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Answer
is:30
METHOD
1: Strategy: Count in an organized
way.
One vertex is on one line and the other two are on the other
line.
Suppose
one vertex is on line 1 and two vertices are on line 2. There
are 6 such triangles using vertex P: PST, PSU, PSV, PTU, PTV,
and PUV. Likewise, there are 6 such triangles using vertex Q
and 6 more using vertex R, for a total of 18.
Now
suppose one vertex is on line 2 and two vertices are on line
1. There are 3 such triangles using vertex S: SPQ, SPR, and
SQR. Likewise there are 3 such triangles using vertex T, 3 more
using vertex U, and 3 more using vertex V, for a total of 12.
In all, 30 triangles can be formed using any three of the points
as vertices.
METHOD
2: Strategy (for more advanced students):
Use combinatorics.
For each of the 3 points on line 1, there are 4C2 = 6 pairs
of points on line 1. This gives us 18 triangles. For each of
the 4 points on line 2, there are 3C2 = 3 pairs of points on
line 1. This gives us 12 more triangles. Thus, a total of 30
triangles can be formed.
FOLLOW-UPS:
(1) Suppose all 7 points are on a circle. How many triangles
can be formed using any 3 of the points as vertices? [35] (2)
See pages 111-114 in Creative Problem Solving in School Mathematics,
2nd Edition.
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