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Of all polygons inscribed in a given circle which one has the maximum sum of squares of side lengths?
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My son presented me with an interesting problem:
Of all possible polygons inscribed in a circle of radius $R$, find the
one that has the sum $S$ of squared side lengths maximized:
$S=a_1^2+a_2^2+dots+a_n^2$, with $a_i$ representing the length of the
$i$-th side. The number of sides is not fixed, you should consider all
triangles, quadrilaterals, pentagons...
It's not that complicated, at least in the beginning. It's easy to show that the optimal polygon (with $n>3$) cannot have obtuse ($>90^circ$) angles. For example, if such an angle $A_i-1A_iA_i+1$ exists, by cosine theorem:
$$|A_i-1A_i|^2+|A_iA_i+1|^2<|A_i-1A_i+1|^2$$
So if you drop vertex $A_i$, you get a polygon with a bigger $S$. This quickly eliminates all polygons with $n>4$.
All candidate polygons with $n=4$ must be rectangles and if their sides are $a$ and $b$, the sum $S$ is $2a^2+2b^2=8R^2$. So with respect to $S$ all rectangles inscribed in the circle are equivalent. In fact, a right triangle with sides $a$, $b$ and $2R$ has the same $S$ as any inscribed rectangle.
But maybe there is an inscribed triangle with $S>8R^2$. I was able to show that for an inscribed triangle with sides $a,b,c$ and $bne c$, an isosceles triangle with all acute angles and base $a$ has better value of $S$. So the optimal triangle must be isosceles. Looking from all three sides, the only possible solution is the equilateral triangle and the sum $S$ in that case is $9R^2$.
However, to prove that fact I had to use trigonometry which is not so complicated (and I can present it here if you want so), but it leaves impression that there has to be some simpler explanation why the equilateral triangle is the best choice. My trigonometry proof takes a few lines of text, I want something more elegant.
Just an idea: if you draw lines through the center of the circle perpendicular to the sides of a triangle and denote the pedal lengths with $h_a,h_b,h_c$, it's easy to show that in order to maximize $a^2+b^2+c^2$ you have to minimize $h_a^2+h_b^2+h_c^2$. And then - what?
EDIT: I want to present the part of the proof that I don't like. Take an arbitrary triangle $ABC$ with sides $a,b,c$ inscribed in a circle. Consider side $a$ fixed and play with angle $gamma$ to get different values of $b,c$. I want to prove that isosceles triangle $BCA_1$ has bigger $S$ than any other triangle with one side equal to $a$.
$$b=2Rsinfracpi-alpha+gamma2=2Rcos(fracalpha2-fracgamma2)$$
$$c=2Rsinfracpi-alpha-gamma2=2Rcos(fracalpha2+fracgamma2)$$
$$b^2=4R^2cos^2(fracalpha2-fracgamma2)=2R^2(1+cos(alpha-gamma))$$
$$c^2=4R^2cos^2(fracalpha2+fracgamma2)=2R^2(1+cos(alpha+gamma))$$
$$b^2+c^2=4R^2+2R^2(cos(alpha-gamma)+cos(alpha+gamma))=4R^2(1+cosalphacosgamma)$$
And this sum achieves maximum obviously for $gamma=0$, or for $Aequiv A_1$. So for any given side $a$, $b$ and $c$ must be of equal. But you can look at the optimal triangle from sides $b$ and $c$ as well. The only triangle which has no better option is equilateral triangle.
euclidean-geometry triangle
asked 2 hours ago
Oldboy
3,1211319
add a comment |Â
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5
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My son presented me with an interesting problem:
Of all possible polygons inscribed in a circle of radius $R$, find the
one that has the sum $S$ of squared side lengths maximized:
$S=a_1^2+a_2^2+dots+a_n^2$, with $a_i$ representing the length of the
$i$-th side. The number of sides is not fixed, you should consider all
triangles, quadrilaterals, pentagons...
It's not that complicated, at least in the beginning. It's easy to show that the optimal polygon (with $n>3$) cannot have obtuse ($>90^circ$) angles. For example, if such an angle $A_i-1A_iA_i+1$ exists, by cosine theorem:
$$|A_i-1A_i|^2+|A_iA_i+1|^2<|A_i-1A_i+1|^2$$
So if you drop vertex $A_i$, you get a polygon with a bigger $S$. This quickly eliminates all polygons with $n>4$.
All candidate polygons with $n=4$ must be rectangles and if their sides are $a$ and $b$, the sum $S$ is $2a^2+2b^2=8R^2$. So with respect to $S$ all rectangles inscribed in the circle are equivalent. In fact, a right triangle with sides $a$, $b$ and $2R$ has the same $S$ as any inscribed rectangle.
But maybe there is an inscribed triangle with $S>8R^2$. I was able to show that for an inscribed triangle with sides $a,b,c$ and $bne c$, an isosceles triangle with all acute angles and base $a$ has better value of $S$. So the optimal triangle must be isosceles. Looking from all three sides, the only possible solution is the equilateral triangle and the sum $S$ in that case is $9R^2$.
However, to prove that fact I had to use trigonometry which is not so complicated (and I can present it here if you want so), but it leaves impression that there has to be some simpler explanation why the equilateral triangle is the best choice. My trigonometry proof takes a few lines of text, I want something more elegant.
Just an idea: if you draw lines through the center of the circle perpendicular to the sides of a triangle and denote the pedal lengths with $h_a,h_b,h_c$, it's easy to show that in order to maximize $a^2+b^2+c^2$ you have to minimize $h_a^2+h_b^2+h_c^2$. And then - what?
EDIT: I want to present the part of the proof that I don't like. Take an arbitrary triangle $ABC$ with sides $a,b,c$ inscribed in a circle. Consider side $a$ fixed and play with angle $gamma$ to get different values of $b,c$. I want to prove that isosceles triangle $BCA_1$ has bigger $S$ than any other triangle with one side equal to $a$.
$$b=2Rsinfracpi-alpha+gamma2=2Rcos(fracalpha2-fracgamma2)$$
$$c=2Rsinfracpi-alpha-gamma2=2Rcos(fracalpha2+fracgamma2)$$
$$b^2=4R^2cos^2(fracalpha2-fracgamma2)=2R^2(1+cos(alpha-gamma))$$
$$c^2=4R^2cos^2(fracalpha2+fracgamma2)=2R^2(1+cos(alpha+gamma))$$
$$b^2+c^2=4R^2+2R^2(cos(alpha-gamma)+cos(alpha+gamma))=4R^2(1+cosalphacosgamma)$$
And this sum achieves maximum obviously for $gamma=0$, or for $Aequiv A_1$. So for any given side $a$, $b$ and $c$ must be of equal. But you can look at the optimal triangle from sides $b$ and $c$ as well. The only triangle which has no better option is equilateral triangle.
euclidean-geometry triangle
asked 2 hours ago
Oldboy
3,1211319
add a comment |Â
up vote
5
down vote
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up vote
5
down vote
favorite
My son presented me with an interesting problem:
Of all possible polygons inscribed in a circle of radius $R$, find the
one that has the sum $S$ of squared side lengths maximized:
$S=a_1^2+a_2^2+dots+a_n^2$, with $a_i$ representing the length of the
$i$-th side. The number of sides is not fixed, you should consider all
triangles, quadrilaterals, pentagons...
It's not that complicated, at least in the beginning. It's easy to show that the optimal polygon (with $n>3$) cannot have obtuse ($>90^circ$) angles. For example, if such an angle $A_i-1A_iA_i+1$ exists, by cosine theorem:
$$|A_i-1A_i|^2+|A_iA_i+1|^2<|A_i-1A_i+1|^2$$
So if you drop vertex $A_i$, you get a polygon with a bigger $S$. This quickly eliminates all polygons with $n>4$.
All candidate polygons with $n=4$ must be rectangles and if their sides are $a$ and $b$, the sum $S$ is $2a^2+2b^2=8R^2$. So with respect to $S$ all rectangles inscribed in the circle are equivalent. In fact, a right triangle with sides $a$, $b$ and $2R$ has the same $S$ as any inscribed rectangle.
But maybe there is an inscribed triangle with $S>8R^2$. I was able to show that for an inscribed triangle with sides $a,b,c$ and $bne c$, an isosceles triangle with all acute angles and base $a$ has better value of $S$. So the optimal triangle must be isosceles. Looking from all three sides, the only possible solution is the equilateral triangle and the sum $S$ in that case is $9R^2$.
However, to prove that fact I had to use trigonometry which is not so complicated (and I can present it here if you want so), but it leaves impression that there has to be some simpler explanation why the equilateral triangle is the best choice. My trigonometry proof takes a few lines of text, I want something more elegant.
Just an idea: if you draw lines through the center of the circle perpendicular to the sides of a triangle and denote the pedal lengths with $h_a,h_b,h_c$, it's easy to show that in order to maximize $a^2+b^2+c^2$ you have to minimize $h_a^2+h_b^2+h_c^2$. And then - what?
EDIT: I want to present the part of the proof that I don't like. Take an arbitrary triangle $ABC$ with sides $a,b,c$ inscribed in a circle. Consider side $a$ fixed and play with angle $gamma$ to get different values of $b,c$. I want to prove that isosceles triangle $BCA_1$ has bigger $S$ than any other triangle with one side equal to $a$.
$$b=2Rsinfracpi-alpha+gamma2=2Rcos(fracalpha2-fracgamma2)$$
$$c=2Rsinfracpi-alpha-gamma2=2Rcos(fracalpha2+fracgamma2)$$
$$b^2=4R^2cos^2(fracalpha2-fracgamma2)=2R^2(1+cos(alpha-gamma))$$
$$c^2=4R^2cos^2(fracalpha2+fracgamma2)=2R^2(1+cos(alpha+gamma))$$
$$b^2+c^2=4R^2+2R^2(cos(alpha-gamma)+cos(alpha+gamma))=4R^2(1+cosalphacosgamma)$$
And this sum achieves maximum obviously for $gamma=0$, or for $Aequiv A_1$. So for any given side $a$, $b$ and $c$ must be of equal. But you can look at the optimal triangle from sides $b$ and $c$ as well. The only triangle which has no better option is equilateral triangle.
euclidean-geometry triangle
asked 2 hours ago
Oldboy
3,1211319
My son presented me with an interesting problem:
Of all possible polygons inscribed in a circle of radius $R$, find the
one that has the sum $S$ of squared side lengths maximized:
$S=a_1^2+a_2^2+dots+a_n^2$, with $a_i$ representing the length of the
$i$-th side. The number of sides is not fixed, you should consider all
triangles, quadrilaterals, pentagons...
It's not that complicated, at least in the beginning. It's easy to show that the optimal polygon (with $n>3$) cannot have obtuse ($>90^circ$) angles. For example, if such an angle $A_i-1A_iA_i+1$ exists, by cosine theorem:
$$|A_i-1A_i|^2+|A_iA_i+1|^2<|A_i-1A_i+1|^2$$
So if you drop vertex $A_i$, you get a polygon with a bigger $S$. This quickly eliminates all polygons with $n>4$.
All candidate polygons with $n=4$ must be rectangles and if their sides are $a$ and $b$, the sum $S$ is $2a^2+2b^2=8R^2$. So with respect to $S$ all rectangles inscribed in the circle are equivalent. In fact, a right triangle with sides $a$, $b$ and $2R$ has the same $S$ as any inscribed rectangle.
But maybe there is an inscribed triangle with $S>8R^2$. I was able to show that for an inscribed triangle with sides $a,b,c$ and $bne c$, an isosceles triangle with all acute angles and base $a$ has better value of $S$. So the optimal triangle must be isosceles. Looking from all three sides, the only possible solution is the equilateral triangle and the sum $S$ in that case is $9R^2$.
However, to prove that fact I had to use trigonometry which is not so complicated (and I can present it here if you want so), but it leaves impression that there has to be some simpler explanation why the equilateral triangle is the best choice. My trigonometry proof takes a few lines of text, I want something more elegant.
Just an idea: if you draw lines through the center of the circle perpendicular to the sides of a triangle and denote the pedal lengths with $h_a,h_b,h_c$, it's easy to show that in order to maximize $a^2+b^2+c^2$ you have to minimize $h_a^2+h_b^2+h_c^2$. And then - what?
EDIT: I want to present the part of the proof that I don't like. Take an arbitrary triangle $ABC$ with sides $a,b,c$ inscribed in a circle. Consider side $a$ fixed and play with angle $gamma$ to get different values of $b,c$. I want to prove that isosceles triangle $BCA_1$ has bigger $S$ than any other triangle with one side equal to $a$.
$$b=2Rsinfracpi-alpha+gamma2=2Rcos(fracalpha2-fracgamma2)$$
$$c=2Rsinfracpi-alpha-gamma2=2Rcos(fracalpha2+fracgamma2)$$
$$b^2=4R^2cos^2(fracalpha2-fracgamma2)=2R^2(1+cos(alpha-gamma))$$
$$c^2=4R^2cos^2(fracalpha2+fracgamma2)=2R^2(1+cos(alpha+gamma))$$
$$b^2+c^2=4R^2+2R^2(cos(alpha-gamma)+cos(alpha+gamma))=4R^2(1+cosalphacosgamma)$$
And this sum achieves maximum obviously for $gamma=0$, or for $Aequiv A_1$. So for any given side $a$, $b$ and $c$ must be of equal. But you can look at the optimal triangle from sides $b$ and $c$ as well. The only triangle which has no better option is equilateral triangle.
euclidean-geometry triangle
euclidean-geometry triangle
asked 2 hours ago
Oldboy
3,1211319
asked 2 hours ago
Oldboy
3,1211319
edited 1 min ago
asked 2 hours ago
Oldboy
3,1211319
asked 2 hours ago
Oldboy
3,1211319
asked 2 hours ago
Oldboy
3,1211319
3,1211319
add a comment |Â
add a comment |Â
2 Answers
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active
oldest
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up vote
3
down vote
Yes, the maximal sum is the one of the equilateral triangle, that is $9R^2$.
Since Prove that in any triangle $ABC$, $cos^2A+cos^2B+cos^2Cgeqfrac34$ then
$$
sin^2 A+sin^2 B+sin^2 C=3-cos^2 A-cos^2 B-cos^2 Cleq frac94
$$
where $A$, $B$ and $C$ are non negative numbers such that $A+B+C=pi$. Hence, for any inscribed triangle, the sum of the squares of the sides is
$$(2Rsin A)^2+(2Rsin B)^2+(2Rsin C)^2leq 9R^2.$$
answered 1 hour ago
Robert Z
85.7k1055123
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up vote
0
down vote
This problem can be stated as
$$
max_nsum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2
$$
s. t.
$$
sum_k=1^ntheta_k = 2pi
$$
but
$$
sum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2ge nleft(2^2nr^2nprod_k=1^nsin^2left(fractheta_k2right)right)^frac 1n
$$
assuming $theta_1=cdots=theta_n$ we have
$$
max_nsum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2ge nleft(2^2nr^2nsin^2nleft(fracpinright)right)^frac 1n = n2^2r^2sin^2left(fracpinright)
$$
Now calling
$$
f(n) = nsin^2left(fracpinright)
$$
we have clearly a maximum about $n = 3$ as can be depicted in the attached plot
answered 18 mins ago
Cesareo
5,9922413
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Yes, the maximal sum is the one of the equilateral triangle, that is $9R^2$.
Since Prove that in any triangle $ABC$, $cos^2A+cos^2B+cos^2Cgeqfrac34$ then
$$
sin^2 A+sin^2 B+sin^2 C=3-cos^2 A-cos^2 B-cos^2 Cleq frac94
$$
where $A$, $B$ and $C$ are non negative numbers such that $A+B+C=pi$. Hence, for any inscribed triangle, the sum of the squares of the sides is
$$(2Rsin A)^2+(2Rsin B)^2+(2Rsin C)^2leq 9R^2.$$
answered 1 hour ago
Robert Z
85.7k1055123
add a comment |Â
up vote
3
down vote
Yes, the maximal sum is the one of the equilateral triangle, that is $9R^2$.
Since Prove that in any triangle $ABC$, $cos^2A+cos^2B+cos^2Cgeqfrac34$ then
$$
sin^2 A+sin^2 B+sin^2 C=3-cos^2 A-cos^2 B-cos^2 Cleq frac94
$$
where $A$, $B$ and $C$ are non negative numbers such that $A+B+C=pi$. Hence, for any inscribed triangle, the sum of the squares of the sides is
$$(2Rsin A)^2+(2Rsin B)^2+(2Rsin C)^2leq 9R^2.$$
answered 1 hour ago
Robert Z
85.7k1055123
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Yes, the maximal sum is the one of the equilateral triangle, that is $9R^2$.
Since Prove that in any triangle $ABC$, $cos^2A+cos^2B+cos^2Cgeqfrac34$ then
$$
sin^2 A+sin^2 B+sin^2 C=3-cos^2 A-cos^2 B-cos^2 Cleq frac94
$$
where $A$, $B$ and $C$ are non negative numbers such that $A+B+C=pi$. Hence, for any inscribed triangle, the sum of the squares of the sides is
$$(2Rsin A)^2+(2Rsin B)^2+(2Rsin C)^2leq 9R^2.$$
answered 1 hour ago
Robert Z
85.7k1055123
Yes, the maximal sum is the one of the equilateral triangle, that is $9R^2$.
Since Prove that in any triangle $ABC$, $cos^2A+cos^2B+cos^2Cgeqfrac34$ then
$$
sin^2 A+sin^2 B+sin^2 C=3-cos^2 A-cos^2 B-cos^2 Cleq frac94
$$
where $A$, $B$ and $C$ are non negative numbers such that $A+B+C=pi$. Hence, for any inscribed triangle, the sum of the squares of the sides is
$$(2Rsin A)^2+(2Rsin B)^2+(2Rsin C)^2leq 9R^2.$$
answered 1 hour ago
Robert Z
85.7k1055123
edited 1 hour ago
answered 1 hour ago
Robert Z
85.7k1055123
answered 1 hour ago
Robert Z
85.7k1055123
answered 1 hour ago
Robert Z
85.7k1055123
85.7k1055123
add a comment |Â
add a comment |Â
up vote
0
down vote
This problem can be stated as
$$
max_nsum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2
$$
s. t.
$$
sum_k=1^ntheta_k = 2pi
$$
but
$$
sum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2ge nleft(2^2nr^2nprod_k=1^nsin^2left(fractheta_k2right)right)^frac 1n
$$
assuming $theta_1=cdots=theta_n$ we have
$$
max_nsum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2ge nleft(2^2nr^2nsin^2nleft(fracpinright)right)^frac 1n = n2^2r^2sin^2left(fracpinright)
$$
Now calling
$$
f(n) = nsin^2left(fracpinright)
$$
we have clearly a maximum about $n = 3$ as can be depicted in the attached plot
answered 18 mins ago
Cesareo
5,9922413
add a comment |Â
up vote
0
down vote
This problem can be stated as
$$
max_nsum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2
$$
s. t.
$$
sum_k=1^ntheta_k = 2pi
$$
but
$$
sum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2ge nleft(2^2nr^2nprod_k=1^nsin^2left(fractheta_k2right)right)^frac 1n
$$
assuming $theta_1=cdots=theta_n$ we have
$$
max_nsum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2ge nleft(2^2nr^2nsin^2nleft(fracpinright)right)^frac 1n = n2^2r^2sin^2left(fracpinright)
$$
Now calling
$$
f(n) = nsin^2left(fracpinright)
$$
we have clearly a maximum about $n = 3$ as can be depicted in the attached plot
answered 18 mins ago
Cesareo
5,9922413
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This problem can be stated as
$$
max_nsum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2
$$
s. t.
$$
sum_k=1^ntheta_k = 2pi
$$
but
$$
sum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2ge nleft(2^2nr^2nprod_k=1^nsin^2left(fractheta_k2right)right)^frac 1n
$$
assuming $theta_1=cdots=theta_n$ we have
$$
max_nsum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2ge nleft(2^2nr^2nsin^2nleft(fracpinright)right)^frac 1n = n2^2r^2sin^2left(fracpinright)
$$
Now calling
$$
f(n) = nsin^2left(fracpinright)
$$
we have clearly a maximum about $n = 3$ as can be depicted in the attached plot
answered 18 mins ago
Cesareo
5,9922413
This problem can be stated as
$$
max_nsum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2
$$
s. t.
$$
sum_k=1^ntheta_k = 2pi
$$
but
$$
sum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2ge nleft(2^2nr^2nprod_k=1^nsin^2left(fractheta_k2right)right)^frac 1n
$$
assuming $theta_1=cdots=theta_n$ we have
$$
max_nsum_k=1^nleft(2rsinleft(fractheta_k2right)right)^2ge nleft(2^2nr^2nsin^2nleft(fracpinright)right)^frac 1n = n2^2r^2sin^2left(fracpinright)
$$
Now calling
$$
f(n) = nsin^2left(fracpinright)
$$
we have clearly a maximum about $n = 3$ as can be depicted in the attached plot
answered 18 mins ago
Cesareo
5,9922413
answered 18 mins ago
Cesareo
5,9922413
answered 18 mins ago
Cesareo
5,9922413
answered 18 mins ago
Cesareo
5,9922413
5,9922413
add a comment |Â
add a comment |Â
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