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Question Number 120569 by prakash jain last updated on 01/Nov/20

Commented by prakash jain last updated on 01/Nov/20

$$\mathrm{There}\mathrm{are}4\mathrm{equilateral}\mathrm{triangles}\mathrm{in}$$$$\mathrm{figure}.\mathrm{Find}\mathrm{ratio}\mathrm{of}\mathrm{colored}\mathrm{region}$$$$\mathrm{to}\mathrm{the}\mathrm{largest}\mathrm{triangle}$$

Commented by MJS_new last updated on 01/Nov/20

$$\mathrm{I}\mathrm{started}\mathrm{with}\mathrm{the}\mathrm{blue}\mathrm{triangle}\mathrm{letting}a=1$$$$\mathrm{we}\mathrm{are}\mathrm{free}\mathrm{to}\mathrm{choose}\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{red}$$$$\mathrm{triangle}\mathrm{with}0<b⩽1$$$$\Rightarrow \mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{red}\mathrm{circle}\mathrm{is}$$$$r=\frac{\sqrt{b^2+b+1}}{\sqrt{3}}$$$$\Rightarrow \mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{yellow}\mathrm{triangle}\mathrm{is}\sqrt{b}\mathrm{and}$$$$\mathrm{the}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{greatest}\mathrm{one}\mathrm{is}2\sqrt{b^2+b+1}$$$$\Rightarrow \mathrm{the}\mathrm{ratio}\mathrm{is}1:4$$

Commented by prakash jain last updated on 01/Nov/20

$$\mathrm{How}\mathrm{did}\mathrm{you}\mathrm{calculate}\mathrm{radius}\mathrm{of}$$$$\mathrm{circle}=\frac{\sqrt{b^2+b+1}}{\sqrt{3}}?$$$$\mathrm{I}\mathrm{dont}\mathrm{most}\mathrm{of}\mathrm{geometry}\mathrm{theorem}$$$$\mathrm{so}\mathrm{want}\mathrm{to}\mathrm{understand}.$$

Commented by MJS_new last updated on 01/Nov/20

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{circumcircle}\mathrm{of}\mathrm{a}\mathrm{triangle}\mathrm{with}\mathrm{sides}$$$$(b+1),1\mathrm{and}\sqrt{b^2+b+1}(\mathrm{blue}\mathrm{and}\mathrm{red}\mathrm{ones}$$$$\mathrm{together})$$$$\mathrm{the}\mathrm{last}\mathrm{is}\sqrt{h^2+{(b+\frac{1}{2})}^2}\mathrm{with}h=\frac{\sqrt{3}}{2}$$$$R=\frac{ABC}{\sqrt{(A+B+C)(-A+B+C)(A-B+C)(A+B-C)}}$$$$\mathrm{here}\mathrm{we}\mathrm{get}R=\frac{\sqrt{b^2+b+1}}{\sqrt{3}}$$

Commented by prakash jain last updated on 01/Nov/20

$$\mathrm{Thanks}.$$

Answered by mr W last updated on 01/Nov/20

Commented by prakash jain last updated on 01/Nov/20

$$\mathrm{Three}\mathrm{sides}\mathrm{of}\mathrm{triangle}\mathrm{are}a,b,c.$$$$\mathrm{Four}\mathrm{point}\mathrm{on}\mathrm{circle}\mathrm{are}\mathrm{then}\mathrm{given}\mathrm{by}$$$$\mathrm{A}=(0,0)$$$$\mathrm{C}=(a+c,0)$$$$D=(\frac{a}{2},-\frac{\sqrt{3}a}{2})$$$$B=(a+\frac{b}{2},-\frac{\sqrt{3}b}{2})$$$$\mathrm{Circle}:x^2+y^2+2fx+2gy=0$$$$C$$$${(a+c)}^2+2f(a+c)=0\Rightarrow f=-(a+c)/2$$$$D$$$$\frac{a^2}{4}+\frac{3a^2}{4}-(a+c)\frac{a}{2}-\sqrt{3}ga=0$$$$\Rightarrow g=\frac{a-c}{2\sqrt{3}}$$$$r^2=f^2+g^2$$$$=\frac{1}{12}(3{(a+c)}^2+{(a-c)}^2)$$$$12r^2=4a^2+4c^2+4ac$$$$a^2+c^2+ac=3r^2$$$$\mathrm{We}\mathrm{need}\mathrm{to}\mathrm{get}$$$$a^2+b^2+c^2\mathrm{in}\mathrm{terms}\mathrm{of}{\mathrm{r}}^2$$$${(a+\frac{b}{2})}^2+\frac{3b^2}{4}-(a+c)(a+\frac{b}{2})$$$$-2\times \left(\frac{a-c}{2\sqrt{3}}\right)\times \frac{\sqrt{3}b}{2}=0$$$$a^2+\frac{b^2}{4}+ab-a^2-ac-\frac{ab}{2}-\frac{bc}{2}+\frac{bc}{2}-\frac{ab}{2}=0$$$$b^2=ac$$$$a^2+b^2+c^2=3r^2$$$$\mathrm{Area}\mathrm{of}\mathrm{colored}\mathrm{region}$$$$=\frac{\sqrt{3}}{4}\times 3r^2=\frac{3\sqrt{3}}{4}{r}^2$$$$\mathrm{Area}=\frac{\sqrt{3}}{4}\times {\left(2\sqrt{3}\mathrm{r}\right)}^2=3\sqrt{3}{r}^2$$$$\mathrm{Ratio}=4$$

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