Amc 10 a amc 12 a problem 21 we divide into cases

2010 AMC 10 A, Problem #21 —
2010 AMC 12 A, Problem #21 —
“We divide into cases according to the three possible values of n ,
namely n = 1 , 2 , or 3 .”
this case. The total number of solutions for all cases is 4.
Difficulty: Hard
NCTM Standard: Number and Operations Standard for Grades 9–12: understand meanings of
operations and how they relate to one another.

Three distinct vertices are selected at random from a
regular 14-sided polygon (tetradecagon) and used as
vertices to form a triangle. What is the probability
that the triangle is a right triangle?
13
2010 AMC 10 A, Problem #22 —
“Circumscribe the regular tetradecagon with a circle.
The triangle will be a right triangle if and only if two
of the vertices are antipodal points of the circle.”

Triangle ABC has vertices at A (0 , 0) , B (8 , 14) and
C (20 , 0) . The line y = mx divides △ ABC into two
triangles of equal area. What is m ?
2010 AMC 10 A, Problem #23 —
“The line y = mx passes through vertex A .”