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Question Number 164305 by ajfour last updated on 16/Jan/22

Commented by ajfour last updated on 16/Jan/22

$$Find\frac{r}{R},ifallcoloredregions$$$$(blue,brown,maroon)have$$$$equalareas.Takeh=7.$$

Answered by mr W last updated on 16/Jan/22

Commented by mr W last updated on 16/Jan/22

$$areaofeachregionis\frac{h\times 1}{3}=\frac{h}{3}$$$${\left(\frac{R+\sqrt{R^2-\frac{b^2}{4}}-1}{R+\sqrt{R^2-\frac{b^2}{4}}}\right)}^2=\frac{1}{2}$$$$\frac{R+\sqrt{R^2-\frac{b^2}{4}}-1}{R+\sqrt{R^2-\frac{b^2}{4}}}=\frac{1}{\sqrt{2}}$$$$\frac{1}{R+\sqrt{R^2-\frac{b^2}{4}}}=1-\frac{1}{\sqrt{2}}=\frac{\sqrt{2}-1}{\sqrt{2}}$$$$R+\sqrt{R^2-\frac{b^2}{4}}=\frac{\sqrt{2}}{\sqrt{2}-1}=2+\sqrt{2}$$$$\sqrt{R^2-\frac{b^2}{4}}=2+\sqrt{2}-R$$$$R^2-\frac{b^2}{4}={(2+\sqrt{2})}^2+R^2-2(2+\sqrt{2})R$$$${(2+\sqrt{2})}^2+\frac{b^2}{4}=2(2+\sqrt{2})R$$$$\Rightarrow R=1+\frac{\sqrt{2}}{2}+\frac{(2-\sqrt{2})b^2}{16}...\left(i\right)$$$$\frac{b}{2}\times (R+\sqrt{R^2-\frac{b^2}{4}})=2\times \frac{h}{3}$$$$\Rightarrow b(R+\sqrt{R^2-\frac{b^2}{4}})=\frac{4h}{3}...\left(ii\right)$$$$$$$$\frac{R+\sqrt{R^2-\frac{b^2}{4}}-1-r}{r}=\frac{2\sqrt{{[R+\sqrt{R^2-\frac{b^2}{4}}]}^2+\frac{b^2}{4}}}{b}$$$$\frac{R+\sqrt{R^2-\frac{b^2}{4}}-1}{r}=1+\frac{2\sqrt{2R(R+\sqrt{R^2-\frac{b^2}{4}})}}{b}$$$$\Rightarrow r=\frac{R+\sqrt{R^2-\frac{b^2}{4}}-1}{1+\frac{2\sqrt{2R(R+\sqrt{R^2-\frac{b^2}{4}})}}{b}}...\left(iii\right)$$$$$$$$example:h=7$$$$\Rightarrow b\approx 2.7337$$$$\Rightarrow R\approx 1.9807$$$$\Rightarrow r\approx 0.6542$$

Commented by mr W last updated on 16/Jan/22

Commented by mr W last updated on 16/Jan/22

$$check:$$$$b=2.7337$$$$R=1.9807$$$$\sqrt{R^2-{\left(\frac{b}{2}\right)}^2}=1.4335$$$$areaofbigtriangle$$$$=\frac{2.7337\times (1.9807+1.4335)}{2}=4.6667$$$$=\frac{7}{3}\times 2✓$$$$areaofsmalltriangle$$$$=4.6667\times {\left(\frac{1.9807+1.4335-1}{1.9807+1.4335}\right)}^2=2.3333$$$$=\frac{7}{3}✓$$

Commented by ajfour last updated on 16/Jan/22

$$Yeah,ialsogotitsir.$$

Commented by Tawa11 last updated on 16/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

Answered by ajfour last updated on 16/Jan/22

Commented by ajfour last updated on 16/Jan/22

$$area=a=\frac{h}{3}=pq$$$$y_B=b=\frac{q}{p}(p+1)$$$$a_{maroon}=b+q=pq$$$$\Rightarrow p+1=p(p-1)$$$$p^2-2p-1=0$$$$p=1+\sqrt{2}$$$$q=\frac{h(\sqrt{2}-1)}{3}$$$$\tan \theta =\frac{q}{p}=\frac{h}{3}\left(\frac{\sqrt{2}-1}{\sqrt{2}+1}\right)=\lambda$$$$\sin \theta =\frac{r}{p-r}\Rightarrow r=\frac{p\sin \theta }{1+\sin \theta }$$$$r=\frac{(\sqrt{2}+1)\lambda }{\lambda +\sqrt{{\lambda }^2+1}}$$$$\tan 2\theta =\frac{b}{p+1-R}$$$$\Rightarrow$$$$R=p+1-\frac{b}{\tan 2\theta }$$$$R=2+\sqrt{2}-\frac{(1-{\lambda }^2)(2+\sqrt{2})}{2}$$$$=(1+\frac{1}{\sqrt{2}})(1+{\lambda }^2)$$$$exampleh=7$$$$2b\approx 2.73$$$$r\approx 0.654$$$$R\approx 1.98$$

Commented by mr W last updated on 16/Jan/22

$$perfect!betterthanmysolutionwhich$$$$deliversonlyapproximatedresult.$$

Commented by Tawa11 last updated on 16/Jan/22

$$\mathrm{Great}\mathrm{sir}$$

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