Difference between revisions of "1989 AIME Problems/Problem 7"
m |
(solution, box) |
||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | + | Call the terms of the [[arithmetic progression]] <math>a,\ a + d,\ a + 2d</math>, making their squares <math>a^2,\ a^2 + 2ad + d^2,\ a^2 + 4ad + 4d^2</math>. | |
| + | |||
| + | We know that <math>\displaystyle a^2 = 36 + k</math> and <math>\displaystyle (a + d)^2 = 300 + k</math>, and subtracting these two we get <math>\displaystyle 264 = 2ad + d^2</math> (1). Similarly, using <math>\displaystyle (a + d)^2 = 300 + k</math> and <math>\displaystyle (a + 2d)^2 = 596 + k</math>, subtraction yields <math>\displaystyle 296 = 2ad + 3d^2</math> (2). | ||
| + | |||
| + | Subtracting the first equation from the second, we get <math>\displaystyle 2d^2 = 32</math>, so <math>\displaystyle d = 4</math>. Substituting backwards yields that <math>\displaystyle a = 31</math> and <math>\displaystyle k = 925</math>. | ||
== See also == | == See also == | ||
| − | + | {{AIME box|year=1989|num-b=6|num-a=8}} | |
| − | |||
| − | |||
Revision as of 16:47, 28 February 2007
Problem
If the integer 
is added to each of the numbers 
, 
, and 
, one obtains the squares of three consecutive terms of an arithmetic series. Find .
Solution
Call the terms of the arithmetic progression 
, making their squares 
.
We know that 
and 
, and subtracting these two we get 
(1). Similarly, using and 
, subtraction yields 
(2).
Subtracting the first equation from the second, we get 
, so 
. Substituting backwards yields that 
and 
.
See also
| 1989 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
