
C D A B
... positive integer solution of 123b 2014a = 1 with b < 2014. Suppose that there is another solution 123b0 2014a0 = 1. Then subtracting these equations we would get 123(b b0 ) = 2014(a0 a). Thus 123(b b0 ) is divisible by 2014, and since 123 is relatively prime to 2014, we get that 2014 divides b b0 . ...
... positive integer solution of 123b 2014a = 1 with b < 2014. Suppose that there is another solution 123b0 2014a0 = 1. Then subtracting these equations we would get 123(b b0 ) = 2014(a0 a). Thus 123(b b0 ) is divisible by 2014, and since 123 is relatively prime to 2014, we get that 2014 divides b b0 . ...
Euclidean Geometry
... This project introdeced us the theorems about cyclic quadrilaterals. According to these three theorems, we could find out some interesting results about the cyclic quadrilaterals. And the key point of all the proofs is based on the attributes of circumscribed triangles: If two triangles have a same ...
... This project introdeced us the theorems about cyclic quadrilaterals. According to these three theorems, we could find out some interesting results about the cyclic quadrilaterals. And the key point of all the proofs is based on the attributes of circumscribed triangles: If two triangles have a same ...
Geometry - Hamilton School District
... solving ability, using a set standard of materials to reason through, and then logically find step by step solutions to more complex exercises. Students will use formulas for problem solving in plane geometrical figures. 2. COURSE OBJECTIVES: The objective of this class is for students to discover a ...
... solving ability, using a set standard of materials to reason through, and then logically find step by step solutions to more complex exercises. Students will use formulas for problem solving in plane geometrical figures. 2. COURSE OBJECTIVES: The objective of this class is for students to discover a ...