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Question

A polyhedron $(3-	ext { polytope) is called regular if all its facets }$ are congruent regular polygons and all the angles at the vertices are equal. Supply the details in the following proof that there are only five regular polyhedra. 
a. Suppose that a regular polyhedron has $r$ facets, each of which is a $k$ -sided regular polygon, and that $s$ edges meet at each vertex. Letting $v$ and $e$ denote the numbers of vertices and edges in the polyhedron, explain why $k r=2 e$ and $s v=2 e$ 
b. Use Euler's formula to show that $\frac{1}{s}+\frac{1}{k}=\frac{1}{2}+\frac{1}{e}$ 
c. Find all the integral solutions of the equation in part (b) that satisfy the geometric constraints of the problem. (How small can $k$ and $s$ be? $)$

Chris Trentman · Accepted Answer

were asked to prove that there are only five regular Polly Hydra. And in part a begin the proof. Suppose that a regular Paula Hadron has our facets. So our faces, each of which is a que cited regular polygon and that s edges meat at each vertex of the poly Headroom were asked. Okay, let V and e b the number of Vergis season edges in the Pali Hadron. Whereas to explain why K R is equal to two e and S V is equal to two ia&#x27;s well, so consider that each edge belongs to two facets. So it follows that twice the number of edges is going to be the number of sides. Okay, times the number of facets are so reach facet has K sides. Yeah, this is going to be twice the number of edges. Now each edge has to vergis ease my definition of an edge. So it follows the number of edges at each vertex. There s edges that meet each vertex times. The number of vergis is which is V. This is going be twice the number of total edges. So to e, this is what we wanted to prove for part a now in part B brass to use Oilers formula to show at one of her s plus one over K is equal to one of her two plus one over e. So by Oilers formula for a Poly Hadron we have the number of vergis is minus the number of edges plus the number of facets which in this case is our This is going to be equal to one plus native and squared or two. So manipulating this equation, we have that to e we substitute in two year over s minus e plus two year over K equals two or, in other words, one over S plus one of her K. If we divide both sides by two e after adding it to both sides, this is equal to one over e plus one over to exactly as we wanted to prove for Part B, now, in part, C were asked to find all the integral solutions of the equation in part B that satisfy the geometric constraints of the problem. So first of all, let&#x27;s determine what the geometric constraints are. So we know a polygon must have at least three sides. So the number of sides for the facet number of edges K is going to be greater than or equal to three. That&#x27;s one constraint. We also know that at least three edges meat at each vertex. Since the facets are regular polygons so it follows. That s must be greater than or equal to three. But we cannot have that Both K and s are greater than three. Because if we suppose that que is greater than three and s is greater than three and since they&#x27;re integers that follows the K is greater than or equal to four and s is greater than or equal to four. So it follows that one over K would be less than or equal to 1/4 when over s would be less than or equal toe 1/4 and by part B, we would have that one of her s plus whenever kay but we have one over E plus 1/2 would be less than or equal to one half, so that one over E would be less than or equal to zero. This is, of course, a contradiction have to have a finite number of edges. The edges have to be non negative. So it follows that both K and s have to be three or no, they can&#x27;t both be greater than three. So I would say that K equals three or s equals three. Mm. Now consider the case where K equals three. Then we have that one of our s minus. When sixth equals one over E so s must be equal to three, four or five since s specific. Greater than or equal to three. And it can&#x27;t be six or anything greater than six. Now, if S is three we get that e is six if s is four. We get that he is 12 and if S is five. B a t e is 30. So when s is three and e is six. This corresponds. So it&#x27;s tetrahedron. When S is four and e is 12. This corresponds to a octahedron when S is five and e is 30. So we have 30 edges with five edges. 30 edges with five edges meeting at each vertex. Well, this is going to be an e Cosa hadron and you can look this up to see exactly what this looks like. If you need to now consider the case. Where s equals three And we have that one over. K minus 1/6 is one e. So we have that once again cake and be 345 And we have the e once again will be 6. 12 or 18 depending on the value of K started 6. 12 or 30 depending on the value of K. Now in the case or cake was three and e equals six. Well, k equals three tells us that there are going to be see is the number of edges for each facet. It facet has three edges. We have that at each vertex three edges meat and that there are six edges total. This is going to be a tetrahedron priority saw before. Now consider the case where K is for and he is 12 and ss three. So we have three edges meeting at each vertex we have Each facet has four edges and there are 12 edges. This is going to be a cube And now consider when s three k is five is 30. This means we have three inches meeting each Vertex. We have five edges on each facet and we have 30 edges total. Well, this is going to be the Dodecahedron. And once again, if you need to see what this looks like, you can look it up on Google. And so we have found all the integral solutions to the equation from Part B. And we have shown that there are only five regular Paula Hydra. There are the Tetrahedron, the Octahedron, the I Cosa Hadron, the Cube and the Dodecahedron. This is different from the number of regular polygons. Because there are an infinite number of regular polygons. There&#x27;s one for each natural number.

Vikash Ranjan · Answer

and the ghost unit is given. Basked, What is this? Plus faces is equal to. Is this glass to therefore? What? This is Maya&#x27;s age class. Face is a quality too. This is the important formula on this formula. But, sir, this fight the problem given in the question Now the answer to the question is, Do they had done? Let me draw the figure. The people can be drawn like this, Right? Is that, uh what this is? Okay, all this effort, this is no leveraging the color This Okay? This&#x27;d is the signs where no, we can see the number off. What? This is it You want to 20 Number off. It just is equal to 30 on DDE faces. Is it? What did? Well, you can see from the figure itself? No, The do deca drone has vortices gente is this tactic and faces that answer

Chris Trentman · Answer

were given statements from rest to determine if they are true or false and to explain so in part a or given the statement that Upali Tope is the convex hole of a finite set of points, this is true. Polito is the Comdex hole of a finite set of points. In fact, this is true by definition, a bit Polly tube Now in part B statement is let P being extreme point of a convex set s if you and the lying this convicts at S and P lives on the line segment between U and V and P is not equal to you. Thampy must be equal to V. This statement is true. So to see why consider the definition of an extreme point. So first of all, let s be a convicts set a point p in the convicts. That s is an extreme point of s. If P is not in the interior of any line segment that lies in s. So we&#x27;re told that p lies on a line segment which is inside s since U and V are both in s and this is a complex set and we&#x27;re told that P is not the end point you. Well, if he was not the endpoint V then P would lie on the interior of a line segment that goes through s and so P would not be an extreme point to be a contradiction. So this is also true by definition, now, in part, C statement is, if s is a non empty convex subset of r n. Then s is the convex hull of its profile. This is false. See why consider the&#x27;re, um 15. This says that if S is a non empty, compact compact set, then s is the convex whole of its profile. So this would only be true will not only be true, but it would be true if s were also compact. So here a simple counter example will be, say, a line segment. Hmm, not a line segment, but a line in our n. Well, a line is a non empty convex subset of our end since every line segment on online lies on the line. So it lies in the set. However, the profile of a line is the empty set. So it follows that the convex whole of its profile is also empty. So s is not the convex all of its profile. There are other counter examples you give for this question. Now, in part D were given the statement. He four dimensional simplex has exactly five facets, each of which is a three dimensional tetrahedron. This statement is true. In fact, this was discussed in the book after figure seven. So we see that figure seven the four dimensional simplex projected onto our to and we can see just by looking they&#x27;re going to be five Tetra, he drel facets. So there are There&#x27;s a facet for every vertex in the four dimensional simplex, there are five courtesies.

Chris Trentman · Answer

in part a were asked to find a formula for the number of Vergis is of an N pyramid in terms of the number of J faces of and minus one. Polito que sir Call it the and Pyramid is formed by taking the conduct hole of a N minus one Polito que and a point X, which is not in a fine space of cute. So in part, a pretty clear that f zero of P v n This is the number of Vergis is we&#x27;re really only adding when we&#x27;re Vertex, which is that vortex outside the Afghans base of Q. This is going to be as zero of Q Plus one now in Part B, were asked to find the number of K faces of an end pyramid where K ranges from one all the way up to end minus two. So consider the number of two faces or the number of square faces. I guess you could start with number of one faces thes air number of edges of A N pyramid. Well, we have all the edges from Polito P Q. And then, in addition to that, or each Vertex Nepali Tokyu. The point. Add exactly one mawr edge in the and pyramid, so it follows that F one of PV en is equal to F one of Q plus F zero of Q. Now consider the number of square faces of P to the end. Well, this is going to be the number of square faces in the Polito Que, in addition to that for each edge in the Polly took. Q. We&#x27;re going to be adding exactly one more square face in P the end. So the edge in that third point outside the Afghan space of Q forms a new square face. So we have that F two of PDM is equal to F one of Q I&#x27;m sorry, f two of Q Plus F one of cute, and in fact, you might see a pattern here. So you might guess that in general, SK of P to the end is going to be f k sq plus F K minus one of cute. And to see why, well, we have that the number of K faces of P to the end. This is going to be We have all the K faces from probably to a Q and then for each of the K minus one faces of Q. The addition of the new point and p the N leads to exactly one more que face in PDM. So we have one more K face for each K minus one face in Q. So the total number of K Faces and Pugh the end is going to be f k of Q plus F K minus one of Q then and this only goes from K equals 12 k equals end minus two. And this makes sense because trying to find the number of end minus one faces of an N minus one dimensional Polito pop doesn&#x27;t really work. Okay, now, in part C were asked to find the number of n minus one dimensional facets of P to the ends. This is what I was referring to in previous part. So f n minus one of P to the end. So think about the polito que that were given Now que has no end minus one faces, but it does have n minus two faces, and we know that for each end, minus two face of Q with additional point that we add in p v n, we form an n minus one face in ppm. So you know the number of n minus one faces impede the end is at least going to be the number of end minus two faces in Q. So this is equal to end minus F of end minus two of Q. But then we also have to consider that Q itself in P d. N. This copy of Q is going to be a n minus one face of Peter the End. So you have to add one more. And this is the total number of M minus one faces of PGN. It&#x27;s number of end minus two faces of Q Plus one.

Chris Trentman · Answer

We&#x27;re told that if we&#x27;re given a simple polygon with Enver theses listed so that consecutive Verdecia Zehr connected by an edge and first and last Vergis ease by an edge. And we&#x27;re told that the definition of an ear is when the Vertex is such that the line segment kicking that to Vergis he&#x27;s adjacent to it is an interior diagonal of this simple polygon. And we have the two years are called non overlapping. If the interiors of the triangles with the Vertex V i and it&#x27;s two adjacent courtesies and VJ it&#x27;s to reason Vergis, ease, Don&#x27;t Intersect were asked to prove that every simple polygon with at least four various years has at least two non overlapping years. So to prove this statement first, understand the statement is a simple polygon. P, with Enver theses has at least two non overlapping years on the basis Step is one and is equal before. So in this case, let P be a simple polygon with four burgess ease, B one, b two, b three and before and therefore follows that he is a quadrilateral. Every vertex of a quadrilateral isn&#x27;t here, along with its two adjacent Vergis ease you know that? V one t three r tonight overlapping because the trying with Verdecia zvi 12 and four in the tranquil Vergis ease be three before the to I only have the dying will be to before in a common and so not anything of the interior. So this proves that P four is true. The induction step. I suppose that p for all the way through PK is true for some K greater than or equal four. You want to show that PK plus one is also true simple, Paula, Government and Ferguson said, keep Cavor diseases at least two non ever Latin years to prove that peak of this one is true. So let P be polygon with K plus one Vergis ease and let DB a diagonal of P with endpoints B, i and d j so d divides P into to Paul eons p one and p two with less vergis ease each and then because P four thorpey care True, it follows that p one and P two have at least to non overlapping years, and then since V i N. B. J cannot be a pair of non overlapping years, since they are adjacent. At least one of the non overlapping years except P one is also a near of P and at least one of the non ever 11 years. Why of P one is also in your a p. Sorry, this should be non overlapping year and we have that at least one of the non overlapping years. Why of P two is also in non overlapping here. P. And this implies that the Vergis he&#x27;s X and y art years of P, while X and I need to be not overlapping years of p by. So I should just say, instead of overlapping years, easier just years just in year p and this is and year of you So x and y our years a p while X and Y are non overlapping years of p and this is because X and y and that two adjacent Vergis ease Rex and why are in a different polygon. And so we&#x27;ve shown that p k plus one is true. And so by strong induction he follows, that&#x27;s PN is true pearl and greater than or equal to four

Kara Merfeld · Answer

right. Doral Gate employees dream about trying along cup triangles on bottom and then connected like this. Right? You calculate number. Found the receives, faces and edges. So start with poverty&#x27;s 0123456 for me. He wants it. No help Edges. We have 123 456 789 So Adria&#x27;s called nine. No, your faces one Do free 45 So Walt five. So Oilers formula V minus B plus f equals two. We&#x27;ll check six minus nine. What&#x27;s five. You, too, Um, in the truth.

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David C. Lay, Steven R. Lay, Judi J. McDonaldLinear Algebra and Its Applications

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David C. Lay, Steven R. Lay, Judi J. McDonald

Linear Algebra and Its Applications