Download this year's Mathcounts handbook here. (It's free.)
#5 1993 Mathcounts National Target : Find the probability that four randomly selected points on the geoboard below will be the vertices of a square? Express your answer as a common fraction.
#5 2004 AMC 10A: A set of three points is chosen randomly from the grid shown. Each three-point (same image as the below question) set has the same probability of being chosen. What is the probability that the points lie on the same straight line?
2007 Mathcounts Chapter Sprint #29 : The points of this 3-by-3 grid are equally spaced
horizontally and vertically. How many different sets of three points of this grid can be the three
vertices of an isosceles triangle?
Solution:
#5 National Target: There are 16C4 = \(\dfrac {16\times 15\times 14\times 13} {4\times 3\times 2\times 1}\)= 1820 ways to select 4 points on the geoboard.
There
are 3 x 3 = 9 one by one squares and 2 x 2 = 4 two by two squares and 1
x 1 = 1 three by three squares. (Do you see the pattern?)
There are 4 other squares that have side length of √ 2
and 2 other larger squares that have side length of √ 5.
9 + 4 + 1 + 4 + 2 = 20 and \(\dfrac {20} {1820}=\dfrac {1} {91}\)
#5: Solution:
AMC-10A: There are 9C3 = \(\dfrac {9\times 8\times 7} {3\times 2\times 1}\)= 84 ways to chose the three dots and 8 of the lines connecting the three dots will form straight lines. (Three verticals, three horizontals and two diagonals.) so
#29: Solution:
Use the length of the two congruent legs to solve this problem systematically.
There are 16 1 - 1 - \(\sqrt {2}\) isosceles triangles.
There are 8 \(\sqrt {2}\) by \(\sqrt {2}\) by 2 isosceles triangles. (See that ?)
There are 4 2 - 2 - \(2\sqrt {2}\) isosceles triangles.
There are 4 \(\sqrt {5}\) by \(\sqrt {5}\) by 2 isosceles triangles.
Finally, there are 4 \(\sqrt {5}\) by \(\sqrt {5}\) by \(\sqrt {2}\) isosceles triangles.
Add them up and the answer is 36.
