Difference between revisions of "1993 AHSME Problems/Problem 28"
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== Solution == | == Solution == | ||
| − | <math>\fbox{D}</math> | + | The vertices of the triangles are limited to a <math>4\times4</math> grid, with <math>16</math> points total. Every triangle is determined by <math>3</math> points chosen from these <math>16</math> for a total of <math>\binom{16}{3}=560</math>. However, triangles formed by collinear points do not have positive area. For each column or row, there are <math>\binom{4}{3}=4</math> such degenerate triangles. There are <math>8</math> total columns and rows, contributing <math>32</math> invalid triangles. There are also <math>4</math> for both of the diagonals and <math>1</math> for each of the <math>4</math> shorter diagonals. There are a total of <math>32+8+4=44</math> invalid triangles counted in the <math>560</math>, so the answer is <math>560-44=516</math> or answer choice <math>\fbox{D}</math>. |
== See also == | == See also == | ||
Latest revision as of 14:00, 1 August 2016
Problem
How many triangles with positive area are there whose vertices are points in the 
-plane whose coordinates are integers 
satisfying 
and 
?
Solution
The vertices of the triangles are limited to a 
grid, with 
points total. Every triangle is determined by 
points chosen from these for a total of 
. However, triangles formed by collinear points do not have positive area. For each column or row, there are 
such degenerate triangles. There are 
total columns and rows, contributing 
invalid triangles. There are also 
for both of the diagonals and 
for each of the shorter diagonals. There are a total of 
invalid triangles counted in the 
, so the answer is 
or answer choice 
.
See also
| 1993 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 27 |
Followed by Problem 29 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
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