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Question Number 210090 by Tawa11 last updated on 30/Jul/24

Commented by mr W last updated on 30/Jul/24

are there 3 times letter p in your picture?

$${are}\:{there}\:\mathrm{3}\:{times}\:{letter}\:{p}\:{in}\:{your}\: \\ $$$${picture}? \\ $$

Commented by Tawa11 last updated on 30/Jul/24

two times sir.

$$\mathrm{two}\:\mathrm{times}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 30/Jul/24

2 times p, a 5 and a 4? then the question is not so interesting as with 3 times p.

$$\mathrm{2}\:{times}\:{p},\:{a}\:\mathrm{5}\:{and}\:{a}\:\mathrm{4}? \\ $$$${then}\:{the}\:{question}\:{is}\:{not}\:{so} \\ $$$${interesting}\:{as}\:{with}\:\mathrm{3}\:{times}\:{p}. \\ $$

Commented by Tawa11 last updated on 30/Jul/24

Sir, please do it as 2 times and 3 times p. The question did not give any information. only the diagram and find p

$$\mathrm{Sir},\:\mathrm{please}\:\mathrm{do}\:\mathrm{it}\:\mathrm{as}\:\mathrm{2}\:\mathrm{times}\:\mathrm{and}\:\mathrm{3}\:\mathrm{times}\:\mathrm{p}. \\ $$$$\mathrm{The}\:\mathrm{question}\:\mathrm{did}\:\mathrm{not}\:\mathrm{give}\:\mathrm{any}\:\mathrm{information}. \\ $$$$\mathrm{only}\:\mathrm{the}\:\mathrm{diagram}\:\mathrm{and}\:\mathrm{find}\:\mathrm{p} \\ $$

Commented by mr W last updated on 30/Jul/24

Commented by mr W last updated on 30/Jul/24

4×9=a(a+2b) p^2 −b^2 =9^2 −(a+b)^2 ⇒9^2 −p^2 =a(a+2b)=4×9 ⇒p^2 =9^2 −4×9 ⇒p=3(√5)

$$\mathrm{4}×\mathrm{9}={a}\left({a}+\mathrm{2}{b}\right) \\ $$$${p}^{\mathrm{2}} −{b}^{\mathrm{2}} =\mathrm{9}^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}^{\mathrm{2}} −{p}^{\mathrm{2}} ={a}\left({a}+\mathrm{2}{b}\right)=\mathrm{4}×\mathrm{9} \\ $$$$\Rightarrow{p}^{\mathrm{2}} =\mathrm{9}^{\mathrm{2}} −\mathrm{4}×\mathrm{9}\:\Rightarrow{p}=\mathrm{3}\sqrt{\mathrm{5}} \\ $$

Commented by mr W last updated on 30/Jul/24

Commented by mr W last updated on 30/Jul/24

p^2 −b^2 =(p+5)^2 −(a+b)^2 ⇒a(a+2b)=10p+25 p(p+5)=a(a+2b)=10p+25 ⇒p^2 −5p−25=0 ⇒p=((5(1+(√5)))/2)=5ϕ (ϕ=golden ratio)

$${p}^{\mathrm{2}} −{b}^{\mathrm{2}} =\left({p}+\mathrm{5}\right)^{\mathrm{2}} −\left({a}+{b}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{a}\left({a}+\mathrm{2}{b}\right)=\mathrm{10}{p}+\mathrm{25} \\ $$$${p}\left({p}+\mathrm{5}\right)={a}\left({a}+\mathrm{2}{b}\right)=\mathrm{10}{p}+\mathrm{25} \\ $$$$\Rightarrow{p}^{\mathrm{2}} −\mathrm{5}{p}−\mathrm{25}=\mathrm{0} \\ $$$$\Rightarrow{p}=\frac{\mathrm{5}\left(\mathrm{1}+\sqrt{\mathrm{5}}\right)}{\mathrm{2}}=\mathrm{5}\varphi\:\:\:\left(\varphi={golden}\:{ratio}\right) \\ $$

Commented by Tawa11 last updated on 30/Jul/24

Thanks so much sir. I really appreciate.

$$\mathrm{Thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir}. \\ $$$$\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}. \\ $$

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