NOT a general method. While this does correctly inscribe a triangle, square, or hexagon in a circle, it does not work for other regular polygons. It does produce remarkably close results for a pentagon (.065% angle error), heptagon (.17% error), and octagon (.42% error), but the error continues to grow as the sides increase. By the time it reaches a 25-gon, the error is large enough to not even fit 24 sides in the circle. It's fine if you want to draw an n-gon approximation with a small number of sides, but don't use this on a math test.
@@Goofayball To inscribe an exact regular n-gon in a circle with straight-edge and compass? It's not possible because most regular n-gons cannot be constructed with straight-edge and compass. What you can do is inscribe a regular p-gon into a circle where p is a Fermat prime (a prime of the form 1 + 2^{2^n}), take an inscribed m-gon and n-gon and inscribe an LCM(m,n)-gon, and double the sides of an inscribed n-gon to get an inscribed 2n-gon. Combining the above allows the construction of a k-gon when k = 2ⁿ·p₁·p₂·p₃..., where the pᵢ values are 0 or more distinct Fermat primes. No other n-gon can be constructed. Note that there are only 5 known Fermat primes (3, 5, 17, 257, and 65537). So it's possible to construct an 8-gon (octagon) and a 60-gon, since 8 = 2³ and 60 = 2²·3·5, but not a 7-gon (heptagon) or 9-gon (nonagon) because 7 is a non-Fermat odd prime and 9 = 3² is the result of multiplying two non-distinct Fermat primes (3 with itself). Doubling a regular polygon just involves bisecting a side or angle and intersecting it with the circumscribed circle. Inscribing an LCM(m,n)-gon involves inscribing an m- and n-gon in a circle with a common vertex, then finding a pair of the nearest non-common vertexes between the m- and n-gon and copying that around the circle. Inscribing a Fermat p-gon in a circle can be done by memorizing the known cases-it's unlikely you'll need anything beyond the triangle and pentagon in practice. I'm sure there is a marvelous method to construct any Fermat p-gon which this comment is too narrow to contain.
Can you please explain me the reason that how that first side you get as one fifth of the circumference? I cannot get it don't know which property is used here .. please tell...
Dividing the diameter of a circle into five equal parts, as illustrated, results in a distance between point A and 2-dash that is approximately equivalent to the length of one side of a pentagon. This constructional approach is employed to approximate the general shape of a pentagon. For precise length calculations, the formula 2Rsin36(degree) can be utilized, where R represents the diameter of the outer circle.
bruh complicating it too much, cant you just divide 360 by number of edges for polygon and just draw in circle using those angles to define the points, finally connect them and get it over with?
My students will surely shed tears 😂
Hi sir I didn't shed any tears I did do it
As a student not from your university, will confirm that we are not shedding tears, but shedding waterfalls
U really hit my expectations
Thanks you so much sir
Thank you so much sir ❤
nice one ADTW Study
Why is it 2 that the external point connects to?
Can we draw 6 sided polygon with this method
Yup
How do I make a regular pentagon of side 2.5 cm.??
NOT a general method.
While this does correctly inscribe a triangle, square, or hexagon in a circle, it does not work for other regular polygons. It does produce remarkably close results for a pentagon (.065% angle error), heptagon (.17% error), and octagon (.42% error), but the error continues to grow as the sides increase. By the time it reaches a 25-gon, the error is large enough to not even fit 24 sides in the circle.
It's fine if you want to draw an n-gon approximation with a small number of sides, but don't use this on a math test.
we use it all over spain, not as an exact methow but as aproximate method
So what’s the best method?
@@Goofayball To inscribe an exact regular n-gon in a circle with straight-edge and compass? It's not possible because most regular n-gons cannot be constructed with straight-edge and compass.
What you can do is inscribe a regular p-gon into a circle where p is a Fermat prime (a prime of the form 1 + 2^{2^n}), take an inscribed m-gon and n-gon and inscribe an LCM(m,n)-gon, and double the sides of an inscribed n-gon to get an inscribed 2n-gon.
Combining the above allows the construction of a k-gon when k = 2ⁿ·p₁·p₂·p₃..., where the pᵢ values are 0 or more distinct Fermat primes. No other n-gon can be constructed. Note that there are only 5 known Fermat primes (3, 5, 17, 257, and 65537).
So it's possible to construct an 8-gon (octagon) and a 60-gon, since 8 = 2³ and 60 = 2²·3·5, but not a 7-gon (heptagon) or 9-gon (nonagon) because 7 is a non-Fermat odd prime and 9 = 3² is the result of multiplying two non-distinct Fermat primes (3 with itself).
Doubling a regular polygon just involves bisecting a side or angle and intersecting it with the circumscribed circle.
Inscribing an LCM(m,n)-gon involves inscribing an m- and n-gon in a circle with a common vertex, then finding a pair of the nearest non-common vertexes between the m- and n-gon and copying that around the circle.
Inscribing a Fermat p-gon in a circle can be done by memorizing the known cases-it's unlikely you'll need anything beyond the triangle and pentagon in practice. I'm sure there is a marvelous method to construct any Fermat p-gon which this comment is too narrow to contain.
The first metho, this is method 2@@Goofayball
can you draw a 7 sided regular polygon with this method?
Yes, divide the horizontal line into 7 parts and draw a vertical line passing through point 2 on the horizontal line...
@@ADTWstudy pffffffftttttttt
@@lanceraltria No you cannot. It will be an approximation but it will be slightly off. Try it yourself.
@@ADTWstudy No you cannot. it will be a close approximation but not correct
Can you please explain me the reason that how that first side you get as one fifth of the circumference?
I cannot get it don't know which property is used here .. please tell...
Dividing the diameter of a circle into five equal parts, as illustrated, results in a distance between point A and 2-dash that is approximately equivalent to the length of one side of a pentagon. This constructional approach is employed to approximate the general shape of a pentagon. For precise length calculations, the formula 2Rsin36(degree) can be utilized, where R represents the diameter of the outer circle.
Ok
bruh complicating it too much, cant you just divide 360 by number of edges for polygon and just draw in circle using those angles to define the points, finally connect them and get it over with?
I HATE YOU MAKING ME DOING SOMETHING WRONG WITH MY WORK
Hindi nya fault kung uto tuto ka