Question-74747

Question Number 74747 by mr W last updated on 30/Nov/19

Answered by mind is power last updated on 30/Nov/19

let p,q,r side of red squar,a,b,c blue α,β,ς angle thetriangl inside red squar a^2 =p^2 +q^2 +2pqcos(α) b^2 =p^2 +r^2 +2prcos(β) c^2 =r^2 +q^2 +2rqcos(ς) cos(α)=((p^2 +q^2 −r^2 )/(2pq)),cos(β)=((p^2 +r^2 −q^2 )/(2pr)) cos(ς)=((q^2 +r^2 −p^2 )/(2qr)) ⇒a^2 =2p^2 +2q^2 −r^2 ,b^2 =2p^2 +2r^2 −q^2 ,c^2 =2r^2 +2q^2 −p^2 ⇒a^2 +b^2 +c^2 =3(p^2 +q^2 +r^2 ) ⇒((a^2 +b^2 +c^2 )/(p^2 +q^2 +r^2 ))=3

$$\mathrm{let}\:\mathrm{p},\mathrm{q},\mathrm{r}\:\mathrm{side}\:\mathrm{of}\:\mathrm{red}\:\mathrm{squar},\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{blue} \\ $$$$\alpha,\beta,\varsigma\:\:\mathrm{angle}\:\mathrm{thetriangl}\:\mathrm{inside}\:\mathrm{red}\:\mathrm{squar} \\ $$$$\mathrm{a}^{\mathrm{2}} =\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{2pqcos}\left(\alpha\right) \\ $$$$\mathrm{b}^{\mathrm{2}} =\mathrm{p}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} +\mathrm{2prcos}\left(\beta\right) \\ $$$$\mathrm{c}^{\mathrm{2}} =\mathrm{r}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{2rqcos}\left(\varsigma\right) \\ $$$$\mathrm{cos}\left(\alpha\right)=\frac{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} }{\mathrm{2pq}},\mathrm{cos}\left(\beta\right)=\frac{\mathrm{p}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} −\mathrm{q}^{\mathrm{2}} }{\mathrm{2pr}} \\ $$$$\mathrm{cos}\left(\varsigma\right)=\frac{\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} }{\mathrm{2qr}} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} =\mathrm{2p}^{\mathrm{2}} +\mathrm{2q}^{\mathrm{2}} −\mathrm{r}^{\mathrm{2}} ,\mathrm{b}^{\mathrm{2}} =\mathrm{2p}^{\mathrm{2}} +\mathrm{2r}^{\mathrm{2}} −\mathrm{q}^{\mathrm{2}} ,\mathrm{c}^{\mathrm{2}} =\mathrm{2r}^{\mathrm{2}} +\mathrm{2q}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} =\mathrm{3}\left(\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\frac{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} }{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{r}^{\mathrm{2}} }=\mathrm{3} \\ $$$$ \\ $$

Commented by mr W last updated on 30/Nov/19

thanks sir! i got the same result in Q74726.

$${thanks}\:{sir}! \\ $$$${i}\:{got}\:{the}\:{same}\:{result}\:{in}\:{Q}\mathrm{74726}. \\ $$

Commented by mind is power last updated on 30/Nov/19

$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom}\: \\ $$

Facebook Tweet Pin

Leave a Reply Cancel reply