Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com