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Question Number 216180 by mr W last updated on 29/Jan/25

Answered by AntonCWX last updated on 29/Jan/25

Commented by AntonCWX last updated on 29/Jan/25

$${FG}^2=FC\times FB$$$$6^2=4(CB+4)\Rightarrow CB=5$$$$CE=CB=5$$$$$$$$GE=a$$$$EH=b$$$$ED=2r-a$$$$$$$$GE\times ED=CE\times EH$$$$a(2r-a)=5b$$$$2r-a=\frac{5b}{a}$$$$2r=\frac{5b+a^2}{a}$$$$r=\frac{5b+a^2}{2a}$$$$$$$$From\mathrm{\Delta }FCE,$$$${FE}^2={FC}^2+{CE}^2$$$${FE}^2=4^2+5^2\Rightarrow {FE}^2=41$$$$$$$$From\mathrm{\Delta }FGE,$$$${FE}^2={FG}^2+{GE}^2$$$$41=6^2+a^2$$$$a^2=5\Rightarrow a=\sqrt{5}$$$$$$$$r=\frac{5b+5}{2\sqrt{5}}$$$$$$$$From\mathrm{\Delta }CBH,$$$${BH}^2={BC}^2+{CH}^2$$$${\left(2r\right)}^2=5^2+{(5+b)}^2$$$$4{\left(\frac{5b+5}{2\sqrt{5}}\right)}^2=25+25+10b+b^2$$$$\frac{25b^2+50b+25}{5}=50+10b+b^2$$$$25b^2+50b+25=250+50b+5b^2$$$$20b^2=225\Rightarrow b=\sqrt{\frac{45}{4}}=\frac{3\sqrt{5}}{2}$$$$$$$$r=\frac{5\left(\frac{3\sqrt{5}}{2}\right)+5}{2\sqrt{5}}=\frac{2\sqrt{5}+15}{4}$$

Commented by mr W last updated on 29/Jan/25

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Answered by A5T last updated on 29/Jan/25

Commented by AntonCWX last updated on 30/Jan/25

$$Ithinkyoumistakenlytypedinthe1there...$$$$\frac{2\sqrt{5}+15}{41}$$$$Removethe1anditsdone.$$

Commented by A5T last updated on 30/Jan/25

$$\mathrm{y}=\mathrm{ED}=\mathrm{EF}\Rightarrow 6^2=4(4+\mathrm{y})\Rightarrow \mathrm{y}=5$$$$6^2+{\mathrm{AD}}^2=4^2+5^2\Rightarrow \mathrm{AD}=\sqrt{5}\Rightarrow \mathrm{CD}=\sqrt{41}$$$$\left(\mathrm{AD}\times \mathrm{CE}\right)+\left(\mathrm{AC}\times \mathrm{DE}\right)=\mathrm{CD}\times \mathrm{EA}$$$$\Rightarrow \mathrm{EA}=\frac{4\sqrt{5}+30}{\sqrt{41}}=\mathrm{z}$$$$2\mathrm{r}\left(\sqrt{5}\right)(2\mathrm{r}-\sqrt{5})+5^2\left(2\mathrm{r}\right)=({4\mathrm{r}}^2-{\mathrm{EA}}^2)\sqrt{5}+{\mathrm{EA}}^2(2\mathrm{r}-\sqrt{5})$$$$\Rightarrow \mathrm{r}({\mathrm{EA}}^2-20)={\mathrm{EA}}^2\sqrt{5}\Rightarrow \mathrm{r}=\frac{{\mathrm{EA}}^2\sqrt{5}}{{\mathrm{EA}}^2-20}$$$$\Rightarrow \mathrm{r}=\frac{49\sqrt{5}+60}{8+12\sqrt{5}}=\frac{2\sqrt{5}+15}{4}\approx 4.868$$

Commented by A5T last updated on 30/Jan/25

$$\mathrm{It}\mathrm{was}\mathrm{a}\mathrm{typo},\mathrm{thanks}.$$

Commented by mr W last updated on 29/Jan/25

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Answered by mr W last updated on 29/Jan/25

Commented by mr W last updated on 29/Jan/25

$$4(4+x)=6^2$$$$\Rightarrow x=5$$$$6^2+y^2=4^2+5^2$$$$\Rightarrow y=\sqrt{5}$$$${(R-\sqrt{5})}^2={\left(\frac{x}{2}\right)}^2+{(x-\sqrt{R^2-{\left(\frac{x}{2}\right)}^2})}^2$$$$10+R\sqrt{5}=5\sqrt{R^2-2.5^2}$$$$R^2-\sqrt{5}R-\frac{205}{16}=0$$$$\Rightarrow R=\frac{\sqrt{5}+\sqrt{5+\frac{205}{4}}}{2}=\frac{2\sqrt{5}+15}{4}\approx 4.868$$

Answered by A5T last updated on 29/Jan/25

Commented by A5T last updated on 29/Jan/25

$$4\times (4+\mathrm{EF})=6^2\Rightarrow \mathrm{EF}=\mathrm{ED}=5$$$$6^2+{\mathrm{AD}}^2={\mathrm{CE}}^2+{\mathrm{ED}}^2\Rightarrow \mathrm{AD}=\sqrt{5}$$$$\mathrm{AD}\times \mathrm{DH}=\mathrm{ED}\times \mathrm{DG}\Rightarrow \sqrt{5}(2\mathrm{r}-\sqrt{5})=5\mathrm{DG}...\left(\mathrm{i}\right)$$$$\mathrm{In}△\mathrm{EFG};{\mathrm{GF}}^2={\mathrm{EF}}^2+{\mathrm{EG}}^2$$$$\Rightarrow {4\mathrm{r}}^2=25+{(5+\mathrm{DG})}^2\Rightarrow {4\mathrm{r}}^2={\mathrm{DG}}^2+10\mathrm{DG}...\left(\mathrm{ii}\right)$$$$\left(\mathrm{i}\right)\mathrm{in}\left(\mathrm{ii}\right)$$$$\Rightarrow {4\mathrm{r}}^2=25+{(5+\frac{2\mathrm{r}-\sqrt{5}}{\sqrt{5}})}^2\Rightarrow {4\mathrm{r}}^2=\frac{125+{(2\mathrm{r}+4\sqrt{5})}^2}{5}$$$$\Rightarrow {16\mathrm{r}}^2-16\sqrt{5}\mathrm{r}-205=0$$$$\Rightarrow \mathrm{r}=\frac{16\sqrt{5}\underset{-}{+}\sqrt{{16}^2\left(5\right)-4\left(16\right)(-205)}}{32}=\frac{16\sqrt{5}\underset{-}{+}120}{32}>0$$$$\Rightarrow \mathrm{r}=\frac{2\sqrt{5}+15}{4}$$

Commented by AntonCWX last updated on 31/Jan/25

$$Sameideaasmine\underset{―}{\underbrace{⋚}}$$

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