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Put the centers of the circles at corners of a square with side length r * sqrt (2).
Each of the four vesica piscis shapes is made of two circle segments.
One circle segment is difference between area of quarter circle and area of an isos right triangle with legs r.
https://www.desmos.com/calculator/nljf3lte8l
Therefore:
Overlap = 4 * 2 * [ (1/4) π r^2 – (1/2) r^2 ]
This simplifies to
Overlap = 2 π r^2 – 4 r^2
Overlap = (2π – 4) r^2
I do. well done!
Here is mine, https://www.geogebra.org/calculator/kptuhaz3
I don't follow normal Dimensions because it doesn't make sense to me, not that I don't understand it; but rather, it doesn't follow geometric harmony. For example. A cube doesn't just exist on its own, because the point of the cube are determined by another geometry. A sphere. My thoughts any way. lol 😂 Also, this is a great website to make some geometry!
I used Geogebra to give you a decimal answer. It's possible to write out an expression using square roots and phi, but so cumbersome I didn't do it. How important is it for you to have an expression for this length?
1 + x = 1.403 476 252 ...
phi = 1.618 033 989 ...
That ratio is 0.867396, and it is closer to the exact heptagon value 2 sin pi/7 than the Dürer approximation.
Durer ratio side to radius: 0.866 025 ...
Your ratio side to radius: 0.867 396 ...
Actual ratio side to radius: 0.867 767 ...
For a radius of 1 m, the Dürer heptagon side would be off by 1.74 mm. For a radius of 1 m, your construction's heptagon side would be off by 0.37 mm.
Your problem intrigued me and I did a few more cross sections to see if there would be any example where they would be equiangular (even if not equilateral) shapes.
Interestingly, I find the answer is no (except for top and bottom faces). As you let the cross section plane rise upward, you get the following shapes:
(1) a 3-gon with 3 angles of 60° (bottom face)
(2) 9-gons with 3 angles about 135.52249° and 6 angles about 142.23876° (only one of these is equilateral)
(3) a 6-gon with 3 angles of 135.52249° and 3 angles of 104.47751° (this happens when plane intersects three icosahedron vertices)
(4) 12-gons with 6 angles of 135.52249° and 6 angles of 164.47751°
(5) equilateral 12-gon occurs when the plane passes through the center of the icosahedron, but it is not regular because it has those 6 angles of 135.52249° and 6 angles of 164.47751°
Rest of the way up just reverses those types.
This reveals there isn't any cross section that is a 9-gon that has 9 angles of 140° nor any 12-gon that has 12 angles of 150°. In other words, none of the 9-gons, 6-gon, or 12-gons are equiangular.
Spin the model here:
https://www.geogebra.org/3d/peeu23hz
I suppose I shouldn't have been surprised that the 135.52249° angles are constant; it's because the cross section planes are parallel. But these things are very hard for me to visualize.
Thanks for the interesting query!
I constructed this in GeoGebra.org,
I started with a circle. I made a segment that intersected the circle in two places (a chord of the circle.)
Then I made another chord that was parallel to the first chord. Then I connected the endpoints to make a trapezoid.
If you find the perpendicular bisector of two contiguous sides, they intersect at the center of the circle.
You can move the blue vertices around on the circle and see how the center of the circle moves depending on the size of the angles. It looks like the center is on the trapezoid when the greater angles are 130. I'm not sure why! And now I'm curious.
Here is a link to the construction in GeoGebra.org text
If you think of the segment AC as an angle bisector, this means that any point on AC is the same distance from the sides of the angles. Since points D and B are on the sides of both angle D and DCB, the point E must be the same distance from both and the lengths are 1:1. But if you go to the GeoGebra construction, you can change the side lengths of the square and see the lengths DE and BE are always the same.
So I constructed it by making congruent top angles. If there are two pairs of congruent angles, then there must be three pairs (third angles theorem). So angle A = angle D. BM = DM. The two triangles are similar, not congruent unless angle B = angle A. Sorry, I dont know how to upload my image. Maybe this link will work. Https://www.geogebra.org/classic/jsm7kgvt
If you can get to this, you can pull the vertices and see how the triangles react.
Here is the solution
Password is 41AB
I made a triangular prism in Desmos, encoded with the ability to slide and rotate the 2D slice around. Check it out:
https://www.desmos.com/calculator/1mmbj9339n
use 'a' to translate in 3D (a=0 is midsection) , 'b,c' to rotate on two independent planes of rotation. You can find the 2 different diagonal slices you're looking for by setting b=1.57,c=0.47 and the other is b=0.785,c=1.57
I haven't made a rectangular prism yet, but a cube, also in Desmos:
https://www.desmos.com/calculator/fgqzkxuntu
This one starts off with the cube angled to pass through corner -first.
Hello guys!
I have a real-world problem I need solved, and I humbly bow my head and request help!
A cymbal with center point A has an edge crack that a cymbal repair-guy (me) needs to cut away safely, illustrated in the figure as segment BP. I know from experience that all material 45 degrees on either side of the crack needs to be removed, and also that the cut-away must leave gentle curve that transitions smoothly from the bottom of the crack to an imaginary line tangent to the edge of the cymbal. I have drawn a diagram depicting a theoretical "perfect" edge-crack repair, where the orange line depicts where I must cut in order to cut away the crack, and smoothly transition into the tangent of the circle at points J and K.
https://www.geogebra.org/geometry/hcbenszh (please excuse my poor geogebra skills, I could not constrain everything so it maintained the relationships- my end goal)
What I need to know for this to be of use to me practically is what the relationship is between BP, BC, UH and IV so that the shape of the curve is always maintained? If I could figure out how to simply use this diagram to create real-world templates, I could theoretically use this template to fix any sized edge crack for any sized cymbal.
​
Thank you so much, any help would be greatly appreciated!
I suppose I should explain my thinking to invite an argument.
So a Sphere is a perfect object, its geometry is even and something i would call harmonious. 1 sphere or circle has a center location and then the edge. The point is what determines a spacial position. So 1 circle = 1 D. A Second circle adds another location or point, which you can draw a line; thus 2 D. so on and so forth. However, i use what appears to be 7 circles which is more like 8 in "3d" So, 1s=1d, 2s=2d, 3s=3d, 8s=4d, 16s=5d or so you would think. However, I believe there is more to it. https://www.geogebra.org/calculator/xv2y7k62
Geogebra has a useful 3d option.
Apple's OS X used to have an inbuilt 2d/3d grapher called "Grapher" but it's been years since I've used that operating system so I don't know if it's still there.
Gnuplot is also useful. It has probably the most flexibility but harder to learn than other options and lacks a GUI.
https://www.geogebra.org/classic/g6kwcxne
Here is a more visual interpretation of how to solve the problem. Drag point"I" along line to see the path point "J" makes.
Synthetic proof
Diagram using Geogebra:
https://www.geogebra.org/graphing/eb9havh2
As shown in the diagram, label right triangle ABC with right angle at A(0, 0), B(2, 0), and C(0, 2√3).
One point of equilateral triangle is M(1, 0) and another is point N as yet not known.
Rotate blue right triangle MAN exactly 60° clockwise around center M to form red triangle MA'N' where N' is the third vertex of the equilateral triangle.
AM = A'M = 1, and angle AMA' = 60°, so drop perpendicular from A' to AB at foot F. Then notice that triangle MFA' is a 30-60-90 triangle with hypotenuse 1. Therefore FM = 1/2 and FA' = √3/2.
Angle AMA' = 60° and angle ABC = 60°, so MA' and BC are parallel. Also A'N' is perpendicular to A'M because triangle N'A'M is image of right triangle NAM. Therefore angle A'N'B = 90° because if a transversal A'N' is perpendicular to one of two parallel lines it is perpendicular to the other.
Notice also that A'F = √3/2 and FM = 1/2 and MB = 1, so FB = 3/2. Therefore triangle A'FB is a 30-60-90 triangle with angle FBA' = 30°. But since angle ABC = 60° this means angle A'BN' = 30° and angle BA'N = 60°.
By Angle-Side-Angle congruence, triangle A'F'B is congruent to A'N'B. That means that A'F = A'N' = √3/2.
Since A'N' =√3/2, you can conclude AN = √3/2, since AN is the preimage of A'N'.
Now we know AN = √3/2 and AM = 1 in right triangle MAN. Hypotenuse MN must therefore by Pythagorean theorem be √[ (√3/2)^2 + 1^2 ] = √[ 3/4 + 4/4 ] = √[7/4] = √7/√4 = √7/2.
The perimeter of the region is the length of the thick black line, right? This line has 3 sections of 6 cm each plus 3 curves (see the GRH arc on the drawing), each of which is a third part of the circle perimeter/circumference.
Do you need the reasoning for why that is 1/3 of the circumference?
You got me looking through my old work now and making me nostalgic for it. No lie, I don't think I'll have it done anytime soon because I've got a lot of other projects on my plate. But I'd be happy to share what I have now since it still might be enlightening to look at especially for teaching (and I use some ideas in my own tutoring too). If you want, drop your email in a DM and I can shoot it to you, along with some other resources.
Fair warning, it is focused more on the Greek translation, at least this beginning part, and presupposes a strong familiarity with Hilbert's project. It's also a little chicken-scratchy but you could sift through and figure it out eventually.
That all said, what it seems you're looking for is most specifically "a complete overview of Hilbert's vindication of Euclid." Hartshorne is, as far as I know, the canonical text on the matter, though it isn't necessarily "complete" and I've got other issues with it. I'm sure there's a PDF of it lying around somewhere on the Interwebz...
You should tell teacher to fix the diagram. The angle QAK must be 50°, not 20°, based on the given information.
Reasoning: Solve equal angles to show x = 11. Then angle QPR = 70, angle PRQ = 80, so angle RQP = 30 and angle AQK = 15. Looking at triangle QPK with angle P = 35, angle Q = 30, we conclude angle QKP = 115. Looking at triangle QAK, if angle KQA = 15 and angle QKA = 115, then angle QAK must be 50.