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A conjecture about the sum of the areas of three triangles built on the sides of any given triangle
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Given any triangle $triangle ABC$, and given one of its side, we can draw two lines perpendicular to that side passing through its two vertices. If we do this construction for each side, we obtain the points $D,E,F$ where two of these perpendicular lines meet at the minimum distance to each side.
These three points can be used to build three triangles on each side of the starting triangle.
The conjecture is that
The sum of the areas of the triangles $triangle AFB$, $triangle BDC$, and $triangle CEA$ is equal to the area of $triangle ABC$.
This is likely an obvious and very well known result. But I cannot find an easy proof of this. Therefore I apologize for possible triviality, and I thank you for any suggestion.
geometry euclidean-geometry triangle geometric-construction
asked 54 mins ago
Andrea Prunotto
1,704728
add a comment |
up vote
4
down vote
favorite
Given any triangle $triangle ABC$, and given one of its side, we can draw two lines perpendicular to that side passing through its two vertices. If we do this construction for each side, we obtain the points $D,E,F$ where two of these perpendicular lines meet at the minimum distance to each side.
These three points can be used to build three triangles on each side of the starting triangle.
The conjecture is that
The sum of the areas of the triangles $triangle AFB$, $triangle BDC$, and $triangle CEA$ is equal to the area of $triangle ABC$.
This is likely an obvious and very well known result. But I cannot find an easy proof of this. Therefore I apologize for possible triviality, and I thank you for any suggestion.
geometry euclidean-geometry triangle geometric-construction
asked 54 mins ago
Andrea Prunotto
1,704728
I am not too sure but did you consider that there might be a point inside the triangle, $G$ such that $AFBG$, $BDCG$ and $AECG$ are parallelograms?
– Raptor
43 mins ago
I agree with Raptor, and would guess it is the center of the inner circle. i.e. intersection of angle symmetrals. however, have you also tried it for triangles with an obtuse angle? there should be weird stuff happening there, since one of you orthogonals actually will go into the triangle!
– Enkidu
27 mins ago
The heights are meeting at that point
– Moti
27 mins ago
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Given any triangle $triangle ABC$, and given one of its side, we can draw two lines perpendicular to that side passing through its two vertices. If we do this construction for each side, we obtain the points $D,E,F$ where two of these perpendicular lines meet at the minimum distance to each side.
These three points can be used to build three triangles on each side of the starting triangle.
The conjecture is that
The sum of the areas of the triangles $triangle AFB$, $triangle BDC$, and $triangle CEA$ is equal to the area of $triangle ABC$.
This is likely an obvious and very well known result. But I cannot find an easy proof of this. Therefore I apologize for possible triviality, and I thank you for any suggestion.
geometry euclidean-geometry triangle geometric-construction
asked 54 mins ago
Andrea Prunotto
1,704728
Given any triangle $triangle ABC$, and given one of its side, we can draw two lines perpendicular to that side passing through its two vertices. If we do this construction for each side, we obtain the points $D,E,F$ where two of these perpendicular lines meet at the minimum distance to each side.
These three points can be used to build three triangles on each side of the starting triangle.
The conjecture is that
The sum of the areas of the triangles $triangle AFB$, $triangle BDC$, and $triangle CEA$ is equal to the area of $triangle ABC$.
This is likely an obvious and very well known result. But I cannot find an easy proof of this. Therefore I apologize for possible triviality, and I thank you for any suggestion.
geometry euclidean-geometry triangle geometric-construction
geometry euclidean-geometry triangle geometric-construction
asked 54 mins ago
Andrea Prunotto
1,704728
asked 54 mins ago
Andrea Prunotto
1,704728
asked 54 mins ago
Andrea Prunotto
1,704728
asked 54 mins ago
Andrea Prunotto
1,704728
asked 54 mins ago
Andrea Prunotto
1,704728
1,704728
I am not too sure but did you consider that there might be a point inside the triangle, $G$ such that $AFBG$, $BDCG$ and $AECG$ are parallelograms?
– Raptor
43 mins ago
I agree with Raptor, and would guess it is the center of the inner circle. i.e. intersection of angle symmetrals. however, have you also tried it for triangles with an obtuse angle? there should be weird stuff happening there, since one of you orthogonals actually will go into the triangle!
– Enkidu
27 mins ago
The heights are meeting at that point
– Moti
27 mins ago
add a comment |
I am not too sure but did you consider that there might be a point inside the triangle, $G$ such that $AFBG$, $BDCG$ and $AECG$ are parallelograms?
– Raptor
43 mins ago
I agree with Raptor, and would guess it is the center of the inner circle. i.e. intersection of angle symmetrals. however, have you also tried it for triangles with an obtuse angle? there should be weird stuff happening there, since one of you orthogonals actually will go into the triangle!
– Enkidu
27 mins ago
The heights are meeting at that point
– Moti
27 mins ago
I am not too sure but did you consider that there might be a point inside the triangle, $G$ such that $AFBG$, $BDCG$ and $AECG$ are parallelograms?
– Raptor
43 mins ago
I am not too sure but did you consider that there might be a point inside the triangle, $G$ such that $AFBG$, $BDCG$ and $AECG$ are parallelograms?
– Raptor
43 mins ago
I agree with Raptor, and would guess it is the center of the inner circle. i.e. intersection of angle symmetrals. however, have you also tried it for triangles with an obtuse angle? there should be weird stuff happening there, since one of you orthogonals actually will go into the triangle!
– Enkidu
27 mins ago
I agree with Raptor, and would guess it is the center of the inner circle. i.e. intersection of angle symmetrals. however, have you also tried it for triangles with an obtuse angle? there should be weird stuff happening there, since one of you orthogonals actually will go into the triangle!
– Enkidu
27 mins ago
The heights are meeting at that point
– Moti
27 mins ago
The heights are meeting at that point
– Moti
27 mins ago
add a comment |
1 Answer
1
active
oldest
votes
up vote
5
down vote
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered 23 mins ago
Moti
1,094510
Nice and easy solution!
– YiFan
10 mins ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered 23 mins ago
Moti
1,094510
Nice and easy solution!
– YiFan
10 mins ago
add a comment |
up vote
5
down vote
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered 23 mins ago
Moti
1,094510
Nice and easy solution!
– YiFan
10 mins ago
add a comment |
up vote
5
down vote
up vote
5
down vote
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered 23 mins ago
Moti
1,094510
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered 23 mins ago
Moti
1,094510
answered 23 mins ago
Moti
1,094510
answered 23 mins ago
Moti
1,094510
answered 23 mins ago
Moti
1,094510
1,094510
Nice and easy solution!
– YiFan
10 mins ago
add a comment |
Nice and easy solution!
– YiFan
10 mins ago
Nice and easy solution!
– YiFan
10 mins ago
Nice and easy solution!
– YiFan
10 mins ago
add a comment |
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I am not too sure but did you consider that there might be a point inside the triangle, $G$ such that $AFBG$, $BDCG$ and $AECG$ are parallelograms?
– Raptor
43 mins ago
I agree with Raptor, and would guess it is the center of the inner circle. i.e. intersection of angle symmetrals. however, have you also tried it for triangles with an obtuse angle? there should be weird stuff happening there, since one of you orthogonals actually will go into the triangle!
– Enkidu
27 mins ago
The heights are meeting at that point
– Moti
27 mins ago