Math 451: Euclidean and Non-Euclidean Geometry MWF 3pm, Gasson 204 Homework 3 Solutions

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1 Math 451: Euclidea ad No-Euclidea Geometry MWF 3pm, Gasso 204 Homework 3 Solutios Exercises from 1.4 ad 1.5 of the otes: 4.3, 4.10, 4.12, 4.14, 4.15, 5.3, 5.4, 5.5 Exercise 4.3. Explai why Hp, q) = {x R 2 x p) q p) < q p 1 q p }, ad use this to give a algebraic proof that Hp, q) is covex, by showig explicitly that if x, y Hp, q) ad t [0, 1], the 2 tx + 1 t)y Hp, q) as well. Solutio. The set Hp, q) is the set of poits of R 2 that are i the compoet of the complemet of the perpedicular bisector of the segmet pq cotaiig p. This is precisely the set of poits x so that the projectio of x to the lie through p ad q lies closer to p tha to q alog this lie. The set of poits alog this lie closer to p tha to q is just {p + tq p) t < 1/2}. The projectio of x to this lie ca be doe by addig p to the projectio of the vector x p to q p, i.e. p + proj q p x p). Recallig projectio of vectors see Exercise 1.4), we have proj q p x p) = x p) q p) q p 2 q p), so that the coditio x Hp, q) is equivalet to x p) q p) q p 2 < 1 2. Multiplyig both sides of this iequality by q p we obtai the characterizatio { Hp, q) = x R 2 x p) q p) < 1 } q p 2 q p. Now suppose that x, y Hp, q), so that x p) q p) q p 2 < 1 2 ad 1 y p) q p) q p 2 < 1 2.

2 2 we have tx + 1 t)y p) q p) tx + 1 t)y t + 1 t)p) q p) = q p 2 q p 2 tx p) + 1 t)y p)) q p) = q p 2 tx p) q p) + 1 t)y p) q p) = q p 2 x p) q p) = t ) < t t) q p 2 ) t) = Thus tx + 1 t)y Hp, q), so that Hp, q) is covex. y p) q p) q p 2 Exercise Caratheodory s Theorem i R 2 ). Show that if X R 2, the every poit i hullx) is a covex combiatio of three poits i X. Hit: Argue geometrically that ay poit i the covex hull of four poits i R 2 lies i the covex hull of three of the poits, the use iductio. Solutio. We will show by iductio o k that a covex combiatio of k poits from X is is actually a covex combiatio of three poits from X. By Theorem 4.8, the covex hull hullx) is the set of all covex combiatios of poits i X, so we will have show that every poit of hullx) is the covex combiatio of three poits from X. We ivestigate the base case k = 4. Note that we could more easily check the case k = 3 a tautology), but the case k = 4 will be ecessary below i applyig iductio. So we suppose that x is a covex combiatio of four poits from X. By Theorem 4.8, x is therefore i the covex hull of these four poits, which is either a lie segmet, triagle, or a covex quadrilateral by Problem A4. I the first case, the lie segmet is the set of covex combiatios of the edpoits; i the secod case, a triagle is the covex hull of its vertices. I either of these cases by Theorem 4.8 x is a covex combiatio of three poits from X. I the last case, choose a triagulatio of the quadrilateral. Because x is i the quadrilateral, it is i oe of the triagles of the triagulatio, which agai implies that x is a covex combiatio of three poits from X. We suppose ow that ay covex combiatio of k poits from X is also a covex combiatio of three poits from X, ad let x X be

3 a covex combiatio of k + 1 poits from X, i.e. there exist poits x 1,..., x k+1 X ad umbers λ i 0 with λ λ k+1 = 1 satisfyig x = λ 1 x λ k+1 x k+1. Rewritig this slightly, we have ) λ1 x λ k x k x = λ λ k ) + λ k+1 x k+1. λ λ k The term i paretheses ca be rewritte λ 1 x λ k x k λ 1 λ k = x x k, λ λ k λ λ k λ λ k which is evidetly a covex combiatio of the poits x 1,..., x k X. By the iductive hypothesis, this covex combiatio of k poits ca be rewritte as a covex combiatio of three poits, i.e. there exist poits x 1, x 2, x 3 X ad o-egative umbers λ 1, λ 2, λ 3 whose sum is oe, ad so that λ 1 x λ k x k = λ λ λ 1x 1 + λ 2x 2 + λ 3x 3. k Lettig s = λ λ k, we have x = sλ 1x 1 + sλ 2x 2 + sλ 3x 3 + λ k+1 x k+1, a covex combiatio of the four poits x 1, x 2, x 3, x k+1 X. As we demostrated i the base case above, this implies that x is the covex combiatio of three poits of X. Alterate Solutio. Pick x hullx). By Theorem 4.8, x is give by a covex combiatio of some poits x 1,..., x X, ad hece also i the covex hull hull{x 1,..., x }). By Problem A4, we have hull{x 1,..., x }) is a covex k-go, for some k. If k = 2 we are doe, as x is a covex combiatio of the two edpoits so trivially a covex combiatio of three poits of X). Choose a triagulatio of this polygo. As x is i this polygo, it lies i some triagle T of this triagulatio. Ivokig Theorem 4.8 oce more, T is the set of covex combiatios of its three vertices, so x is a covex combiatio of the three poits of X that form the vertices of T. Exercise If f : R R, let C f = {x 1, x 2 ) R 2 x2 fx 1 )}. Show that if f is covex, the C f is a covex subset of R 2. 3 Solutio. Suppose that x 1, y 1 ) ad x 2, y 2 ) are poits of C f. I order to show that C f is covex, we must show that 1 t)x 1, y 1 ) + tx 2, y 2 ) which we deote by p t ) is i C f, for ay t [0, 1]. Addig

4 4 the terms defiig p t, the x-coordiate is give by 1 t)x 1 + tx 2. By covexity of f ad the defiitio of C f we have f1 t)x 1 + tx 2 ) 1 t)fx 1 ) + tfx 2 ) 1 t)y 1 + ty 2, which is the y-coordiate of p t. We coclude that p t C f t [0, 1]. Thus C f is covex. for each Exercise Suppose that f : R R is covex. If we have x 1,..., x R ad t 1,..., t 0, 1) with t t = 1, show that ) t i fx i ) f t i x i. If f is strictly covex, show that this ca be a equality oly if x 1 =... = x. Solutio. We proceed by iductio o. The base case = 2 is precisely the defiitio of covexity of f. Suppose that the statemet holds for, ad suppose that x 1,..., x +1 R ad t 1,..., t +1 0, 1) with t t +1 = 1. We rewrite +1 t i x i = s s x t ) s x + t +1 x +1, where s = t t. As f is covex, f s s x t ) ) s x + t +1 x +1 sf s x t ) s x + t +1 fx +1 ). Sice t 1 s t = 1, the iductive hypothesis applies, ad we have s f s x t ) s x t 1 s fx 1) t s fx ). Puttig these together, we coclude that +1 ) f t i x i s s fx 1) t ) s fx +1 ) +t +1 fx +1 ) = t i fx i ), as desired. If f is strictly covex ad equality holds, we demostrate by iductio agai that x 1 =... = x. The case = 2 is the defiitio of strict

5 covexity. The case with +1 poits proceeds similarly as above: Suppose that t 1 x t +1 x +1 is a covex combiatio of the poits x 1,..., x +1 satisfyig f +1 ) +1 t i x i = t i fx i ). Lettig s = t t ad usig the covexity of f we have +1 t i fx i ) = f s s x t ) ) s x + t +1 x +1 sf s x t ) s x + t +1 fx +1 ). Subtractig t +1 fx +1 ) from each side ad usig the covexity of f, we obtai t i fx i ) sf s x t ) s x t i s s fx i) = t i fx i ). The left ad right sides of this chai of iequalities are equal, so it implies that each iequality is actually a equality. Thus we have sf s x t ) s x t i = s s fx i). Dividig both sides by s ote that s 0 sice t i 0, 1)), we obtai the equality f s x t ) s x t i = s fx i). Sice t 1 s x t x s is a covex combiatio of x 1,..., x X, the iductive hypothesis applies, ad the equality above implies that x 1 =... = x. Returig to the origial equality, we obtai f sx 1 + t +1 x +1 ) = f +1 ) +1 t i x i = t i fx i ) = sfx 1 )+t +1 fx +1 ). The strict covexity of f implies that x 1 = x +1, so that x 1 =... = x +1, as desired. Exercise 4.15 Iequality of meas). If x 1,..., x R show that x x x 1... x, 5

6 6 with equality if ad oly if x 1 =... = x. Hit: use the previous exercise ad the fact that l is a decreasig, strictly covex fuctio. Solutio. Clearly, 1 x x is a covex combiatio of the poits x 1,..., x. Sice l is covex see *) below) Exercise 4.14 implies that we have 1 l x ) x 1 l x 1) l x ). We multiply by 1 ad use the properties of the logarithm to rewrite this, so that 1 l x ) x 1 l x 1... x = l x 1... x. As the expoetial fuctio e x is a icreasig fuctio sice it has positive derivative), a b implies that e a e b. Thus we have e l x x ) e l x 1... x. Sice l ad the expoetial fuctio are iverse of each other, this reduces to the desired iequality: x x x 1... x. I case we have the equality x x = x 1... x, takig l of both sides ad multiplyig by 1 we obtai 1 l x ) x = 1 l x 1) l x ). The strict covexity see *) below) of l via Exercise 4.14 implies that x 1 =... = x. *): Though we may take this as give i the problem, i fact the strict covexity of l follows from Exercise 4.13, sice we have for all x. l x) = 1 x ) = 1 x 2 > 0 Exercise 5.3. Verify that the legth of a lie segmet is the distace betwee its edpoits. Hit: Use the triagle iequality.

7 7 Solutio. Let γ : [0, 1] R be the liearly parameterized lie segmet from p to q, i.e. γt) = 1 t)p + tq. We will show that for ay partitio {0 = t 0 <... < t = 1} of [0, 1] the approximatio to the legth of γ satisfies 1 dγt i ), γt i+1 )) = dp, q) i=0 by iductio o the, the umber of pieces of the partitio. The statemet is obvious if = 1, but we will perform iductio startig with = 2 as we will oly apply the iductive step for values 2. So suppose that our partitio is give by 0 = t 0 < t 1 < t 2 = 1. The triagle iequality Corollary 1.6) states that dp, q) = dγt 0 ), γt 2 )) dγt 0 ), γt 1 )) + dγt 1 ), γt 2 )), with equality exactly whe γt 0 ), γt 1 ), γt 2 ) are colliear. Evidetly, they are i this case, so we obtai the desired equality. Now assume the statemet holds for partitios with pieces, for 2, ad suppose that {0 = t 0 <... < t +1 = 1} is a partitio of [0, 1]. Note that the partitio obtaied by droppig t is {0 = t 0 <... < t 1 < t +1 = 1}, which has pieces. The iductive hypothesis implies that dp, q) = dγt 0 ), γt 1 ))+...+dγt 2 ), γt 1 ))+dγt 1 ), γt +1 )). The triagle iequality oce more applies, with the poits γt 1 ),γt ), ad γt +1 ) colliear. Thus dγt 1 ), γt +1 )) = dγt 1 ), γt )) + dγt, t +1 )), ad combiig this with the previous gives dp, q) = dγt 0 ), γt 1 )) dγt 2 ), γt 1 )) + dγt 1 ), γt )) + dγt ), γt +1 )). This is exactly the equality for the legth approximatio we eeded. We ve thus demostrated that, for ay choice of partitio {0 = t 0 <... < t = 1} of [0, 1], we have 1 dp, q) = dγt i ), γt i+1 )). i=0 The legth of γ is the least upper boud of these approximatios, each of which is equal to dp, q), so the legth is also dp, q). Ay other upper boud for {dp, q)} would be greater tha dp, q), ad therefore ot least).

8 8 Exercise 5.4. Regular -gos ca be used as a piecewise liear approximatios of the uit circle, which we expect has legth 2π. Show that the perimeters of regular -gos whose vertices lie o the uit circle do ideed coverge to 2π as. Hit: use the law of cosies. You might also fid l Hôpital s rule useful. Solutio. Draw the radial segmets from the ceter of the uit circle to the vertices of the regular -go. This divides the -go ito triagles, each of which has sides give by oe edge of the -go ad two radii. By SSS) all of these triagles are cogruet, ad the agles at the ceter of the circle add to 2π. Sice there are of them, each has measure 2π/ radias. The law of cosies allows us to compute the o-radial side which is a side of the -go): Deote this legth by x. We have x 2 = cos 2π = 2 2 cos 2π, so that x = 2 2 cos 2π. Let s deote the perimeter of the -go by P ). Sice each side has legth 2 2 cos 2π, we have P ) = 2 2 cos 2π. Our goal is to compute lim P ). We use the Taylor series for cos x: so that cos x = 1 x2 2 + x4 4!..., P ) = 2 2 cos 2π ) = π/)2 + 2π/) ! 4π 2 = 2 16π = 4π 2 4π Note that the... terms above all have a deomiator with a positive power of, so it is ow clear that lim P ) = 4π 2 = 2π,

9 9 as desired. Exercise 5.5. Fid a formula for the legth of K see p. 24 for the costructio of K ), ad show that it goes to as icreases. Solutio. The costructio of the curves K is doe iductively; i order to obtai K +1 from K we take each of the lie segmets of K ad replace them by four lie segmets, where each of these four ew lie segmets is oe-third of the legth of the previous. Sice K +1 is obtaied by doig this to each of the lie segmets of K, the legth of K +1 is 4/3 of the legth of K. Moreover, the legth of K 1 is 1 by defiitio. This implies that the legth of K is 4/3) 1. This ca be see by iductio. It s evidetly true for = 1. Now the legth of K +1 is 4/3 that of K, which is 4/3) 1 by iductio. Thus the legth of K +1 is 4/3) 4/3) 1 = 4/3), completig the iductive step. Fially, we ca ow see that the legth of K goes to with : lim ) 1 4 =, 3 sice the commo ratio 4/3 is greater tha 1. Problem A5. Cosider the followig proof by iductio. Lemma. All horses have the same color. Proof. We prove this by iductio o, the umber of horses there are. The base case = 1 holds; if there were oe horse, it would be true that all horses have the same color. Suppose ow that the statemet if there are horses, the they all have the same color holds for the umber, ad cosider + 1 horses. List the horses i order, ad choose the first of them. Note that by iductio they all have the same color. Now choose the last of them, ad agai ote they have the same color. Because the first horse has the same color as a horse i the middle, ad a horse i the middle has the same color as the last horse, all of the + 1 horses have the same color. What s wrog with this proof?

10 10 Solutio. The iductive step oly applies for 2: Cosider the case = 1, i which we assume that ay set of horses of size oe all have the same color, ad the iductive step aims to show that ay set of two horses have the same color. Let s ame the horses {h, k}. It s true that the first of these the set {h}) all have the same color ad the last of these the set {k}) all have the same color by the the iductive step), but sice these sets do t overlap we caot coclude that {h, k} have the same color. Of course, this is a obvious statemet: We kow that there ca be o way to prove that ay pair of horses have the same color sice there exist a pair of horses of differet colors! Google horse images for more detail.)

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