Triangles have a Magic Highway - Numberphile
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Date: 2022-04-08
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Comments and reviews: 9
Venkat
Sin waves are the real creators of the ratios. That's why you have the ratios. Each thing like centroid etc are special angles of sin. A triangle is a special planar vortex. They can always be split to two right. Equi is quantum duality. Resonance is a equal side match of any shape triangles. A D in the formation is where all center meet. That's why most creatures are like worms. The shape of BH. Equality is the chance of meeting or superficial. BH are the male aspects of the cosmos. Croissants. A cocktail of Eclipses of moon. Do universes flip yes sometimes. Maybe once in a blue moon. They are just quantum predictions.
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Sin waves are the real creators of the ratios. That's why you have the ratios. Each thing like centroid etc are special angles of sin. A triangle is a special planar vortex. They can always be split to two right. Equi is quantum duality. Resonance is a equal side match of any shape triangles. A D in the formation is where all center meet. That's why most creatures are like worms. The shape of BH. Equality is the chance of meeting or superficial. BH are the male aspects of the cosmos. Croissants. A cocktail of Eclipses of moon. Do universes flip yes sometimes. Maybe once in a blue moon. They are just quantum predictions.
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Kurtlane
That was my school geometry:
Given any triangle, prove that angle bysectors, medians and heights of its 3 angles always intersect at one point, and so do the perpendicular bycenters. (The 4 points of intersections do not have to be the same)
In other words, prove that in enter, medicenter, orthocenter and circumcenter always exist for any triangle.
Anyone can point me to the proofs.
We didn't learn about the Euler line, but the proof of that would be interesting too.
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That was my school geometry:
Given any triangle, prove that angle bysectors, medians and heights of its 3 angles always intersect at one point, and so do the perpendicular bycenters. (The 4 points of intersections do not have to be the same)
In other words, prove that in enter, medicenter, orthocenter and circumcenter always exist for any triangle.
Anyone can point me to the proofs.
We didn't learn about the Euler line, but the proof of that would be interesting too.
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Star
now that i have revisited the video, i'm also astounded by the method to proof centroid separates the line 2: 1
because the center triangle forms a parallelogram & its centroid remains the same
therefore repeatedly add center triangles will get the longer side equals 1-0. 5+0. 25. = 1/(1+0. 5) = 2/3
& so the ratio is (2/3): (1/3) which is 2: 1
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now that i have revisited the video, i'm also astounded by the method to proof centroid separates the line 2: 1
because the center triangle forms a parallelogram & its centroid remains the same
therefore repeatedly add center triangles will get the longer side equals 1-0. 5+0. 25. = 1/(1+0. 5) = 2/3
& so the ratio is (2/3): (1/3) which is 2: 1
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ApresSavant
Using euclidean tools, and draw circles based upon the centers of the three sides, and then draw a circle that encloses and touches all three circles, one would presume that the center of that larger circle would also be on that line. But it isn't. Do we know why? Is this another case of the In-center?
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Using euclidean tools, and draw circles based upon the centers of the three sides, and then draw a circle that encloses and touches all three circles, one would presume that the center of that larger circle would also be on that line. But it isn't. Do we know why? Is this another case of the In-center?
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Generic
Amazing video! Two questions:
1. Can you find the point that represents the smallest total summed distance from itself to each of the points? (Smallest value for XA+XB+XC)
2. Can these nuggets of wisdom be extrapolated to any other n-sided shape? Does it work for shapes that contain concave angles?
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Amazing video! Two questions:
1. Can you find the point that represents the smallest total summed distance from itself to each of the points? (Smallest value for XA+XB+XC)
2. Can these nuggets of wisdom be extrapolated to any other n-sided shape? Does it work for shapes that contain concave angles?
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Rhydian
If you apply a projective transformation to the equilateral triangle case how is it that one point is split into three if under projection distinct points are mapped to distinct points? Bisecting the Euler line gives the centre of the Feuerbach 9-point ciecle. Nice animations. (Dr) Rhydian Harker.
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If you apply a projective transformation to the equilateral triangle case how is it that one point is split into three if under projection distinct points are mapped to distinct points? Bisecting the Euler line gives the centre of the Feuerbach 9-point ciecle. Nice animations. (Dr) Rhydian Harker.
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David
Question: Given the three -centers- is it possible to determine the triangle that generated them? If not, what is the class of triangles that may have generated them? What is the situation in the degenerative cases where two or all three of the -centers- coincide?
Any thoughts?
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Question: Given the three -centers- is it possible to determine the triangle that generated them? If not, what is the class of triangles that may have generated them? What is the situation in the degenerative cases where two or all three of the -centers- coincide?
Any thoughts?
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Ye
Wow! I never knew that the three centers lie on the same line! I thought that my education system beat Euclidean geometry to death and I guess not. I am here after watching all the -clown- videos (wink wink. I am so happy that I didn't skip this video.
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Wow! I never knew that the three centers lie on the same line! I thought that my education system beat Euclidean geometry to death and I guess not. I am here after watching all the -clown- videos (wink wink. I am so happy that I didn't skip this video.
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education
thanks so much, brilliant subject and brilliantly explained. i dont want to exagerate, but so many answers to so many misteries in life (from piramids to perpetum mobile, to the masonic logo) may be answered with this explanation
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thanks so much, brilliant subject and brilliantly explained. i dont want to exagerate, but so many answers to so many misteries in life (from piramids to perpetum mobile, to the masonic logo) may be answered with this explanation
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