Today I took a book with English grammar which I used at school just to refresh my knowledge. And I found a home task in geometry which was given to us by our mathematics teacher Loboda Oleksiy Anatoliyovych at 10th grade. Since it is very interesting and dear for me in some way I've decided to translate it into English and post it here for everybody interested and for me to remember and not to loose. Here it is.
Home Task 11, geometry., 10th grade, to 04.12.00. Vectors
1. The points A,B,C and D are given. Points M and N - are the centers of the segments AB and CD, the point O is a center of the segment MN ( if a segment degenerates into a point then the point is the center). Prove that for any point S the following equation is valid:
SA + SB + SC + SD = 4SO.
2. The points M and N are the centers of the edges AB and A1D1 of the parallelepiped ABCDA1B1C1D1. The plane CMN intersects
the lines B1C1 and DB1 respectively at the points P and Q. Express the vectors AP and AQ in terms of the vectors AB = a, AD = b, AA1 = c.
(My answer: AP = a - 1/2 b + c; AQ = ( 4a + 5b + 4c ) / 9)
3. Different points A, B, C and D are given. The points M and N are taken at AC and BD respectively and in a way that the following relations are true: AM = λ⋅AC, BN = λ⋅BD. Prove that the vectors AB, CD and MN are coplanar and express MN in terms of AB and CD.
(My answer: MN = λ⋅CD + (1-λ)⋅AB )
4. Different points A, B, C and D are given. The points M and N are the centers of the segments BC and AD respectively. Prove that if the equation MN = 0.5⋅(AB + CD) is valid, then AB || CD.
5. The skew lines a and b are given and respectively the points A1, A2, A3 and B1, B2, B3 are at the lines (A2 is between A1 and A3, B2 is between B1 and B3 ). And also given that A1A2:A2A3 = B1B2:B2B3. Prove that the centers of the segments A1B1, A2B2, A3B3 belong to the same line.
6. ABCDA1B1C1D1 is a parallelepiped. Let's draw a line which intersects the lines AA1, BC and C1D1 respectively at points M, N and P in a way that MN:MP = 2. What is the value of the relation BN:BC (find all solutions)?
(My answer: BN:BC = 2, could not find any additional cases yet).
6H (Example of how a problem difficulty can increase almost to infinity by changing one word:), I don't think this problem has sufficient data in the statement). ABCDA1B1C1D1 is a parallelepiped. Let's draw a plane which intersects the lines AA1, BC and C1D1 respectively at points M, N and P in a way that MN:MP = 2. What is the value of the relation BN:BC (find all solutions)?
OK, it took a little longer than I thought, so the rest will appear here later ... I hope soon.
