A conjecture about the sum of the areas of three triangles built on the sides of any given triangle
up vote
21
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
add a comment |
up vote
21
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
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
Nov 9 at 7:58
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
Nov 9 at 8:14
The heights are meeting at that point
– Moti
Nov 9 at 8:14
Thanks all for your comments!
– user559615
Nov 9 at 9:38
@Raptor@Enkidu. Working on it!
– user559615
Nov 9 at 9:52
add a comment |
up vote
21
down vote
favorite
up vote
21
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
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 Nov 9 at 7:47
user559615
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
Nov 9 at 7:58
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
Nov 9 at 8:14
The heights are meeting at that point
– Moti
Nov 9 at 8:14
Thanks all for your comments!
– user559615
Nov 9 at 9:38
@Raptor@Enkidu. Working on it!
– user559615
Nov 9 at 9:52
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
Nov 9 at 7:58
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
Nov 9 at 8:14
The heights are meeting at that point
– Moti
Nov 9 at 8:14
Thanks all for your comments!
– user559615
Nov 9 at 9:38
@Raptor@Enkidu. Working on it!
– user559615
Nov 9 at 9:52
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
Nov 9 at 7:58
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
Nov 9 at 7:58
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
Nov 9 at 8:14
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
Nov 9 at 8:14
The heights are meeting at that point
– Moti
Nov 9 at 8:14
The heights are meeting at that point
– Moti
Nov 9 at 8:14
Thanks all for your comments!
– user559615
Nov 9 at 9:38
Thanks all for your comments!
– user559615
Nov 9 at 9:38
@Raptor@Enkidu. Working on it!
– user559615
Nov 9 at 9:52
@Raptor@Enkidu. Working on it!
– user559615
Nov 9 at 9:52
add a comment |
2 Answers
2
active
oldest
votes
up vote
29
down vote
accepted
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered Nov 9 at 8:18
Moti
1,325712
Nice and easy solution!
– YiFan
Nov 9 at 8:31
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 Nov 9 at 11:32
Francesco Iovine
29115
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
Nov 9 at 11:51
2
Signed areas do, indeed, save the result.
– Blue
Nov 9 at 11:54
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
Nov 9 at 12:57
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
29
down vote
accepted
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered Nov 9 at 8:18
Moti
1,325712
Nice and easy solution!
– YiFan
Nov 9 at 8:31
add a comment |
up vote
29
down vote
accepted
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered Nov 9 at 8:18
Moti
1,325712
Nice and easy solution!
– YiFan
Nov 9 at 8:31
add a comment |
up vote
29
down vote
accepted
up vote
29
down vote
accepted
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered Nov 9 at 8:18
Moti
1,325712
Draw the orthocenter. You get three parallelograms which immediately provide the answer.
answered Nov 9 at 8:18
Moti
1,325712
answered Nov 9 at 8:18
Moti
1,325712
answered Nov 9 at 8:18
Moti
1,325712
answered Nov 9 at 8:18
Moti
1,325712
1,325712
Nice and easy solution!
– YiFan
Nov 9 at 8:31
add a comment |
Nice and easy solution!
– YiFan
Nov 9 at 8:31
Nice and easy solution!
– YiFan
Nov 9 at 8:31
Nice and easy solution!
– YiFan
Nov 9 at 8:31
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 Nov 9 at 11:32
Francesco Iovine
29115
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
Nov 9 at 11:51
2
Signed areas do, indeed, save the result.
– Blue
Nov 9 at 11:54
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
Nov 9 at 12:57
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 Nov 9 at 11:32
Francesco Iovine
29115
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
Nov 9 at 11:51
2
Signed areas do, indeed, save the result.
– Blue
Nov 9 at 11:54
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
Nov 9 at 12:57
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 Nov 9 at 11:32
Francesco Iovine
29115
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 Nov 9 at 11:32
Francesco Iovine
29115
answered Nov 9 at 11:32
Francesco Iovine
29115
answered Nov 9 at 11:32
Francesco Iovine
29115
answered Nov 9 at 11:32
Francesco Iovine
29115
29115
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
Nov 9 at 11:51
2
Signed areas do, indeed, save the result.
– Blue
Nov 9 at 11:54
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
Nov 9 at 12:57
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
Nov 9 at 11:51
2
Signed areas do, indeed, save the result.
– Blue
Nov 9 at 11:54
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
Nov 9 at 12:57
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
Nov 9 at 11:51
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
Nov 9 at 11:51
2
2
Signed areas do, indeed, save the result.
– Blue
Nov 9 at 11:54
Signed areas do, indeed, save the result.
– Blue
Nov 9 at 11:54
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
Nov 9 at 12:57
@FrancescoIovine That the sum of the oriented areas is equal to the original triangle's area is also easily shown using the orthocenter.
– Vaelus
Nov 9 at 12:57
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2991075%2fa-conjecture-about-the-sum-of-the-areas-of-three-triangles-built-on-the-sides-of%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
Nov 9 at 7:58
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
Nov 9 at 8:14
The heights are meeting at that point
– Moti
Nov 9 at 8:14
Thanks all for your comments!
– user559615
Nov 9 at 9:38
@Raptor@Enkidu. Working on it!
– user559615
Nov 9 at 9:52