Distances to line passing through the centroid of triangle The Next CEO of Stack OverflowIf a line through the centroid $G$ of $triangle ABC$ meets $AB$ in $M$ and $AC$ in $N$, then how to prove that $ANcdot MB+AMcdot NC=AMcdot AN$ .A line through the centroid G of $triangle ABC$ intersects the sides at points X, Y, Z.A triangle ABC with vertex $C(4,3)$. The bisector and the median line equation drawn from the same vertex are given. Find the vertices A & B.Where to put angle ending on right triangle, only using variables.Largest of the smallest angles of incidence from arbitrary point to tetrahedron vertex/centroid lineIf the equation of side BC is $2x-y=10$,then find the possible coordinates of vertex A.Constructing an isosceles triangle inside an acute angle.In triangle ABC, ∠B>90∘ Let H is point on side AC, AH=BH ⊥ BH and D and E is midpoint of AB and BC respectively…Constructing a line segment through a point so it has a ratio $1:2$A Geometry Problem from Indian RMO(Regional Mathematical Olympiad) 2016.Prove that the vertex C of a certain family of equilateral triangles ABC lies on a line
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Distances to line passing through the centroid of triangle
The Next CEO of Stack OverflowIf a line through the centroid $G$ of $triangle ABC$ meets $AB$ in $M$ and $AC$ in $N$, then how to prove that $ANcdot MB+AMcdot NC=AMcdot AN$ .A line through the centroid G of $triangle ABC$ intersects the sides at points X, Y, Z.A triangle ABC with vertex $C(4,3)$. The bisector and the median line equation drawn from the same vertex are given. Find the vertices A & B.Where to put angle ending on right triangle, only using variables.Largest of the smallest angles of incidence from arbitrary point to tetrahedron vertex/centroid lineIf the equation of side BC is $2x-y=10$,then find the possible coordinates of vertex A.Constructing an isosceles triangle inside an acute angle.In triangle ABC, ∠B>90∘ Let H is point on side AC, AH=BH ⊥ BH and D and E is midpoint of AB and BC respectively…Constructing a line segment through a point so it has a ratio $1:2$A Geometry Problem from Indian RMO(Regional Mathematical Olympiad) 2016.Prove that the vertex C of a certain family of equilateral triangles ABC lies on a line
$begingroup$
Let $p$ be a line that pass through the centroid of a triangle $ABC$. Unless the line pass through one vertex, then $2$ verices are one side of the line, while the third one is on the other side. Without loss of generality, let vertices $A$ and $C$ be one side, while the vertex $B$ be on the other side of the line. Let $x,y,z$ be the distances from the vertices $A,B,C$ to the line $p$. Prove that $x + z = y$
I've tried something, but I didn't make any progress. If we multiply both sides of the equality by $fracoverlineXY2$, then we need to prove:
$$P_BXY = P_AXY + P_CXY iff fracXB cdot BY cdot sin angle XBY2 = fracXA cdot AY cdot sin angle XAY2 + fracXC cdot CY cdot sin angle XCY2 iff XB cdot BY = XA cdot AY + XC cdot CY$$
But this is the most that I manage to get.
geometry triangles
edited Mar 20 at 7:46
Glorfindel
3,41581830
asked Jan 30 '14 at 18:43
Stefan4024Stefan4024
30.6k63579
$endgroup$
add a comment |
$begingroup$
Let $p$ be a line that pass through the centroid of a triangle $ABC$. Unless the line pass through one vertex, then $2$ verices are one side of the line, while the third one is on the other side. Without loss of generality, let vertices $A$ and $C$ be one side, while the vertex $B$ be on the other side of the line. Let $x,y,z$ be the distances from the vertices $A,B,C$ to the line $p$. Prove that $x + z = y$
I've tried something, but I didn't make any progress. If we multiply both sides of the equality by $fracoverlineXY2$, then we need to prove:
$$P_BXY = P_AXY + P_CXY iff fracXB cdot BY cdot sin angle XBY2 = fracXA cdot AY cdot sin angle XAY2 + fracXC cdot CY cdot sin angle XCY2 iff XB cdot BY = XA cdot AY + XC cdot CY$$
But this is the most that I manage to get.
geometry triangles
edited Mar 20 at 7:46
Glorfindel
3,41581830
asked Jan 30 '14 at 18:43
Stefan4024Stefan4024
30.6k63579
$endgroup$
$begingroup$
You haven't used the fact that $p$ passes through $M$ yet, although you will need it somewhere. What are some nice properties of the centroid that may be useful here?
$endgroup$
– Ragnar
Jan 30 '14 at 19:00
add a comment |
$begingroup$
Let $p$ be a line that pass through the centroid of a triangle $ABC$. Unless the line pass through one vertex, then $2$ verices are one side of the line, while the third one is on the other side. Without loss of generality, let vertices $A$ and $C$ be one side, while the vertex $B$ be on the other side of the line. Let $x,y,z$ be the distances from the vertices $A,B,C$ to the line $p$. Prove that $x + z = y$
I've tried something, but I didn't make any progress. If we multiply both sides of the equality by $fracoverlineXY2$, then we need to prove:
$$P_BXY = P_AXY + P_CXY iff fracXB cdot BY cdot sin angle XBY2 = fracXA cdot AY cdot sin angle XAY2 + fracXC cdot CY cdot sin angle XCY2 iff XB cdot BY = XA cdot AY + XC cdot CY$$
But this is the most that I manage to get.
geometry triangles
edited Mar 20 at 7:46
Glorfindel
3,41581830
asked Jan 30 '14 at 18:43
Stefan4024Stefan4024
30.6k63579
$endgroup$
Let $p$ be a line that pass through the centroid of a triangle $ABC$. Unless the line pass through one vertex, then $2$ verices are one side of the line, while the third one is on the other side. Without loss of generality, let vertices $A$ and $C$ be one side, while the vertex $B$ be on the other side of the line. Let $x,y,z$ be the distances from the vertices $A,B,C$ to the line $p$. Prove that $x + z = y$
I've tried something, but I didn't make any progress. If we multiply both sides of the equality by $fracoverlineXY2$, then we need to prove:
$$P_BXY = P_AXY + P_CXY iff fracXB cdot BY cdot sin angle XBY2 = fracXA cdot AY cdot sin angle XAY2 + fracXC cdot CY cdot sin angle XCY2 iff XB cdot BY = XA cdot AY + XC cdot CY$$
But this is the most that I manage to get.
geometry triangles
geometry triangles
edited Mar 20 at 7:46
Glorfindel
3,41581830
asked Jan 30 '14 at 18:43
Stefan4024Stefan4024
30.6k63579
edited Mar 20 at 7:46
Glorfindel
3,41581830
asked Jan 30 '14 at 18:43
Stefan4024Stefan4024
30.6k63579
edited Mar 20 at 7:46
Glorfindel
3,41581830
edited Mar 20 at 7:46
Glorfindel
3,41581830
edited Mar 20 at 7:46
Glorfindel
3,41581830
3,41581830
asked Jan 30 '14 at 18:43
Stefan4024Stefan4024
30.6k63579
asked Jan 30 '14 at 18:43
Stefan4024Stefan4024
30.6k63579
asked Jan 30 '14 at 18:43
Stefan4024Stefan4024
30.6k63579
30.6k63579
$begingroup$
You haven't used the fact that $p$ passes through $M$ yet, although you will need it somewhere. What are some nice properties of the centroid that may be useful here?
$endgroup$
– Ragnar
Jan 30 '14 at 19:00
add a comment |
$begingroup$
You haven't used the fact that $p$ passes through $M$ yet, although you will need it somewhere. What are some nice properties of the centroid that may be useful here?
$endgroup$
– Ragnar
Jan 30 '14 at 19:00
$begingroup$
You haven't used the fact that $p$ passes through $M$ yet, although you will need it somewhere. What are some nice properties of the centroid that may be useful here?
$endgroup$
– Ragnar
Jan 30 '14 at 19:00
$begingroup$
You haven't used the fact that $p$ passes through $M$ yet, although you will need it somewhere. What are some nice properties of the centroid that may be useful here?
$endgroup$
– Ragnar
Jan 30 '14 at 19:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $F$ the the midpoint of line segment $AC$. We know that $F$ lies on $BM$, because $M$ is the centroid. Let $G$ be the point of intersection of $p$ and the perpendicular to $p$ through $F$. It's not hard to see that
$$
|FG|=fracCJ2
$$
although I can't see a nice simple proof of it right now. (Found one, see below)
Because $M$ is the centroid, we know that $|BM|=2|MF|$. Now, we have that $triangle BMHsimtriangle FMG$, because of equal angles en parallel lines. It follows that
$$
fracBH=frac=frac 12
$$
Combining this with the earlier found equation, we get that $|AI|+|CJ|=|BH|$ indeed.
EDIT
To show the first equality, you can look at the triangles with sides $AF$ and $CF$, parts of the line through $F$ parallel to $p$ and the corresponding parts of the perpendicular through $A$ and $C$ on $p$. The two triangles you obtain are congruent, so the lengths of the excess at $A$ and the shortage at $C$ are the same.
answered Jan 30 '14 at 19:06
RagnarRagnar
5,9721120
$endgroup$
1
$begingroup$
Quite nice proof. I tried to use the fact the M is te centroid and splits $BF$ in ratio $2:1$, but I couldn't make any use of it.
$endgroup$
– Stefan4024
Jan 30 '14 at 19:12
add a comment |
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$begingroup$
Let $F$ the the midpoint of line segment $AC$. We know that $F$ lies on $BM$, because $M$ is the centroid. Let $G$ be the point of intersection of $p$ and the perpendicular to $p$ through $F$. It's not hard to see that
$$
|FG|=fracCJ2
$$
although I can't see a nice simple proof of it right now. (Found one, see below)
Because $M$ is the centroid, we know that $|BM|=2|MF|$. Now, we have that $triangle BMHsimtriangle FMG$, because of equal angles en parallel lines. It follows that
$$
fracBH=frac=frac 12
$$
Combining this with the earlier found equation, we get that $|AI|+|CJ|=|BH|$ indeed.
EDIT
To show the first equality, you can look at the triangles with sides $AF$ and $CF$, parts of the line through $F$ parallel to $p$ and the corresponding parts of the perpendicular through $A$ and $C$ on $p$. The two triangles you obtain are congruent, so the lengths of the excess at $A$ and the shortage at $C$ are the same.
answered Jan 30 '14 at 19:06
RagnarRagnar
5,9721120
$endgroup$
1
$begingroup$
Quite nice proof. I tried to use the fact the M is te centroid and splits $BF$ in ratio $2:1$, but I couldn't make any use of it.
$endgroup$
– Stefan4024
Jan 30 '14 at 19:12
add a comment |
$begingroup$
Let $F$ the the midpoint of line segment $AC$. We know that $F$ lies on $BM$, because $M$ is the centroid. Let $G$ be the point of intersection of $p$ and the perpendicular to $p$ through $F$. It's not hard to see that
$$
|FG|=fracCJ2
$$
although I can't see a nice simple proof of it right now. (Found one, see below)
Because $M$ is the centroid, we know that $|BM|=2|MF|$. Now, we have that $triangle BMHsimtriangle FMG$, because of equal angles en parallel lines. It follows that
$$
fracBH=frac=frac 12
$$
Combining this with the earlier found equation, we get that $|AI|+|CJ|=|BH|$ indeed.
EDIT
To show the first equality, you can look at the triangles with sides $AF$ and $CF$, parts of the line through $F$ parallel to $p$ and the corresponding parts of the perpendicular through $A$ and $C$ on $p$. The two triangles you obtain are congruent, so the lengths of the excess at $A$ and the shortage at $C$ are the same.
answered Jan 30 '14 at 19:06
RagnarRagnar
5,9721120
$endgroup$
1
$begingroup$
Quite nice proof. I tried to use the fact the M is te centroid and splits $BF$ in ratio $2:1$, but I couldn't make any use of it.
$endgroup$
– Stefan4024
Jan 30 '14 at 19:12
add a comment |
$begingroup$
Let $F$ the the midpoint of line segment $AC$. We know that $F$ lies on $BM$, because $M$ is the centroid. Let $G$ be the point of intersection of $p$ and the perpendicular to $p$ through $F$. It's not hard to see that
$$
|FG|=fracCJ2
$$
although I can't see a nice simple proof of it right now. (Found one, see below)
Because $M$ is the centroid, we know that $|BM|=2|MF|$. Now, we have that $triangle BMHsimtriangle FMG$, because of equal angles en parallel lines. It follows that
$$
fracBH=frac=frac 12
$$
Combining this with the earlier found equation, we get that $|AI|+|CJ|=|BH|$ indeed.
EDIT
To show the first equality, you can look at the triangles with sides $AF$ and $CF$, parts of the line through $F$ parallel to $p$ and the corresponding parts of the perpendicular through $A$ and $C$ on $p$. The two triangles you obtain are congruent, so the lengths of the excess at $A$ and the shortage at $C$ are the same.
answered Jan 30 '14 at 19:06
RagnarRagnar
5,9721120
$endgroup$
Let $F$ the the midpoint of line segment $AC$. We know that $F$ lies on $BM$, because $M$ is the centroid. Let $G$ be the point of intersection of $p$ and the perpendicular to $p$ through $F$. It's not hard to see that
$$
|FG|=fracCJ2
$$
although I can't see a nice simple proof of it right now. (Found one, see below)
Because $M$ is the centroid, we know that $|BM|=2|MF|$. Now, we have that $triangle BMHsimtriangle FMG$, because of equal angles en parallel lines. It follows that
$$
fracBH=frac=frac 12
$$
Combining this with the earlier found equation, we get that $|AI|+|CJ|=|BH|$ indeed.
EDIT
To show the first equality, you can look at the triangles with sides $AF$ and $CF$, parts of the line through $F$ parallel to $p$ and the corresponding parts of the perpendicular through $A$ and $C$ on $p$. The two triangles you obtain are congruent, so the lengths of the excess at $A$ and the shortage at $C$ are the same.
answered Jan 30 '14 at 19:06
RagnarRagnar
5,9721120
edited Jan 30 '14 at 19:16
answered Jan 30 '14 at 19:06
RagnarRagnar
5,9721120
answered Jan 30 '14 at 19:06
RagnarRagnar
5,9721120
answered Jan 30 '14 at 19:06
RagnarRagnar
5,9721120
5,9721120
1
$begingroup$
Quite nice proof. I tried to use the fact the M is te centroid and splits $BF$ in ratio $2:1$, but I couldn't make any use of it.
$endgroup$
– Stefan4024
Jan 30 '14 at 19:12
add a comment |
1
$begingroup$
Quite nice proof. I tried to use the fact the M is te centroid and splits $BF$ in ratio $2:1$, but I couldn't make any use of it.
$endgroup$
– Stefan4024
Jan 30 '14 at 19:12
1
1
$begingroup$
Quite nice proof. I tried to use the fact the M is te centroid and splits $BF$ in ratio $2:1$, but I couldn't make any use of it.
$endgroup$
– Stefan4024
Jan 30 '14 at 19:12
$begingroup$
Quite nice proof. I tried to use the fact the M is te centroid and splits $BF$ in ratio $2:1$, but I couldn't make any use of it.
$endgroup$
– Stefan4024
Jan 30 '14 at 19:12
add a comment |
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$begingroup$
You haven't used the fact that $p$ passes through $M$ yet, although you will need it somewhere. What are some nice properties of the centroid that may be useful here?
$endgroup$
– Ragnar
Jan 30 '14 at 19:00