Menu Close

Determine-all-triangle-with-1-The-lengths-of-sides-positive-integers-and-at-least-one-is-prime-number-2-The-semiperimetr-is-positive-integer-and-area-is-equal-with-perimetr-

Tinku Tara 0 Comments
Pin




Question Number 155100 by mathdanisur last updated on 25/Sep/21
Determine all triangle with: 1.The lengths of sides positive integers and at least one is prime number. 2.The semiperimetr is positive integer and area is equal with perimetr.
$$\mathrm{Determine}\:\mathrm{all}\:\mathrm{triangle}\:\mathrm{with}: \\ $$$$\mathrm{1}.\mathrm{The}\:\mathrm{lengths}\:\mathrm{of}\:\mathrm{sides}\:\mathrm{positive}\:\mathrm{integers} \\ $$$$\:\:\:\:\:\mathrm{and}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{is}\:\mathrm{prime}\:\mathrm{number}. \\ $$$$\mathrm{2}.\mathrm{The}\:\mathrm{semiperimetr}\:\mathrm{is}\:\mathrm{positive}\:\mathrm{integer} \\ $$$$\:\:\:\:\:\mathrm{and}\:\mathrm{area}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{with}\:\mathrm{perimetr}. \\ $$
Commented by MJS_new last updated on 25/Sep/21
if a<b<c there are only 4 I think 5/12/13 6/25/29 7/15/20 9/10/17 [there′s a 5^(th) one without primes: 6/8/10]
$$\mathrm{if}\:{a}<{b}<{c}\:\mathrm{there}\:\mathrm{are}\:\mathrm{only}\:\mathrm{4}\:\mathrm{I}\:\mathrm{think} \\ $$$$\mathrm{5}/\mathrm{12}/\mathrm{13} \\ $$$$\mathrm{6}/\mathrm{25}/\mathrm{29} \\ $$$$\mathrm{7}/\mathrm{15}/\mathrm{20} \\ $$$$\mathrm{9}/\mathrm{10}/\mathrm{17} \\ $$$$ \\ $$$$\:\:\:\:\:\left[\mathrm{there}'\mathrm{s}\:\mathrm{a}\:\mathrm{5}^{\mathrm{th}} \:\mathrm{one}\:\mathrm{without}\:\mathrm{primes}:\:\mathrm{6}/\mathrm{8}/\mathrm{10}\right] \\ $$
Commented by mathdanisur last updated on 25/Sep/21
perfect my dear thankyou
$$\mathrm{perfect}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{thankyou} \\ $$
Answered by Rasheed.Sindhi last updated on 26/Sep/21
•a,b,c,s=((a+b+c)/2)∈Z^+ ∧ a(say)∈P •(√(s(s−a)(s−b)(s−c))) =a+b+c_(where s=(a+b+c)/2) ((a+b+c)/2)∈Z^+ ⇒a+b+c∈E ⇒ { ((a,b,c all are eve even⇒prime is 2)),((Two of a,b,c are odd,the third is even.)) :} C-1: a,b,c∈E⇒a(prime)=2 b=2m,c=2n ; m,n∈Z^+ s=(2+2m+2n)/2=m+n+1 s−a=1+m+n−2=m+n−1 s−b=1+m+n−2m=−m+n+1 s−c=1+m+n−2n=m−n+1 (√(s(s−a)(s−b)(s−c)))=a+b+c ▶(√((m+n+1)(m+n−1)(−m+n+1)(m−n+1))) =2m+2n+2 ▶ (m+n+1)(m+n−1)(−m+n+1)(m−n+1) =4(m+n+1)^2 ▶ (m+n−1)(−m+n+1)(m−n+1) =4(m+n+1) (o,o)-case:m,n∈O determinant ((((o,o)-case:m,n∈O))) RHS is clearly even. LHS:(o+o−o)(−o+o+o)(o−o+o) =(e−o)(e+o)(e+o) =(o)(o)(o)=o LHS is odd. Contradiction. determinant ((((e,e)-case:m,n∈E))) RHS=even LHS:(e+e−o)(−e+e+o)(e−e+o) (e−o)(e+o)(e+o) (o)(o)(o)=o Contradiction. determinant ((((e,o) or (o,e)-case:m∈E,n∈O))) Continue
$$\:\bullet{a},{b},{c},{s}=\frac{{a}+{b}+{c}}{\mathrm{2}}\in\mathbb{Z}^{+} \wedge\:{a}\left({say}\right)\in\mathbb{P}\: \\ $$$$\:\bullet\underset{{where}\:{s}=\left({a}+{b}+{c}\right)/\mathrm{2}} {\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}\:={a}+{b}+{c}} \\ $$$$\frac{{a}+{b}+{c}}{\mathrm{2}}\in\mathbb{Z}^{+} \Rightarrow{a}+{b}+{c}\in\mathbb{E} \\ $$$$\Rightarrow\begin{cases}{{a},{b},{c}\:{all}\:{are}\:{eve}\:{even}\Rightarrow{prime}\:{is}\:\mathrm{2}}\\{\mathcal{T}{wo}\:{of}\:{a},{b},{c}\:{are}\:{odd},{the}\:{third}\:{is}\:{even}.}\end{cases} \\ $$$$\mathrm{C}-\mathrm{1}:\:{a},{b},{c}\in\mathbb{E}\Rightarrow{a}\left({prime}\right)=\mathrm{2} \\ $$$$\:\:\:\:\:\:\:{b}=\mathrm{2}{m},{c}=\mathrm{2}{n}\:;\:{m},{n}\in\mathbb{Z}^{+} \\ $$$$\:\:\:{s}=\left(\mathrm{2}+\mathrm{2}{m}+\mathrm{2}{n}\right)/\mathrm{2}={m}+{n}+\mathrm{1} \\ $$$$\:\:\:{s}−{a}=\mathrm{1}+{m}+{n}−\mathrm{2}={m}+{n}−\mathrm{1} \\ $$$$\:\:\:{s}−{b}=\mathrm{1}+{m}+{n}−\mathrm{2}{m}=−{m}+{n}+\mathrm{1} \\ $$$$\:\:\:{s}−{c}=\mathrm{1}+{m}+{n}−\mathrm{2}{n}={m}−{n}+\mathrm{1} \\ $$$$\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}={a}+{b}+{c} \\ $$$$\blacktriangleright\sqrt{\left({m}+{n}+\mathrm{1}\right)\left({m}+{n}−\mathrm{1}\right)\left(−{m}+{n}+\mathrm{1}\right)\left({m}−{n}+\mathrm{1}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{m}+\mathrm{2}{n}+\mathrm{2} \\ $$$$\blacktriangleright\:\left({m}+{n}+\mathrm{1}\right)\left({m}+{n}−\mathrm{1}\right)\left(−{m}+{n}+\mathrm{1}\right)\left({m}−{n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\left({m}+{n}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\blacktriangleright\:\left({m}+{n}−\mathrm{1}\right)\left(−{m}+{n}+\mathrm{1}\right)\left({m}−{n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}\left({m}+{n}+\mathrm{1}\right) \\ $$$$\:\:\:\:\left({o},{o}\right)-{case}:{m},{n}\in\mathbb{O} \\ $$$$\:\:\:\:\:\begin{array}{|c|}{\left({o},{o}\right)-{case}:{m},{n}\in\mathbb{O}}\\\hline\end{array}\: \\ $$$$\:\:\:\:\mathrm{RHS}\:{is}\:{clearly}\:\boldsymbol{{even}}. \\ $$$$\:\:\:\:\:\mathrm{LHS}:\left({o}+{o}−{o}\right)\left(−{o}+{o}+{o}\right)\left({o}−{o}+{o}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({e}−{o}\right)\left({e}+{o}\right)\left({e}+{o}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:=\left({o}\right)\left({o}\right)\left({o}\right)={o} \\ $$$$\:\:\:\:\:\:\mathrm{LHS}\:{is}\:\boldsymbol{{odd}}. \\ $$$$\:\:\:\:\:\:{Contradiction}. \\ $$$$\:\:\:\:\:\begin{array}{|c|}{\left({e},{e}\right)-{case}:{m},{n}\in\mathbb{E}}\\\hline\end{array}\: \\ $$$$\:\:\:\:\:\:\mathrm{RHS}=\boldsymbol{{even}} \\ $$$$\:\:\:\:\mathrm{LHS}:\left({e}+{e}−{o}\right)\left(−{e}+{e}+{o}\right)\left({e}−{e}+{o}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({e}−{o}\right)\left({e}+{o}\right)\left({e}+{o}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({o}\right)\left({o}\right)\left({o}\right)={o} \\ $$$$\:\:\:\:\:\:{Contradiction}. \\ $$$$\:\:\:\:\:\begin{array}{|c|}{\left({e},{o}\right)\:{or}\:\left({o},{e}\right)-{case}:{m}\in\mathbb{E},{n}\in\mathbb{O}}\\\hline\end{array}\: \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${Continue} \\ $$
Commented by talminator2856791 last updated on 26/Sep/21
how did you put it in a box?
$$\:\mathrm{how}\:\mathrm{did}\:\mathrm{you}\:\mathrm{put}\:\mathrm{it}\:\mathrm{in}\:\mathrm{a}\:\mathrm{box}? \\ $$
Commented by Rasheed.Sindhi last updated on 26/Sep/21
The manu that appears on click of matrix- button contains table with borders.I′ve deleted its row/s except one which I′ve used as box.
$$\:\:{The}\:\:{manu}\:{that}\:{appears}\:{on}\:{click} \\ $$$${of}\:{matrix}-\:{button}\:{contains}\:\boldsymbol{{table}} \\ $$$$\boldsymbol{{with}}\:\boldsymbol{{borders}}.{I}'{ve}\:{deleted}\:{its}\:{row}/{s} \\ $$$${except}\:{one}\:{which}\:{I}'{ve}\:{used}\:{as}\:{box}. \\ $$
Commented by talminator2856791 last updated on 05/Oct/21
how to delete the rows?
$$\:\mathrm{how}\:\mathrm{to}\:\mathrm{delete}\:\mathrm{the}\:\mathrm{rows}? \\ $$
Commented by Rasheed.Sindhi last updated on 05/Oct/21
When you are in a row (that you want to delete)open side mau and click ′delete row′.
$${When}\:{you}\:{are}\:{in}\:{a}\:{row}\:\left({that}\:{you}\:{want}\right. \\ $$$$\left.{to}\:{delete}\right){open}\:{side}\:{mau}\:{and}\:{click} \\ $$$$'{delete}\:{row}'. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *