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Question Number 186627 by Mingma last updated on 07/Feb/23

Commented by Frix last updated on 07/Feb/23

As always: The answer is 42 • We have 2 equations for 6 unknowns ⇒ We can choose 4 of them and solve for the remaining 2. • We can set ((a+b+c)/(x+y+z))=42 ⇒ We now have 3 equations for 6 unknowns ⇒ We can choose 3 of them and solve for the remaining 3

$$\mathrm{As}\:\mathrm{always}:\:\mathrm{The}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{42} \\ $$$$\bullet\:\mathrm{We}\:\mathrm{have}\:\mathrm{2}\:\mathrm{equations}\:\mathrm{for}\:\mathrm{6}\:\mathrm{unknowns}\:\Rightarrow \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{choose}\:\mathrm{4}\:\mathrm{of}\:\mathrm{them}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:\mathrm{the} \\ $$$$\mathrm{remaining}\:\mathrm{2}. \\ $$$$\bullet\:\mathrm{We}\:\mathrm{can}\:\mathrm{set}\:\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\mathrm{42}\:\Rightarrow\:\mathrm{We}\:\mathrm{now}\:\mathrm{have} \\ $$$$\mathrm{3}\:\mathrm{equations}\:\mathrm{for}\:\mathrm{6}\:\mathrm{unknowns}\:\Rightarrow\:\mathrm{We}\:\mathrm{can} \\ $$$$\mathrm{choose}\:\mathrm{3}\:\mathrm{of}\:\mathrm{them}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:\mathrm{the}\:\mathrm{remaining}\:\mathrm{3} \\ $$

Answered by ajfour last updated on 07/Feb/23

z=qx, y=px x(a+bp+cq)=20 y((a/p)+b+((cq)/p))=20 z((a/q)+((bp)/q)+c)=20 ax+by+cz=Σ((20a)/(a+bp+cq))=20 ⇒ a((a/p)+b+((cq)/p))((a/q)+((bp)/q)+c) +b((a/q)+((bp)/q)+c)(a+bp+cq) +c(a+bp+cq)((a/p)+b+((cq)/p))=1 let ?=(1/Q) ⇒ x+y+z=Q(a+b+c) ⇒ (1/((a+bp+cq)))+(1/(((a/p)+b+((cq)/p)))) +(1/(((a/q)+((bp)/q)+c)))= ((Q(a+b+c))/(20)) ⇒ ((1+p+q)/s)=((Q(a+b+c))/(20)) & ((as^2 )/(pq))+((bs^2 )/q)+((cs^2 )/p)=1 ⇒ s^3 =pq ⇒ 20(1+p+q)=Q(a+b+c)(pq)^(1/3) (a+b+c)^2 =16−2(ab+bc+ca) Q=((20(1+p+q))/((pq)^(1/3) (a+b+c))) Q=((20(1+(y/x)+(z/x)))/((((yz)/x^2 ))^(1/3) (a+b+c)))=((20Q)/((xyz)^(1/3) )) ⇒ (xyz)^(1/3) =20 by symmetry a^2 =b^2 =c^2 =((16)/3) x+y+z=20+20+20 (1/?)=Q=((x+y+z)/(a+b+c)) Q=±((60)/(3((4/( (√3))))))=±5(√3) ?=(1/Q)=±(1/(5(√3)))

$${z}={qx},\:\:{y}={px} \\ $$$${x}\left({a}+{bp}+{cq}\right)=\mathrm{20} \\ $$$${y}\left(\frac{{a}}{{p}}+{b}+\frac{{cq}}{{p}}\right)=\mathrm{20} \\ $$$${z}\left(\frac{{a}}{{q}}+\frac{{bp}}{{q}}+{c}\right)=\mathrm{20} \\ $$$${ax}+{by}+{cz}=\Sigma\frac{\mathrm{20}{a}}{{a}+{bp}+{cq}}=\mathrm{20} \\ $$$$\Rightarrow\:{a}\left(\frac{{a}}{{p}}+{b}+\frac{{cq}}{{p}}\right)\left(\frac{{a}}{{q}}+\frac{{bp}}{{q}}+{c}\right) \\ $$$$+{b}\left(\frac{{a}}{{q}}+\frac{{bp}}{{q}}+{c}\right)\left({a}+{bp}+{cq}\right) \\ $$$$+{c}\left({a}+{bp}+{cq}\right)\left(\frac{{a}}{{p}}+{b}+\frac{{cq}}{{p}}\right)=\mathrm{1} \\ $$$${let}\:\:\:\:?=\frac{\mathrm{1}}{{Q}} \\ $$$$\Rightarrow\:\:\:{x}+{y}+{z}={Q}\left({a}+{b}+{c}\right) \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}}{\left({a}+{bp}+{cq}\right)}+\frac{\mathrm{1}}{\left(\frac{{a}}{{p}}+{b}+\frac{{cq}}{{p}}\right)} \\ $$$$\:\:\:+\frac{\mathrm{1}}{\left(\frac{{a}}{{q}}+\frac{{bp}}{{q}}+{c}\right)}=\:\:\frac{{Q}\left({a}+{b}+{c}\right)}{\mathrm{20}} \\ $$$$\Rightarrow\:\:\frac{\mathrm{1}+{p}+{q}}{{s}}=\frac{{Q}\left({a}+{b}+{c}\right)}{\mathrm{20}} \\ $$$$\&\:\:\:\frac{{as}^{\mathrm{2}} }{{pq}}+\frac{{bs}^{\mathrm{2}} }{{q}}+\frac{{cs}^{\mathrm{2}} }{{p}}=\mathrm{1} \\ $$$$\Rightarrow\:\:{s}^{\mathrm{3}} ={pq} \\ $$$$\Rightarrow\:\:\:\mathrm{20}\left(\mathrm{1}+{p}+{q}\right)={Q}\left({a}+{b}+{c}\right)\left({pq}\right)^{\mathrm{1}/\mathrm{3}} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} =\mathrm{16}−\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$${Q}=\frac{\mathrm{20}\left(\mathrm{1}+{p}+{q}\right)}{\left({pq}\right)^{\mathrm{1}/\mathrm{3}} \left({a}+{b}+{c}\right)} \\ $$$${Q}=\frac{\mathrm{20}\left(\mathrm{1}+\frac{{y}}{{x}}+\frac{{z}}{{x}}\right)}{\left(\frac{{yz}}{{x}^{\mathrm{2}} }\right)^{\mathrm{1}/\mathrm{3}} \left({a}+{b}+{c}\right)}=\frac{\mathrm{20}{Q}}{\left({xyz}\right)^{\mathrm{1}/\mathrm{3}} } \\ $$$$\Rightarrow\:\:\left({xyz}\right)^{\mathrm{1}/\mathrm{3}} =\mathrm{20} \\ $$$${by}\:{symmetry} \\ $$$${a}^{\mathrm{2}} ={b}^{\mathrm{2}} ={c}^{\mathrm{2}} =\frac{\mathrm{16}}{\mathrm{3}} \\ $$$${x}+{y}+{z}=\mathrm{20}+\mathrm{20}+\mathrm{20} \\ $$$$\frac{\mathrm{1}}{?}={Q}=\frac{{x}+{y}+{z}}{{a}+{b}+{c}} \\ $$$$\:\:\:\:\:\:{Q}=\pm\frac{\mathrm{60}}{\mathrm{3}\left(\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}\right)}=\pm\mathrm{5}\sqrt{\mathrm{3}} \\ $$$$?=\frac{\mathrm{1}}{{Q}}=\pm\frac{\mathrm{1}}{\mathrm{5}\sqrt{\mathrm{3}}} \\ $$$$ \\ $$$$ \\ $$

Answered by ajfour last updated on 07/Feb/23

or simply if at all answer is unique then we can get the same as a special case too a=b=c a(x+y+z)=20 a^2 =((16)/3) a=±(4/( (√3))) ((a+b+c)/(x+y+z))=((3a)/(20/a))=((3a^2 )/(20))=((16)/(20))=(4/5)

$${or}\:{simply}\:\:{if}\:\:{at}\:{all}\:\:{answer}\:{is}\:{unique} \\ $$$${then}\:\:\:{we}\:{can}\:{get}\:{the}\:{same}\:\:{as}\:{a}\:{special} \\ $$$${case}\:{too}\:\:\:{a}={b}={c} \\ $$$${a}\left({x}+{y}+{z}\right)=\mathrm{20} \\ $$$${a}^{\mathrm{2}} =\frac{\mathrm{16}}{\mathrm{3}} \\ $$$${a}=\pm\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}} \\ $$$$\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{\mathrm{3}{a}}{\mathrm{20}/{a}}=\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{20}}=\frac{\mathrm{16}}{\mathrm{20}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$

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