side}. \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{triangles}\:\mathrm{with}\:\mathrm{integer}\:\mathrm{sides}\:\mathrm{are} \\ $$$$\left(\mathrm{1},\mathrm{1},\mathrm{c}\right) \\ $$$$\:\:\:\mathrm{c}\leqslant\mathrm{4}\:\wedge\:\mathrm{c}<\mathrm{1}+\mathrm{1}=\mathrm{2}\Rightarrow\mathrm{c}<\mathrm{2} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}=\mathrm{1}\:\left(\mathrm{1}\:\mathrm{triangle}\right) \\ $$$$\left(\mathrm{2},\mathrm{2},\mathrm{c}\right) \\ $$$$\:\:\:\:\mathrm{c}\leqslant\mathrm{4}\:\wedge\:\mathrm{c}<\mathrm{2}+\mathrm{2}=\mathrm{4}\Rightarrow\mathrm{c}<\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}=\mathrm{1},\mathrm{2},\mathrm{3}\:\left(\mathrm{3}\:\mathrm{triangles}\right) \\ $$$$\left(\mathrm{3},\mathrm{3},\mathrm{c}\right) \\ $$$$\:\:\mathrm{c}\leqslant\mathrm{4}\:\wedge\:\mathrm{c}<\mathrm{3}+\mathrm{3}=\mathrm{6}\Rightarrow\mathrm{c}\leqslant\mathrm{4} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\:\:\left(\mathrm{4}\:\mathrm{triangles}\right) \\ $$$$\left(\mathrm{4},\mathrm{4},\mathrm{c}\right) \\ $$$$\:\:\mathrm{c}\leqslant\mathrm{4}\:\wedge\:\mathrm{c}<\mathrm{4}+\mathrm{4}=\mathrm{8}\Rightarrow\mathrm{c}\leqslant\mathrm{4}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}=\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\:\:\left(\mathrm{4}\:\mathrm{triangles}\right) \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{required}\:\mathrm{triangles} \\ $$$$\:\:\:\:\:\:\:\:\:=\mathrm{1}+\mathrm{3}+\mathrm{4}+\mathrm{4}=\mathrm{12}\:\mathrm{triangles}. \\ $$$$ \\ $$

Commented by Tinkutara last updated on 11/May/18

Thank you very much Sir! I got the answer. ��������

Answered by Rasheed.Sindhi last updated on 21/May/18

Solution by hint given in question. •At least two sides are equal and any of the sides is not greater than 4. •All sides are positive integer. •Sum of any two sides>Third side. Let p≤q≤r , p,q,r being sides of the triangle. • For a fix r , r+1≤p+q≤2r _(−) r maybe 1,2,3,4 r=4, 5≤p+q≤8 p+q=5 : (1,4,4) p+q=6 : (2,4,4) (3,3,4) p+q=7 : (3,4,4) p+q=8 : (4,4,4) −−5 triangles r=3, 4≤p+q≤6 p+q=4 : (1,3,3)(2,2,3) p+q=5 : (2,3,3) p+q=6 : (3,3,3) −−4 triangles r=2, 3≤p+q≤4 p+q=3 : (1,2,2) p+q=4 : (2,2,2) −−2 triangles r=1 , 2≤p+q≤2 p+q=2 : (1,1,1) −−1 triangle Number of triplets(triangles)= 5+4+2+1=12

$$\mathrm{Solution}\:\mathrm{by}\:\mathrm{hint}\:\mathrm{given}\:\mathrm{in}\:\mathrm{question}. \\ $$$$\:\bullet\mathrm{At}\:\mathrm{least}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{and}\:\mathrm{any} \\ $$$$\:\:\mathrm{of}\:\:\mathrm{the}\:\:\mathrm{sides}\:\mathrm{is}\:\mathrm{not}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{4}. \\ $$$$\bullet\mathrm{All}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{positive}\:\mathrm{integer}. \\ $$$$\bullet\mathrm{Sum}\:\mathrm{of}\:\mathrm{any}\:\mathrm{two}\:\mathrm{sides}>\mathrm{Third}\:\mathrm{side}. \\ $$$$ \\ $$$$\:\mathrm{Let}\:\mathrm{p}\leqslant\mathrm{q}\leqslant\mathrm{r}\:,\:\mathrm{p},\mathrm{q},\mathrm{r}\:\mathrm{being}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{the}\:\mathrm{triangle}. \\ $$$$\:\:\:\bullet\:\mathrm{For}\:\mathrm{a}\:\mathrm{fix}\:\:\mathrm{r}\:,\:\:\underset{−} {\:\:\:\:\:\:\mathrm{r}+\mathrm{1}\leqslant\mathrm{p}+\mathrm{q}\leqslant\mathrm{2r}\:\:\:\:} \\ $$$$\:\:\:\:\:\:\:\mathrm{r}\:\mathrm{maybe}\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4} \\ $$$$\mathrm{r}=\mathrm{4},\:\:\:\:\:\:\mathrm{5}\leqslant\mathrm{p}+\mathrm{q}\leqslant\mathrm{8} \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{5}\::\:\left(\mathrm{1},\mathrm{4},\mathrm{4}\right)\: \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{6}\::\:\left(\mathrm{2},\mathrm{4},\mathrm{4}\right)\:\left(\mathrm{3},\mathrm{3},\mathrm{4}\right)\: \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{7}\::\:\left(\mathrm{3},\mathrm{4},\mathrm{4}\right)\: \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{8}\::\:\left(\mathrm{4},\mathrm{4},\mathrm{4}\right)\:−−\mathrm{5}\:\mathrm{triangles} \\ $$$$\mathrm{r}=\mathrm{3},\:\:\mathrm{4}\leqslant\mathrm{p}+\mathrm{q}\leqslant\mathrm{6} \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{4}\::\:\left(\mathrm{1},\mathrm{3},\mathrm{3}\right)\left(\mathrm{2},\mathrm{2},\mathrm{3}\right) \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{5}\::\:\left(\mathrm{2},\mathrm{3},\mathrm{3}\right) \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{6}\::\:\left(\mathrm{3},\mathrm{3},\mathrm{3}\right)\:−−\mathrm{4}\:\mathrm{triangles} \\ $$$$\mathrm{r}=\mathrm{2},\:\:\mathrm{3}\leqslant\mathrm{p}+\mathrm{q}\leqslant\mathrm{4} \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{3}\::\:\left(\mathrm{1},\mathrm{2},\mathrm{2}\right) \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{4}\::\:\left(\mathrm{2},\mathrm{2},\mathrm{2}\right)\:−−\mathrm{2}\:\mathrm{triangles} \\ $$$$\mathrm{r}=\mathrm{1}\:,\:\:\mathrm{2}\leqslant\mathrm{p}+\mathrm{q}\leqslant\mathrm{2} \\ $$$$\mathrm{p}+\mathrm{q}=\mathrm{2}\::\:\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)\:−−\mathrm{1}\:\mathrm{triangle} \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{triplets}\left(\mathrm{triangles}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{5}+\mathrm{4}+\mathrm{2}+\mathrm{1}=\mathrm{12} \\ $$

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