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Question Number 34870 by Tinkutara last updated on 12/May/18

Commented by Tinkutara last updated on 15/May/18

please help

Commented by Tinkutara last updated on 16/May/18

Anyone please see it.

Commented by Tinkutara last updated on 16/May/18

Commented by Tinkutara last updated on 16/May/18

Actually this was the whole comprehension.

Answered by Rasheed.Sindhi last updated on 20/May/18

Pattern approach Using strategy of Q#34760,we can determine ′number of triangles′ for various values of c:(In the following N denotes number of triangle.) ((c,N),(1,3),(2,(12)),(3,(27)),(4,(48)),(5,(75)),(⋮,⋮) ) Number of triangles(N) make following sequence : 3,12,27,48,75,108,.... We can easily determine general term: [(c,1,2,3,4,5,6,(..)),(N,3,(12),(27),(48),(75),(108),(..)),(,(3.1),(3.4),(3.9),(3.16),(3.25),(3.36),(..)),(,(3.1^2 ),(3.2^2 ),(3.3^2 ),(3.4^2 ),(3.5^2 ),(3.6^2 ),(..)) ] Hence general term is 3c^2 Number of triangles : 3c^2

$$\mathrm{Pattern}\:\mathrm{approach} \\ $$$$\mathrm{Using}\:\mathrm{strategy}\:\mathrm{of}\:\mathrm{Q}#\mathrm{34760},\mathrm{we}\:\mathrm{can} \\ $$$$\mathrm{determine}\:'\mathrm{number}\:\mathrm{of}\:\mathrm{triangles}'\:\mathrm{for} \\ $$$$\mathrm{various}\:\mathrm{values}\:\mathrm{of}\:\mathrm{c}:\left(\mathrm{In}\:\mathrm{the}\:\mathrm{following}\:\mathrm{N}\right. \\ $$$$\left.\mathrm{denotes}\:\mathrm{number}\:\mathrm{of}\:\mathrm{triangle}.\right) \\ $$$$\begin{pmatrix}{\mathrm{c}}&{\mathrm{N}}\\{\mathrm{1}}&{\mathrm{3}}\\{\mathrm{2}}&{\mathrm{12}}\\{\mathrm{3}}&{\mathrm{27}}\\{\mathrm{4}}&{\mathrm{48}}\\{\mathrm{5}}&{\mathrm{75}}\\{\vdots}&{\vdots}\end{pmatrix} \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{triangles}\left(\mathrm{N}\right)\:\mathrm{make}\:\mathrm{following}\: \\ $$$$\:\mathrm{sequence}\:: \\ $$$$\:\mathrm{3},\mathrm{12},\mathrm{27},\mathrm{48},\mathrm{75},\mathrm{108},.... \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{easily}\:\mathrm{determine}\:\mathrm{general}\:\mathrm{term}: \\ $$$$\begin{bmatrix}{\mathrm{c}}&{\mathrm{1}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{4}}&{\mathrm{5}}&{\mathrm{6}}&{..}\\{\mathrm{N}}&{\mathrm{3}}&{\mathrm{12}}&{\mathrm{27}}&{\mathrm{48}}&{\mathrm{75}}&{\mathrm{108}}&{..}\\{}&{\mathrm{3}.\mathrm{1}}&{\mathrm{3}.\mathrm{4}}&{\mathrm{3}.\mathrm{9}}&{\mathrm{3}.\mathrm{16}}&{\mathrm{3}.\mathrm{25}}&{\mathrm{3}.\mathrm{36}}&{..}\\{}&{\mathrm{3}.\mathrm{1}^{\mathrm{2}} }&{\mathrm{3}.\mathrm{2}^{\mathrm{2}} }&{\mathrm{3}.\mathrm{3}^{\mathrm{2}} }&{\mathrm{3}.\mathrm{4}^{\mathrm{2}} }&{\mathrm{3}.\mathrm{5}^{\mathrm{2}} }&{\mathrm{3}.\mathrm{6}^{\mathrm{2}} }&{..}\end{bmatrix} \\ $$$$\mathrm{Hence}\:\mathrm{general}\:\mathrm{term}\:\mathrm{is}\:\mathrm{3c}^{\mathrm{2}} \\ $$$$\mathrm{Number}\:\mathrm{of}\:\mathrm{triangles}\::\:\mathrm{3c}^{\mathrm{2}} \\ $$

Answered by Tinkutara last updated on 22/May/18

Commented by Tinkutara last updated on 22/May/18

This was the solution. Can you please elaborate it? I can't understand it.

Answered by Rasheed.Sindhi last updated on 24/May/18

A try to elaborate the answer given with question in the book. (See ans by Mr Tinkutara here) (a,a,b):A triangle, (i) with integer sides and (ii) at least two sides are equal with the possibility a=b. By triangle inequality, a+a>b⇒ 2a−1≥b Also b≥1 So, 2a−1≥b≥1...............A 1≤a,b≤2c (Given).......B Case-1: 1≤a≤c 1≤b≤2c ∧ b≤2a−1 ⇒1≤b≤2a−1 ∗ i-e there are 2a−1 choices for b when a is fixed. For a=c → 2c−1 choices for b For a=c−1→ 2(c−1)−1=2c−3 choices for b For a=c−2→ 2(c−2)−1=2c−5 choices for b ...... ..... For a=2→ 2(2)−1=3 choices for b For a=1→ 2(1)−1=1 choices for b Total choices for this case (2c−1)+(2c−3)+(2c+5)...+3+1_(−−−−−−−−−−−−c terms−−−−−−) =c^2 Notice that above is the sum of c consecutive odd numbers.(Treate it as AP) Case-2: c+1≤a≤2c (c values) b≤2c ∧ 1≤b≤2a−1 ⇒1≤b≤2c ∗∗ i-e b takes 2c values. ∵ a takes c values from c+1 to 2c and for each of these values b takes 2c vales ∴ (a,a,b) takes c.2c=2c^2 values Case-1 & Case-2: c^2 +2c^2 =3c^2 Number of (a,a,b)=3c^2

$$\mathrm{A}\:\mathrm{try}\:\mathrm{to}\:\mathrm{elaborate}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{given} \\ $$$$\mathrm{with}\:\mathrm{question}\:\mathrm{in}\:\mathrm{the}\:\mathrm{book}. \\ $$$$\left(\mathrm{See}\:\mathrm{ans}\:\mathrm{by}\:\mathrm{Mr}\:\mathrm{Tinkutara}\:\mathrm{here}\right) \\ $$$$\left({a},{a},{b}\right):\mathrm{A}\:\mathrm{triangle},\:\left(\mathrm{i}\right)\:\mathrm{with}\:\mathrm{integer}\:\mathrm{sides} \\ $$$$\mathrm{and}\:\left(\mathrm{ii}\right)\:\mathrm{at}\:\mathrm{least}\:\mathrm{two}\:\mathrm{sides}\:\mathrm{are}\:\mathrm{equal}\:\mathrm{with} \\ $$$$\mathrm{the}\:\mathrm{possibility}\:{a}={b}. \\ $$$$\mathrm{By}\:\mathrm{triangle}\:\mathrm{inequality}, \\ $$$$\:\:\:\:\:\:\:{a}+{a}>{b}\Rightarrow\:\:\mathrm{2}{a}−\mathrm{1}\geqslant{b} \\ $$$$\mathrm{Also}\:\:\:\:{b}\geqslant\mathrm{1} \\ $$$$\mathrm{So}, \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{a}−\mathrm{1}\geqslant{b}\geqslant\mathrm{1}...............\mathrm{A} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}\leqslant{a},{b}\leqslant\mathrm{2c}\:\:\left(\mathrm{Given}\right).......\mathrm{B} \\ $$$$\mathrm{Case}-\mathrm{1}:\:\:\:\:\:\:\:\mathrm{1}\leqslant{a}\leqslant\mathrm{c}\: \\ $$$$\:\:\:\mathrm{1}\leqslant{b}\leqslant\mathrm{2}{c}\:\wedge\:{b}\leqslant\mathrm{2}{a}−\mathrm{1}\:\Rightarrow\mathrm{1}\leqslant{b}\leqslant\mathrm{2}{a}−\mathrm{1}\:\ast \\ $$$$\:\:\mathrm{i}-\mathrm{e}\:\mathrm{there}\:\mathrm{are}\:\mathrm{2}{a}−\mathrm{1}\:\mathrm{choices}\:\mathrm{for}\:{b}\:\mathrm{when} \\ $$$$\:\:{a}\:\mathrm{is}\:\mathrm{fixed}. \\ $$$$\mathrm{For}\:{a}={c}\:\rightarrow\:\mathrm{2}{c}−\mathrm{1}\:{choices}\:{for}\:{b} \\ $$$$\mathrm{For}\:{a}={c}−\mathrm{1}\rightarrow\:\mathrm{2}\left({c}−\mathrm{1}\right)−\mathrm{1}=\mathrm{2}{c}−\mathrm{3}\:{choices}\:{for}\:{b} \\ $$$$\mathrm{For}\:{a}={c}−\mathrm{2}\rightarrow\:\mathrm{2}\left({c}−\mathrm{2}\right)−\mathrm{1}=\mathrm{2}{c}−\mathrm{5}\:{choices}\:{for}\:{b} \\ $$$$\:\:\:\:\:\:\:...... \\ $$$$\:\:\:\:\:\:\:\:..... \\ $$$$\mathrm{For}\:{a}=\mathrm{2}\rightarrow\:\mathrm{2}\left(\mathrm{2}\right)−\mathrm{1}=\mathrm{3}\:{choices}\:{for}\:{b} \\ $$$$\mathrm{For}\:{a}=\mathrm{1}\rightarrow\:\mathrm{2}\left(\mathrm{1}\right)−\mathrm{1}=\mathrm{1}\:{choices}\:{for}\:{b} \\ $$$$\mathrm{Total}\:\mathrm{choices}\:\mathrm{for}\:\mathrm{this}\:\mathrm{case} \\ $$$$\underset{−−−−−−−−−−−−{c}\:{terms}−−−−−−} {\left(\mathrm{2c}−\mathrm{1}\right)+\left(\mathrm{2c}−\mathrm{3}\right)+\left(\mathrm{2c}+\mathrm{5}\right)...+\mathrm{3}+\mathrm{1}}={c}^{\mathrm{2}} \\ $$$$\mathrm{Notice}\:\mathrm{that}\:\mathrm{above}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\:\mathrm{of}\:\mathrm{c}\:\mathrm{consecutive} \\ $$$$\mathrm{odd}\:\mathrm{numbers}.\left(\mathrm{Treate}\:\mathrm{it}\:\mathrm{as}\:\mathrm{AP}\right) \\ $$$$\mathrm{Case}-\mathrm{2}:\:\:\:\:\:\:\mathrm{c}+\mathrm{1}\leqslant{a}\leqslant\mathrm{2c}\:\left(\mathrm{c}\:\mathrm{values}\right) \\ $$$$\:\:\:\:{b}\leqslant\mathrm{2c}\:\wedge\:\mathrm{1}\leqslant{b}\leqslant\mathrm{2}{a}−\mathrm{1}\:\Rightarrow\mathrm{1}\leqslant{b}\leqslant\mathrm{2}{c}\:\ast\ast \\ $$$$\mathrm{i}-\mathrm{e}\:{b}\:\mathrm{takes}\:\:\mathrm{2c}\:\mathrm{values}. \\ $$$$\because\:{a}\:\mathrm{takes}\:{c}\:\mathrm{values}\:\mathrm{from}\:\mathrm{c}+\mathrm{1}\:\mathrm{to}\:\mathrm{2c} \\ $$$$\:\:\:\:\mathrm{and}\:\mathrm{for}\:\mathrm{each}\:\mathrm{of}\:\mathrm{these}\:\mathrm{values}\: \\ $$$$\:\:\:\:\:{b}\:\mathrm{takes}\:\mathrm{2}{c}\:\mathrm{vales} \\ $$$$\therefore\:\left({a},{a},{b}\right)\:\mathrm{takes}\:{c}.\mathrm{2}{c}=\mathrm{2}{c}^{\mathrm{2}} \:\mathrm{values} \\ $$$$\mathrm{Case}-\mathrm{1}\:\:\&\:\:\mathrm{Case}-\mathrm{2}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{c}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} =\mathrm{3}{c}^{\mathrm{2}} \\ $$$${Number}\:{of}\:\left({a},{a},{b}\right)=\mathrm{3}{c}^{\mathrm{2}} \\ $$

Commented by Rasheed.Sindhi last updated on 24/May/18

∗ For 1≤a≤c b≤2c ∧ b≤2a−1 ⇒b≤2a−1 [ b≤2c ∧ b≤2a−1⇒b≤min(2c,2a−1) 1≤a≤c⇒2≤2a−1≤2c−1 ⇒2a−1≤2c So 2a−1 is always less than 2c ] ∗∗ For c+1≤a≤2c b≤2c ∧ b≤2a−1 ⇒b≤2c [ b≤2c ∧ b≤2a−1⇒b≤min(2c,2a−1) c+1≤a≤2c⇒2c≤2a−1≤4c−1 So 2c always less than 2a−1

$$\:\ast\:\mathrm{For}\:\mathrm{1}\leqslant{a}\leqslant{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\leqslant\mathrm{2}{c}\:\wedge\:{b}\leqslant\mathrm{2}{a}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{b}\leqslant\mathrm{2}{a}−\mathrm{1} \\ $$$$\left[\:\:{b}\leqslant\mathrm{2}{c}\:\wedge\:{b}\leqslant\mathrm{2}{a}−\mathrm{1}\Rightarrow{b}\leqslant\mathrm{min}\left(\mathrm{2}{c},\mathrm{2}{a}−\mathrm{1}\right)\right. \\ $$$$\mathrm{1}\leqslant{a}\leqslant{c}\Rightarrow\mathrm{2}\leqslant\mathrm{2}{a}−\mathrm{1}\leqslant\mathrm{2}{c}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\mathrm{2}{a}−\mathrm{1}\leqslant\mathrm{2}{c} \\ $$$$\left.\mathrm{So}\:\mathrm{2}{a}−\mathrm{1}\:\mathrm{is}\:\mathrm{always}\:\mathrm{less}\:\mathrm{than}\:\mathrm{2}{c}\:\right] \\ $$$$\ast\ast\:\mathrm{For}\:\:\:{c}+\mathrm{1}\leqslant{a}\leqslant\mathrm{2}{c} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:{b}\leqslant\mathrm{2}{c}\:\wedge\:{b}\leqslant\mathrm{2}{a}−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow{b}\leqslant\mathrm{2}{c} \\ $$$$\left[\:\:{b}\leqslant\mathrm{2}{c}\:\wedge\:{b}\leqslant\mathrm{2}{a}−\mathrm{1}\Rightarrow{b}\leqslant\mathrm{min}\left(\mathrm{2}{c},\mathrm{2}{a}−\mathrm{1}\right)\right. \\ $$$$\:{c}+\mathrm{1}\leqslant{a}\leqslant\mathrm{2}{c}\Rightarrow\mathrm{2}{c}\leqslant\mathrm{2}{a}−\mathrm{1}\leqslant\mathrm{4}{c}−\mathrm{1} \\ $$$$\mathrm{So}\:\mathrm{2}{c}\:\mathrm{always}\:\mathrm{less}\:\mathrm{than}\:\mathrm{2}{a}−\mathrm{1} \\ $$$$ \\ $$

Commented by Tinkutara last updated on 24/May/18

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

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