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# Circles to make a pentagon

## Question

Using only a compass to make circles of any size you choose, find a method to locate five points that if connected would make a regular pentagon.

Is it possible to make a heptagon using a compass and circles?

## 17 answers to this question

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You'll need a straight edge (ruler) also.

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Why would you need a straight edge? I just want points.

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Why would you need a straight edge? I just want points.

One of the points used in the standard construction is the midpoint of a radius.

I can't think offhand of a compass-only construction that bisects a line.

But now I'll try to find one.

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Why would you need a straight edge? I just want points.

One of the points used in the standard construction is the midpoint of a radius.

I can't think offhand of a compass-only construction that bisects a line.

But now I'll try to find one.

But remember, I do not need a line. I just want the vertices that make the desired shapes. There is no request to actually connect the vertices.

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Why would you need a straight edge? I just want points.

One of the points used in the standard construction is the midpoint of a radius.

I can't think offhand of a compass-only construction that bisects a line.

But now I'll try to find one.

But remember, I do not need a line. I just want the vertices that make the desired shapes. There is no request to actually connect the vertices.

It seems you need the use of a straight edge (ruler) to bisect a line segment.

That's why your solution must be different from the standard construction.

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In the sense of needing to bisect a line....yes.

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Make 7 dots in a row by constructing equilateral tri-points ,then erase upper circles.

On 5th dot from left construct the 4 radius circle (green).

Back 3 dots to 2nd dot and construct the 5 radius circle (red). The intersection point (blue) form a right triangle (3,4,5) w/ red & green centers.

Construct circle from green circle radius 'mid point' (white) to blue dot to intersect the horizontal dia. of green circle (not visible).at (-), giving the side length of the pentagon: (-) to blue. .but since straight lines are out we alternatively pick intersection (+) .for the side length.

Erase all white circles and construct the 2 side lenght radius circles to complete 5 vertices.

Edited by TimeSpaceLightForce

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Make 7 dots in a row by constructing equilateral tri-points ,then erase upper circles.

On 5th dot from left construct the 4 radius circle (green).

Back 3 dots to 2nd dot and construct the 5 radius circle (red). The intersection point (blue) form a right triangle (3,4,5) w/ red & green centers.

Construct circle from green circle radius 'mid point' (white) to blue dot to intersect the horizontal dia. of green circle (not visible).at (-), giving the side length of the pentagon: (-) to blue. .but since straight lines are out we alternatively pick intersection (+) .for the side length.

Erase all white circles and construct the 2 side lenght radius circles to complete 5 vertices.

well done. Now about that heptagon.

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make 3 dots in a row
do the cyan circles
do the white and continue
for the 4 vertices

@BMAD my pentagon solution seems inaccurate.I found out the side length point at (+) being off . I would like to request your better solution for im having trouble with finding the last point of side length.

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tslf, you have me worried. For I have done it much the same way. Let me re-examine my own solution as well.

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heptacle.JPG

make 3 dots in a row

do the cyan circles

do the white and continue

for the 4 vertices

@BMAD my pentagon solution seems inaccurate.I found out the side length point at (+) being off . I would like to request your better solution for im having trouble with finding the last point of side length.

A ruler is used in the construction, as it true for all the solutions that I know about.

But if a ruler is allowed, this link gives the prettiest description that I know of.

I'm interested to learn how/whether a ruler may not be necessary.

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Remember @Bonanova, we can only use a compass.

Given two points A and B,
C1 = A(B),
C2 = B(A) to intersect at C and D,
C3 = C(D) to intersect at E and C2 at F,
C4 = A(F),
C5 = B(E) to intersect at G and H.
C6 = GÂ© to intersect at I and J,
C7 = HÂ© to intersect at K and L.
Then DIKLJ is a regular pentagon.
Edited by BMAD
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Thanks!

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lol

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Can you construct a heptagon without assuming you can make several linear points?

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well constructing linear points is a valid part of a compass.

simply construct a circle, and mark the point you left off on, or if you can't do that,

pick a point on the circle, draw a circle keeping the compass the same radius.

then from one of the points of intersection, construct another circle. the two newest circles meet in the middle.

and then most new and the first circle meet, you can then construct another point using a similar process

for a pentagon, you can use the golden ratio, (1+sqrt(5))/2. which is easy enough to construct.

with a heptagon, its impossible. see http://en.wikipedia.org/wiki/Heptagon

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I was under the same impression Phil regarding the Heptagon, which is why i asked in the op (a possible and impossible solution). Thanks for the link and thank you for explaining the construction of linear points. I missed that part .

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