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Question Number 153638 by mathdanisur last updated on 08/Sep/21
Commented by EDWIN88 last updated on 09/Sep/21
b×c=12a ⇒a^2 =b^2 +c^2 ⇒a^2 =(b+c)^2 −24a ⇒a^2 +24a−(b+c)^2 =0 ⇒(a+12)^2 =144+(b+c)^2 ⇒(a+12+b+c)(a+12−b−c)=144 ⇒(a+b+c+12)(a−b−c+12)=144 factor 144={1,2,3,4,6,8,9,12,16,18,24,36,48,72,144} a+b+c+12>0 it follows that case(1)a+b+c+12=16 ∧a−b−c+12=9 ⇒a+b+c=4∧a−b−c=−3 (no solution) case(2)a+b+c+12=18 ∧a−b−c+12=8 ⇒a+b+c=6∧a−b−c=−4 (no solution) case(3)a+b+c+12=24∧a−b−c+12=6 ⇒a+b+c=12∧a−b−c=−6 ⇒a=3 ∧ { ((b+c=9)),((b×c=36)) :}⇒no solution case(4)a+b+c+12=36∧a−b−c+12=4 ⇒a+b+c=24∧a−b−c=−8 ⇒a=8∧ { ((b+c=16)),((b×c=96)) :}⇒no solution case(5)a+b+c+12=48∧a−b−c+12=3 ⇒a+b+c=36∧a−b−c=−9 ⇒no solution case(6)a+b+c+12=72∧a−b−c+12=2 ⇒a+b+c=60∧a−b−c=−10 ⇒a=25→ { ((b+c=35)),((b×c=300)) :} ⇒c=35−b⇒b×(35−b)=300 ⇒b^2 −35b+300=0 ⇒b=((35+5)/2)=20 ∧c=15 solution (a,b,c)= { (((25,20,15))),(((25,15,20))) :} case(7)a+b+c+12=144∧a−b−c+12=1 ⇒a+b+c=132∧a−b−c=−11 no solution
$${b}×{c}=\mathrm{12}{a} \\ $$$$\Rightarrow{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} =\left({b}+{c}\right)^{\mathrm{2}} −\mathrm{24}{a} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +\mathrm{24}{a}−\left({b}+{c}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({a}+\mathrm{12}\right)^{\mathrm{2}} =\mathrm{144}+\left({b}+{c}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\left({a}+\mathrm{12}+{b}+{c}\right)\left({a}+\mathrm{12}−{b}−{c}\right)=\mathrm{144} \\ $$$$\Rightarrow\left({a}+{b}+{c}+\mathrm{12}\right)\left({a}−{b}−{c}+\mathrm{12}\right)=\mathrm{144} \\ $$$${factor}\:\mathrm{144}=\left\{\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4},\mathrm{6},\mathrm{8},\mathrm{9},\mathrm{12},\mathrm{16},\mathrm{18},\mathrm{24},\mathrm{36},\mathrm{48},\mathrm{72},\mathrm{144}\right\} \\ $$$${a}+{b}+{c}+\mathrm{12}>\mathrm{0}\:{it}\:{follows}\:{that} \\ $$$${case}\left(\mathrm{1}\right){a}+{b}+{c}+\mathrm{12}=\mathrm{16}\:\wedge{a}−{b}−{c}+\mathrm{12}=\mathrm{9} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{4}\wedge{a}−{b}−{c}=−\mathrm{3}\:\left({no}\:{solution}\right) \\ $$$${case}\left(\mathrm{2}\right){a}+{b}+{c}+\mathrm{12}=\mathrm{18}\:\wedge{a}−{b}−{c}+\mathrm{12}=\mathrm{8} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{6}\wedge{a}−{b}−{c}=−\mathrm{4}\:\left({no}\:{solution}\right) \\ $$$${case}\left(\mathrm{3}\right){a}+{b}+{c}+\mathrm{12}=\mathrm{24}\wedge{a}−{b}−{c}+\mathrm{12}=\mathrm{6} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{12}\wedge{a}−{b}−{c}=−\mathrm{6} \\ $$$$\Rightarrow{a}=\mathrm{3}\:\wedge\begin{cases}{{b}+{c}=\mathrm{9}}\\{{b}×{c}=\mathrm{36}}\end{cases}\Rightarrow{no}\:{solution} \\ $$$${case}\left(\mathrm{4}\right){a}+{b}+{c}+\mathrm{12}=\mathrm{36}\wedge{a}−{b}−{c}+\mathrm{12}=\mathrm{4} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{24}\wedge{a}−{b}−{c}=−\mathrm{8} \\ $$$$\Rightarrow{a}=\mathrm{8}\wedge\begin{cases}{{b}+{c}=\mathrm{16}}\\{{b}×{c}=\mathrm{96}}\end{cases}\Rightarrow{no}\:{solution} \\ $$$${case}\left(\mathrm{5}\right){a}+{b}+{c}+\mathrm{12}=\mathrm{48}\wedge{a}−{b}−{c}+\mathrm{12}=\mathrm{3} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{36}\wedge{a}−{b}−{c}=−\mathrm{9} \\ $$$$\Rightarrow{no}\:{solution} \\ $$$${case}\left(\mathrm{6}\right){a}+{b}+{c}+\mathrm{12}=\mathrm{72}\wedge{a}−{b}−{c}+\mathrm{12}=\mathrm{2} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{60}\wedge{a}−{b}−{c}=−\mathrm{10} \\ $$$$\Rightarrow{a}=\mathrm{25}\rightarrow\begin{cases}{{b}+{c}=\mathrm{35}}\\{{b}×{c}=\mathrm{300}}\end{cases} \\ $$$$\Rightarrow{c}=\mathrm{35}−{b}\Rightarrow{b}×\left(\mathrm{35}−{b}\right)=\mathrm{300} \\ $$$$\Rightarrow{b}^{\mathrm{2}} −\mathrm{35}{b}+\mathrm{300}=\mathrm{0} \\ $$$$\Rightarrow{b}=\frac{\mathrm{35}+\mathrm{5}}{\mathrm{2}}=\mathrm{20}\:\wedge{c}=\mathrm{15} \\ $$$${solution}\:\left({a},{b},{c}\right)=\begin{cases}{\left(\mathrm{25},\mathrm{20},\mathrm{15}\right)}\\{\left(\mathrm{25},\mathrm{15},\mathrm{20}\right)}\end{cases} \\ $$$${case}\left(\mathrm{7}\right){a}+{b}+{c}+\mathrm{12}=\mathrm{144}\wedge{a}−{b}−{c}+\mathrm{12}=\mathrm{1} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{132}\wedge{a}−{b}−{c}=−\mathrm{11} \\ $$$${no}\:{solution} \\ $$
Commented by mathdanisur last updated on 11/Sep/21
thanks ser nice
$$\mathrm{thanks}\:\mathrm{ser}\:\mathrm{nice} \\ $$
Answered by MJS_new last updated on 09/Sep/21
a^2 =b^2 +c^2 a=((bc)/(12)) ⇒ b^2 c^2 =144(b^2 +c^2 ) c^2 =((144b^2 )/(b^2 −144)) ⇒ b>12∧c>12 (symmetry)) b=15∧c=20 ∨ b=20∧c=15 ⇒ a=25
$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \\ $$$${a}=\frac{{bc}}{\mathrm{12}} \\ $$$$\Rightarrow\:{b}^{\mathrm{2}} {c}^{\mathrm{2}} =\mathrm{144}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\left.{c}^{\mathrm{2}} =\frac{\mathrm{144}{b}^{\mathrm{2}} }{{b}^{\mathrm{2}} −\mathrm{144}}\:\Rightarrow\:{b}>\mathrm{12}\wedge{c}>\mathrm{12}\:\left(\mathrm{symmetry}\right)\right) \\ $$$${b}=\mathrm{15}\wedge{c}=\mathrm{20}\:\vee\:{b}=\mathrm{20}\wedge{c}=\mathrm{15}\:\Rightarrow\:{a}=\mathrm{25} \\ $$
Commented by Rasheed.Sindhi last updated on 09/Sep/21
Sir , the given is: ((b∗c)/a)=12⇒((b×c)/a)=12⇒a=(1/(12))(b×c)
$$\mathcal{S}{ir}\:,\:{the}\:{given}\:{is}: \\ $$$$\frac{{b}\ast{c}}{{a}}=\mathrm{12}\Rightarrow\frac{{b}×{c}}{{a}}=\mathrm{12}\Rightarrow{a}=\frac{\mathrm{1}}{\mathrm{12}}\left({b}×{c}\right) \\ $$
Commented by MJS_new last updated on 09/Sep/21
ok. my eyes...
$$\mathrm{ok}.\:\mathrm{my}\:\mathrm{eyes}… \\ $$
Commented by MJS_new last updated on 09/Sep/21
corrected my post. thank you!
$$\mathrm{corrected}\:\mathrm{my}\:\mathrm{post}.\:\mathrm{thank}\:\mathrm{you}! \\ $$

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