Difference between revisions of "1993 AHSME Problems/Problem 28"
(→Solution) |
|||
(One intermediate revision by one other user not shown) | |||
Line 1: | Line 1: | ||
− | How many triangles | + | == Problem == |
+ | How many triangles with positive area are there whose vertices are points in the <math>xy</math>-plane whose coordinates are integers <math>(x,y)</math> satisfying <math>1\le x\le 4</math> and <math>1\le y\le 4</math>? | ||
+ | |||
+ | <math>\text{(A) } 496\quad | ||
+ | \text{(B) } 500\quad | ||
+ | \text{(C) } 512\quad | ||
+ | \text{(D) } 516\quad | ||
+ | \text{(E) } 560</math> | ||
+ | |||
+ | == Solution == | ||
+ | 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 == | ||
+ | {{AHSME box|year=1993|num-b=27|num-a=29}} | ||
+ | |||
+ | [[Category: Intermediate Combinatorics Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.