Question-77390

Question Number 77390 by Maclaurin Stickker last updated on 05/Jan/20

Commented by Maclaurin Stickker last updated on 05/Jan/20

$${find}\:{the}\:{radius}\:{of}\:{the}\:{shaded}\: \\ $$$${circumference}\:{as}\:{a}\:{function}\:{of} \\ $$$${side}\:{a}\:{of}\:{the}\:{square}. \\ $$

Answered by jagoll last updated on 06/Jan/20

$$\mathrm{r}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:=\:\mathrm{a}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$$$\mathrm{r}\:=\:\mathrm{a}\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$

Commented by mr W last updated on 06/Jan/20

$${it}'{s}\:{not}\:{correct}\:{sir}! \\ $$$${the}\:{center}\:{of}\:{the}\:{small}\:{circle}\:{doesn}'{t} \\ $$$${lie}\:{on}\:{the}\:{diagonal}\:{of}\:{the}\:{square}! \\ $$

Commented by jagoll last updated on 06/Jan/20

$$\:\:{o}\:{yes}\:{sir}\:{i}\:{agree} \\ $$

Answered by mr W last updated on 06/Jan/20

Commented by mr W last updated on 06/Jan/20

AC=a−r BC=a+r sin β=((DC)/(AC))=(r/(a−r))=cos α BC^2 =AC^2 +AB^2 −2×AC×AB×cos α (a+r)^2 =(a−r)^2 +a^2 −2(a−r)a×(r/(a−r)) a^2 +r^2 +2ar=a^2 −2ar+r^2 +a^2 −2ar 0=a^2 −6ar ⇒r=(a/6)

$${AC}={a}−{r} \\ $$$${BC}={a}+{r} \\ $$$$\mathrm{sin}\:\beta=\frac{{DC}}{{AC}}=\frac{{r}}{{a}−{r}}=\mathrm{cos}\:\alpha \\ $$$${BC}^{\mathrm{2}} ={AC}^{\mathrm{2}} +{AB}^{\mathrm{2}} −\mathrm{2}×{AC}×{AB}×\mathrm{cos}\:\alpha \\ $$$$\left({a}+{r}\right)^{\mathrm{2}} =\left({a}−{r}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}\left({a}−{r}\right){a}×\frac{{r}}{{a}−{r}} \\ $$$${a}^{\mathrm{2}} +{r}^{\mathrm{2}} +\mathrm{2}{ar}={a}^{\mathrm{2}} −\mathrm{2}{ar}+{r}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{ar} \\ $$$$\mathrm{0}={a}^{\mathrm{2}} −\mathrm{6}{ar} \\ $$$$\Rightarrow{r}=\frac{{a}}{\mathrm{6}} \\ $$

Commented by mr W last updated on 06/Jan/20

or: AD^2 =AC^2 −DC^2 =(a−r)^2 −r^2 =a^2 −2ar BC^2 =AD^2 +(AB−DC)^2 (a+r)^2 =a^2 −2ar+(a−r)^2 ⇒r=(a/6)

$${or}: \\ $$$${AD}^{\mathrm{2}} ={AC}^{\mathrm{2}} −{DC}^{\mathrm{2}} =\left({a}−{r}\right)^{\mathrm{2}} −{r}^{\mathrm{2}} ={a}^{\mathrm{2}} −\mathrm{2}{ar} \\ $$$${BC}^{\mathrm{2}} ={AD}^{\mathrm{2}} +\left({AB}−{DC}\right)^{\mathrm{2}} \\ $$$$\left({a}+{r}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} −\mathrm{2}{ar}+\left({a}−{r}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{{a}}{\mathrm{6}} \\ $$

Commented by jagoll last updated on 06/Jan/20