Finding the smallest possible perimeter of a 4-gon inscribed in a square
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"Take the square $ABCD$. Let $E_1$ be a point on the side $AB$ such that $AE_1 : E_1B = 2 : 1$. Find the points $E_2, E_3, E_4$ on the sides $BC, CD$, and $DB$ respectively such that the sum $|E_1E_2| + |E_2E_3| + |E_3E_4| + |E_4E_1|$ is the smallest. Find this sum."
My Attempt:
I first took the two triangles (formed by the diagonal $DB$) individually. I reflected $E_1$ in each of the sides and considered the line joining these two reflected points. Where that line hit the triangle on $DB$, I labelled as $E_4$. I did the same for the second triangle and formed a path.
I don't know if this method is correct and, if it is, I do not know how to find the perimeter of this shape, as the question requires. Any help?
geometry euclidean-geometry
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add a comment |
$begingroup$
"Take the square $ABCD$. Let $E_1$ be a point on the side $AB$ such that $AE_1 : E_1B = 2 : 1$. Find the points $E_2, E_3, E_4$ on the sides $BC, CD$, and $DB$ respectively such that the sum $|E_1E_2| + |E_2E_3| + |E_3E_4| + |E_4E_1|$ is the smallest. Find this sum."
My Attempt:
I first took the two triangles (formed by the diagonal $DB$) individually. I reflected $E_1$ in each of the sides and considered the line joining these two reflected points. Where that line hit the triangle on $DB$, I labelled as $E_4$. I did the same for the second triangle and formed a path.
I don't know if this method is correct and, if it is, I do not know how to find the perimeter of this shape, as the question requires. Any help?
geometry euclidean-geometry
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$begingroup$
Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
$endgroup$
– KReiser
Dec 17 '18 at 5:14
add a comment |
$begingroup$
"Take the square $ABCD$. Let $E_1$ be a point on the side $AB$ such that $AE_1 : E_1B = 2 : 1$. Find the points $E_2, E_3, E_4$ on the sides $BC, CD$, and $DB$ respectively such that the sum $|E_1E_2| + |E_2E_3| + |E_3E_4| + |E_4E_1|$ is the smallest. Find this sum."
My Attempt:
I first took the two triangles (formed by the diagonal $DB$) individually. I reflected $E_1$ in each of the sides and considered the line joining these two reflected points. Where that line hit the triangle on $DB$, I labelled as $E_4$. I did the same for the second triangle and formed a path.
I don't know if this method is correct and, if it is, I do not know how to find the perimeter of this shape, as the question requires. Any help?
geometry euclidean-geometry
$endgroup$
"Take the square $ABCD$. Let $E_1$ be a point on the side $AB$ such that $AE_1 : E_1B = 2 : 1$. Find the points $E_2, E_3, E_4$ on the sides $BC, CD$, and $DB$ respectively such that the sum $|E_1E_2| + |E_2E_3| + |E_3E_4| + |E_4E_1|$ is the smallest. Find this sum."
My Attempt:
I first took the two triangles (formed by the diagonal $DB$) individually. I reflected $E_1$ in each of the sides and considered the line joining these two reflected points. Where that line hit the triangle on $DB$, I labelled as $E_4$. I did the same for the second triangle and formed a path.
I don't know if this method is correct and, if it is, I do not know how to find the perimeter of this shape, as the question requires. Any help?
geometry euclidean-geometry
geometry euclidean-geometry
edited Dec 17 '18 at 5:14
KReiser
9,72721435
9,72721435
asked Dec 17 '18 at 5:12
helpneededhelpneeded
897
897
$begingroup$
Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
$endgroup$
– KReiser
Dec 17 '18 at 5:14
add a comment |
$begingroup$
Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
$endgroup$
– KReiser
Dec 17 '18 at 5:14
$begingroup$
Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
$endgroup$
– KReiser
Dec 17 '18 at 5:14
$begingroup$
Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
$endgroup$
– KReiser
Dec 17 '18 at 5:14
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Whatever $E_2$, $E_3$, $E_4$ you choose,
Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.
The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.
The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.
The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.
The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot.
In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (frac23,0)$ and $H_1 = (frac83,2)$. In general, we always have $|E_1H_1| = 2sqrt{2}|AB|$.
If one stare at above diagram long enough, one will notice
$|E_2E_3| = |E_2F_3|$,
$|E_3E_4| = |F_3F_4| = |F_3G_4|$,
$|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.
From this, we can deduce
$$begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\
= &; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\
ge &; |E_1H_1| = 2sqrt{2}|AB|
end{align}$$
This means the perimeter we seek is bounded from below by $2sqrt{2}|AB|$.
To see this is the actual minimum perimeter, we can
repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct
the "optimal" quadrilateral. We find when
$$BE_2 : E_2C = DE_4 : E_4A = 1 : 2quadtext{ and }quad CE_3 : E_3D = 2 : 1$$
the quadrilateral $E_1E_2E_3E_4$ does have $2sqrt{2}|AB|$ as perimeter.
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$begingroup$
Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
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– Quang Hoang
Dec 17 '18 at 16:04
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@Quang Hoang I think, my solution is the same.
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– Michael Rozenberg
Dec 17 '18 at 17:34
2
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Yes, exactly, but the picture helps clarify the proof a lot though.
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– Quang Hoang
Dec 17 '18 at 17:35
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@Quang Hoang I need to learn how to draw pictures.
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– Michael Rozenberg
Dec 17 '18 at 17:36
1
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@helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
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– achille hui
Dec 19 '18 at 0:09
|
show 2 more comments
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Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares
and $E_1'in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$
Now, let $E_1E_1'cap BC'={E_2'},$ $E_1E_1'cap D''C'={E_3'}$ and $E_1E_1'cap A'''D''={E_4'}.$
Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and
$$BE_2:E_2C=BE_2':E_2C'=1:2,$$
$$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and
$$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$
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add a comment |
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Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.
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Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
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– helpneeded
Dec 17 '18 at 5:54
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I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
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– Andrei
Dec 17 '18 at 6:01
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Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
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– helpneeded
Dec 17 '18 at 6:04
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Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
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– Andrei
Dec 17 '18 at 6:23
add a comment |
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Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that
$|PB|=|PQ|cosangle QPB$
$|QB|=|PQ|sinangle QPB$
$dfrac{|PQ|}{|PB|+|QB|}=dfrac{1}{cosangle QPB+sinangle QPB}ge dfrac{1}{sqrt{2}}$
We know that $cosangle QPB+sinangle QPB lesqrt{2}$ because $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)=2$.
Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $ge 1/sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).
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add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Whatever $E_2$, $E_3$, $E_4$ you choose,
Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.
The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.
The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.
The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.
The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot.
In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (frac23,0)$ and $H_1 = (frac83,2)$. In general, we always have $|E_1H_1| = 2sqrt{2}|AB|$.
If one stare at above diagram long enough, one will notice
$|E_2E_3| = |E_2F_3|$,
$|E_3E_4| = |F_3F_4| = |F_3G_4|$,
$|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.
From this, we can deduce
$$begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\
= &; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\
ge &; |E_1H_1| = 2sqrt{2}|AB|
end{align}$$
This means the perimeter we seek is bounded from below by $2sqrt{2}|AB|$.
To see this is the actual minimum perimeter, we can
repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct
the "optimal" quadrilateral. We find when
$$BE_2 : E_2C = DE_4 : E_4A = 1 : 2quadtext{ and }quad CE_3 : E_3D = 2 : 1$$
the quadrilateral $E_1E_2E_3E_4$ does have $2sqrt{2}|AB|$ as perimeter.
$endgroup$
$begingroup$
Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
$endgroup$
– Quang Hoang
Dec 17 '18 at 16:04
$begingroup$
@Quang Hoang I think, my solution is the same.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:34
2
$begingroup$
Yes, exactly, but the picture helps clarify the proof a lot though.
$endgroup$
– Quang Hoang
Dec 17 '18 at 17:35
$begingroup$
@Quang Hoang I need to learn how to draw pictures.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:36
1
$begingroup$
@helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
$endgroup$
– achille hui
Dec 19 '18 at 0:09
|
show 2 more comments
$begingroup$
Whatever $E_2$, $E_3$, $E_4$ you choose,
Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.
The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.
The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.
The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.
The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot.
In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (frac23,0)$ and $H_1 = (frac83,2)$. In general, we always have $|E_1H_1| = 2sqrt{2}|AB|$.
If one stare at above diagram long enough, one will notice
$|E_2E_3| = |E_2F_3|$,
$|E_3E_4| = |F_3F_4| = |F_3G_4|$,
$|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.
From this, we can deduce
$$begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\
= &; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\
ge &; |E_1H_1| = 2sqrt{2}|AB|
end{align}$$
This means the perimeter we seek is bounded from below by $2sqrt{2}|AB|$.
To see this is the actual minimum perimeter, we can
repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct
the "optimal" quadrilateral. We find when
$$BE_2 : E_2C = DE_4 : E_4A = 1 : 2quadtext{ and }quad CE_3 : E_3D = 2 : 1$$
the quadrilateral $E_1E_2E_3E_4$ does have $2sqrt{2}|AB|$ as perimeter.
$endgroup$
$begingroup$
Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
$endgroup$
– Quang Hoang
Dec 17 '18 at 16:04
$begingroup$
@Quang Hoang I think, my solution is the same.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:34
2
$begingroup$
Yes, exactly, but the picture helps clarify the proof a lot though.
$endgroup$
– Quang Hoang
Dec 17 '18 at 17:35
$begingroup$
@Quang Hoang I need to learn how to draw pictures.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:36
1
$begingroup$
@helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
$endgroup$
– achille hui
Dec 19 '18 at 0:09
|
show 2 more comments
$begingroup$
Whatever $E_2$, $E_3$, $E_4$ you choose,
Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.
The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.
The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.
The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.
The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot.
In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (frac23,0)$ and $H_1 = (frac83,2)$. In general, we always have $|E_1H_1| = 2sqrt{2}|AB|$.
If one stare at above diagram long enough, one will notice
$|E_2E_3| = |E_2F_3|$,
$|E_3E_4| = |F_3F_4| = |F_3G_4|$,
$|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.
From this, we can deduce
$$begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\
= &; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\
ge &; |E_1H_1| = 2sqrt{2}|AB|
end{align}$$
This means the perimeter we seek is bounded from below by $2sqrt{2}|AB|$.
To see this is the actual minimum perimeter, we can
repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct
the "optimal" quadrilateral. We find when
$$BE_2 : E_2C = DE_4 : E_4A = 1 : 2quadtext{ and }quad CE_3 : E_3D = 2 : 1$$
the quadrilateral $E_1E_2E_3E_4$ does have $2sqrt{2}|AB|$ as perimeter.
$endgroup$
Whatever $E_2$, $E_3$, $E_4$ you choose,
Reflect square $ABCD$ wrt side $BC$ to get square $A'BCD'$.
The quadrilateral $E_1E_2E_3E_4$ get mapped to quadrilateral $F_1E_2F_3F_4$.Reflect square $A'BCD'$ wrt side $CD'$ to get square $A''B'CD'$.
The quadrilateral $F_1E_2F_3F_4$ get mapped to quadrilateral $G_1G_2F_3G_4$.Reflect square $A''B'CD'$ wrt side $A''D'$ to get square $A''B''C'D'$.
The quadrilateral $G_1G_2F_3G_4$ get mapped to quadrilateral $H_1H_2H_3G_4$.
The key is independent of choice of $E_2,E_3,E_4$, the point $H_1$ always mapped to same spot.
In fact, if one choose a coordinate system so that $A = (0,0), B = (1,0), C = (1,1)$ and $D = (0,1)$, one find $E_1 = (frac23,0)$ and $H_1 = (frac83,2)$. In general, we always have $|E_1H_1| = 2sqrt{2}|AB|$.
If one stare at above diagram long enough, one will notice
$|E_2E_3| = |E_2F_3|$,
$|E_3E_4| = |F_3F_4| = |F_3G_4|$,
$|E_4E_1| = |F_4F_1| = |G_4G_1| = |G_4H_1|$.
From this, we can deduce
$$begin{align} &|E_1E_2|+|E_2E_3|+|E_3E_4|+|E_4E_1|\
= &; |E_1E_2| + |E_2F_3| + |F_3G_4|+|G_4H_1|\
ge &; |E_1H_1| = 2sqrt{2}|AB|
end{align}$$
This means the perimeter we seek is bounded from below by $2sqrt{2}|AB|$.
To see this is the actual minimum perimeter, we can
repeatedly "fold" segment $E_1H_1$ back to square $ABC$ to construct
the "optimal" quadrilateral. We find when
$$BE_2 : E_2C = DE_4 : E_4A = 1 : 2quadtext{ and }quad CE_3 : E_3D = 2 : 1$$
the quadrilateral $E_1E_2E_3E_4$ does have $2sqrt{2}|AB|$ as perimeter.
edited Dec 17 '18 at 18:38
answered Dec 17 '18 at 14:37
achille huiachille hui
96k5132259
96k5132259
$begingroup$
Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
$endgroup$
– Quang Hoang
Dec 17 '18 at 16:04
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@Quang Hoang I think, my solution is the same.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:34
2
$begingroup$
Yes, exactly, but the picture helps clarify the proof a lot though.
$endgroup$
– Quang Hoang
Dec 17 '18 at 17:35
$begingroup$
@Quang Hoang I need to learn how to draw pictures.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:36
1
$begingroup$
@helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
$endgroup$
– achille hui
Dec 19 '18 at 0:09
|
show 2 more comments
$begingroup$
Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
$endgroup$
– Quang Hoang
Dec 17 '18 at 16:04
$begingroup$
@Quang Hoang I think, my solution is the same.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:34
2
$begingroup$
Yes, exactly, but the picture helps clarify the proof a lot though.
$endgroup$
– Quang Hoang
Dec 17 '18 at 17:35
$begingroup$
@Quang Hoang I need to learn how to draw pictures.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:36
1
$begingroup$
@helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
$endgroup$
– achille hui
Dec 19 '18 at 0:09
$begingroup$
Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
$endgroup$
– Quang Hoang
Dec 17 '18 at 16:04
$begingroup$
Nice solution, and it shows that the ratio $AE_1:E_1B$ is not really needed here. Given any point $E_1$, we can construct a rectangle $E_1E_2E_3E_4$ with edges parallels to the diagonals of $ABCD$ and that $E_1E_2E_3E_4$ has the minimum perimeter.
$endgroup$
– Quang Hoang
Dec 17 '18 at 16:04
$begingroup$
@Quang Hoang I think, my solution is the same.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:34
$begingroup$
@Quang Hoang I think, my solution is the same.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:34
2
2
$begingroup$
Yes, exactly, but the picture helps clarify the proof a lot though.
$endgroup$
– Quang Hoang
Dec 17 '18 at 17:35
$begingroup$
Yes, exactly, but the picture helps clarify the proof a lot though.
$endgroup$
– Quang Hoang
Dec 17 '18 at 17:35
$begingroup$
@Quang Hoang I need to learn how to draw pictures.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:36
$begingroup$
@Quang Hoang I need to learn how to draw pictures.
$endgroup$
– Michael Rozenberg
Dec 17 '18 at 17:36
1
1
$begingroup$
@helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
$endgroup$
– achille hui
Dec 19 '18 at 0:09
$begingroup$
@helpneeded you can use the fact the points are constructed by reflections, so $|E_1B| = |H_1B'|$. Since $AB$ is parallel to $A''B''$, you get $|E_1H_1| = |BB''|$ so it is twice the diagonal of any one of the squares. Since the "optimal" quadrilateral can be constructed by folding the segment $E_1H_1$ back into square $ABCD$, its perimeter is the length of segment $E_1H_1$.
$endgroup$
– achille hui
Dec 19 '18 at 0:09
|
show 2 more comments
$begingroup$
Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares
and $E_1'in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$
Now, let $E_1E_1'cap BC'={E_2'},$ $E_1E_1'cap D''C'={E_3'}$ and $E_1E_1'cap A'''D''={E_4'}.$
Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and
$$BE_2:E_2C=BE_2':E_2C'=1:2,$$
$$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and
$$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$
$endgroup$
add a comment |
$begingroup$
Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares
and $E_1'in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$
Now, let $E_1E_1'cap BC'={E_2'},$ $E_1E_1'cap D''C'={E_3'}$ and $E_1E_1'cap A'''D''={E_4'}.$
Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and
$$BE_2:E_2C=BE_2':E_2C'=1:2,$$
$$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and
$$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$
$endgroup$
add a comment |
$begingroup$
Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares
and $E_1'in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$
Now, let $E_1E_1'cap BC'={E_2'},$ $E_1E_1'cap D''C'={E_3'}$ and $E_1E_1'cap A'''D''={E_4'}.$
Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and
$$BE_2:E_2C=BE_2':E_2C'=1:2,$$
$$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and
$$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$
$endgroup$
Let $ABC'D'$, $A''BC'D''$, $A'''B''C'D''$ and $A'''B'''C'''D''$ be squares
and $E_1'in A'''B'''$ such that $A'''E_1':E_1'B'''=2:1.$
Now, let $E_1E_1'cap BC'={E_2'},$ $E_1E_1'cap D''C'={E_3'}$ and $E_1E_1'cap A'''D''={E_4'}.$
Now, easy to see that $$E_1E_2+E_2E_3+E_3E_4+E_4E_1geq E_1E_1'=E_1E_2'+E_2'E_3'+E_3'E_4'+E_4'E_1'$$ and
$$BE_2:E_2C=BE_2':E_2C'=1:2,$$
$$CE_3:E_3D=C'E_3':E_3'D''=2:1$$ and
$$DE_4:E_4A=D''E_4':E_4'A'''=1:2.$$
answered Dec 17 '18 at 13:27
Michael RozenbergMichael Rozenberg
105k1892197
105k1892197
add a comment |
add a comment |
$begingroup$
Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.
$endgroup$
$begingroup$
Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
$endgroup$
– helpneeded
Dec 17 '18 at 5:54
$begingroup$
I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
$endgroup$
– Andrei
Dec 17 '18 at 6:01
$begingroup$
Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
$endgroup$
– helpneeded
Dec 17 '18 at 6:04
$begingroup$
Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
$endgroup$
– Andrei
Dec 17 '18 at 6:23
add a comment |
$begingroup$
Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.
$endgroup$
$begingroup$
Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
$endgroup$
– helpneeded
Dec 17 '18 at 5:54
$begingroup$
I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
$endgroup$
– Andrei
Dec 17 '18 at 6:01
$begingroup$
Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
$endgroup$
– helpneeded
Dec 17 '18 at 6:04
$begingroup$
Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
$endgroup$
– Andrei
Dec 17 '18 at 6:23
add a comment |
$begingroup$
Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.
$endgroup$
Hint: Start by drawing the $ABCD$ square, and the point $E_1$ on $AB$. Now let's choose a random point $E_3$ on $CD$. How do you get the shortest path from $E_1$ to $E_3$ so that it touches $BC$ in some $E_2$? Now repeat the procedure for the $AD$ side as well. Once you have these steps, there is just one more (similar) step to find the final answer.
answered Dec 17 '18 at 5:44
AndreiAndrei
12.4k21128
12.4k21128
$begingroup$
Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
$endgroup$
– helpneeded
Dec 17 '18 at 5:54
$begingroup$
I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
$endgroup$
– Andrei
Dec 17 '18 at 6:01
$begingroup$
Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
$endgroup$
– helpneeded
Dec 17 '18 at 6:04
$begingroup$
Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
$endgroup$
– Andrei
Dec 17 '18 at 6:23
add a comment |
$begingroup$
Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
$endgroup$
– helpneeded
Dec 17 '18 at 5:54
$begingroup$
I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
$endgroup$
– Andrei
Dec 17 '18 at 6:01
$begingroup$
Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
$endgroup$
– helpneeded
Dec 17 '18 at 6:04
$begingroup$
Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
$endgroup$
– Andrei
Dec 17 '18 at 6:23
$begingroup$
Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
$endgroup$
– helpneeded
Dec 17 '18 at 5:54
$begingroup$
Do you reflect the point $E_1$ in $BC$ and then join that point with $E_3$ to get $E_2$?
$endgroup$
– helpneeded
Dec 17 '18 at 5:54
$begingroup$
I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
$endgroup$
– Andrei
Dec 17 '18 at 6:01
$begingroup$
I did reflect $E_3$, but it's similar. Now calculate the distance between the reflections (these are along one side of the square. If you reflect $E_3$, let's call these points $E_{3C}$ and $E_{3D}$. What's the distance between them? What's the shortest distance also touching $AB$?
$endgroup$
– Andrei
Dec 17 '18 at 6:01
$begingroup$
Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
$endgroup$
– helpneeded
Dec 17 '18 at 6:04
$begingroup$
Here is where I am getting stuck. I have no idea how to calculate the distances between them. All I know is that reflecting provides the shortest distance.
$endgroup$
– helpneeded
Dec 17 '18 at 6:04
$begingroup$
Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
$endgroup$
– Andrei
Dec 17 '18 at 6:23
$begingroup$
Ok. So in my notation, say $E_3$ is at some distance $x$ from $C$. Where is $E_{3C}$? If the side of the square is $a$, $E_3$ is at $a-x$ from $D$. Where is $E_{3D}$? Do you have the picture?
$endgroup$
– Andrei
Dec 17 '18 at 6:23
add a comment |
$begingroup$
Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that
$|PB|=|PQ|cosangle QPB$
$|QB|=|PQ|sinangle QPB$
$dfrac{|PQ|}{|PB|+|QB|}=dfrac{1}{cosangle QPB+sinangle QPB}ge dfrac{1}{sqrt{2}}$
We know that $cosangle QPB+sinangle QPB lesqrt{2}$ because $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)=2$.
Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $ge 1/sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).
$endgroup$
add a comment |
$begingroup$
Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that
$|PB|=|PQ|cosangle QPB$
$|QB|=|PQ|sinangle QPB$
$dfrac{|PQ|}{|PB|+|QB|}=dfrac{1}{cosangle QPB+sinangle QPB}ge dfrac{1}{sqrt{2}}$
We know that $cosangle QPB+sinangle QPB lesqrt{2}$ because $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)=2$.
Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $ge 1/sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).
$endgroup$
add a comment |
$begingroup$
Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that
$|PB|=|PQ|cosangle QPB$
$|QB|=|PQ|sinangle QPB$
$dfrac{|PQ|}{|PB|+|QB|}=dfrac{1}{cosangle QPB+sinangle QPB}ge dfrac{1}{sqrt{2}}$
We know that $cosangle QPB+sinangle QPB lesqrt{2}$ because $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)=2$.
Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $ge 1/sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).
$endgroup$
Let $A,B,C$ be three corners of the square in rotational order. Draw a linecsegment $PQ$ from point $P$ on $AB$ to point $Q$ on $BC$. Using SOH CAH TOA on right triangle $PQB$ infer that
$|PB|=|PQ|cosangle QPB$
$|QB|=|PQ|sinangle QPB$
$dfrac{|PQ|}{|PB|+|QB|}=dfrac{1}{cosangle QPB+sinangle QPB}ge dfrac{1}{sqrt{2}}$
We know that $cosangle QPB+sinangle QPB lesqrt{2}$ because $(costheta+sintheta)^2+(costheta-sintheta)^2=2(cos^2theta+sin^2theta)=2$.
Applying this result to all four sides of the inscribed quadrilateral leads to the perimeter of this quadrilateral being $ge 1/sqrt{2}$ times that of the square. That this bound is sharp is proved from the simple case of the sides of the inscribed quadrilateral being parallel to the angle bisectors of the square (always possible given the symmetries involved).
edited Dec 17 '18 at 15:36
answered Dec 17 '18 at 15:03
Oscar LanziOscar Lanzi
12.8k12136
12.8k12136
add a comment |
add a comment |
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$begingroup$
Please read tags before using them. The algebraic-geometry tag specifically mentions it is not for use in elementary problems involving both algebra and geometry.
$endgroup$
– KReiser
Dec 17 '18 at 5:14