Simple geometry question involving the circumradius of a triangle.
$begingroup$
Prove that
$sin(2A)overrightarrow{OA}+sin(2B)overrightarrow{OB}+sin(2C)overrightarrow{OC} =0 $
Where $O$ is the circumradius of the $ABC$ triangle
How to approach this type of problem?
How do we demonstrate that for P ( a point in the triangle ) we get :
$overrightarrow{PO}=frac{sin(2A)overrightarrow{PA}+sin(2B)overrightarrow{PB}+sin(2C)overrightarrow{PC}}{sin(2A)+sin(2B)+sin(2C)} =0 $ only if O = P
triangle
asked Dec 1 '18 at 15:10
SADBOYSSADBOYS
4288
$endgroup$
add a comment |
$begingroup$
Prove that
$sin(2A)overrightarrow{OA}+sin(2B)overrightarrow{OB}+sin(2C)overrightarrow{OC} =0 $
Where $O$ is the circumradius of the $ABC$ triangle
How to approach this type of problem?
How do we demonstrate that for P ( a point in the triangle ) we get :
$overrightarrow{PO}=frac{sin(2A)overrightarrow{PA}+sin(2B)overrightarrow{PB}+sin(2C)overrightarrow{PC}}{sin(2A)+sin(2B)+sin(2C)} =0 $ only if O = P
triangle
asked Dec 1 '18 at 15:10
SADBOYSSADBOYS
4288
$endgroup$
1
$begingroup$
I think there is a mistake in your problem. The side should not appear in the sine function. Suppose that it's true for a particular triangle. If I increase the size of the triangle by a factor of $2$, the $OA, OB, OC$ increase by a factor of two as well. Which is fine in the first equation. But the argument of sines wil double, and I assume it might change the equation. The only valid solution would be $O=A=B=C$
$endgroup$
– Andrei
Dec 1 '18 at 15:22
$begingroup$
The mistake was that the angles were written in lowercase and I thought those were the sides
$endgroup$
– SADBOYS
Dec 1 '18 at 15:26
add a comment |
$begingroup$
Prove that
$sin(2A)overrightarrow{OA}+sin(2B)overrightarrow{OB}+sin(2C)overrightarrow{OC} =0 $
Where $O$ is the circumradius of the $ABC$ triangle
How to approach this type of problem?
How do we demonstrate that for P ( a point in the triangle ) we get :
$overrightarrow{PO}=frac{sin(2A)overrightarrow{PA}+sin(2B)overrightarrow{PB}+sin(2C)overrightarrow{PC}}{sin(2A)+sin(2B)+sin(2C)} =0 $ only if O = P
triangle
asked Dec 1 '18 at 15:10
SADBOYSSADBOYS
4288
$endgroup$
Prove that
$sin(2A)overrightarrow{OA}+sin(2B)overrightarrow{OB}+sin(2C)overrightarrow{OC} =0 $
Where $O$ is the circumradius of the $ABC$ triangle
How to approach this type of problem?
How do we demonstrate that for P ( a point in the triangle ) we get :
$overrightarrow{PO}=frac{sin(2A)overrightarrow{PA}+sin(2B)overrightarrow{PB}+sin(2C)overrightarrow{PC}}{sin(2A)+sin(2B)+sin(2C)} =0 $ only if O = P
triangle
triangle
asked Dec 1 '18 at 15:10
SADBOYSSADBOYS
4288
asked Dec 1 '18 at 15:10
SADBOYSSADBOYS
4288
edited Dec 1 '18 at 15:24
SADBOYS
asked Dec 1 '18 at 15:10
SADBOYSSADBOYS
4288
asked Dec 1 '18 at 15:10
SADBOYSSADBOYS
4288
asked Dec 1 '18 at 15:10
SADBOYSSADBOYS
4288
4288
1
$begingroup$
I think there is a mistake in your problem. The side should not appear in the sine function. Suppose that it's true for a particular triangle. If I increase the size of the triangle by a factor of $2$, the $OA, OB, OC$ increase by a factor of two as well. Which is fine in the first equation. But the argument of sines wil double, and I assume it might change the equation. The only valid solution would be $O=A=B=C$
$endgroup$
– Andrei
Dec 1 '18 at 15:22
$begingroup$
The mistake was that the angles were written in lowercase and I thought those were the sides
$endgroup$
– SADBOYS
Dec 1 '18 at 15:26
add a comment |
1
$begingroup$
I think there is a mistake in your problem. The side should not appear in the sine function. Suppose that it's true for a particular triangle. If I increase the size of the triangle by a factor of $2$, the $OA, OB, OC$ increase by a factor of two as well. Which is fine in the first equation. But the argument of sines wil double, and I assume it might change the equation. The only valid solution would be $O=A=B=C$
$endgroup$
– Andrei
Dec 1 '18 at 15:22
$begingroup$
The mistake was that the angles were written in lowercase and I thought those were the sides
$endgroup$
– SADBOYS
Dec 1 '18 at 15:26
1
1
$begingroup$
I think there is a mistake in your problem. The side should not appear in the sine function. Suppose that it's true for a particular triangle. If I increase the size of the triangle by a factor of $2$, the $OA, OB, OC$ increase by a factor of two as well. Which is fine in the first equation. But the argument of sines wil double, and I assume it might change the equation. The only valid solution would be $O=A=B=C$
$endgroup$
– Andrei
Dec 1 '18 at 15:22
$begingroup$
I think there is a mistake in your problem. The side should not appear in the sine function. Suppose that it's true for a particular triangle. If I increase the size of the triangle by a factor of $2$, the $OA, OB, OC$ increase by a factor of two as well. Which is fine in the first equation. But the argument of sines wil double, and I assume it might change the equation. The only valid solution would be $O=A=B=C$
$endgroup$
– Andrei
Dec 1 '18 at 15:22
$begingroup$
The mistake was that the angles were written in lowercase and I thought those were the sides
$endgroup$
– SADBOYS
Dec 1 '18 at 15:26
$begingroup$
The mistake was that the angles were written in lowercase and I thought those were the sides
$endgroup$
– SADBOYS
Dec 1 '18 at 15:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The exact trilinear coordinates of the circumcenter $O$ are $[Rcos A;Rcos B;Rcos C]$, hence the trilinear coordinates are $[cos A;cos B;cos C]$ and the barycentric coordinates are $[acos A; bcos B; ccos C]$ or $[sin(2A);sin(2B);sin(2C)]$. It follows that
$$vec{O}=frac{sin(2A)vec{A}+sin(2B)vec{B}+sin(2C)vec{C}}{sin(2A)+sin(2B)+sin(2C)}$$
and if the reference system is centered at $O$
$$0=frac{sin(2A)vec{OA}+sin(2B)vec{OB}+sin(2C)vec{OC}}{sin(2A)+sin(2B)+sin(2C)}.$$
answered Dec 1 '18 at 15:36
Jack D'AurizioJack D'Aurizio
288k33280659
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add a comment |
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1 Answer
1
active
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1 Answer
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active
oldest
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$begingroup$
The exact trilinear coordinates of the circumcenter $O$ are $[Rcos A;Rcos B;Rcos C]$, hence the trilinear coordinates are $[cos A;cos B;cos C]$ and the barycentric coordinates are $[acos A; bcos B; ccos C]$ or $[sin(2A);sin(2B);sin(2C)]$. It follows that
$$vec{O}=frac{sin(2A)vec{A}+sin(2B)vec{B}+sin(2C)vec{C}}{sin(2A)+sin(2B)+sin(2C)}$$
and if the reference system is centered at $O$
$$0=frac{sin(2A)vec{OA}+sin(2B)vec{OB}+sin(2C)vec{OC}}{sin(2A)+sin(2B)+sin(2C)}.$$
answered Dec 1 '18 at 15:36
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
The exact trilinear coordinates of the circumcenter $O$ are $[Rcos A;Rcos B;Rcos C]$, hence the trilinear coordinates are $[cos A;cos B;cos C]$ and the barycentric coordinates are $[acos A; bcos B; ccos C]$ or $[sin(2A);sin(2B);sin(2C)]$. It follows that
$$vec{O}=frac{sin(2A)vec{A}+sin(2B)vec{B}+sin(2C)vec{C}}{sin(2A)+sin(2B)+sin(2C)}$$
and if the reference system is centered at $O$
$$0=frac{sin(2A)vec{OA}+sin(2B)vec{OB}+sin(2C)vec{OC}}{sin(2A)+sin(2B)+sin(2C)}.$$
answered Dec 1 '18 at 15:36
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
add a comment |
$begingroup$
The exact trilinear coordinates of the circumcenter $O$ are $[Rcos A;Rcos B;Rcos C]$, hence the trilinear coordinates are $[cos A;cos B;cos C]$ and the barycentric coordinates are $[acos A; bcos B; ccos C]$ or $[sin(2A);sin(2B);sin(2C)]$. It follows that
$$vec{O}=frac{sin(2A)vec{A}+sin(2B)vec{B}+sin(2C)vec{C}}{sin(2A)+sin(2B)+sin(2C)}$$
and if the reference system is centered at $O$
$$0=frac{sin(2A)vec{OA}+sin(2B)vec{OB}+sin(2C)vec{OC}}{sin(2A)+sin(2B)+sin(2C)}.$$
answered Dec 1 '18 at 15:36
Jack D'AurizioJack D'Aurizio
288k33280659
$endgroup$
The exact trilinear coordinates of the circumcenter $O$ are $[Rcos A;Rcos B;Rcos C]$, hence the trilinear coordinates are $[cos A;cos B;cos C]$ and the barycentric coordinates are $[acos A; bcos B; ccos C]$ or $[sin(2A);sin(2B);sin(2C)]$. It follows that
$$vec{O}=frac{sin(2A)vec{A}+sin(2B)vec{B}+sin(2C)vec{C}}{sin(2A)+sin(2B)+sin(2C)}$$
and if the reference system is centered at $O$
$$0=frac{sin(2A)vec{OA}+sin(2B)vec{OB}+sin(2C)vec{OC}}{sin(2A)+sin(2B)+sin(2C)}.$$
answered Dec 1 '18 at 15:36
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 1 '18 at 15:36
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 1 '18 at 15:36
Jack D'AurizioJack D'Aurizio
288k33280659
answered Dec 1 '18 at 15:36
Jack D'AurizioJack D'Aurizio
288k33280659
288k33280659
add a comment |
add a comment |
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$begingroup$
I think there is a mistake in your problem. The side should not appear in the sine function. Suppose that it's true for a particular triangle. If I increase the size of the triangle by a factor of $2$, the $OA, OB, OC$ increase by a factor of two as well. Which is fine in the first equation. But the argument of sines wil double, and I assume it might change the equation. The only valid solution would be $O=A=B=C$
$endgroup$
– Andrei
Dec 1 '18 at 15:22
$begingroup$
The mistake was that the angles were written in lowercase and I thought those were the sides
$endgroup$
– SADBOYS
Dec 1 '18 at 15:26