Question-203018

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Question Number 203018 by mr W last updated on 07/Jan/24

Commented by mr W last updated on 07/Jan/24

$$\left[{an}\:{unsolved}\:{old}\:{question}\:{Q}\mathrm{202864}\right] \\ $$

Commented by esmaeil last updated on 07/Jan/24

Commented by MathematicalUser2357 last updated on 09/Jan/24

$${an}\:{unquolved}\:\mathrm{202864} \\ $$

Answered by mr W last updated on 07/Jan/24

Commented by mr W last updated on 07/Jan/24

we can see that the three green lines are parallel to each other. let α=(q/p), β=(r/p) 1+((2825+B)/(690+A))=(((q+r)/p))^2 =(α+β)^2 ⇒A=((2825+B)/((α+β)^2 −1))−690 1+(B/(46))=((p/q))^2 =(1/α^2 ) ⇒α=(1/( (√(1+(B/(46)))))) (B/(2825))=((p+q)/r)=((1+α)/β) ⇒β=((2825)/B)(1+(1/( (√(1+(B/(46))))))) 1+((2825×2+B)/A)=(((q+2r)/p))^2 =(α+2β)^2 ⇒A=((2825×2+B)/((α+2β)^2 −1)) ((2825+B)/((α+β)^2 −1))−690=((2825×2+B)/((α+2β)^2 −1)) ⇒B≈1593.2072 ⇒A≈412.5313 ?=A+B≈2005.7385 ✓

$${we}\:{can}\:{see}\:{that}\:{the}\:{three}\:{green}\:{lines} \\ $$$${are}\:{parallel}\:{to}\:{each}\:{other}. \\ $$$${let}\:\alpha=\frac{{q}}{{p}},\:\beta=\frac{{r}}{{p}} \\ $$$$\mathrm{1}+\frac{\mathrm{2825}+{B}}{\mathrm{690}+{A}}=\left(\frac{{q}+{r}}{{p}}\right)^{\mathrm{2}} =\left(\alpha+\beta\right)^{\mathrm{2}} \\ $$$$\Rightarrow{A}=\frac{\mathrm{2825}+{B}}{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{1}}−\mathrm{690} \\ $$$$\mathrm{1}+\frac{{B}}{\mathrm{46}}=\left(\frac{{p}}{{q}}\right)^{\mathrm{2}} =\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:\Rightarrow\alpha=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{B}}{\mathrm{46}}}} \\ $$$$\frac{{B}}{\mathrm{2825}}=\frac{{p}+{q}}{{r}}=\frac{\mathrm{1}+\alpha}{\beta}\:\Rightarrow\beta=\frac{\mathrm{2825}}{{B}}\left(\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{B}}{\mathrm{46}}}}\right) \\ $$$$\mathrm{1}+\frac{\mathrm{2825}×\mathrm{2}+{B}}{{A}}=\left(\frac{{q}+\mathrm{2}{r}}{{p}}\right)^{\mathrm{2}} =\left(\alpha+\mathrm{2}\beta\right)^{\mathrm{2}} \\ $$$$\Rightarrow{A}=\frac{\mathrm{2825}×\mathrm{2}+{B}}{\left(\alpha+\mathrm{2}\beta\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{\mathrm{2825}+{B}}{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{1}}−\mathrm{690}=\frac{\mathrm{2825}×\mathrm{2}+{B}}{\left(\alpha+\mathrm{2}\beta\right)^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{B}\approx\mathrm{1593}.\mathrm{2072} \\ $$$$\Rightarrow{A}\approx\mathrm{412}.\mathrm{5313} \\ $$$$?={A}+{B}\approx\mathrm{2005}.\mathrm{7385}\:\checkmark \\ $$

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