Mixing
Hexagon not regular
$begingroup$
Plotting the set:
$$ small
left{
(x,,y) in mathbb{R}^2 :
x + frac{3}{2} le y le x + 2, ;
3,x le y le 3,x + 1, ;
-2,x + 3 le y le -2,x + 4
right} $$
the following hexagon is obtained:
which is characterized by sides that are two to two parallel, but not congruent.
I noticed that the lines that connect the midpoints of the parallel sides pass through a single point. I'm thinking how to proof this fact with synthetic geometry, but nothing comes to mind of elementary. Ideas?
geometry
asked Jan 18 at 23:27
TeMTeM
459316
$endgroup$
add a comment |
$begingroup$
Plotting the set:
$$ small
left{
(x,,y) in mathbb{R}^2 :
x + frac{3}{2} le y le x + 2, ;
3,x le y le 3,x + 1, ;
-2,x + 3 le y le -2,x + 4
right} $$
the following hexagon is obtained:
which is characterized by sides that are two to two parallel, but not congruent.
I noticed that the lines that connect the midpoints of the parallel sides pass through a single point. I'm thinking how to proof this fact with synthetic geometry, but nothing comes to mind of elementary. Ideas?
geometry
asked Jan 18 at 23:27
TeMTeM
459316
$endgroup$
add a comment |
$begingroup$
Plotting the set:
$$ small
left{
(x,,y) in mathbb{R}^2 :
x + frac{3}{2} le y le x + 2, ;
3,x le y le 3,x + 1, ;
-2,x + 3 le y le -2,x + 4
right} $$
the following hexagon is obtained:
which is characterized by sides that are two to two parallel, but not congruent.
I noticed that the lines that connect the midpoints of the parallel sides pass through a single point. I'm thinking how to proof this fact with synthetic geometry, but nothing comes to mind of elementary. Ideas?
geometry
asked Jan 18 at 23:27
TeMTeM
459316
$endgroup$
Plotting the set:
$$ small
left{
(x,,y) in mathbb{R}^2 :
x + frac{3}{2} le y le x + 2, ;
3,x le y le 3,x + 1, ;
-2,x + 3 le y le -2,x + 4
right} $$
the following hexagon is obtained:
which is characterized by sides that are two to two parallel, but not congruent.
I noticed that the lines that connect the midpoints of the parallel sides pass through a single point. I'm thinking how to proof this fact with synthetic geometry, but nothing comes to mind of elementary. Ideas?
geometry
geometry
asked Jan 18 at 23:27
TeMTeM
459316
asked Jan 18 at 23:27
TeMTeM
459316
asked Jan 18 at 23:27
TeMTeM
459316
asked Jan 18 at 23:27
TeMTeM
459316
asked Jan 18 at 23:27
TeMTeM
459316
459316
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
This is an old problem. It appeared e.g. in Prasolov's book as problem 5.80.
Below I present a proof I came up with once; it assumes familiarity with some projective geometry though.
Let $ABCDEF$ be a hexagon whose pairs of opposite sides are parallel. Note that $X:=AB cap DE, Y:=BC cap EF, Z:=CD cap FA$ are points at infinity, and therefore are collinear (they lie on the line in the infinity). By Pascal theorem, $ABCDEF$ is inscribed in a conic.
Denoting the midpoints of $AB$ and $DE$ by $K$ and $L$, respectively, we see that quadruples $(A,B,X,K)$, $(D,E,X,L)$ are harmonic. It follows that $KL$ is the polar line of $X$ with respect to the circumconic of $ABCDEF$. Hence, the pole of the line in infinity lies on $KL$. An analogous argument shows that the other two lines determined by the midpoints of opposite sides of $ABCDEF$ pass through the pole of the line in infinity, which solves the problem.
answered Jan 19 at 0:08
timon92timon92
4,4171826
$endgroup$
$begingroup$
I heartily thank you, but unfortunately my level of geometry does not allow me to understand this demonstration! Is it not that you also know a less elegant demonstration that uses only elementary Euclidean geometry?
$endgroup$
– TeM
Jan 19 at 9:07
1
$begingroup$
@TeM Check out my edited answer. I have added a source of this problem, you can find an elementary solution there.
$endgroup$
– timon92
Jan 19 at 14:13
$begingroup$
proof of exquisite workmanship! ^_^
$endgroup$
– TeM
Jan 19 at 14:41
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
This is an old problem. It appeared e.g. in Prasolov's book as problem 5.80.
Below I present a proof I came up with once; it assumes familiarity with some projective geometry though.
Let $ABCDEF$ be a hexagon whose pairs of opposite sides are parallel. Note that $X:=AB cap DE, Y:=BC cap EF, Z:=CD cap FA$ are points at infinity, and therefore are collinear (they lie on the line in the infinity). By Pascal theorem, $ABCDEF$ is inscribed in a conic.
Denoting the midpoints of $AB$ and $DE$ by $K$ and $L$, respectively, we see that quadruples $(A,B,X,K)$, $(D,E,X,L)$ are harmonic. It follows that $KL$ is the polar line of $X$ with respect to the circumconic of $ABCDEF$. Hence, the pole of the line in infinity lies on $KL$. An analogous argument shows that the other two lines determined by the midpoints of opposite sides of $ABCDEF$ pass through the pole of the line in infinity, which solves the problem.
answered Jan 19 at 0:08
timon92timon92
4,4171826
$endgroup$
$begingroup$
I heartily thank you, but unfortunately my level of geometry does not allow me to understand this demonstration! Is it not that you also know a less elegant demonstration that uses only elementary Euclidean geometry?
$endgroup$
– TeM
Jan 19 at 9:07
1
$begingroup$
@TeM Check out my edited answer. I have added a source of this problem, you can find an elementary solution there.
$endgroup$
– timon92
Jan 19 at 14:13
$begingroup$
proof of exquisite workmanship! ^_^
$endgroup$
– TeM
Jan 19 at 14:41
add a comment |
$begingroup$
This is an old problem. It appeared e.g. in Prasolov's book as problem 5.80.
Below I present a proof I came up with once; it assumes familiarity with some projective geometry though.
Let $ABCDEF$ be a hexagon whose pairs of opposite sides are parallel. Note that $X:=AB cap DE, Y:=BC cap EF, Z:=CD cap FA$ are points at infinity, and therefore are collinear (they lie on the line in the infinity). By Pascal theorem, $ABCDEF$ is inscribed in a conic.
Denoting the midpoints of $AB$ and $DE$ by $K$ and $L$, respectively, we see that quadruples $(A,B,X,K)$, $(D,E,X,L)$ are harmonic. It follows that $KL$ is the polar line of $X$ with respect to the circumconic of $ABCDEF$. Hence, the pole of the line in infinity lies on $KL$. An analogous argument shows that the other two lines determined by the midpoints of opposite sides of $ABCDEF$ pass through the pole of the line in infinity, which solves the problem.
answered Jan 19 at 0:08
timon92timon92
4,4171826
$endgroup$
$begingroup$
I heartily thank you, but unfortunately my level of geometry does not allow me to understand this demonstration! Is it not that you also know a less elegant demonstration that uses only elementary Euclidean geometry?
$endgroup$
– TeM
Jan 19 at 9:07
1
$begingroup$
@TeM Check out my edited answer. I have added a source of this problem, you can find an elementary solution there.
$endgroup$
– timon92
Jan 19 at 14:13
$begingroup$
proof of exquisite workmanship! ^_^
$endgroup$
– TeM
Jan 19 at 14:41
add a comment |
$begingroup$
This is an old problem. It appeared e.g. in Prasolov's book as problem 5.80.
Below I present a proof I came up with once; it assumes familiarity with some projective geometry though.
Let $ABCDEF$ be a hexagon whose pairs of opposite sides are parallel. Note that $X:=AB cap DE, Y:=BC cap EF, Z:=CD cap FA$ are points at infinity, and therefore are collinear (they lie on the line in the infinity). By Pascal theorem, $ABCDEF$ is inscribed in a conic.
Denoting the midpoints of $AB$ and $DE$ by $K$ and $L$, respectively, we see that quadruples $(A,B,X,K)$, $(D,E,X,L)$ are harmonic. It follows that $KL$ is the polar line of $X$ with respect to the circumconic of $ABCDEF$. Hence, the pole of the line in infinity lies on $KL$. An analogous argument shows that the other two lines determined by the midpoints of opposite sides of $ABCDEF$ pass through the pole of the line in infinity, which solves the problem.
answered Jan 19 at 0:08
timon92timon92
4,4171826
$endgroup$
This is an old problem. It appeared e.g. in Prasolov's book as problem 5.80.
Below I present a proof I came up with once; it assumes familiarity with some projective geometry though.
Let $ABCDEF$ be a hexagon whose pairs of opposite sides are parallel. Note that $X:=AB cap DE, Y:=BC cap EF, Z:=CD cap FA$ are points at infinity, and therefore are collinear (they lie on the line in the infinity). By Pascal theorem, $ABCDEF$ is inscribed in a conic.
Denoting the midpoints of $AB$ and $DE$ by $K$ and $L$, respectively, we see that quadruples $(A,B,X,K)$, $(D,E,X,L)$ are harmonic. It follows that $KL$ is the polar line of $X$ with respect to the circumconic of $ABCDEF$. Hence, the pole of the line in infinity lies on $KL$. An analogous argument shows that the other two lines determined by the midpoints of opposite sides of $ABCDEF$ pass through the pole of the line in infinity, which solves the problem.
answered Jan 19 at 0:08
timon92timon92
4,4171826
edited Jan 19 at 14:11
answered Jan 19 at 0:08
timon92timon92
4,4171826
answered Jan 19 at 0:08
timon92timon92
4,4171826
answered Jan 19 at 0:08
timon92timon92
4,4171826
4,4171826
$begingroup$
I heartily thank you, but unfortunately my level of geometry does not allow me to understand this demonstration! Is it not that you also know a less elegant demonstration that uses only elementary Euclidean geometry?
$endgroup$
– TeM
Jan 19 at 9:07
1
$begingroup$
@TeM Check out my edited answer. I have added a source of this problem, you can find an elementary solution there.
$endgroup$
– timon92
Jan 19 at 14:13
$begingroup$
proof of exquisite workmanship! ^_^
$endgroup$
– TeM
Jan 19 at 14:41
add a comment |
$begingroup$
I heartily thank you, but unfortunately my level of geometry does not allow me to understand this demonstration! Is it not that you also know a less elegant demonstration that uses only elementary Euclidean geometry?
$endgroup$
– TeM
Jan 19 at 9:07
1
$begingroup$
@TeM Check out my edited answer. I have added a source of this problem, you can find an elementary solution there.
$endgroup$
– timon92
Jan 19 at 14:13
$begingroup$
proof of exquisite workmanship! ^_^
$endgroup$
– TeM
Jan 19 at 14:41
$begingroup$
I heartily thank you, but unfortunately my level of geometry does not allow me to understand this demonstration! Is it not that you also know a less elegant demonstration that uses only elementary Euclidean geometry?
$endgroup$
– TeM
Jan 19 at 9:07
$begingroup$
I heartily thank you, but unfortunately my level of geometry does not allow me to understand this demonstration! Is it not that you also know a less elegant demonstration that uses only elementary Euclidean geometry?
$endgroup$
– TeM
Jan 19 at 9:07
1
1
$begingroup$
@TeM Check out my edited answer. I have added a source of this problem, you can find an elementary solution there.
$endgroup$
– timon92
Jan 19 at 14:13
$begingroup$
@TeM Check out my edited answer. I have added a source of this problem, you can find an elementary solution there.
$endgroup$
– timon92
Jan 19 at 14:13
$begingroup$
proof of exquisite workmanship! ^_^
$endgroup$
– TeM
Jan 19 at 14:41
$begingroup$
proof of exquisite workmanship! ^_^
$endgroup$
– TeM
Jan 19 at 14:41
add a comment |
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