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An equilateral triangle is constructed on each side of a square with side length $2,$ as shown below. The four outer vertices are then joined to form a large square. Find the side length of the large square.

 

https://artofproblemsolving.com/texer/edxnhgjk - for image 

 Mar 6, 2023
 #1
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[asy]
size(4cm);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((0,0)--(0,-1)--(1,-3^.5)--(2,-1)--(2,0));
draw((2,0)--(3,1)--(2,2)--(1,3)--(0,2));
draw((0,2)--(-1,1)--(-3^.5,0)--(-1,-1)--(0,-2));
draw((0,-2)--(1,-3)--(2,-2)--(3,-1)--(2,0),dashed);

[/asy]

Let $O$ be the center of the square. Since the side length of the square is 2, the distance from $O$ to any vertex of the square is $\sqrt{2^2 + 2^2} = 2\sqrt{2}.$

Let $A,$ $B,$ $C,$ $D$ be the vertices of the square, in that order. Let $E,$ $F,$ $G,$ $H$ be the centers of the equilateral triangles with $\overline{AB},$ $\overline{BC},$ $\overline{CD},$ $\overline{DA}$ as bases, respectively.

Let $X$ be the intersection of $\overline{EF}$ and $\overline{OG},$ and let $Y$ be the intersection of $\overline{GH}$ and $\overline{OF}.$

[asy]
size(6cm);
draw((0,0)--(2,0)--(2,2)--(0,2)--cycle);
draw((0,0)--(0,-1)--(1,-3^.5)--(2,-1)--(2,0));
draw((2,0)--(3,1)--(2,2)--(1,3)--(0,2));
draw((0,2)--(-1,1)--(-3^.5,0)--(-1,-1)--(0,-2));
draw((0,-2)--(1,-3)--(2,-2)--(3,-1)--(2,0),dashed);
draw((0,0)--(1,-3^.5)--(2,0),dashed);
draw((0,2)--(-1,1)--(-3^.5,0)--(0,-1)--(0,0),dashed);
draw((2,0)--(3,1)--(2,2)--(1,1)--(1,0),dashed);
label("$A$",(0,2),NW);
label("$B$",(2,2),NE);
label("$C$",(2,0),SE);
label("$D$",(0,0),SW);
label("$E$",(1,-3^.5),S);
label("$F$",(2,0),E);
label("$G$",(-1,1),W);
label("$H$",(0,2),N);
label("$O$",(1,1),NW);
label("$X$",(5/3,-1/3),S);
label("$Y$",(4/3,4/3),E);
draw((1,1)--(5/3,-1/3),dashed);
draw((1,1)--(4/3,4/3),dashed);
[/asy]

Since $\triangle ABE$ is equilateral, $AE = BE = 2.$ Since $O$ is the center of the square, $OE = OD - DE = 2 - 1 = 1.$ Therefore, $OX = OE + EX = 1 + \frac{2}{\sqrt{3}}.$ Similarly, $OY = OE + EY = 1 + \frac{2}{\sqrt{3}}.$

Then, the side length of the large square is $XY = OX + OY = 2 + \frac{4}{\sqrt{3}}.$

Therefore, the answer is $\boxed{2 + \frac{4}{\sqrt{3}}}.$

 Mar 6, 2023
 #2
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0

The side length of the square is 3*sqrt(2).

 Mar 6, 2023

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