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 ). L R Proof. Clearly R  L. Wnop* X  R  X  L. B =  b i A i,  b i = 1 (B  Aff (A 1, A 2, …, A k )) X = b B +  a n A n, b +  a n =1 (X  Aff (B,A 1, A 2, …, A k )) = b  b i A i +  a n A n =  (b b i + a i ) A i,  (b b i + a i ) = b  b i +  a i = b +  a i = 1  X  L. Remark. In the convex case is only additional b, b i,, a n  0. * We need only to prove

2. Exercise 2-2’ Part 1 Prove that Aff /Conv(A 1, A 2, …, A k ) contains the line AB / segment [AB] with each pair of its points A,B. Proof. A,B  Aff / Conv (A 1, A 2, …, A k )  A =  a n A n, B =  b n A n,  a n =1,  b n =1, a n, b m  0 in conv. case. X  AB  [AB] in conv. case. X =  A + (1 -  ) B = 1    0 in conv. case. =  (  a n A n ) + (1 -  )  b n A n =  (  a n + (1 -  ) b n ) A n,  (  a n + (1 -  ) b n ) =  (  a n ) + (1 -  )  b n =  + (1 -  ) =1, In conv. case we only assert that  a n + (1 -  ) b n  0.

3. Exercise 2-2’ Part 2. Prove that A = Aff /Conv(A 1, A 2, …, A k ) is the minimal set containing {A 1, A 2, …, 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 B has this property, and X =  a n A n, (a n  0 in the convex case),  a n = 1, is a point from A. Wnop: X  B. Choose some i such that a i  0,1. Then y =  n  i a n  0,1 and (1) Aff /Conv(A 1, …, A i-1, A i+1 …, A k )  B by induction. (2) Y =  n  i (a n / y) A n  Aff /Conv(A 1, …, A i-1, A i+1 …, A k )  B since  n  i a n / y = (  n  i a n ) / y =1, and a n / y ≥ 0 in the convex case. (3) Hence B, acc. to the assumption, contains line A i Y ( segment [A i Y]). (4) X is a point from the line YA i (segment [YA i ]) because: yY + (1- y) A i = y  n  i (a n / y) A n + (1-  n  i a n ) A i =  n  i a n A n + a i A i = X (in convex case we check y=  n  i a n > 0 and 1 - y = a i > 0). Finally, (3) and (4) imply X  B. (1)

4. Exercise 3-3’ Prove that Aff / Conv (A 1,, A 2, …, A k ) is ndependent on the transformation of coordinates, i.e. that (1) M Aff (A 1,,…, A k ) = Aff ( M (A 1 ),,…, M( A k )), and (2) O + Aff (A 1,,…, A k )= Aff (O + (A 1 ),,…, O + ( A k )) Proof 1. (1)X  L iff* X = M (  a n A n ) =  a n M (A n ) iff X  R.  a n =1 (a n  0 in the convex case) (2) X  L iff X = O +  a n A n = (  a n ) O +  a n A n =  a n O +  a n A n =  a n (O + A n ) iff X  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 4 If a convex set S contains the vertices A 1, A 2, …, A k of a polygon P=A 1 A 2 …A k, it contains the polygon P. Hint: Interior point property. Proof. (1) By the Exercise 2‘ is [ A 1 A 2 ], …, [A k-1 A k ]  S. Hence wnop that S contains the interior points of P. (2) An interior point Z of a polygon P = A 1 A 2 …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 [ A 1 A 2 ], …, [A k-1 A k ] in points X,Y which are in S (  1). Since Z  [XY] using 2‘ again we conclude Z  S. Z X Y 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 A has the desired property. Than in (2) we prove the minimallity. (1)Let M,N  A. Than M =  a n 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 C r }= {  (  e r + (1-  )f r ) C r }  A because  e r + (1-  )f r =  e r + (1-  )  f r =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 is a point of A. Wnop M  X. M =  a n 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))={  a n A n :  a n =1,(a n > 0),for all finite subsets A i  S}. def

7. Exercise 6-6’ Prove that: Aff/Conv*(A 1,A 2, …,A k ) = Aff/Conv(A 1,Aff (A 2, …, A k )) Prove. Since S = {A 1 }  Aff (A 2, …, A k )  {A 1,A 2, …, 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 (A 2, …, A k ) (1) or A 1 = B 1 and {B m, m  1}  Aff (A 2, …, A k ) (2). (1) Aff(A 1,A 2, …, A k )  Aff(A 2, …, A k ) = Aff(B 1,…,B r,A 2, …, A k )  Aff(B 1,…,B r ) and M  Aff(B 1,…,B r )  M  L. (2) M =  b m B m =b 1 A 1 +  m  1 b m B m = b 1 A 1 + (1-b 1 )  m  1 b m / (1-b 1 ) B m = b 1 A 1 + (1-b 1 )B, B =  m  1 b m / (1-b 1 ) B m. (1)  B  L (  m  1 b m / (1-b 1 ) = 1 easy to check)  M  L (since M is on the line A 1 B= {b 1 A 1 + (1-b 1 )B, b 1 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 {A 1, A 2,…, A k } is affinely independent iff the set of vectors {A 1 A 2 …, A 1 A k } is linearly independent. Lemma {A 1, A 2,…, A k } are affinely dependent iff  m a m A m =0 for some a m such that  m a m = 0 and not all a m are 0. Proof of Lemma. (  ) Suppose {A 1, A 2,…, A k } are affinely dependent. Than i.e. A i =  m  i a m A m,  m  i a m = 1   a m A m = 0, a i = -1,  m  i a m = 1. (  ) Suppose  a m A m =0 for some {a m } such that  a m = 0 and that a i  0. Hence, A i =  m  i –a m / a i A m. Using Lemma we get:  a m A m =0,  a m = 0 iff  a m A m -  a m A 1 =  a m A 1 A m = 0, what is a very well known characteristic of linear dependency for vectors.

9. Exercise 12 Polarity 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 X therefore holds: C= A + (1- ) B = ( a 1 + (1- )b 1, a 2 + (1- )b 2, a 3 + (1- )b 3 ),  R. After applying polarity: A   : a 1 X + a 2 Y = a 3 + Z, B   : b 1 X + b 2 Y = b 3 + Z, (1) C   : ( a 1 + (1- )b 1 )X + ( a 2 + (1- )b 2 )Y=( a 3 + (1- )b 3 ) + Z. (2) Now, (1) and (2) prove the statement since   {  + (1- ) ,  R}. Exercise 11 Polarity 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=(m a, m b, m c ) be a point of it. This simply means : a m a + b m b = c + m c (1) After applying polarity, one obtains:   (a,b,c), M  m a X + m b Y = m c + Z (2) Now, (1) and (2) prove the statement.

10. Exercise 13 Polarity maps points of P to the planes tangent to P. Proof. Let A be a point of P, i.e. let a 1 2 +a 2 2 = 2 a 3. (1) The tangent plane to z = f( x, y) at A has the equation: z - a 3 = f x (a 1,a 2 ) (x-a 1 ) + f y (a 1, a 2 ) (y-a 2 ). (2) Since P has the equation z = (x 2 +y 2 )/2, the tangent plane at A is z - a 3 = a 1 (x-a 1 ) + a 2 (y-a 2 ). (3) Wnop that (3) and f P (A): a 1 x + a 2 y = z + a 3 are the same. After the substitution - a 1 2 - a 2 2 = - 2 a 3 (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 = r 2 in  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 r 2 /2 to the plane f P (S), S=(a,b). (a) Let C(c 1,c 2 ) be a point of , i.e. let (c 1 -a) 2 + (c 2 -b) 2 = r 2, (1) i.e. let c 1 a + c 2 b - (c 1 2 + c 2 2 )/2 = (a 2 +b 2 - r 2 )/2. (2) Since (c 1,c 2,(c 1 2 + c 2 2 )/2) is exactly point  (C)=C, (2) implies that  (C) is in the fixed* plane X a + Y b - Z = (a 2 +b 2 - r 2 )/2. Another notation for it: Z = - X a - Y b + (a 2 +b 2 - r 2 )/2. (3) (b) For interior points (1) and (2) turn into inequalities respectively: c 1 a + c 2 b - (c 1 2 + c 2 2 )/2 > (a 2 +b 2 - r 2 )/2 (2’) i.e. for them is Z < - X a - Y b+(a 2 +b 2 - r 2 )/2. Comparing to (3) we prove (b). (c) Point S =  (S) has coordinates (a,b,(a 2 +b 2 )/2) and the tangent plane at S is X a +Y b = Z + (a 2 +b 2 )/2. The vertical projection of C onto it is C = (c 1,c 2, c 1 a + c 2 b - (a 1 2 + a 2 2 )/2 ). The z-distance of C to C is (c 1 2 + c 2 2 )/2 – (c 1 a + c 2 b - (a 1 2 + a 2 2 )/2) **= r 2 * S=(a,b) is the center of k; **see (2) ~~