How to maximize area of a square inscribed in a equilateral triangle?Find the maximum area possible of equilateral triangle that inside the given squareArea of an equilateral triangleHow to find the sides of an equilateral triangle given all angles.Triangle perimeter and areaSquare with equilateral triangle drawn it it, find area of the triangle.Equilateral triangle touching three sides of a squareHow fast is the area increasing for an equilateral triangle under the given conditions?Maximize area of triangleMaximal area of equilateral triangle inside rectangle.
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How to maximize area of a square inscribed in a equilateral triangle?
Find the maximum area possible of equilateral triangle that inside the given squareArea of an equilateral triangleHow to find the sides of an equilateral triangle given all angles.Triangle perimeter and areaSquare with equilateral triangle drawn it it, find area of the triangle.Equilateral triangle touching three sides of a squareHow fast is the area increasing for an equilateral triangle under the given conditions?Maximize area of triangleMaximal area of equilateral triangle inside rectangle.
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We have an equilateral triangle and want to inscribe a square, but want to do so in the way that maximizes the area of the square.
I sketched two possible ways, not to scale and not perfect. Note I am not sure if the second way will really have all square corners touching the triangle sides.
The second case appears to have bigger side-lengths of the square, so bigger area. But I do not know how to determine the angles involved. How to solve this?
geometry euclidean-geometry triangles area
edited Aug 13 at 3:50
Michael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
asked Aug 13 at 2:38
DrZ214DrZ214
1,1695 silver badges14 bronze badges
$endgroup$
|
show 4 more comments
$begingroup$
We have an equilateral triangle and want to inscribe a square, but want to do so in the way that maximizes the area of the square.
I sketched two possible ways, not to scale and not perfect. Note I am not sure if the second way will really have all square corners touching the triangle sides.
The second case appears to have bigger side-lengths of the square, so bigger area. But I do not know how to determine the angles involved. How to solve this?
geometry euclidean-geometry triangles area
edited Aug 13 at 3:50
Michael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
asked Aug 13 at 2:38
DrZ214DrZ214
1,1695 silver badges14 bronze badges
$endgroup$
$begingroup$
@TobyMak This isn't a duplicate, since OP is asking which of two specific configurations is better.
$endgroup$
– Parcly Taxel
Aug 13 at 2:53
3
$begingroup$
How do you define "inscribed"? In your first sketch, one corner of the square is actually floating, not touching the triangle. I think it is safe to assume that there is only one way to inscribe a square in an equilateral triangle, and it is when one side of the square lays exactly on one side of the triangle (as in your second sketch). As a result, the problem of maximization is a non-problem.
$endgroup$
– virolino
Aug 13 at 11:04
$begingroup$
@ParclyTaxel Are they? Because that's not the question asked in the title.
$endgroup$
– Jack M
Aug 13 at 11:22
$begingroup$
@JackM It is in the consideration of the title problem that the real question has been asked.
$endgroup$
– Parcly Taxel
Aug 13 at 11:24
$begingroup$
A very nice problem indeed! Do you consider the first case to be symmetric also, like the answer below? Or just in general, with just 3 of the tips of a square, touching the triangle and the fourth just somewhere inside the area of the triangle?
$endgroup$
– dmtri
Aug 18 at 9:43
|
show 4 more comments
$begingroup$
We have an equilateral triangle and want to inscribe a square, but want to do so in the way that maximizes the area of the square.
I sketched two possible ways, not to scale and not perfect. Note I am not sure if the second way will really have all square corners touching the triangle sides.
The second case appears to have bigger side-lengths of the square, so bigger area. But I do not know how to determine the angles involved. How to solve this?
geometry euclidean-geometry triangles area
edited Aug 13 at 3:50
Michael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
asked Aug 13 at 2:38
DrZ214DrZ214
1,1695 silver badges14 bronze badges
$endgroup$
We have an equilateral triangle and want to inscribe a square, but want to do so in the way that maximizes the area of the square.
I sketched two possible ways, not to scale and not perfect. Note I am not sure if the second way will really have all square corners touching the triangle sides.
The second case appears to have bigger side-lengths of the square, so bigger area. But I do not know how to determine the angles involved. How to solve this?
geometry euclidean-geometry triangles area
geometry euclidean-geometry triangles area
edited Aug 13 at 3:50
Michael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
asked Aug 13 at 2:38
DrZ214DrZ214
1,1695 silver badges14 bronze badges
edited Aug 13 at 3:50
Michael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
asked Aug 13 at 2:38
DrZ214DrZ214
1,1695 silver badges14 bronze badges
edited Aug 13 at 3:50
Michael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
edited Aug 13 at 3:50
Michael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
edited Aug 13 at 3:50
Michael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
133k25 gold badges108 silver badges216 bronze badges
asked Aug 13 at 2:38
DrZ214DrZ214
1,1695 silver badges14 bronze badges
asked Aug 13 at 2:38
DrZ214DrZ214
1,1695 silver badges14 bronze badges
asked Aug 13 at 2:38
DrZ214DrZ214
1,1695 silver badges14 bronze badges
1,1695 silver badges14 bronze badges
$begingroup$
@TobyMak This isn't a duplicate, since OP is asking which of two specific configurations is better.
$endgroup$
– Parcly Taxel
Aug 13 at 2:53
3
$begingroup$
How do you define "inscribed"? In your first sketch, one corner of the square is actually floating, not touching the triangle. I think it is safe to assume that there is only one way to inscribe a square in an equilateral triangle, and it is when one side of the square lays exactly on one side of the triangle (as in your second sketch). As a result, the problem of maximization is a non-problem.
$endgroup$
– virolino
Aug 13 at 11:04
$begingroup$
@ParclyTaxel Are they? Because that's not the question asked in the title.
$endgroup$
– Jack M
Aug 13 at 11:22
$begingroup$
@JackM It is in the consideration of the title problem that the real question has been asked.
$endgroup$
– Parcly Taxel
Aug 13 at 11:24
$begingroup$
A very nice problem indeed! Do you consider the first case to be symmetric also, like the answer below? Or just in general, with just 3 of the tips of a square, touching the triangle and the fourth just somewhere inside the area of the triangle?
$endgroup$
– dmtri
Aug 18 at 9:43
|
show 4 more comments
$begingroup$
@TobyMak This isn't a duplicate, since OP is asking which of two specific configurations is better.
$endgroup$
– Parcly Taxel
Aug 13 at 2:53
3
$begingroup$
How do you define "inscribed"? In your first sketch, one corner of the square is actually floating, not touching the triangle. I think it is safe to assume that there is only one way to inscribe a square in an equilateral triangle, and it is when one side of the square lays exactly on one side of the triangle (as in your second sketch). As a result, the problem of maximization is a non-problem.
$endgroup$
– virolino
Aug 13 at 11:04
$begingroup$
@ParclyTaxel Are they? Because that's not the question asked in the title.
$endgroup$
– Jack M
Aug 13 at 11:22
$begingroup$
@JackM It is in the consideration of the title problem that the real question has been asked.
$endgroup$
– Parcly Taxel
Aug 13 at 11:24
$begingroup$
A very nice problem indeed! Do you consider the first case to be symmetric also, like the answer below? Or just in general, with just 3 of the tips of a square, touching the triangle and the fourth just somewhere inside the area of the triangle?
$endgroup$
– dmtri
Aug 18 at 9:43
$begingroup$
@TobyMak This isn't a duplicate, since OP is asking which of two specific configurations is better.
$endgroup$
– Parcly Taxel
Aug 13 at 2:53
$begingroup$
@TobyMak This isn't a duplicate, since OP is asking which of two specific configurations is better.
$endgroup$
– Parcly Taxel
Aug 13 at 2:53
3
3
$begingroup$
How do you define "inscribed"? In your first sketch, one corner of the square is actually floating, not touching the triangle. I think it is safe to assume that there is only one way to inscribe a square in an equilateral triangle, and it is when one side of the square lays exactly on one side of the triangle (as in your second sketch). As a result, the problem of maximization is a non-problem.
$endgroup$
– virolino
Aug 13 at 11:04
$begingroup$
How do you define "inscribed"? In your first sketch, one corner of the square is actually floating, not touching the triangle. I think it is safe to assume that there is only one way to inscribe a square in an equilateral triangle, and it is when one side of the square lays exactly on one side of the triangle (as in your second sketch). As a result, the problem of maximization is a non-problem.
$endgroup$
– virolino
Aug 13 at 11:04
$begingroup$
@ParclyTaxel Are they? Because that's not the question asked in the title.
$endgroup$
– Jack M
Aug 13 at 11:22
$begingroup$
@ParclyTaxel Are they? Because that's not the question asked in the title.
$endgroup$
– Jack M
Aug 13 at 11:22
$begingroup$
@JackM It is in the consideration of the title problem that the real question has been asked.
$endgroup$
– Parcly Taxel
Aug 13 at 11:24
$begingroup$
@JackM It is in the consideration of the title problem that the real question has been asked.
$endgroup$
– Parcly Taxel
Aug 13 at 11:24
$begingroup$
A very nice problem indeed! Do you consider the first case to be symmetric also, like the answer below? Or just in general, with just 3 of the tips of a square, touching the triangle and the fourth just somewhere inside the area of the triangle?
$endgroup$
– dmtri
Aug 18 at 9:43
$begingroup$
A very nice problem indeed! Do you consider the first case to be symmetric also, like the answer below? Or just in general, with just 3 of the tips of a square, touching the triangle and the fourth just somewhere inside the area of the triangle?
$endgroup$
– dmtri
Aug 18 at 9:43
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Let $a_1$ and $a_2$ be the side lengths of the two squares. To determine which one is larger, we simply look at their ratio below.
With the angles in the diagram,
$$d_1=frac12tan 30a_1=fracsqrt32a_1$$
$$d_2=fracsin 15sin 30a_2=frac12cos 15a_2$$
Assume both equilateral triangles have unit height.
$$1=a_1+d_1=left(1+fracsqrt32right)a_1=frac12(2+sqrt3)a_1$$
$$1=sqrt2a_2+d_2=left(sqrt2+frac12cos 15right)a_2=frac12(sqrt2+sqrt6)a_2$$
So, their ratio is
$$fraca_1a_2= fracsqrt2+sqrt62+sqrt3 =left(frac8+4sqrt37+4sqrt3right)^frac12 > 1$$
answered Aug 13 at 3:55
QuantoQuanto
12.9k1 gold badge5 silver badges21 bronze badges
$endgroup$
add a comment
|
$begingroup$
The second configuration (square has edge contact with triangle) indeed has a bigger inscribed square. If the square has unit sides, the triangle's side is $1+frac2sqrt3$:
The symmetric first configuration may be resolved as follows. Set the unit square's bottom corner as $(0,0)$, so that the top corner is $(0,sqrt2)$. Let the side length of the triangle be $r$. Then we have, by similar triangles,
$$frac(sqrt3/2)r-sqrt2/2sqrt2/2=sqrt3$$
$$r=(1+sqrt3)sqrtfrac23=2.230dots$$
and this is greater than $1+frac2sqrt3=2.154dots$, so the first configuration has a smaller inscribed square than the second.
answered Aug 13 at 2:53
Parcly TaxelParcly Taxel
53.4k13 gold badges81 silver badges120 bronze badges
$endgroup$
$begingroup$
Maybe, you mean the second, in your the sentence before last?
$endgroup$
– dmtri
Aug 18 at 9:51
1
$begingroup$
@dmtri I got my words right there, checking again.
$endgroup$
– Parcly Taxel
Aug 18 at 9:52
$begingroup$
Sorry , you are right , you are talking about the sides of the triangle not the square....
$endgroup$
– dmtri
Aug 20 at 18:36
add a comment
|
$begingroup$
Let sides-lengths of the equilateral triangle be equal to $1$.
Let $x$ be sides-lengths of the square in the first configuration.
Thus, by law of sines we obtain:
$$fracxsin60^circ=fracfrac12sin75^circ$$ or
$$fracxfracsqrt32=fracfrac12frac1+sqrt32sqrt2$$ or
$$x=fracsqrt3sqrt2(1+sqrt3)$$ and for the area of the square we obtain:
$$frac32(4+2sqrt3)=frac34(2-sqrt3).$$
Let $y$ be sides-lengths of the square in the second configuration.
Thus, by similarity we obtain:
$$fracy1=fracfracsqrt32-yfracsqrt32$$ or
$$y=sqrt3(2-sqrt3)$$
and for the area of the square we obtain:
$$3(7-4sqrt3),$$ which is a bit of greater.
answered Aug 13 at 3:45
Michael RozenbergMichael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
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add a comment
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $a_1$ and $a_2$ be the side lengths of the two squares. To determine which one is larger, we simply look at their ratio below.
With the angles in the diagram,
$$d_1=frac12tan 30a_1=fracsqrt32a_1$$
$$d_2=fracsin 15sin 30a_2=frac12cos 15a_2$$
Assume both equilateral triangles have unit height.
$$1=a_1+d_1=left(1+fracsqrt32right)a_1=frac12(2+sqrt3)a_1$$
$$1=sqrt2a_2+d_2=left(sqrt2+frac12cos 15right)a_2=frac12(sqrt2+sqrt6)a_2$$
So, their ratio is
$$fraca_1a_2= fracsqrt2+sqrt62+sqrt3 =left(frac8+4sqrt37+4sqrt3right)^frac12 > 1$$
answered Aug 13 at 3:55
QuantoQuanto
12.9k1 gold badge5 silver badges21 bronze badges
$endgroup$
add a comment
|
$begingroup$
Let $a_1$ and $a_2$ be the side lengths of the two squares. To determine which one is larger, we simply look at their ratio below.
With the angles in the diagram,
$$d_1=frac12tan 30a_1=fracsqrt32a_1$$
$$d_2=fracsin 15sin 30a_2=frac12cos 15a_2$$
Assume both equilateral triangles have unit height.
$$1=a_1+d_1=left(1+fracsqrt32right)a_1=frac12(2+sqrt3)a_1$$
$$1=sqrt2a_2+d_2=left(sqrt2+frac12cos 15right)a_2=frac12(sqrt2+sqrt6)a_2$$
So, their ratio is
$$fraca_1a_2= fracsqrt2+sqrt62+sqrt3 =left(frac8+4sqrt37+4sqrt3right)^frac12 > 1$$
answered Aug 13 at 3:55
QuantoQuanto
12.9k1 gold badge5 silver badges21 bronze badges
$endgroup$
add a comment
|
$begingroup$
Let $a_1$ and $a_2$ be the side lengths of the two squares. To determine which one is larger, we simply look at their ratio below.
With the angles in the diagram,
$$d_1=frac12tan 30a_1=fracsqrt32a_1$$
$$d_2=fracsin 15sin 30a_2=frac12cos 15a_2$$
Assume both equilateral triangles have unit height.
$$1=a_1+d_1=left(1+fracsqrt32right)a_1=frac12(2+sqrt3)a_1$$
$$1=sqrt2a_2+d_2=left(sqrt2+frac12cos 15right)a_2=frac12(sqrt2+sqrt6)a_2$$
So, their ratio is
$$fraca_1a_2= fracsqrt2+sqrt62+sqrt3 =left(frac8+4sqrt37+4sqrt3right)^frac12 > 1$$
answered Aug 13 at 3:55
QuantoQuanto
12.9k1 gold badge5 silver badges21 bronze badges
$endgroup$
Let $a_1$ and $a_2$ be the side lengths of the two squares. To determine which one is larger, we simply look at their ratio below.
With the angles in the diagram,
$$d_1=frac12tan 30a_1=fracsqrt32a_1$$
$$d_2=fracsin 15sin 30a_2=frac12cos 15a_2$$
Assume both equilateral triangles have unit height.
$$1=a_1+d_1=left(1+fracsqrt32right)a_1=frac12(2+sqrt3)a_1$$
$$1=sqrt2a_2+d_2=left(sqrt2+frac12cos 15right)a_2=frac12(sqrt2+sqrt6)a_2$$
So, their ratio is
$$fraca_1a_2= fracsqrt2+sqrt62+sqrt3 =left(frac8+4sqrt37+4sqrt3right)^frac12 > 1$$
answered Aug 13 at 3:55
QuantoQuanto
12.9k1 gold badge5 silver badges21 bronze badges
edited Aug 16 at 17:30
answered Aug 13 at 3:55
QuantoQuanto
12.9k1 gold badge5 silver badges21 bronze badges
answered Aug 13 at 3:55
QuantoQuanto
12.9k1 gold badge5 silver badges21 bronze badges
answered Aug 13 at 3:55
QuantoQuanto
12.9k1 gold badge5 silver badges21 bronze badges
12.9k1 gold badge5 silver badges21 bronze badges
add a comment
|
add a comment
|
$begingroup$
The second configuration (square has edge contact with triangle) indeed has a bigger inscribed square. If the square has unit sides, the triangle's side is $1+frac2sqrt3$:
The symmetric first configuration may be resolved as follows. Set the unit square's bottom corner as $(0,0)$, so that the top corner is $(0,sqrt2)$. Let the side length of the triangle be $r$. Then we have, by similar triangles,
$$frac(sqrt3/2)r-sqrt2/2sqrt2/2=sqrt3$$
$$r=(1+sqrt3)sqrtfrac23=2.230dots$$
and this is greater than $1+frac2sqrt3=2.154dots$, so the first configuration has a smaller inscribed square than the second.
answered Aug 13 at 2:53
Parcly TaxelParcly Taxel
53.4k13 gold badges81 silver badges120 bronze badges
$endgroup$
$begingroup$
Maybe, you mean the second, in your the sentence before last?
$endgroup$
– dmtri
Aug 18 at 9:51
1
$begingroup$
@dmtri I got my words right there, checking again.
$endgroup$
– Parcly Taxel
Aug 18 at 9:52
$begingroup$
Sorry , you are right , you are talking about the sides of the triangle not the square....
$endgroup$
– dmtri
Aug 20 at 18:36
add a comment
|
$begingroup$
The second configuration (square has edge contact with triangle) indeed has a bigger inscribed square. If the square has unit sides, the triangle's side is $1+frac2sqrt3$:
The symmetric first configuration may be resolved as follows. Set the unit square's bottom corner as $(0,0)$, so that the top corner is $(0,sqrt2)$. Let the side length of the triangle be $r$. Then we have, by similar triangles,
$$frac(sqrt3/2)r-sqrt2/2sqrt2/2=sqrt3$$
$$r=(1+sqrt3)sqrtfrac23=2.230dots$$
and this is greater than $1+frac2sqrt3=2.154dots$, so the first configuration has a smaller inscribed square than the second.
answered Aug 13 at 2:53
Parcly TaxelParcly Taxel
53.4k13 gold badges81 silver badges120 bronze badges
$endgroup$
$begingroup$
Maybe, you mean the second, in your the sentence before last?
$endgroup$
– dmtri
Aug 18 at 9:51
1
$begingroup$
@dmtri I got my words right there, checking again.
$endgroup$
– Parcly Taxel
Aug 18 at 9:52
$begingroup$
Sorry , you are right , you are talking about the sides of the triangle not the square....
$endgroup$
– dmtri
Aug 20 at 18:36
add a comment
|
$begingroup$
The second configuration (square has edge contact with triangle) indeed has a bigger inscribed square. If the square has unit sides, the triangle's side is $1+frac2sqrt3$:
The symmetric first configuration may be resolved as follows. Set the unit square's bottom corner as $(0,0)$, so that the top corner is $(0,sqrt2)$. Let the side length of the triangle be $r$. Then we have, by similar triangles,
$$frac(sqrt3/2)r-sqrt2/2sqrt2/2=sqrt3$$
$$r=(1+sqrt3)sqrtfrac23=2.230dots$$
and this is greater than $1+frac2sqrt3=2.154dots$, so the first configuration has a smaller inscribed square than the second.
answered Aug 13 at 2:53
Parcly TaxelParcly Taxel
53.4k13 gold badges81 silver badges120 bronze badges
$endgroup$
The second configuration (square has edge contact with triangle) indeed has a bigger inscribed square. If the square has unit sides, the triangle's side is $1+frac2sqrt3$:
The symmetric first configuration may be resolved as follows. Set the unit square's bottom corner as $(0,0)$, so that the top corner is $(0,sqrt2)$. Let the side length of the triangle be $r$. Then we have, by similar triangles,
$$frac(sqrt3/2)r-sqrt2/2sqrt2/2=sqrt3$$
$$r=(1+sqrt3)sqrtfrac23=2.230dots$$
and this is greater than $1+frac2sqrt3=2.154dots$, so the first configuration has a smaller inscribed square than the second.
answered Aug 13 at 2:53
Parcly TaxelParcly Taxel
53.4k13 gold badges81 silver badges120 bronze badges
answered Aug 13 at 2:53
Parcly TaxelParcly Taxel
53.4k13 gold badges81 silver badges120 bronze badges
answered Aug 13 at 2:53
Parcly TaxelParcly Taxel
53.4k13 gold badges81 silver badges120 bronze badges
answered Aug 13 at 2:53
Parcly TaxelParcly Taxel
53.4k13 gold badges81 silver badges120 bronze badges
53.4k13 gold badges81 silver badges120 bronze badges
$begingroup$
Maybe, you mean the second, in your the sentence before last?
$endgroup$
– dmtri
Aug 18 at 9:51
1
$begingroup$
@dmtri I got my words right there, checking again.
$endgroup$
– Parcly Taxel
Aug 18 at 9:52
$begingroup$
Sorry , you are right , you are talking about the sides of the triangle not the square....
$endgroup$
– dmtri
Aug 20 at 18:36
add a comment
|
$begingroup$
Maybe, you mean the second, in your the sentence before last?
$endgroup$
– dmtri
Aug 18 at 9:51
1
$begingroup$
@dmtri I got my words right there, checking again.
$endgroup$
– Parcly Taxel
Aug 18 at 9:52
$begingroup$
Sorry , you are right , you are talking about the sides of the triangle not the square....
$endgroup$
– dmtri
Aug 20 at 18:36
$begingroup$
Maybe, you mean the second, in your the sentence before last?
$endgroup$
– dmtri
Aug 18 at 9:51
$begingroup$
Maybe, you mean the second, in your the sentence before last?
$endgroup$
– dmtri
Aug 18 at 9:51
1
1
$begingroup$
@dmtri I got my words right there, checking again.
$endgroup$
– Parcly Taxel
Aug 18 at 9:52
$begingroup$
@dmtri I got my words right there, checking again.
$endgroup$
– Parcly Taxel
Aug 18 at 9:52
$begingroup$
Sorry , you are right , you are talking about the sides of the triangle not the square....
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– dmtri
Aug 20 at 18:36
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Sorry , you are right , you are talking about the sides of the triangle not the square....
$endgroup$
– dmtri
Aug 20 at 18:36
add a comment
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$begingroup$
Let sides-lengths of the equilateral triangle be equal to $1$.
Let $x$ be sides-lengths of the square in the first configuration.
Thus, by law of sines we obtain:
$$fracxsin60^circ=fracfrac12sin75^circ$$ or
$$fracxfracsqrt32=fracfrac12frac1+sqrt32sqrt2$$ or
$$x=fracsqrt3sqrt2(1+sqrt3)$$ and for the area of the square we obtain:
$$frac32(4+2sqrt3)=frac34(2-sqrt3).$$
Let $y$ be sides-lengths of the square in the second configuration.
Thus, by similarity we obtain:
$$fracy1=fracfracsqrt32-yfracsqrt32$$ or
$$y=sqrt3(2-sqrt3)$$
and for the area of the square we obtain:
$$3(7-4sqrt3),$$ which is a bit of greater.
answered Aug 13 at 3:45
Michael RozenbergMichael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
$endgroup$
add a comment
|
$begingroup$
Let sides-lengths of the equilateral triangle be equal to $1$.
Let $x$ be sides-lengths of the square in the first configuration.
Thus, by law of sines we obtain:
$$fracxsin60^circ=fracfrac12sin75^circ$$ or
$$fracxfracsqrt32=fracfrac12frac1+sqrt32sqrt2$$ or
$$x=fracsqrt3sqrt2(1+sqrt3)$$ and for the area of the square we obtain:
$$frac32(4+2sqrt3)=frac34(2-sqrt3).$$
Let $y$ be sides-lengths of the square in the second configuration.
Thus, by similarity we obtain:
$$fracy1=fracfracsqrt32-yfracsqrt32$$ or
$$y=sqrt3(2-sqrt3)$$
and for the area of the square we obtain:
$$3(7-4sqrt3),$$ which is a bit of greater.
answered Aug 13 at 3:45
Michael RozenbergMichael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
$endgroup$
add a comment
|
$begingroup$
Let sides-lengths of the equilateral triangle be equal to $1$.
Let $x$ be sides-lengths of the square in the first configuration.
Thus, by law of sines we obtain:
$$fracxsin60^circ=fracfrac12sin75^circ$$ or
$$fracxfracsqrt32=fracfrac12frac1+sqrt32sqrt2$$ or
$$x=fracsqrt3sqrt2(1+sqrt3)$$ and for the area of the square we obtain:
$$frac32(4+2sqrt3)=frac34(2-sqrt3).$$
Let $y$ be sides-lengths of the square in the second configuration.
Thus, by similarity we obtain:
$$fracy1=fracfracsqrt32-yfracsqrt32$$ or
$$y=sqrt3(2-sqrt3)$$
and for the area of the square we obtain:
$$3(7-4sqrt3),$$ which is a bit of greater.
answered Aug 13 at 3:45
Michael RozenbergMichael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
$endgroup$
Let sides-lengths of the equilateral triangle be equal to $1$.
Let $x$ be sides-lengths of the square in the first configuration.
Thus, by law of sines we obtain:
$$fracxsin60^circ=fracfrac12sin75^circ$$ or
$$fracxfracsqrt32=fracfrac12frac1+sqrt32sqrt2$$ or
$$x=fracsqrt3sqrt2(1+sqrt3)$$ and for the area of the square we obtain:
$$frac32(4+2sqrt3)=frac34(2-sqrt3).$$
Let $y$ be sides-lengths of the square in the second configuration.
Thus, by similarity we obtain:
$$fracy1=fracfracsqrt32-yfracsqrt32$$ or
$$y=sqrt3(2-sqrt3)$$
and for the area of the square we obtain:
$$3(7-4sqrt3),$$ which is a bit of greater.
answered Aug 13 at 3:45
Michael RozenbergMichael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
answered Aug 13 at 3:45
Michael RozenbergMichael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
answered Aug 13 at 3:45
Michael RozenbergMichael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
answered Aug 13 at 3:45
Michael RozenbergMichael Rozenberg
133k25 gold badges108 silver badges216 bronze badges
133k25 gold badges108 silver badges216 bronze badges
add a comment
|
add a comment
|
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@TobyMak This isn't a duplicate, since OP is asking which of two specific configurations is better.
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– Parcly Taxel
Aug 13 at 2:53
3
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How do you define "inscribed"? In your first sketch, one corner of the square is actually floating, not touching the triangle. I think it is safe to assume that there is only one way to inscribe a square in an equilateral triangle, and it is when one side of the square lays exactly on one side of the triangle (as in your second sketch). As a result, the problem of maximization is a non-problem.
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– virolino
Aug 13 at 11:04
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@ParclyTaxel Are they? Because that's not the question asked in the title.
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– Jack M
Aug 13 at 11:22
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@JackM It is in the consideration of the title problem that the real question has been asked.
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– Parcly Taxel
Aug 13 at 11:24
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A very nice problem indeed! Do you consider the first case to be symmetric also, like the answer below? Or just in general, with just 3 of the tips of a square, touching the triangle and the fourth just somewhere inside the area of the triangle?
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– dmtri
Aug 18 at 9:43