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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

There are 4 equilateral triangles in figure. Find ratio of colored region to the largest triangle

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

Commented by MJS_new last updated on 01/Nov/20

I started with the blue triangle letting a=1 we are free to choose the side of the red triangle with 0<b≤1 ⇒ the radius of the red circle is r=((√(b^2 +b+1))/( (√3))) ⇒ the side of the yellow triangle is (√b) and the side of the greatest one is 2(√(b^2 +b+1)) ⇒ the ratio is 1:4

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

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

How did you calculate radius of circle = ((√(b^2 +b+1))/( (√3)))? I dont most of geometry theorem so want to understand.

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

Commented by MJS_new last updated on 01/Nov/20

it′s the circumcircle of a triangle with sides (b+1), 1 and (√(b^2 +b+1)) (blue and red ones together) the last is (√(h^2 +(b+(1/2))^2 )) with h=((√3)/2) R=((ABC)/( (√((A+B+C)(−A+B+C)(A−B+C)(A+B−C))))) here we get R=((√(b^2 +b+1))/( (√3)))

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

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

Thanks.

$$\mathrm{Thanks}. \\ $$

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

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

Three sides of triangle are a,b,c. Four point on circle are then given by A=(0,0) C=(a+c,0) D=((a/2),−(((√3)a)/2)) B=(a+(b/2),−(((√3)b)/2)) Circle: x^2 +y^2 +2fx+2gy=0 C (a+c)^2 +2f(a+c)=0⇒f=−(a+c)/2 D (a^2 /4)+((3a^2 )/4)−(a+c)(a/2)−(√3)ga=0 ⇒g=((a−c)/(2(√3))) r^2 =f^2 +g^2 =(1/(12))(3(a+c)^2 +(a−c)^2 ) 12r^2 =4a^2 +4c^2 +4ac a^2 +c^2 +ac=3r^2 We need to get a^2 +b^2 +c^2 in terms of r^2 (a+(b/2))^2 +((3b^2 )/4)−(a+c)(a+(b/2)) −2×(((a−c)/(2(√3))))×(((√3)b)/2)=0 a^2 +(b^2 /4)+ab−a^2 −ac−((ab)/2)−((bc)/2)+((bc)/2)−((ab)/2)=0 b^2 =ac a^2 +b^2 +c^2 =3r^2 Area of colored region =((√3)/4)×3r^2 =((3(√3))/4)r^2 Area=((√3)/4)×(2(√3)r)^2 =3(√3)r^2 Ratio=4

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

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