Locus of centroid of triangle inside a circle.
A circle $x^2+y^2 = 3$ has a chord $AB$ which subtends an angle of $45^{circ}$ on a moving point $P$ on the circle. Find the locus of centroid of triangle $PAB$.
euclidean-geometry analytic-geometry geometric-transformation
edited 1 hour ago
greedoid
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asked 3 hours ago
Harsh Joshi
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A circle $x^2+y^2 = 3$ has a chord $AB$ which subtends an angle of $45^{circ}$ on a moving point $P$ on the circle. Find the locus of centroid of triangle $PAB$.
euclidean-geometry analytic-geometry geometric-transformation
edited 1 hour ago
greedoid
37.2k114794
asked 3 hours ago
Harsh Joshi
192
New contributor
Harsh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
3 hours ago
Please phrase the question properly. If possible supply a diagram.
– John Mitchell
3 hours ago
add a comment |
A circle $x^2+y^2 = 3$ has a chord $AB$ which subtends an angle of $45^{circ}$ on a moving point $P$ on the circle. Find the locus of centroid of triangle $PAB$.
euclidean-geometry analytic-geometry geometric-transformation
edited 1 hour ago
greedoid
37.2k114794
asked 3 hours ago
Harsh Joshi
192
New contributor
Harsh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
A circle $x^2+y^2 = 3$ has a chord $AB$ which subtends an angle of $45^{circ}$ on a moving point $P$ on the circle. Find the locus of centroid of triangle $PAB$.
euclidean-geometry analytic-geometry geometric-transformation
euclidean-geometry analytic-geometry geometric-transformation
edited 1 hour ago
greedoid
37.2k114794
asked 3 hours ago
Harsh Joshi
192
New contributor
Harsh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago
greedoid
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asked 3 hours ago
Harsh Joshi
192
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Harsh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 1 hour ago
greedoid
37.2k114794
edited 1 hour ago
greedoid
37.2k114794
edited 1 hour ago
greedoid
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37.2k114794
asked 3 hours ago
Harsh Joshi
192
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Harsh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked 3 hours ago
Harsh Joshi
192
asked 3 hours ago
Harsh Joshi
192
192
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Harsh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
Harsh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Harsh Joshi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
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Please phrase the question properly. If possible supply a diagram.
– John Mitchell
3 hours ago
add a comment |
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
3 hours ago
Please phrase the question properly. If possible supply a diagram.
– John Mitchell
3 hours ago
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
3 hours ago
Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
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Please phrase the question properly. If possible supply a diagram.
– John Mitchell
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Please phrase the question properly. If possible supply a diagram.
– John Mitchell
3 hours ago
add a comment |
2 Answers
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Radius of the circle is $R=sqrt3$. Introduce the following coordinates:
$A(-frac{sqrt{6}}{2}, frac{sqrt{6}}{2})$, $B(frac{sqrt{6}}{2}, frac{sqrt{6}}{2})$, $P(x_P,y_P)$.
Coordinates of the centroid are:
$$x_T=frac{x_A+x_B+x_P}{3}=frac {x_P}3$$
$$y_T=frac{y_A+y_B+y_P}{3}=frac {sqrt 6+y_P}3$$
or:
$$3 x_T={x_P}$$
$$3y_T-sqrt6=y_P$$
$$9 x_T^2+(3y_T-sqrt6)^2=x_P^2+y_P^2=3$$
$$9 x_T^2+9(y_T-frac{sqrt6}{3})^2=3$$
$$x_T^2+(y_T-frac{sqrt6}{3})^2=(frac{sqrt3}3)^2$$
...which is a circle with center $(0, frac{sqrt{6}}{3})$ and radius $frac{sqrt3}3$
answered 3 hours ago
Oldboy
6,5401730
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Anyway, this could be done also with synthetic (Euclidian) geometry.
Let $S= (0,0)$ be a center of a given circle.
Let $M$ be a midpoint of a chord $AB$. Then $vec{MT} = {1over 3}vec{MP}$ for every $P$. So we get $T$ after performing a homothety with center at $M$ and dilatation coefficient $k= {1over 3}$ on $P$. Since $P$ is on a fixed circle and homothety takes a circle to a circle we see that $T$ describe a circle with center at $C$ which is on ${1over 3}MS$ and with radius $r' = {rover 3} = {sqrt{3}over 3}$.
So if we take a picture of oldboy we have $M=(0,{sqrt{6}over 2})$ so $C = (0,{sqrt{6}over 3})$ and thus $$x^2+(y-{sqrt{6}over 3})^2= {1over 3}$$
answered 1 hour ago
greedoid
37.2k114794
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
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oldest
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active
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active
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votes
Radius of the circle is $R=sqrt3$. Introduce the following coordinates:
$A(-frac{sqrt{6}}{2}, frac{sqrt{6}}{2})$, $B(frac{sqrt{6}}{2}, frac{sqrt{6}}{2})$, $P(x_P,y_P)$.
Coordinates of the centroid are:
$$x_T=frac{x_A+x_B+x_P}{3}=frac {x_P}3$$
$$y_T=frac{y_A+y_B+y_P}{3}=frac {sqrt 6+y_P}3$$
or:
$$3 x_T={x_P}$$
$$3y_T-sqrt6=y_P$$
$$9 x_T^2+(3y_T-sqrt6)^2=x_P^2+y_P^2=3$$
$$9 x_T^2+9(y_T-frac{sqrt6}{3})^2=3$$
$$x_T^2+(y_T-frac{sqrt6}{3})^2=(frac{sqrt3}3)^2$$
...which is a circle with center $(0, frac{sqrt{6}}{3})$ and radius $frac{sqrt3}3$
answered 3 hours ago
Oldboy
6,5401730
add a comment |
Radius of the circle is $R=sqrt3$. Introduce the following coordinates:
$A(-frac{sqrt{6}}{2}, frac{sqrt{6}}{2})$, $B(frac{sqrt{6}}{2}, frac{sqrt{6}}{2})$, $P(x_P,y_P)$.
Coordinates of the centroid are:
$$x_T=frac{x_A+x_B+x_P}{3}=frac {x_P}3$$
$$y_T=frac{y_A+y_B+y_P}{3}=frac {sqrt 6+y_P}3$$
or:
$$3 x_T={x_P}$$
$$3y_T-sqrt6=y_P$$
$$9 x_T^2+(3y_T-sqrt6)^2=x_P^2+y_P^2=3$$
$$9 x_T^2+9(y_T-frac{sqrt6}{3})^2=3$$
$$x_T^2+(y_T-frac{sqrt6}{3})^2=(frac{sqrt3}3)^2$$
...which is a circle with center $(0, frac{sqrt{6}}{3})$ and radius $frac{sqrt3}3$
answered 3 hours ago
Oldboy
6,5401730
add a comment |
Radius of the circle is $R=sqrt3$. Introduce the following coordinates:
$A(-frac{sqrt{6}}{2}, frac{sqrt{6}}{2})$, $B(frac{sqrt{6}}{2}, frac{sqrt{6}}{2})$, $P(x_P,y_P)$.
Coordinates of the centroid are:
$$x_T=frac{x_A+x_B+x_P}{3}=frac {x_P}3$$
$$y_T=frac{y_A+y_B+y_P}{3}=frac {sqrt 6+y_P}3$$
or:
$$3 x_T={x_P}$$
$$3y_T-sqrt6=y_P$$
$$9 x_T^2+(3y_T-sqrt6)^2=x_P^2+y_P^2=3$$
$$9 x_T^2+9(y_T-frac{sqrt6}{3})^2=3$$
$$x_T^2+(y_T-frac{sqrt6}{3})^2=(frac{sqrt3}3)^2$$
...which is a circle with center $(0, frac{sqrt{6}}{3})$ and radius $frac{sqrt3}3$
answered 3 hours ago
Oldboy
6,5401730
Radius of the circle is $R=sqrt3$. Introduce the following coordinates:
$A(-frac{sqrt{6}}{2}, frac{sqrt{6}}{2})$, $B(frac{sqrt{6}}{2}, frac{sqrt{6}}{2})$, $P(x_P,y_P)$.
Coordinates of the centroid are:
$$x_T=frac{x_A+x_B+x_P}{3}=frac {x_P}3$$
$$y_T=frac{y_A+y_B+y_P}{3}=frac {sqrt 6+y_P}3$$
or:
$$3 x_T={x_P}$$
$$3y_T-sqrt6=y_P$$
$$9 x_T^2+(3y_T-sqrt6)^2=x_P^2+y_P^2=3$$
$$9 x_T^2+9(y_T-frac{sqrt6}{3})^2=3$$
$$x_T^2+(y_T-frac{sqrt6}{3})^2=(frac{sqrt3}3)^2$$
...which is a circle with center $(0, frac{sqrt{6}}{3})$ and radius $frac{sqrt3}3$
answered 3 hours ago
Oldboy
6,5401730
answered 3 hours ago
Oldboy
6,5401730
answered 3 hours ago
Oldboy
6,5401730
answered 3 hours ago
Oldboy
6,5401730
6,5401730
add a comment |
add a comment |
Anyway, this could be done also with synthetic (Euclidian) geometry.
Let $S= (0,0)$ be a center of a given circle.
Let $M$ be a midpoint of a chord $AB$. Then $vec{MT} = {1over 3}vec{MP}$ for every $P$. So we get $T$ after performing a homothety with center at $M$ and dilatation coefficient $k= {1over 3}$ on $P$. Since $P$ is on a fixed circle and homothety takes a circle to a circle we see that $T$ describe a circle with center at $C$ which is on ${1over 3}MS$ and with radius $r' = {rover 3} = {sqrt{3}over 3}$.
So if we take a picture of oldboy we have $M=(0,{sqrt{6}over 2})$ so $C = (0,{sqrt{6}over 3})$ and thus $$x^2+(y-{sqrt{6}over 3})^2= {1over 3}$$
answered 1 hour ago
greedoid
37.2k114794
add a comment |
Anyway, this could be done also with synthetic (Euclidian) geometry.
Let $S= (0,0)$ be a center of a given circle.
Let $M$ be a midpoint of a chord $AB$. Then $vec{MT} = {1over 3}vec{MP}$ for every $P$. So we get $T$ after performing a homothety with center at $M$ and dilatation coefficient $k= {1over 3}$ on $P$. Since $P$ is on a fixed circle and homothety takes a circle to a circle we see that $T$ describe a circle with center at $C$ which is on ${1over 3}MS$ and with radius $r' = {rover 3} = {sqrt{3}over 3}$.
So if we take a picture of oldboy we have $M=(0,{sqrt{6}over 2})$ so $C = (0,{sqrt{6}over 3})$ and thus $$x^2+(y-{sqrt{6}over 3})^2= {1over 3}$$
answered 1 hour ago
greedoid
37.2k114794
add a comment |
Anyway, this could be done also with synthetic (Euclidian) geometry.
Let $S= (0,0)$ be a center of a given circle.
Let $M$ be a midpoint of a chord $AB$. Then $vec{MT} = {1over 3}vec{MP}$ for every $P$. So we get $T$ after performing a homothety with center at $M$ and dilatation coefficient $k= {1over 3}$ on $P$. Since $P$ is on a fixed circle and homothety takes a circle to a circle we see that $T$ describe a circle with center at $C$ which is on ${1over 3}MS$ and with radius $r' = {rover 3} = {sqrt{3}over 3}$.
So if we take a picture of oldboy we have $M=(0,{sqrt{6}over 2})$ so $C = (0,{sqrt{6}over 3})$ and thus $$x^2+(y-{sqrt{6}over 3})^2= {1over 3}$$
answered 1 hour ago
greedoid
37.2k114794
Anyway, this could be done also with synthetic (Euclidian) geometry.
Let $S= (0,0)$ be a center of a given circle.
Let $M$ be a midpoint of a chord $AB$. Then $vec{MT} = {1over 3}vec{MP}$ for every $P$. So we get $T$ after performing a homothety with center at $M$ and dilatation coefficient $k= {1over 3}$ on $P$. Since $P$ is on a fixed circle and homothety takes a circle to a circle we see that $T$ describe a circle with center at $C$ which is on ${1over 3}MS$ and with radius $r' = {rover 3} = {sqrt{3}over 3}$.
So if we take a picture of oldboy we have $M=(0,{sqrt{6}over 2})$ so $C = (0,{sqrt{6}over 3})$ and thus $$x^2+(y-{sqrt{6}over 3})^2= {1over 3}$$
answered 1 hour ago
greedoid
37.2k114794
answered 1 hour ago
greedoid
37.2k114794
answered 1 hour ago
greedoid
37.2k114794
answered 1 hour ago
greedoid
37.2k114794
37.2k114794
add a comment |
add a comment |
Harsh Joshi is a new contributor. Be nice, and check out our Code of Conduct.
Harsh Joshi is a new contributor. Be nice, and check out our Code of Conduct.
Harsh Joshi is a new contributor. Be nice, and check out our Code of Conduct.
Harsh Joshi is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– José Carlos Santos
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Please phrase the question properly. If possible supply a diagram.
– John Mitchell
3 hours ago