A conjecture about the sum of the areas of three triangles built on the sides of any given triangle
up vote
18
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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 yesterday
Andrea Prunotto
1,826730
add a comment |
up vote
18
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 yesterday
Andrea Prunotto
1,826730
1
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
yesterday
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
yesterday
The heights are meeting at that point
– Moti
yesterday
Thanks all for your comments!
– Andrea Prunotto
yesterday
@Raptor@Enkidu. Working on it!
– Andrea Prunotto
yesterday
add a comment |
up vote
18
down vote
favorite
up vote
18
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 yesterday
Andrea Prunotto
1,826730
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 yesterday
Andrea Prunotto
1,826730
asked yesterday
Andrea Prunotto
1,826730
asked yesterday
Andrea Prunotto
1,826730
asked yesterday
Andrea Prunotto
1,826730
asked yesterday
Andrea Prunotto
1,826730
1,826730
1
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
yesterday
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
yesterday
The heights are meeting at that point
– Moti
yesterday
Thanks all for your comments!
– Andrea Prunotto
yesterday
@Raptor@Enkidu. Working on it!
– Andrea Prunotto
yesterday
add a comment |
1
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
yesterday
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
yesterday
The heights are meeting at that point
– Moti
yesterday
Thanks all for your comments!
– Andrea Prunotto
yesterday
@Raptor@Enkidu. Working on it!
– Andrea Prunotto
yesterday
1
1
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
yesterday
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
yesterday
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
yesterday
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
yesterday
The heights are meeting at that point
– Moti
yesterday
The heights are meeting at that point
– Moti
yesterday
Thanks all for your comments!
– Andrea Prunotto
yesterday
Thanks all for your comments!
– Andrea Prunotto
yesterday
@Raptor@Enkidu. Working on it!
– Andrea Prunotto
yesterday
@Raptor@Enkidu. Working on it!
– Andrea Prunotto
yesterday
add a comment |
2 Answers
2
active
oldest
votes
up vote
27
down vote
accepted
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered yesterday
Moti
1,299712
Nice and easy solution!
– YiFan
yesterday
add a comment |
up vote
12
down vote
The answer of Moti is perfect.
Note, though, that this also means that this property holds only if the orthocenter is inside the triangle, otherwise the external triangles overlap and the required property does not hold any more.
Internal orthocenter.
External orthocenter.
answered yesterday
Francesco Iovine
30115
2
No, I do not think so. Maybe that we must consider the oriented area of the triangles build with respect to each side. I guess that the sum of the oriented areas of these triangles (some positive some negative) still equals the area of the original triangle.
– Francesco Iovine
yesterday
2
Signed areas do, indeed, save the result.
– Blue
yesterday
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
27
down vote
accepted
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered yesterday
Moti
1,299712
Nice and easy solution!
– YiFan
yesterday
add a comment |
up vote
27
down vote
accepted
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered yesterday
Moti
1,299712
Nice and easy solution!
– YiFan
yesterday
add a comment |
up vote
27
down vote
accepted
up vote
27
down vote
accepted
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered yesterday
Moti
1,299712
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered yesterday
Moti
1,299712
answered yesterday
Moti
1,299712
answered yesterday
Moti
1,299712
answered yesterday
Moti
1,299712
1,299712
Nice and easy solution!
– YiFan
yesterday
add a comment |
Nice and easy solution!
– YiFan
yesterday
Nice and easy solution!
– YiFan
yesterday
Nice and easy solution!
– YiFan
yesterday
add a comment |
up vote
12
down vote
The answer of Moti is perfect.
Note, though, that this also means that this property holds only if the orthocenter is inside the triangle, otherwise the external triangles overlap and the required property does not hold any more.
Internal orthocenter.
External orthocenter.
answered yesterday
Francesco Iovine
30115
2
No, I do not think so. Maybe that we must consider the oriented area of the triangles build with respect to each side. I guess that the sum of the oriented areas of these triangles (some positive some negative) still equals the area of the original triangle.
– Francesco Iovine
yesterday
2
Signed areas do, indeed, save the result.
– Blue
yesterday
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
yesterday
add a comment |
up vote
12
down vote
The answer of Moti is perfect.
Note, though, that this also means that this property holds only if the orthocenter is inside the triangle, otherwise the external triangles overlap and the required property does not hold any more.
Internal orthocenter.
External orthocenter.
answered yesterday
Francesco Iovine
30115
2
No, I do not think so. Maybe that we must consider the oriented area of the triangles build with respect to each side. I guess that the sum of the oriented areas of these triangles (some positive some negative) still equals the area of the original triangle.
– Francesco Iovine
yesterday
2
Signed areas do, indeed, save the result.
– Blue
yesterday
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
yesterday
add a comment |
up vote
12
down vote
up vote
12
down vote
The answer of Moti is perfect.
Note, though, that this also means that this property holds only if the orthocenter is inside the triangle, otherwise the external triangles overlap and the required property does not hold any more.
Internal orthocenter.
External orthocenter.
answered yesterday
Francesco Iovine
30115
The answer of Moti is perfect.
Note, though, that this also means that this property holds only if the orthocenter is inside the triangle, otherwise the external triangles overlap and the required property does not hold any more.
Internal orthocenter.
External orthocenter.
answered yesterday
Francesco Iovine
30115
answered yesterday
Francesco Iovine
30115
answered yesterday
Francesco Iovine
30115
answered yesterday
Francesco Iovine
30115
30115
2
No, I do not think so. Maybe that we must consider the oriented area of the triangles build with respect to each side. I guess that the sum of the oriented areas of these triangles (some positive some negative) still equals the area of the original triangle.
– Francesco Iovine
yesterday
2
Signed areas do, indeed, save the result.
– Blue
yesterday
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
yesterday
add a comment |
2
No, I do not think so. Maybe that we must consider the oriented area of the triangles build with respect to each side. I guess that the sum of the oriented areas of these triangles (some positive some negative) still equals the area of the original triangle.
– Francesco Iovine
yesterday
2
Signed areas do, indeed, save the result.
– Blue
yesterday
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
yesterday
2
2
No, I do not think so. Maybe that we must consider the oriented area of the triangles build with respect to each side. I guess that the sum of the oriented areas of these triangles (some positive some negative) still equals the area of the original triangle.
– Francesco Iovine
yesterday
No, I do not think so. Maybe that we must consider the oriented area of the triangles build with respect to each side. I guess that the sum of the oriented areas of these triangles (some positive some negative) still equals the area of the original triangle.
– Francesco Iovine
yesterday
2
2
Signed areas do, indeed, save the result.
– Blue
yesterday
Signed areas do, indeed, save the result.
– Blue
yesterday
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
yesterday
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
yesterday
add a comment |
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1
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
yesterday
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
yesterday
The heights are meeting at that point
– Moti
yesterday
Thanks all for your comments!
– Andrea Prunotto
yesterday
@Raptor@Enkidu. Working on it!
– Andrea Prunotto
yesterday