Proving the distance between centroid and orthocentre is twice the distance from centroid to circumcentre.
$begingroup$
Let $ABC$ be a triangle with right angle at $C$. First find the circumcentre, $O$, the centroid $G$, and the orthocentre $H$ of triangle ABC. Then prove that these three points are collinear and that $HG = 2GO$. All advice appreciated.
geometry
asked Dec 10 '18 at 12:13
Matlab rookieMatlab rookie
357
$endgroup$
add a comment |
$begingroup$
Let $ABC$ be a triangle with right angle at $C$. First find the circumcentre, $O$, the centroid $G$, and the orthocentre $H$ of triangle ABC. Then prove that these three points are collinear and that $HG = 2GO$. All advice appreciated.
geometry
asked Dec 10 '18 at 12:13
Matlab rookieMatlab rookie
357
$endgroup$
3
$begingroup$
The $M$ be the midpoint of $AB$. Then $H=C$, $O=M$ and $G$ lies at $frac{1}{3}$ of the median $CM$.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 12:15
add a comment |
$begingroup$
Let $ABC$ be a triangle with right angle at $C$. First find the circumcentre, $O$, the centroid $G$, and the orthocentre $H$ of triangle ABC. Then prove that these three points are collinear and that $HG = 2GO$. All advice appreciated.
geometry
asked Dec 10 '18 at 12:13
Matlab rookieMatlab rookie
357
$endgroup$
Let $ABC$ be a triangle with right angle at $C$. First find the circumcentre, $O$, the centroid $G$, and the orthocentre $H$ of triangle ABC. Then prove that these three points are collinear and that $HG = 2GO$. All advice appreciated.
geometry
geometry
asked Dec 10 '18 at 12:13
Matlab rookieMatlab rookie
357
asked Dec 10 '18 at 12:13
Matlab rookieMatlab rookie
357
edited Dec 10 '18 at 12:23
Matlab rookie
asked Dec 10 '18 at 12:13
Matlab rookieMatlab rookie
357
asked Dec 10 '18 at 12:13
Matlab rookieMatlab rookie
357
asked Dec 10 '18 at 12:13
Matlab rookieMatlab rookie
357
357
3
$begingroup$
The $M$ be the midpoint of $AB$. Then $H=C$, $O=M$ and $G$ lies at $frac{1}{3}$ of the median $CM$.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 12:15
add a comment |
3
$begingroup$
The $M$ be the midpoint of $AB$. Then $H=C$, $O=M$ and $G$ lies at $frac{1}{3}$ of the median $CM$.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 12:15
3
3
$begingroup$
The $M$ be the midpoint of $AB$. Then $H=C$, $O=M$ and $G$ lies at $frac{1}{3}$ of the median $CM$.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 12:15
$begingroup$
The $M$ be the midpoint of $AB$. Then $H=C$, $O=M$ and $G$ lies at $frac{1}{3}$ of the median $CM$.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 12:15
add a comment |
1 Answer
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$begingroup$
The midpoints of $AB,,BC,,CA$ are the vertices of the median triangle, which also has centroid $G$. We get from $triangle ABC$ to its median triangle by a $-1/2$ scaling centred on $G$. This maps $H$ to the median triangle's orthocentre. Each side of $triangle ABC$ is a chord of its circmcircle, and their perpendicular bisectors meet at $O$. In particular, the bisection is at the median triangle's vertices. Therefore, $O$ is the median triangle's orthocentre.
answered Dec 10 '18 at 13:32
J.G.J.G.
25.4k22539
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add a comment |
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$begingroup$
The midpoints of $AB,,BC,,CA$ are the vertices of the median triangle, which also has centroid $G$. We get from $triangle ABC$ to its median triangle by a $-1/2$ scaling centred on $G$. This maps $H$ to the median triangle's orthocentre. Each side of $triangle ABC$ is a chord of its circmcircle, and their perpendicular bisectors meet at $O$. In particular, the bisection is at the median triangle's vertices. Therefore, $O$ is the median triangle's orthocentre.
answered Dec 10 '18 at 13:32
J.G.J.G.
25.4k22539
$endgroup$
add a comment |
$begingroup$
The midpoints of $AB,,BC,,CA$ are the vertices of the median triangle, which also has centroid $G$. We get from $triangle ABC$ to its median triangle by a $-1/2$ scaling centred on $G$. This maps $H$ to the median triangle's orthocentre. Each side of $triangle ABC$ is a chord of its circmcircle, and their perpendicular bisectors meet at $O$. In particular, the bisection is at the median triangle's vertices. Therefore, $O$ is the median triangle's orthocentre.
answered Dec 10 '18 at 13:32
J.G.J.G.
25.4k22539
$endgroup$
add a comment |
$begingroup$
The midpoints of $AB,,BC,,CA$ are the vertices of the median triangle, which also has centroid $G$. We get from $triangle ABC$ to its median triangle by a $-1/2$ scaling centred on $G$. This maps $H$ to the median triangle's orthocentre. Each side of $triangle ABC$ is a chord of its circmcircle, and their perpendicular bisectors meet at $O$. In particular, the bisection is at the median triangle's vertices. Therefore, $O$ is the median triangle's orthocentre.
answered Dec 10 '18 at 13:32
J.G.J.G.
25.4k22539
$endgroup$
The midpoints of $AB,,BC,,CA$ are the vertices of the median triangle, which also has centroid $G$. We get from $triangle ABC$ to its median triangle by a $-1/2$ scaling centred on $G$. This maps $H$ to the median triangle's orthocentre. Each side of $triangle ABC$ is a chord of its circmcircle, and their perpendicular bisectors meet at $O$. In particular, the bisection is at the median triangle's vertices. Therefore, $O$ is the median triangle's orthocentre.
answered Dec 10 '18 at 13:32
J.G.J.G.
25.4k22539
answered Dec 10 '18 at 13:32
J.G.J.G.
25.4k22539
answered Dec 10 '18 at 13:32
J.G.J.G.
25.4k22539
answered Dec 10 '18 at 13:32
J.G.J.G.
25.4k22539
25.4k22539
add a comment |
add a comment |
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$begingroup$
The $M$ be the midpoint of $AB$. Then $H=C$, $O=M$ and $G$ lies at $frac{1}{3}$ of the median $CM$.
$endgroup$
– Jack D'Aurizio
Dec 10 '18 at 12:15