Mixing
Elliptical version of Pythagoras’ Theorem?
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up vote
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Consider any right triangle $triangle ABC$.
We focus on one side, $AC$, and we take the midpoint $E$ of this side. Then, we draw the circle with center in $E$ and passing by $A,C$. If we take the perpendicular to $AC$ passing by $E$, we define a point $I$, where the circle intersects the perpendicular line.
Now, we can draw the ellipse with focii in $A$ and $C$ and passing by $I$.
Clearly, we can apply this procedure to both the other sides, obtaining other two ellipses.
My conjecture is that
The sum of the areas of the ellipses constructed on the two catheti is equal to the area of the ellipse constructed on the hypotenuse.
This is probably a very well known result, and I already apologize with the experts.
However, in order to prove the conjecture, I used the formula of the area of the ellipse, $S=pi a b$, where $a$ and $b$ are the lengths of the semi-axes. Although it is easy to prove that, in the case of all our ellipses, one semi-axis is clearly half the side, I am stuck in the attempt to determine the lengths of the other semi-axes, and I suspect however that there should be a very elementary way to prove such claim.
Again, sorry for the naivety, and thank you very much for any help or suggestion for a compact proof.
geometry euclidean-geometry triangle conic-sections geometric-construction
edited Aug 23 at 8:32
Brahadeesh
4,11331550
asked Aug 23 at 8:03
Andrea Prunotto
1,299625
add a comment |Â
up vote
21
down vote
favorite
Consider any right triangle $triangle ABC$.
We focus on one side, $AC$, and we take the midpoint $E$ of this side. Then, we draw the circle with center in $E$ and passing by $A,C$. If we take the perpendicular to $AC$ passing by $E$, we define a point $I$, where the circle intersects the perpendicular line.
Now, we can draw the ellipse with focii in $A$ and $C$ and passing by $I$.
Clearly, we can apply this procedure to both the other sides, obtaining other two ellipses.
My conjecture is that
The sum of the areas of the ellipses constructed on the two catheti is equal to the area of the ellipse constructed on the hypotenuse.
This is probably a very well known result, and I already apologize with the experts.
However, in order to prove the conjecture, I used the formula of the area of the ellipse, $S=pi a b$, where $a$ and $b$ are the lengths of the semi-axes. Although it is easy to prove that, in the case of all our ellipses, one semi-axis is clearly half the side, I am stuck in the attempt to determine the lengths of the other semi-axes, and I suspect however that there should be a very elementary way to prove such claim.
Again, sorry for the naivety, and thank you very much for any help or suggestion for a compact proof.
geometry euclidean-geometry triangle conic-sections geometric-construction
edited Aug 23 at 8:32
Brahadeesh
4,11331550
asked Aug 23 at 8:03
Andrea Prunotto
1,299625
add a comment |Â
up vote
21
down vote
favorite
up vote
21
down vote
favorite
Consider any right triangle $triangle ABC$.
We focus on one side, $AC$, and we take the midpoint $E$ of this side. Then, we draw the circle with center in $E$ and passing by $A,C$. If we take the perpendicular to $AC$ passing by $E$, we define a point $I$, where the circle intersects the perpendicular line.
Now, we can draw the ellipse with focii in $A$ and $C$ and passing by $I$.
Clearly, we can apply this procedure to both the other sides, obtaining other two ellipses.
My conjecture is that
The sum of the areas of the ellipses constructed on the two catheti is equal to the area of the ellipse constructed on the hypotenuse.
This is probably a very well known result, and I already apologize with the experts.
However, in order to prove the conjecture, I used the formula of the area of the ellipse, $S=pi a b$, where $a$ and $b$ are the lengths of the semi-axes. Although it is easy to prove that, in the case of all our ellipses, one semi-axis is clearly half the side, I am stuck in the attempt to determine the lengths of the other semi-axes, and I suspect however that there should be a very elementary way to prove such claim.
Again, sorry for the naivety, and thank you very much for any help or suggestion for a compact proof.
geometry euclidean-geometry triangle conic-sections geometric-construction
edited Aug 23 at 8:32
Brahadeesh
4,11331550
asked Aug 23 at 8:03
Andrea Prunotto
1,299625
Consider any right triangle $triangle ABC$.
We focus on one side, $AC$, and we take the midpoint $E$ of this side. Then, we draw the circle with center in $E$ and passing by $A,C$. If we take the perpendicular to $AC$ passing by $E$, we define a point $I$, where the circle intersects the perpendicular line.
Now, we can draw the ellipse with focii in $A$ and $C$ and passing by $I$.
Clearly, we can apply this procedure to both the other sides, obtaining other two ellipses.
My conjecture is that
The sum of the areas of the ellipses constructed on the two catheti is equal to the area of the ellipse constructed on the hypotenuse.
This is probably a very well known result, and I already apologize with the experts.
However, in order to prove the conjecture, I used the formula of the area of the ellipse, $S=pi a b$, where $a$ and $b$ are the lengths of the semi-axes. Although it is easy to prove that, in the case of all our ellipses, one semi-axis is clearly half the side, I am stuck in the attempt to determine the lengths of the other semi-axes, and I suspect however that there should be a very elementary way to prove such claim.
Again, sorry for the naivety, and thank you very much for any help or suggestion for a compact proof.
geometry euclidean-geometry triangle conic-sections geometric-construction
edited Aug 23 at 8:32
Brahadeesh
4,11331550
asked Aug 23 at 8:03
Andrea Prunotto
1,299625
edited Aug 23 at 8:32
Brahadeesh
4,11331550
edited Aug 23 at 8:32
Brahadeesh
4,11331550
edited Aug 23 at 8:32
Brahadeesh
4,11331550
4,11331550
asked Aug 23 at 8:03
Andrea Prunotto
1,299625
asked Aug 23 at 8:03
Andrea Prunotto
1,299625
asked Aug 23 at 8:03
Andrea Prunotto
1,299625
1,299625
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
16
down vote
accepted
In an ellipse, $a^2 = b^2 + c^2$, where $a$ is the semi-major, $b$ is the semi-minor, and $c$ is half the focal distance. In your case, $b=c$, so that $a=sqrt2b$.
answered Aug 23 at 8:09
what a disgrace
34316
3
Right! Thanks a lot!
– Andrea Prunotto
Aug 23 at 8:12
add a comment |Â
up vote
21
down vote
If you construct similar shapes on the three edges of a right triangle their areas add up as suggested by the Pythagorean theorem.
answered Aug 23 at 8:30
Christian Blatter
165k7109311
1
Sure. I realize that this is really a trivial question. Sorry, but thanks for replying, however!
– Andrea Prunotto
Aug 23 at 8:33
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
16
down vote
accepted
In an ellipse, $a^2 = b^2 + c^2$, where $a$ is the semi-major, $b$ is the semi-minor, and $c$ is half the focal distance. In your case, $b=c$, so that $a=sqrt2b$.
answered Aug 23 at 8:09
what a disgrace
34316
3
Right! Thanks a lot!
– Andrea Prunotto
Aug 23 at 8:12
add a comment |Â
up vote
16
down vote
accepted
In an ellipse, $a^2 = b^2 + c^2$, where $a$ is the semi-major, $b$ is the semi-minor, and $c$ is half the focal distance. In your case, $b=c$, so that $a=sqrt2b$.
answered Aug 23 at 8:09
what a disgrace
34316
3
Right! Thanks a lot!
– Andrea Prunotto
Aug 23 at 8:12
add a comment |Â
up vote
16
down vote
accepted
up vote
16
down vote
accepted
In an ellipse, $a^2 = b^2 + c^2$, where $a$ is the semi-major, $b$ is the semi-minor, and $c$ is half the focal distance. In your case, $b=c$, so that $a=sqrt2b$.
answered Aug 23 at 8:09
what a disgrace
34316
In an ellipse, $a^2 = b^2 + c^2$, where $a$ is the semi-major, $b$ is the semi-minor, and $c$ is half the focal distance. In your case, $b=c$, so that $a=sqrt2b$.
answered Aug 23 at 8:09
what a disgrace
34316
answered Aug 23 at 8:09
what a disgrace
34316
answered Aug 23 at 8:09
what a disgrace
34316
answered Aug 23 at 8:09
what a disgrace
34316
34316
3
Right! Thanks a lot!
– Andrea Prunotto
Aug 23 at 8:12
add a comment |Â
3
Right! Thanks a lot!
– Andrea Prunotto
Aug 23 at 8:12
3
3
Right! Thanks a lot!
– Andrea Prunotto
Aug 23 at 8:12
Right! Thanks a lot!
– Andrea Prunotto
Aug 23 at 8:12
add a comment |Â
up vote
21
down vote
If you construct similar shapes on the three edges of a right triangle their areas add up as suggested by the Pythagorean theorem.
answered Aug 23 at 8:30
Christian Blatter
165k7109311
1
Sure. I realize that this is really a trivial question. Sorry, but thanks for replying, however!
– Andrea Prunotto
Aug 23 at 8:33
add a comment |Â
up vote
21
down vote
If you construct similar shapes on the three edges of a right triangle their areas add up as suggested by the Pythagorean theorem.
answered Aug 23 at 8:30
Christian Blatter
165k7109311
1
Sure. I realize that this is really a trivial question. Sorry, but thanks for replying, however!
– Andrea Prunotto
Aug 23 at 8:33
add a comment |Â
up vote
21
down vote
up vote
21
down vote
If you construct similar shapes on the three edges of a right triangle their areas add up as suggested by the Pythagorean theorem.
answered Aug 23 at 8:30
Christian Blatter
165k7109311
If you construct similar shapes on the three edges of a right triangle their areas add up as suggested by the Pythagorean theorem.
answered Aug 23 at 8:30
Christian Blatter
165k7109311
answered Aug 23 at 8:30
Christian Blatter
165k7109311
answered Aug 23 at 8:30
Christian Blatter
165k7109311
answered Aug 23 at 8:30
Christian Blatter
165k7109311
165k7109311
1
Sure. I realize that this is really a trivial question. Sorry, but thanks for replying, however!
– Andrea Prunotto
Aug 23 at 8:33
add a comment |Â
1
Sure. I realize that this is really a trivial question. Sorry, but thanks for replying, however!
– Andrea Prunotto
Aug 23 at 8:33
1
1
Sure. I realize that this is really a trivial question. Sorry, but thanks for replying, however!
– Andrea Prunotto
Aug 23 at 8:33
Sure. I realize that this is really a trivial question. Sorry, but thanks for replying, however!
– Andrea Prunotto
Aug 23 at 8:33
add a comment |Â
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