A conjecture about the sum of the areas of $3$ triangles built on the sides of any triangle (by means of...
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Given any triangle $triangle ABC$, let us draw its orthocenter $D$. By means of this point, we can draw three circles with centers in $A,B,C$ and passing through $D$.
These circles intersect in the points $E,F,G$, which can be seen as the vertices of three triangles $triangle AFC, triangle CGB$ and $triangle BEA$.
My conjecture is that
The sum of the areas of the triangles $triangle AFC, triangle CGB, triangle BEA$ is equal to the area of the triangle $triangle ABC$.
Furthermore,
If we substitute the orthocenter $D$ with the centroid of $triangle ABC$, then the areas of $triangle AFC, triangle CGB$ and $triangle BEA$ are all equal, and their sum is equal to the area of the triangle $triangle ABC$.
Maybe these are very well known theorems. However, is there a compact proof for such conjectures?
NOTE: These conjectures are very similar to the one exposed in this post.
Thanks for your help, and sorry for imprecision or triviality.
geometry euclidean-geometry triangle circle geometric-construction
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up vote
0
down vote
favorite
Given any triangle $triangle ABC$, let us draw its orthocenter $D$. By means of this point, we can draw three circles with centers in $A,B,C$ and passing through $D$.
These circles intersect in the points $E,F,G$, which can be seen as the vertices of three triangles $triangle AFC, triangle CGB$ and $triangle BEA$.
My conjecture is that
The sum of the areas of the triangles $triangle AFC, triangle CGB, triangle BEA$ is equal to the area of the triangle $triangle ABC$.
Furthermore,
If we substitute the orthocenter $D$ with the centroid of $triangle ABC$, then the areas of $triangle AFC, triangle CGB$ and $triangle BEA$ are all equal, and their sum is equal to the area of the triangle $triangle ABC$.
Maybe these are very well known theorems. However, is there a compact proof for such conjectures?
NOTE: These conjectures are very similar to the one exposed in this post.
Thanks for your help, and sorry for imprecision or triviality.
geometry euclidean-geometry triangle circle geometric-construction
"If we substitute the centroid $D$ with the orthocenter [...], then the areas [...] area all equal ..." Not so! Consider a right triangle (or a very-nearly-right triangle, if you want to restrict yourself to acute figures). Two of the constructed triangles vanish (or are very small); the areas cannot all be equal, nor equal to the area of the original triangle. As greedoid has shown, the sum-of-areas gives the original area regardless of $D$'s location, so each area being equal to original triangle simply never holds.
– Blue
Nov 16 at 10:58
@Blue Sure, sorry, I edit. I meant that the areas are all equal and their sum is equal to the area of the original triangle!
– user559615
Nov 16 at 11:00
@Blue Thanks to point it out.
– user559615
Nov 16 at 11:02
Open problem, the case of the othocenter yields to three equal areas. Why? Is this the only solution?
– user559615
Nov 16 at 11:02
1
That the centroid is the only point making three equal areas is well-known. To see why it must be so, consider: If $triangle ABD$ has one-third the area of $triangle ABC$, then its height relative to base $overline{AB}$ must be one-third that of $triangle ABC$; thus, $D$ is on a line parallel to that base, one-third of the way "up". Likewise, $D$ is on other lines parallel to the other sides. These three (distinct) lines can have at most one point in common; the centroid is that point.
– Blue
Nov 16 at 11:14
|
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given any triangle $triangle ABC$, let us draw its orthocenter $D$. By means of this point, we can draw three circles with centers in $A,B,C$ and passing through $D$.
These circles intersect in the points $E,F,G$, which can be seen as the vertices of three triangles $triangle AFC, triangle CGB$ and $triangle BEA$.
My conjecture is that
The sum of the areas of the triangles $triangle AFC, triangle CGB, triangle BEA$ is equal to the area of the triangle $triangle ABC$.
Furthermore,
If we substitute the orthocenter $D$ with the centroid of $triangle ABC$, then the areas of $triangle AFC, triangle CGB$ and $triangle BEA$ are all equal, and their sum is equal to the area of the triangle $triangle ABC$.
Maybe these are very well known theorems. However, is there a compact proof for such conjectures?
NOTE: These conjectures are very similar to the one exposed in this post.
Thanks for your help, and sorry for imprecision or triviality.
geometry euclidean-geometry triangle circle geometric-construction
Given any triangle $triangle ABC$, let us draw its orthocenter $D$. By means of this point, we can draw three circles with centers in $A,B,C$ and passing through $D$.
These circles intersect in the points $E,F,G$, which can be seen as the vertices of three triangles $triangle AFC, triangle CGB$ and $triangle BEA$.
My conjecture is that
The sum of the areas of the triangles $triangle AFC, triangle CGB, triangle BEA$ is equal to the area of the triangle $triangle ABC$.
Furthermore,
If we substitute the orthocenter $D$ with the centroid of $triangle ABC$, then the areas of $triangle AFC, triangle CGB$ and $triangle BEA$ are all equal, and their sum is equal to the area of the triangle $triangle ABC$.
Maybe these are very well known theorems. However, is there a compact proof for such conjectures?
NOTE: These conjectures are very similar to the one exposed in this post.
Thanks for your help, and sorry for imprecision or triviality.
geometry euclidean-geometry triangle circle geometric-construction
geometry euclidean-geometry triangle circle geometric-construction
edited Nov 16 at 11:13
asked Nov 16 at 10:10
user559615
"If we substitute the centroid $D$ with the orthocenter [...], then the areas [...] area all equal ..." Not so! Consider a right triangle (or a very-nearly-right triangle, if you want to restrict yourself to acute figures). Two of the constructed triangles vanish (or are very small); the areas cannot all be equal, nor equal to the area of the original triangle. As greedoid has shown, the sum-of-areas gives the original area regardless of $D$'s location, so each area being equal to original triangle simply never holds.
– Blue
Nov 16 at 10:58
@Blue Sure, sorry, I edit. I meant that the areas are all equal and their sum is equal to the area of the original triangle!
– user559615
Nov 16 at 11:00
@Blue Thanks to point it out.
– user559615
Nov 16 at 11:02
Open problem, the case of the othocenter yields to three equal areas. Why? Is this the only solution?
– user559615
Nov 16 at 11:02
1
That the centroid is the only point making three equal areas is well-known. To see why it must be so, consider: If $triangle ABD$ has one-third the area of $triangle ABC$, then its height relative to base $overline{AB}$ must be one-third that of $triangle ABC$; thus, $D$ is on a line parallel to that base, one-third of the way "up". Likewise, $D$ is on other lines parallel to the other sides. These three (distinct) lines can have at most one point in common; the centroid is that point.
– Blue
Nov 16 at 11:14
|
show 5 more comments
"If we substitute the centroid $D$ with the orthocenter [...], then the areas [...] area all equal ..." Not so! Consider a right triangle (or a very-nearly-right triangle, if you want to restrict yourself to acute figures). Two of the constructed triangles vanish (or are very small); the areas cannot all be equal, nor equal to the area of the original triangle. As greedoid has shown, the sum-of-areas gives the original area regardless of $D$'s location, so each area being equal to original triangle simply never holds.
– Blue
Nov 16 at 10:58
@Blue Sure, sorry, I edit. I meant that the areas are all equal and their sum is equal to the area of the original triangle!
– user559615
Nov 16 at 11:00
@Blue Thanks to point it out.
– user559615
Nov 16 at 11:02
Open problem, the case of the othocenter yields to three equal areas. Why? Is this the only solution?
– user559615
Nov 16 at 11:02
1
That the centroid is the only point making three equal areas is well-known. To see why it must be so, consider: If $triangle ABD$ has one-third the area of $triangle ABC$, then its height relative to base $overline{AB}$ must be one-third that of $triangle ABC$; thus, $D$ is on a line parallel to that base, one-third of the way "up". Likewise, $D$ is on other lines parallel to the other sides. These three (distinct) lines can have at most one point in common; the centroid is that point.
– Blue
Nov 16 at 11:14
"If we substitute the centroid $D$ with the orthocenter [...], then the areas [...] area all equal ..." Not so! Consider a right triangle (or a very-nearly-right triangle, if you want to restrict yourself to acute figures). Two of the constructed triangles vanish (or are very small); the areas cannot all be equal, nor equal to the area of the original triangle. As greedoid has shown, the sum-of-areas gives the original area regardless of $D$'s location, so each area being equal to original triangle simply never holds.
– Blue
Nov 16 at 10:58
"If we substitute the centroid $D$ with the orthocenter [...], then the areas [...] area all equal ..." Not so! Consider a right triangle (or a very-nearly-right triangle, if you want to restrict yourself to acute figures). Two of the constructed triangles vanish (or are very small); the areas cannot all be equal, nor equal to the area of the original triangle. As greedoid has shown, the sum-of-areas gives the original area regardless of $D$'s location, so each area being equal to original triangle simply never holds.
– Blue
Nov 16 at 10:58
@Blue Sure, sorry, I edit. I meant that the areas are all equal and their sum is equal to the area of the original triangle!
– user559615
Nov 16 at 11:00
@Blue Sure, sorry, I edit. I meant that the areas are all equal and their sum is equal to the area of the original triangle!
– user559615
Nov 16 at 11:00
@Blue Thanks to point it out.
– user559615
Nov 16 at 11:02
@Blue Thanks to point it out.
– user559615
Nov 16 at 11:02
Open problem, the case of the othocenter yields to three equal areas. Why? Is this the only solution?
– user559615
Nov 16 at 11:02
Open problem, the case of the othocenter yields to three equal areas. Why? Is this the only solution?
– user559615
Nov 16 at 11:02
1
1
That the centroid is the only point making three equal areas is well-known. To see why it must be so, consider: If $triangle ABD$ has one-third the area of $triangle ABC$, then its height relative to base $overline{AB}$ must be one-third that of $triangle ABC$; thus, $D$ is on a line parallel to that base, one-third of the way "up". Likewise, $D$ is on other lines parallel to the other sides. These three (distinct) lines can have at most one point in common; the centroid is that point.
– Blue
Nov 16 at 11:14
That the centroid is the only point making three equal areas is well-known. To see why it must be so, consider: If $triangle ABD$ has one-third the area of $triangle ABC$, then its height relative to base $overline{AB}$ must be one-third that of $triangle ABC$; thus, $D$ is on a line parallel to that base, one-third of the way "up". Likewise, $D$ is on other lines parallel to the other sides. These three (distinct) lines can have at most one point in common; the centroid is that point.
– Blue
Nov 16 at 11:14
|
show 5 more comments
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
This is true since for any point in triangle (not just centroid or orthocenter):
$ACF$ is congruent to $ACD$ (sss), ($AD = AF$, $CD=CF$, and common $AC$)
$ABE$ is congruent to $ABD$ (sss) and
$BCG$ is congruent to $BCD$ (sss).
answered Nov 16 at 10:24
greedoid
34.3k114488
Right! Thanks! Very neat. And what about the fact that the areas of the three external triangles are the same in case of the orthocenter? Is this the only case?
– user559615
Nov 16 at 10:28
add a comment |
up vote
0
down vote
Given the arbitrary point D inside the triangle, if you get the symmetrical points with respect to all of the sides, the triangles built outside are congruent to those built inside. And the three of them divide perfeclty the area of the triangle.
So the conjecture is correct.
answered Nov 16 at 11:48
Francesco Iovine
29115
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
This is true since for any point in triangle (not just centroid or orthocenter):
$ACF$ is congruent to $ACD$ (sss), ($AD = AF$, $CD=CF$, and common $AC$)
$ABE$ is congruent to $ABD$ (sss) and
$BCG$ is congruent to $BCD$ (sss).
answered Nov 16 at 10:24
greedoid
34.3k114488
Right! Thanks! Very neat. And what about the fact that the areas of the three external triangles are the same in case of the orthocenter? Is this the only case?
– user559615
Nov 16 at 10:28
add a comment |
up vote
1
down vote
accepted
This is true since for any point in triangle (not just centroid or orthocenter):
$ACF$ is congruent to $ACD$ (sss), ($AD = AF$, $CD=CF$, and common $AC$)
$ABE$ is congruent to $ABD$ (sss) and
$BCG$ is congruent to $BCD$ (sss).
answered Nov 16 at 10:24
greedoid
34.3k114488
Right! Thanks! Very neat. And what about the fact that the areas of the three external triangles are the same in case of the orthocenter? Is this the only case?
– user559615
Nov 16 at 10:28
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
This is true since for any point in triangle (not just centroid or orthocenter):
$ACF$ is congruent to $ACD$ (sss), ($AD = AF$, $CD=CF$, and common $AC$)
$ABE$ is congruent to $ABD$ (sss) and
$BCG$ is congruent to $BCD$ (sss).
answered Nov 16 at 10:24
greedoid
34.3k114488
This is true since for any point in triangle (not just centroid or orthocenter):
$ACF$ is congruent to $ACD$ (sss), ($AD = AF$, $CD=CF$, and common $AC$)
$ABE$ is congruent to $ABD$ (sss) and
$BCG$ is congruent to $BCD$ (sss).
answered Nov 16 at 10:24
greedoid
34.3k114488
answered Nov 16 at 10:24
greedoid
34.3k114488
answered Nov 16 at 10:24
greedoid
34.3k114488
answered Nov 16 at 10:24
greedoid
34.3k114488
34.3k114488
Right! Thanks! Very neat. And what about the fact that the areas of the three external triangles are the same in case of the orthocenter? Is this the only case?
– user559615
Nov 16 at 10:28
add a comment |
Right! Thanks! Very neat. And what about the fact that the areas of the three external triangles are the same in case of the orthocenter? Is this the only case?
– user559615
Nov 16 at 10:28
Right! Thanks! Very neat. And what about the fact that the areas of the three external triangles are the same in case of the orthocenter? Is this the only case?
– user559615
Nov 16 at 10:28
Right! Thanks! Very neat. And what about the fact that the areas of the three external triangles are the same in case of the orthocenter? Is this the only case?
– user559615
Nov 16 at 10:28
add a comment |
up vote
0
down vote
Given the arbitrary point D inside the triangle, if you get the symmetrical points with respect to all of the sides, the triangles built outside are congruent to those built inside. And the three of them divide perfeclty the area of the triangle.
So the conjecture is correct.
answered Nov 16 at 11:48
Francesco Iovine
29115
add a comment |
up vote
0
down vote
Given the arbitrary point D inside the triangle, if you get the symmetrical points with respect to all of the sides, the triangles built outside are congruent to those built inside. And the three of them divide perfeclty the area of the triangle.
So the conjecture is correct.
answered Nov 16 at 11:48
Francesco Iovine
29115
add a comment |
up vote
0
down vote
up vote
0
down vote
Given the arbitrary point D inside the triangle, if you get the symmetrical points with respect to all of the sides, the triangles built outside are congruent to those built inside. And the three of them divide perfeclty the area of the triangle.
So the conjecture is correct.
answered Nov 16 at 11:48
Francesco Iovine
29115
Given the arbitrary point D inside the triangle, if you get the symmetrical points with respect to all of the sides, the triangles built outside are congruent to those built inside. And the three of them divide perfeclty the area of the triangle.
So the conjecture is correct.
answered Nov 16 at 11:48
Francesco Iovine
29115
answered Nov 16 at 11:48
Francesco Iovine
29115
answered Nov 16 at 11:48
Francesco Iovine
29115
answered Nov 16 at 11:48
Francesco Iovine
29115
29115
add a comment |
add a comment |
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"If we substitute the centroid $D$ with the orthocenter [...], then the areas [...] area all equal ..." Not so! Consider a right triangle (or a very-nearly-right triangle, if you want to restrict yourself to acute figures). Two of the constructed triangles vanish (or are very small); the areas cannot all be equal, nor equal to the area of the original triangle. As greedoid has shown, the sum-of-areas gives the original area regardless of $D$'s location, so each area being equal to original triangle simply never holds.
– Blue
Nov 16 at 10:58
@Blue Sure, sorry, I edit. I meant that the areas are all equal and their sum is equal to the area of the original triangle!
– user559615
Nov 16 at 11:00
@Blue Thanks to point it out.
– user559615
Nov 16 at 11:02
Open problem, the case of the othocenter yields to three equal areas. Why? Is this the only solution?
– user559615
Nov 16 at 11:02
1
That the centroid is the only point making three equal areas is well-known. To see why it must be so, consider: If $triangle ABD$ has one-third the area of $triangle ABC$, then its height relative to base $overline{AB}$ must be one-third that of $triangle ABC$; thus, $D$ is on a line parallel to that base, one-third of the way "up". Likewise, $D$ is on other lines parallel to the other sides. These three (distinct) lines can have at most one point in common; the centroid is that point.
– Blue
Nov 16 at 11:14