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Geometry - Proof of Construction of regular pentagon by using compass and straightedge.
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How to prove that the polygon constructed by the method mentioned in the following link is indeed a regular pentagon.
https://www.youtube.com/watch?v=9VceAA2qBPA
geometry geometric-construction
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Mak Man Tung
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How to prove that the polygon constructed by the method mentioned in the following link is indeed a regular pentagon.
https://www.youtube.com/watch?v=9VceAA2qBPA
geometry geometric-construction
asked 2 days ago
Mak Man Tung
6
New contributor
Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
-2
down vote
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up vote
-2
down vote
favorite
How to prove that the polygon constructed by the method mentioned in the following link is indeed a regular pentagon.
https://www.youtube.com/watch?v=9VceAA2qBPA
geometry geometric-construction
asked 2 days ago
Mak Man Tung
6
New contributor
Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How to prove that the polygon constructed by the method mentioned in the following link is indeed a regular pentagon.
https://www.youtube.com/watch?v=9VceAA2qBPA
geometry geometric-construction
geometry geometric-construction
asked 2 days ago
Mak Man Tung
6
New contributor
Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Mak Man Tung
6
New contributor
Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Mak Man Tung
6
New contributor
Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
Mak Man Tung
6
asked 2 days ago
Mak Man Tung
6
6
New contributor
Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mak Man Tung is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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1 Answer
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After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.
By the Pythagorean theorem,
$$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
$$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$
It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
$$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.
Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$
answered 2 days ago
Blue
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.
By the Pythagorean theorem,
$$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
$$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$
It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
$$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.
Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$
answered 2 days ago
Blue
46.3k869146
add a comment |
up vote
1
down vote
After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.
By the Pythagorean theorem,
$$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
$$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$
It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
$$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.
Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$
answered 2 days ago
Blue
46.3k869146
add a comment |
up vote
1
down vote
up vote
1
down vote
After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.
By the Pythagorean theorem,
$$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
$$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$
It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
$$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.
Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$
answered 2 days ago
Blue
46.3k869146
After some setup of perpendicular segments (here, $overline{OA}$ and $overline{OC}$), the first few steps are the standard construction of the midpoint ($B$) of $overline{OA}$, so let's start there; and let's say that $|OB| = |BA| = 1$.
By the Pythagorean theorem,
$$|BC| = sqrt{|OB|^2 + |OC|^2} = sqrt{1+2^2} = sqrt{5}$$
and we construct $bigcirc B$ with that radius, meeting the "horizontal" axis at $D$, so that $|OD| = |BC|-|OB| = sqrt{5}-1$. Thus,
$$|CD| = sqrt{|OD|^2 + |OC|^2} = sqrt{(sqrt{5}-1)^2+2^2} = sqrt{10-2sqrt{5}}$$
It is "known" that the ratio of the side-length of a regular pentagon to its circumradius is
$$sqrt{frac{5-sqrt{5}}{2}}= frac{sqrt{10-2sqrt{5}}}{2}= frac{|CD|}{|OA|}$$ so $|CD|$ is precisely the side-length needed for this regular pentagon.
Then, $bigcirc C$ through $D$ transfers this side-length to $|CE|$, with $E$ on $bigcirc O$. Congruent circles construct additional points at the same distances about the circle, giving the perfect regular pentagon. $square$
answered 2 days ago
Blue
46.3k869146
answered 2 days ago
Blue
46.3k869146
answered 2 days ago
Blue
46.3k869146
answered 2 days ago
Blue
46.3k869146
46.3k869146
add a comment |
add a comment |
Mak Man Tung is a new contributor. Be nice, and check out our Code of Conduct.
Mak Man Tung is a new contributor. Be nice, and check out our Code of Conduct.
Mak Man Tung is a new contributor. Be nice, and check out our Code of Conduct.
Mak Man Tung is a new contributor. Be nice, and check out our Code of Conduct.
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