In this video, I showed the triangle inequality as a basis for determining whether the sides of a triangle are valid. I also described a degenerate triangle as a triangle with zero area.
Excellent explanation, to add to it, one of the used of degenerate cases is they can allow you to use a different set of rules for exploration. The idea of a degenerate triangle cascades forward. A degenerate quadrilateral with one side zero length for instance behaves like a triangle, with two sides zero, the degenerate quadrilateral behaves as a line segment. Why might this be useful? Think about computer animation, allowing for degenerate triangles allows the GPU to process everything as a triangle versus separate processes for segments. Some of y'all might want to read this: en.wikipedia.org/wiki/Degeneracy_(mathematics)
In the primitive-assembly and rasterizer stages of the GPU you usually want to avoid degenerate triangles, because they only waste computing power with no output, and hence cull them as early as possible.
Despite having a Mathematics A level GCE in the UK (in 1961), I had never heard of a degenerate triangle before. I'm pretty sure that it was never covered in the syllabus. Before I watched the video, I did figure out that 7-13-4 was impossible and that 2-3-1 was a straight line. That was just by drawing the triangles in my head. I love mathematics, but having spent 50-odd years as an Engineer, much of it in design, I never found the need to redefine a straight line as a degenerate triangle. "Degenerate" to me is something entirely different! I learn something new every day. At 82 years old, that's a must. Just found your channel and subscribed. Keep it up, mate. I'll be watching.
not having a high level of mathematics but came to the same conclusion (only checked the smaller side) so yeah I had A was "technically" a triangle but didn't know it was called degenerate...
I'm pretty sure the reason it's never mentioned is cause it's irrelevant, save for being a silly bit of trivia. ;) Following that same logic, a triangle is also a degenerate version of every other polygon Taking it to the extreme, a single point is degenerate from of all polygons. Btw, did you know that since two points are enough to define a side, a circle is also an... err, infinito-gon?
@@Slavic_Goblin Excellent reasoning, sir! Because it shows that all polygons are based on the definition of a point. Now all that is needed is for educators to update themselves, rewrite the relevant curricula, & then teach up-to-date stuff right from the start! That leads to the concept of "degenerate" being no longer needed to explain obsoleted curricula. As to a circle being an infinito-gon: YUP!An infinite quantity of infinitely-short lines--which is right-there the definition of "limit"...
Strange. I learned the term in regard to polygons when I was in middle school. Of course that was 1975-ish, so the term might've only recently been applied.
I doubt that. Regardless of which clade of engineering, you always come into a definition of forces/currents/potentials that splits it up into two orthogonal components, giving the components sin(a)F and cos(a)F. Every time this coincidentally only has one component with the other set to 0 (eg, a+j0 or three intersecting rods in a lattice where two of them point in the same direction), you have a degenerate triangle. Top dead centre and Bottom dead centre in a piston engine are degenerate triangles. sin(0) and sin(\pi) denote degenerate triangles. They are in fact incredibly common.
You know what's funny? I was taught all the way to Trigonometry and Calculus AB in High School but never once did I learn what a degenerate triangle was. I didn't even know what the requirements of making a triangle were until today. Thank you. Awesome explanation!!!
You know at the ripe old age of 56, I really like these videos about math. What I realize about all of these is that the theorems and postulates of geometry are woefully not stressed enough. Math teachers should be putting a geometry question on every test to pound home how important knowing this basic information is.
I was going to ask how a degenerate triangle was going to add up to 180 degrees because I thought both of the legs with the hypotenuse were 360 degrees each, but then I realized that a straight line is 180 degrees itself! Geometry can be so fascinating!!
@@derwolf7810 you actually end up with two 0 degree angles along the longest length side, since a 0 degree angle in a triangle represents an immediate turnaround (two sides moving in the same direction away from a vertex). For another perspective consider the supplementary angle, which consists of two edges moving in opposite directions from a vertex, and thus meeting at a 180 degree angle. If an angle's supplement is 180 degrees, the angle itself must measure 0 degrees.
@@sophiastern2719 Yeah, i found my error. Thanks for your info, i deleted that message, round about the time you wrote your message, so i didn't notice you answered. Apologies for that.
@@danquaylesitsspeltpotatoe8307 I agree, your trig functions break down. You start getting complex numbers and divide-by-zero problems, too many things just don't make sense anymore.
It's quicker (and enough) to compare only the longest side with the others. Here is why: Let's call C the longest side, we have automatically C >= A and C >= B, therefore A+C >= B and B+C >= A. We just have to check that A+B >= C.
What is the formula to find the longest side, and the two shorter sides? I assume you could do it if you wanted to use some set theory. Mathematically it seems difficult to describe. it is easier to express ( (A+B)>=C ) & ( (A+C)>=B ) & ( (B+C)>=A ) than to explain how to find the max and the other two sides to use. Keep in mind that 2 or 3 sides could be the same. There is no guarantee that you have one side of maximum length, though if you have multiple sides of max length you will always have a triangle. Just because you can do something intuitively doesn't mean it is easy to describe mathematically.
@@PhilosophicalNonsense-wy9gy While I agree with you, opinions differ. Degenerate triangles may have angles of 0°, 0°, 180°, but that''s a line in my book. Euclid's triangle inequality appears to indicate indicate (A+B)>C. Otherwise, you could argue that if the angle can be zero, the length can be zero, and we get octagons with three sides (five sides have lengths of zero.) IANAM.
Another way to state the triangle inequality is that the shortest path between any points is a straight line. The direct distance from two points will never be longer than the distance between the two points via a third point.
Nice! I love math, though I wouldn't say I'm particularly good at it. I don't know how I haven't stumbled upon your channel before now, but I'm glad I did. I am actually taking a math class this semester (going back to college after almost 20 years, oh boy) and I'm both excited and terrified. I appreciate your calm and methodical demeanor, and when you said, "A degenerate man is still a man," I knew I had to subscribe. Thanks for the great video!
There seems to be different interpretations on the triangle inequality theorem, specifically the difference between "2 sides must have a sum greater than the 3rd" and "2 sides must have a sum greater than OR equal to the 3rd". This is a bit frustrating but I think it should be important to consider what you want for a valid triangle. If you want a triangle that has an area > 0, then you would have to have the sums be greater than the 3rd side, but not equal.
I was going to put exactly the same point. For an actual triangle, no angle can be equal to 0°. If A+B=C, then two angles equal 0° and one angle equals 180°. To me, that's not a triangle.
@@jerry2357 I'm not a geometry expert, so I acknowledge that I could be incorrect, but I strongly agree with you here. I don't think how we are defining a triangle is relevant in this situation because if Sⁿ¹+Sⁿ² = Sⁿ³, then you have a line which is categorically not a triangle which is defined by having 3 sides (a/k/a LINES). The implications of a zero area triangle are pretty ridiculous, since you could just as easily show a triangle and claim it's a pentagon with 2 angles = 0º. Now, I would still answered C since this is the _most_ correct answer, and I could make a steel-man argument by claiming it's possible the length of A¹ equals 1.00000000001 and is reasonably reported as '1' which would create an actual triangle since technically Sⁿ¹+Sⁿ² > Sⁿ³. I know this isn't a great argument (or even a good argument), but it's better than any argument that can be made for the existence of Triangle C.
@@jerry2357 But before you validly can use the word "actual", you must define its meaning. Saying "You know what I mean" is NOT a useful definition in a scientific field. You might require that a triangle must be distinguishable as such. So what must at least 1 angle minimally be? 0.0000000001 degree? Why not 1% of that value? Or even less? Would/should/could ANY value that is not exactly 0 be valid? What is still distinguishable?
@@msskaggs3911 Re "The implications of a zero area triangle are pretty ridiculous, since you could just as easily show a triangle and claim it's a pentagon with 2 angles = 0º": Said claim is perfectly valid & not at all ridiculous. Were ridiculousness a valid argument here, then the concept of limits is absolutely ridiculous. How many of the zeroes in '1.00000000001' can one remove (or insert) and STILL "report it reasonably as '1' "? The idea of reasonableness being allowable is OK for engineers who must BUILD a thing, but it is a ridiculous concept in logic.
@@user-gx1rk8yw6l That rather depends on whether you are looking at a real, physical triangle, or a calculation. For a calculation, the largest angle must be less than 180°. The tolerance depends on the software you are using, or the resolution of the pocket calculator. Or if you're going old school, four figure trig tables worked to a precision of 1 minute of arc. For a real triangle, it rather depends on how you're doing the measurements. With a triangle drawn using a pencil on paper, with the angles measured using a school protractor, then you would be lucky to measure an angle of less than 0.5°. But you could measure a much smaller angle using a theodolite, down to seconds of arc.
Am about the same age. I learned 4 things here. 1: The concept "degenerate triangles". 2: My teacher was not complete in his explanation of triangles. 3: The concept "degenerate" would never have existed (because it would have not been needed) if definitions had been taught -IN-FULL. 4: Even teachers outside the USA's educational system can be not-perfect--even though ever so much better!😄
The shortest distance between two points is a straight line. By this same definition, any path other than a straight line will be longer. Take 🔺️abc If ac is a straight line, then ab+bc must be longer. The sum of any two sides of a triangle will always be *greater* than the third. 1 +2 =3 7 +4
I hope nobody tries to explain degenerate triangles to kids just starting to learn geometry. They will easily see that the sum of any two sides must be greater than the third. They often like to use compasses, so they could prove (or even discover) this rule for themselves.This is about as much knowledge as the average (non-mathematician) person needs to go about their daily life. Still, it’s fascinating to know that so much exists beyond what I learned in school in the 1950’s.
Another way of doing this uses the concept of semiperimeter, which is defined as s = (a+b+c)/2, where a,b and c are the side lengths. It turns out that we always have s >= a, s >= b, s >= c for any triangle. Thus, since the (7, 13, 4) triangle has semiperimeter s = 12, it should not have any sides longer than 12. So the side of length 13 gives a contradiction.
One could argue that other 'Rules of Triangles' needs to be applied. The literal definition: it has to have three (3) non 0° angles between the lines, and the sum of those angles must = 180°. Yes 0° + 0° + 180° = 180°, but you have two angles that are exactly 0°. So applying the other rules for triangles, option 'A' drops out as well because there are angles of 0° between lines 1" & 3" and 2" & 3", and 1" & 2" is 180°, so all three lines literally collapse into a single line in 3D space (i.e. all lines are parallel), and a single line (or three parallel lines no matter the length) does NOT a triangle make.
Informally, A is degenerate or a line segment, B is a pointy isoceles triangle, C is not even a degenerate triangle and its angles cannot be defined, and D is a very flat near-isoseles triangle.
6:03 Triangle A is a degenerate triangle. We can exclude this from discussion of Plane Geometry since it is not unique to one plane. Nice video. You do a wonderful job, as always.
the way i like to think about it is: -take the two shortest sides -connect their ends -align them and start spreading them out -the biggest angle you can get is 180 degrees -therefore the length of the remaining side has to be between 0 and their sum
*flashes back to Cyberchase* Jokes aside, your content is really good, and while I don't think they covered degenerate triangles, I fondly remember the triangle inequality from watching it all those years ago.
Yes! Tou can see this from the thumbnail already. Also, if the sum of the two minor sides is more than the greatest side, that is a sufficient condition that the triangle does exist.
I always thought that the longest side should be smaller than the sum of the other two sides. If the sum was equal to the longest side, it would just lie right on top.
As long as the triangle has an area thats greater than 0 its a triangle, so we can just use herons formula. Find the semi permeter of all three sides then compare if any side is above the semi perimeter , then we know that triangle has no area since we will get either 0 or a complex number as an answer.
Thanks...always i was wondering the relation between the equality of (flat angle ) ir one line angle equality of 180° and the sum of triangle angles 180°
First of all, great video. I'd never heard of degenerate geometry, so this was very interesting to me. I'm still confused on the concept, however. A triangle is defined as having three sides, vertices, and angles, however, if one of those angles is 180 degrees, then it's not conducive to the placement of a vertex. If I can place a vertex on a 180 degree angle, I can place that same vertex anywhere on a straight line with no discernable alteration to the geometry. I could even place multiple vertices on the line and say it's a degenerate square, or hexagon, or n-gon. In short, the provided values produce an exact geometry, but the geometry in isolation does NOT provide exact values. Is there a justification for the inclusion of ambiguous geometries?
My answer to the 'is the 1,2,3 triangle a triangle?' was 'yes' - on the grounds that all triangles would look like a straight line with a point on it if viewed from *the plane of the triangle*. No flatlander would be able to see all three vertices at the same time. It only looks like a three sided shape from our third dimensional, perpendicular to the 2D plane, viewpoint. Triangles on a sphere have interior angles of more than 180 degrees - up to, I just realised, 900 degrees... if you put a degenerate triangle on a sphere with two vertices at the poles and the third on the equator. 360 at each pole and 180 at the equator = 900 - I'm sure there is a HUGE flaw in my thinking here...
You can evaluate this problem by using the law of cosines where you take the longest side as the opposite side. When applying this on the 1-2-3 triangle we find that cos of the angle between the 1 and the 2 is -1 which means 180 degrees which becomes the collapsed degenerate triangle. When applying the same logic on the 7-4-13 triangle we get that cos of the angle between the 4 and the 7 becomes -13/7 but this is impossible because the values of cos range from -1 to 1 when the angle is between 0 and 180 degrees. Hence the 7-4-13 triangle doesn’t exist
Wouldn’t another way of verifying the legitimacy of a triangle’s existence be seeing if there’s an angle on (0,π) corresponding to each corner such that they add up to π? For example I’m not getting real number solns for 4-7-13 and my general sol’n for the 1-2-3 triangle is 2nπ, for which no value of n which fit the angular criteria for a triangle
I did not participate in the pool exactly because of the 1-2-3 case. In my opinion it's a matter of definition, if one considers this case a triangle or not. When I learned about triangles it was considered a special kind of triangle (don't remember exactly if "degenerate" was used to name it). However, I mentioned that presently some definitions seem to differ from how I learned/remember them.
Question: What is the perimeter of a degenerate triangle ? the length of the segment or twice the length of the segment ? I tend to go with twice, because it would be the sum of all sides, but I might be wrong.
I would probably say you would only add it once or not add it at all. If you think of perimeter as 2D surface area then this would mean that the triangle does not have a perimeter since there is no area to be surrounded by. If you think of perimeter as the sum of the lengths of the edges then I guess you could think of the line as one edge and say the perimeter is the length of a segment. While those could be the formal definitions, I would still practically use your definition of the perimeter being 2 times the length of the segment because if you had a set of right triangles with the only change being theta(the theta from the video) then it would be unnatural for the perimeter to suddenly drop in value once theta reaches 0. I feel like this problem is one of those deeply rooted problems in the vocabulary of geometry and it depends on the textbook you read. I wish I had a concrete answer.
I am reasonably certain the perimeter would be twice the line segment length. The perimeter is the sum of the side lengths. If the side lengths are (1,2,3) then the sum is 6. The fact that the three sides are sharing space with each other doesn't matter.
You add them all up. 1+2+3. There are still 3 distinct segments, AB, BC and AC. You do the same when finding the sum of interior angles. If B were the middle segment, angle BAC and BCA would both be 0 degrees, while angle ABC is a 180 degree angle. Interior angles summing up to 180 is one of the definitions of a valid triangle
The area of triangle in the first triangle having sides as 1, 2 & 3 will be 0 This satisfies the other axiom : "the angle of a straight line is 180 degrees" Hence this is triangle
Thanks for the very simple explanation. I have a question with warning that I'm not a math expert and have very basic math education. I came across degenerate triangles when finding out if "in theory" that three points are in the same coordinates, does that count as a triange but also if also counts as an "equilateral triangle"? I found out from Wiki (gain of salt) that three points is a degenerate triangles however each angle is undefined. Would we consider that we can't determine that it's an equilateral triangle because the angles are undefined or do we say that it's possiable but not sure. I'm curious what people think and I'm sorry if my questions doesn't make sense :).
Perhaps in other languages the usage is different but I was taught that the word triangle used alone is always assumed to be a non-degenerate polygon because a triangle is a polygon, polygons are two dimensional objects by definition, (an enclosed area), and a line segment can't be said to be a two-dimensional object.
In my opinion, claiming that a+b may be equal to c means like saying every line is a triangle, and it also applying to other polygons would mean a line is every kind of shape in existence, give you put a set of ∞ number of vertices. Remember ∞={1,2,3,4,...}
7:50 going to dispute this, the probability is not zero. If you can intentionally pick numbers to make it zero, then those same numbers can come from random selection as well. The odds may be small but they can't be zero unless it's impossible to do intentionally as well.
Imagine constructing the triangle with a ruler and a compass. When you draw the longest line with the ruler and try to find the 3rd corner with the compass you have to draw circles that are in sum bigger than the longest side or the circles will never intersect.
Straight line is not a shape it is a line. But mathematically a triangle with 0 area will still work it does not become undefined. Since the system does not blow up then it is part of the solution set.
another method is to explore the terms that make up heron's formula A=√[s(s-a)(s-b)(s-c)] assuming a, b, and c are all positive, if any of (s-a), (s-b) or (s-c) is zero, the triangle is degenerate. if any of (s-a), (s-b), or (s-c) is negative, a triangle cannot be formed with side-lengths a, b, and c. for example, in a=1, b=2, and c=3, s=(a+b+c)/2=(1+2+3)/2=3. thus, we find that s-c=0, and hence the triangle is degenerate in case where a=7, b=4, and c=13, s=(7+4+13)/2=12. in this case we find that s-c is negative, and hence it's impossible to form a triangle
As I asked in a similar video: can't this be explained much more simply by saying that the most you can separate two vertices of a triangle (and remain a triangle) is always less than the sum of the lengths of their respective sides, and therefore the third side must be less than that sum?
Also all the angles of a triangle is 180. Assign letters to each point and you can show it as well. So even with the degenerate triangle is two zeros and one 180 degrees. Just a different proof it is a triangle.
I think the easiest way to recognise that the 1-2-3 is not really a degenerate triangle, is that if it were really a degenerate, we would see behaviour like it wearing leather, doing drugs or playing loud music. Since it is not doing any of these, we can recognise that the shape is not degenerate, and therefore must be considered a straight line (straight being the opposite of degenerate, rather than implying anything about the line's sexuality).
these definitions of mathematical nomenclature and notation differ from country to country. here are some other examples: is 0 a natural number? do you write a vector like [a b] vertically or like (a; b) horizontally? is a square a rectangle?
I assume the 7-13-4 triangle. Any one side cannot be less than the difference between the other two (and this if you had a "0 degree" triangle in the 7/13/4 - that third leg would need to be a least 6)
Answer is A&C. The prime condition for construction of triangle is that the sum of any two sides must be greater than 3rd side. The triangle 4,7,13 can not exist. Also the triangle 1,2,3 can not exist.
6:11 Yes. Three vertices exist (triangle definition does not state that the vertices cannot exist within the edges). Sum of angles is 180: 0,0,180. Three edges. This triangle is undefined as there are infinite number of triangles with the longest side being 10 inches tho. A different classification of triangles that look exactly like straight lines?
Didn't watch the video, just picked what seemed the likeliest answer and skipped to the last 20 seconds to confirm. A triangle is a closed shape so, logically, in order to form a closed shape from three straight lines of lengths x, y and z, the sum of the lengths of any two sides cannot be less than the length of the third side because, if not, the two shortest sides placed end to end would not be able to reach both ends of the longest side, and thus could not create a closed shape of lengths x, y and z. This requirement held true for all but the 7/4/13 option (i.e., 7 + 4 < 13). Was this basically how he explained it?
The total lengths any two sides of a triangle must be greater than the length of the third side in order to form a triangle. So the 1-2-3 triangle isn't really a triangle because the 1-2 sides form a straight line in order to connect with the 3 side. But that all depends on the accepted definition of a triangle. The 4-7-13 triangle can't exist at all because 4-7 adds up to 11, which is too short to connect with the 13 side. The other two triangles are fine.
You are using the standard High School version of the triangle inequality theorem. They lied to you for the sake of simplicity. This is one of the reasons I hated teaching HS geometry.
My exact same thoughts. But after watching the video and seeing him draw the unit circle and showing the trig functions on it, I had to admit there must exist Sin (0) and Cos (90°) and accept the consequences of them. In other words in order for me to accept all the ideas and rules of trigonometry, I must also accept the idea that a triangle with an angle of zero degrees exists. Obviously if it has an angle of zero degrees the two “short” sides must be drawn on top of the long side, so it just looks like a line.
@@christianemden7637under pure Euclidean geometry perhaps, but maths have moved past this. Using >= connects this definition with other metric spaces using the generalized rule |a|+|b| >= |a+b|, e.g., vector spaces. Moreover as he mentioned degenerate triangles are critical to applications of computer graphics. We may not like degenerate triangles, but we need them.
3:49 Nice one. And wouldn't it be more visual too also use circles? At each end of the line 10 you draw a circle or radius 1 and 2. If the the circles do not intersect, obviously the triangle cannot be drawn. Hence the circles must either intersect in 2 points or be tangent.
Well if you look at the definition of a triangle, you can see that degenerate triangles are NOT triangles because they are not even polygons. A triangle is a polygon with 3 sides. A polygon contains a closed polygonal chain where every side of the chain is a side of the polygon. At least this is how it's defined in my country. The sides of a degenerate triangle do not form a closed polygonal chain therefore a degenerate triangle is not a triangle. The sin(theta) argument doesn't make any sense because 1) sin(theta) is defined using the unit circle , the theorem that allows to use sin(x) in right triangle assumes the existence of the triangle in the first place 2) even if we considered a degenerate triangle a triangle it would definitely not be a right one , as it would not have a 90° angle so the theorem would not apply.
Question: what if the triangle was 3,3,0 triangle, where two of the vertices lay on top of each other? Would that make it a right triangle? Or not because there’s no side length?
@@Jeffman978There are operations that are only allowed on triangles, and a triangle can be between the real and imaginary coordinates of a complex number (a+jb). It's the whole Zero-discussion from the medieval ages all over again, when some people couldn't get around the concept that 0 was a number. If you exclude degenerate triangles from triangles, suddenly, half of engineering and maths breaks because there no longer is a 0; and worse, a sinus wave is no longer uninterrupted if \theta=0 suddenly is undefined, as well as every 0-transfer from positive to negative values. You can "define" a degenerate triangle to no longer hold in case of a specific context, but a degenerate triangle behaves like a triangle, can be constructed (using ruler & compass) like any other triangle, and it is necessary for many theoretical and practical engineering solutions.
@andreamiele5842 as you write an answer on the internet, I am very sure that even in your country, a degenerate triangle is a triangle. Otherwise, you wouldn't have AC technology, so no networks, no radio, no AC power grids. What you postulate is in fact that sin and cos waves have undefined gaps whenever the wave transfers from the positive to the negative and vice versa. This obviously cannot be true.
In the beginning you didn't specify that you were using a flat coordinate system. Now consider polar and spherical coordinates. Then consider the triangle as vectors on a spacetime manifold.
Triangle C immediately jumped out at me because its longest side is longer than its other two sides combined, which is impossible. Also, if Triangle A is a real triangle then it has a height of zero, since its longest side (which I'll designate as the base) is the same length as the two others combined. Triangle B is a perfectly possible isosceles triangle, and I see nothing wrong with Triangle D (which is scalene).
Sorry, I can't agree, that a line segment is a triangle. No angle of triangle be either 0, or 180 degrees. And I'm absolutely positive, that the inequalities of triangle are strict, no equal sign is appropriate. By the way there's another part of the inequality |a-b|
No, the triangle inequality in general holds for both the greater than or equal to case. It is only strictly greater than when considering non degenerate triangles.
OK, I admit, I was wrong, but I'm sure in my school we never talked about degenerate triangles, nor have I seen this elsewhere. So that's a surprise for me. Didn't get what it has to flat or not flat space in comments, obviously triangles shown on a sheet of paper, they can't be any curvature? unless it's it mentioned unambiguously
@@XtreeM_FaiL If you step out of Cartesian/Euclidean geometry, you can pretty easily make even the 4/7/13 tri work, but most of mathematics works in an assumed static, non-curving, neither hyperbolic nor hypobolic space unless explicitly stated otherwise. You're not strictly wrong, but the "gotcha!" carp is annoying.
*This has been corrected thanks to an error pointed out in later comments. I had originally misidentified the failure condition as the final answer* I used roughly this principle in a college statistics course I was taking (although I didn't get into degenerate triangles.) I think the problem was: If you take a line segment, and randomly cut it in 2 places, what's the probability you can make a triangle out of the remaining pieces? I solved this, and expanded on it for any n-gon using the principle that all cuts must fall on the same half of the line segment. Looking at the failure condition (all cuts are on the same side) that's 1/2 to make a triangle. 3 cuts, 1/2^2 to make a quadrilateral. 4 cuts, 1/2^3 to make a pentagon, etc. The success condition is 1 those amounts. So the generic formula was the probability of creating a succesful n-gon with this method would be 1-1/(2^(n-2)) While I was doing this, my instructor was showing the class some completely different method that I had trouble following, then assigned solving for a quadrilateral as extra credit. I showed him what I came up with, and verified that my answers were consistent with his. But he wasn't convinced my reasoning was sufficient. He thought all cuts falling on the same half seemed intuitive, but wasn't completely convinced and wanted a proof. Which wasn't something I felt up to.
actually the cuts have to be on opposite halves and the distance between them must be at least half the length, the chance for a triangle ends up being 1/4, very interesting problem though
@@SuperTheNobody Indeed, I did this incorrectly just now. I had established a failure condition in my original comment. So the correct answer is 1-[1/(2^(n-2))] I'll go add an edit. Thank you for the catch!
So degenerate triangles are always isosceles triangles considering its angles (180, 0, 0), but aren’t always isosceles considering its sides (1, 2, 3 for example)?
The 'Triangle' 1 2 3, is not a triangle, it is a straight-line segment. I.E., 1 + 2 = 3. Also, it is not a LINE! It is a line segment. Mathematicians should care. Entertaining. Thank you for your efforts. May you and yours stay well and prosper.
Very nicely done video, thank you for posting. The usefulness of triangles with one angle of 0 degrees is apparent. However, two such triangles with angles of (0,90,90) and (0, 45,135), respectfully, would have the same vertices and appear indistinguishable. In fact, if one specifies such a triangle by listing the vertices, then the only angle that is defined is the one that is zero, so we would in effect have a triangle with only one angle.
I fail to see how you could construct a triangle with angles of (0,45,135) as it would expand into the 2nd dimension. It think the only angles possible for a flat triangle are of (0,90,90) and (0,0,180).
Wouldn't any triangle need to have an area? If one can't differentiate a triangle from a line, is it really a triangle? Couldn't it also be any "degerate" polygon? Maybe it is really a 4 sided polygons, with sides 3,1,1,1. Where do you "draw the line" (pun fully intended")?
