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Exercise 1- 1’

Exercise 1- 1’ Prove that if a point B belongs to the affine / convex hull Aff/Conv ( A 1 , A 2 , …, A k ) of points A 1 , A 2 ,…, A k , then: Aff/Conv ( A 1 , A 2 ,…, A k ) = Aff /Conv( B , A 1 ,…, A k ).

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Exercise 1- 1’

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  1. Exercise 1- 1’ Prove that if a point B belongs to the affine / convex hull Aff/Conv (A1, A2, …, Ak ) of points A1, A2,…, Ak , then: Aff/Conv (A1, A2,…, Ak) = Aff /Conv(B,A1,…, Ak). L R Proof. Clearly R  L. Wnop* X  R  X  L. B =  biAi ,  bi = 1 (B  Aff (A1, A2, …, A k )) X = b • B +  anAn, b +  an=1 (X  Aff (B,A1, A2, …, A k )) = b •  biAi +  anAn =  (b • bi +ai) Ai ,  (b • bi +ai) = b •  bi +  ai = b +  ai = 1  X  L. Remark. In the convex case is only additional b, bi,, an  0. * We need only to prove

  2. Exercise 2-2’ Part 1 Prove that Aff /Conv(A1, A2, …, Ak ) contains the line AB / segment [AB] with each pair of its points A,B. Proof. A,B  Aff / Conv (A1, A2, …, A k )  A =  anAn, B =  bnAn,  an =1,  bn =1 , an, bm  0 in conv. case. X  AB  [AB] in conv. case. X =  A + (1 - ) B = 1    0in conv. case. =  ( anAn) + (1 - )  bnAn=  ( an + (1 - ) bn) An,  ( an + (1 - ) bn) =  ( an ) + (1 - )  bn=  + (1 - ) =1, In conv. case we only assert that  an + (1 - ) bn  0.

  3. Exercise 2-2’ Part 2. Prove that A = Aff /Conv(A1, A2, …, A k ) is the minimal set containing {A1, A2, …, A k } and the line /segment through each pair of its points. Proof. The proof goes by induction. For k=2 the statement is obvious. We inductively assume that the statement holds for k  m and prove that it holds for k=m+1. Suppose set Bhas this property, and X =  anAn , (an  0 in the convex case),  an = 1, is a point from A.Wnop: X  B. Choose some i such that ai  0,1. Then y =  n i an  0,1 and (1) Aff /Conv(A1, … , Ai-1, Ai+1 …, A k )  Bby induction. (2) Y =  n i (an / y) An  Aff /Conv(A1, … , Ai-1, Ai+1 …, A k ) B since  n i an / y = ( n i an)/ y =1, and an / y ≥ 0 in the convex case. (3) Hence B, acc. to the assumption,contains line AiY (segment [Ai Y]). (4) X is a point from the line YAi(segment [YAi]) because: yY + (1- y) Ai= y  n i (an / y) An + (1-  n i an) Ai=  n i anAn + ai Ai = X (in convex case we check y=  n i an > 0 and 1 - y = ai > 0). Finally, (3) and (4) imply X  B. (1)

  4. Exercise 3-3’ • Prove that Aff / Conv (A1,, A2,…, Ak ) is ndependentonthe transformation of coordinates, i.e. that • (1) M •Aff (A1,,…, A k ) = Aff (M(A1),,…, M(A k )), and • (2) O + Aff (A1,,…, A k )= Aff (O +(A1),,…, O + (A k )) • Proof 1. • X  L iff* X = M • ( anAn) =  anM • (An) iff X  R. •  an=1 (an  0 in the convex case) • (2) X  L iff X = O +  anAn = ( an)O +  anAn =  an O +  anAn • =  an (O +An) iffX  R. * if and only if • Proof 2. • By Exercise 2-2’ We need only to remark that after a transformation of coordinates a line remains a line!

  5. Exercise 4If a convex set S contains the vertices A1, A2,…, A k of a polygon P=A1A2…A k , it contains the polygon P.Hint: Interior point property. Proof. (1) By the Exercise 2‘ is [A1A2],…, [A k-1A k]  S. Hence wnop that S contains the interior points of P . (2) An interior point Z of a polygon P = A1A2…A k has the property: Every halfline through it which contains no vertex of P intersects (polygonal line) P in ann odd number of points – consequently at least ones! We „draw“ a line through P which contains no vertex of P. Two halflines of it intersect segments P (  2) in [A1A2],…, [A k-1A k] in points X,Y which are in S (  1). Since Z [XY] using 2‘ again we conclude Z  S. X Y Z S

  6. Exercise 5-5’Prove that A= Aff (S) (Conv ( S))* is the smallest affine (convex) set containing S - the smallest set X which contains S and the line MN (segment [MN]) with each pair of points M,N  X . • Proof. • In 1 we first prove that Ahas the desired property. Than in (2) we prove the minimallity. • Let M,N A. Than M = an A n,N = b m B m for some finite {A n}, {B m} subsets of S. Set {A n}  {B m} is again some finite set {C r} which is a subset of S. Hence, M = e r C r,N = f r C r (some e r / f r are a r / b r,other are 0) and line MN = {M + (1-)N } ={ e r C r + (1-) f r Cr }= • { (e r + (1-)f r) Cr } A because e r + (1-)f r= e r + (1-)fr=1. • (In the convex case is 0    1, and the coefficients e r ,(1-)f r  0.) • (2) Let X is a set having desired property, and let M isa point of A. Wnop M  X . M = an A n for some finite {A i }S. Since S  X, we conclude {A i }  X. Now, Exercise 2-2’ Part2  M  Aff {A i } X. • * Aff (S) (Conv ( S))={ an A n: an =1,(an > 0),for all finite subsetsAi  S}. def

  7. Exercise 6-6’ Prove that: Aff/Conv*(A1,A2,…,A k) = Aff/Conv(A1,Aff (A2,…, A k )) Prove. Since S = {A1} Aff (A2,…, A k )  {A1,A2,…, A k }, R  L. Wnop M  R  M  L. Let hence M =  b m B m for some finite subset {B m} of S. Then {B m}  Aff (A2,…, A k ) (1) or A1 = B1 and {B m, m  1}  Aff (A2,…, A k ) (2). (1) Aff(A1,A2,…, A k )  Aff(A2,…, A k ) = Aff(B1,…,Br ,A2,…, A k )  Aff(B1,…,Br) and M  Aff(B1,…,Br)  M  L. (2) M =  b m B m=b1 A1 + m1 b m B m= b1 A1+ (1-b1) m1 b m / (1-b1) B m= b1 A1+ (1-b1)B, B = m1 b m / (1-b1) B m. (1)  B  L (m1 b m / (1-b1) = 1 easy to check)  M  L (since M is on the line A1B= {b1 A1+ (1-b1)B, b1 a real num.}). *In the convex case it is easy to check that all the coefficients are  0. Ex.1-1‘

  8. Exercise 7 A set {A1, A2,…, A k} is affinely independent iff the set of vectors {A1A2…, A1Ak}is linearly independent. Lemma {A1, A2,…, A k} are affinely dependentiffm amA m=0 for some am such that m am = 0 and not all am are 0. Proof of Lemma. () Suppose {A1, A2,…, A k} are affinely dependent. Than i.e. Ai= m i amA m, m i am = 1   amA m = 0, a i= -1, m i am = 1. () Suppose  amA m=0 for some {am } such that  am = 0 and that a i  0. Hence, Ai= m i –am / a iA m. Using Lemma we get: • amA m=0,  am = 0 iff  amA m-  am A1=  amA1A m = 0, what is a very well known characteristic of linear dependency for vectors.

  9. Exercise 11Polarity maps points of a plane  to the planes through the point f P (). Proof. Let a X + b Y = c + Z be the equation of a plane , and let M=(ma , mb , mc) be a point of it. This simply means : a ma + b m b = c + mc(1) After applying polarity, one obtains:   (a,b,c), M  maX + mbY = mc + Z (2) Now, (1) and (2) prove the statement. Exercise 12Polarity maps points of a line x to the planes through a line (def.) f P (x). Proof. Let X be the line AB. For arbitrary point C of Xtherefore holds: C=  A + (1-  ) B = (a1 + (1-  )b1,a2 + (1-  )b2,a3 + (1-  )b3),  R. Afterapplying polarity: A :a1 X + a2 Y = a3 + Z, B : b1 X + b2 Y = b3 + Z, (1) C  : (a1 + (1-  )b1)X + (a2 + (1-  )b2)Y=(a3 + (1-  )b3) + Z. (2) Now, (1) and (2) prove the statement since  {   + (1-  ) ,  R}.

  10. Exercise 13Polarity maps points of P to the planes tangent to P. Proof. Let A be a point of P, i.e. let a12 +a22= 2 a3. (1) The tangent plane to z = f( x , y) at A has the equation: z - a3 = f x (a1 ,a2 ) (x-a1) + f y (a1 , a2) (y-a2). (2) Since P has the equation z = (x2+y2)/2, the tangent plane at A is z - a3 = a1 (x-a1) + a2 (y-a2). (3) Wnop that (3) and f P (A): a1x + a2 y = z + a3 are the same. After the substitution - a12 - a22= - 2 a3 (from 1) on the right hand side of (3), the proof is done.

  11. Exercise 14a,b,c (a) Points of a circle k: (x-a)2 + (y-b)2 = r2in  are projected by  to the points of some plane . (b)  images of the interior points of k are bellow this plane. (c) C, C k, has z-distance r2/2 to the plane f P (S), S=(a ,b). (a) Let C(c1,c2) be a point of , i.e. let (c1-a)2 + (c2-b)2 = r2,(1) i.e. let c1a + c2b - (c12 + c22)/2 = (a2+b2 - r2)/2. (2) Since (c1,c2,(c12 + c22)/2) is exactly point (C)=C, (2) implies that (C) is in the fixed* plane X a + Y b - Z = (a2+b2 - r2)/2. Another notation for it: Z = - X a - Y b + (a2+b2 - r2)/2. (3) (b) For interior points (1) and (2) turn into inequalities <,> respectively: c1a + c2b - (c12 + c22)/2 > (a2+b2 - r2)/2 (2’) i.e. for them is Z < - X a - Y b+(a2+b2- r2)/2. Comparing to (3) we prove (b). (c) Point S =(S) has coordinates (a ,b,(a2+b2)/2) and the tangent plane at S is X a +Y b = Z + (a2+b2)/2. The vertical projection of C onto it is C = (c1,c2, c1a + c2b - (a12 + a22)/2 ). The z-distance of C to C is (c12 + c22)/2 – (c1a + c2b - (a12 + a22)/2) **= r2 *S=(a ,b) is the center of k; **see (2) ~ ~

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