Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1989 AIME Problems/Problem 7"

m (→‎Solution)
m (→‎Solution 3)
 
(7 intermediate revisions by 4 users not shown)
Line 2: Line 2:
 
If the integer <math>k</math> is added to each of the numbers <math>36</math>, <math>300</math>, and <math>596</math>, one obtains the squares of three consecutive terms of an arithmetic series. Find <math>k</math>.
 
If the integer <math>k</math> is added to each of the numbers <math>36</math>, <math>300</math>, and <math>596</math>, one obtains the squares of three consecutive terms of an arithmetic series. Find <math>k</math>.
  
−
== Solution ==
+
== Solution 1==
 
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>.
 
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>.
  
Line 8: Line 8:
  
 
Subtracting the first equation from the second, we get <math>2d^2 = 32</math>, so <math>d = 4</math>. Substituting backwards yields that <math>a = 31</math> and <math>k = \boxed{925}</math>.
 
Subtracting the first equation from the second, we get <math>2d^2 = 32</math>, so <math>d = 4</math>. Substituting backwards yields that <math>a = 31</math> and <math>k = \boxed{925}</math>.
 +
 +
==Solution 2 (Straighforward, but has big numbers)==
 +
Since terms in an arithmetic progression have constant differences,
 +
<cmath>\sqrt{300+k}-\sqrt{36+k}=\sqrt{596+k}-\sqrt{300+k}</cmath>
 +
<cmath>\implies 2\sqrt{300+k} = \sqrt{596+k}+\sqrt{36+k}</cmath>
 +
<cmath>\implies 4k+1200=596+k+36+k+2\sqrt{(596+k)(36+k)}</cmath>
 +
<cmath>\implies 2k+568=2\sqrt{(596+k)(36+k)}</cmath>
 +
<cmath>\implies k+284=\sqrt{(596+k)(36+k)}</cmath>
 +
<cmath>\implies k^2+568k+80656=k^2+632k+21456</cmath>
 +
<cmath>\implies 568k+80656=632k+21456</cmath>
 +
<cmath>\implies 64k = 59200</cmath>
 +
<cmath>\implies k = \boxed{925}</cmath>
 +
 +
== Solution 3 ==
 +
Let the arithmetic sequence be <math>a-d</math>, <math>a</math>, and <math>a+d</math>. Then <math>(a+d)^2-a^2 = 296</math>, but using the difference of squares, <math>d(2a+d)=296</math>. Also, <math>a^2-(a-d)^2 = 264</math>, and using the difference of squares we get <math>d(2a-d) = 264</math>. Subtracting both equations gives <math>2d^2 = 32</math>, <math>d = 4</math>, and <math>a = 35</math>. Since <math>a = 35</math>, <math>a^2 = 1225 = 300+k</math> and <math>k = \boxed{925}</math>.
 +
 +
~~Disphenoid_lover
 +
 +
== Video Solution by OmegaLearn ==
 +
https://youtu.be/qL0OOYZiaqA?t=251
 +
 +
~ pi_is_3.14
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1989|num-b=6|num-a=8}}
 
{{AIME box|year=1989|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}
 +
[[Category: Introductory Algebra Problems]]

Latest revision as of 14:50, 12 September 2024

Problem

If the integer $k$ is added to each of the numbers $36$, $300$, and $596$, one obtains the squares of three consecutive terms of an arithmetic series. Find $k$.

Solution 1

Call the terms of the arithmetic progression $a,\ a + d,\ a + 2d$, making their squares $a^2,\ a^2 + 2ad + d^2,\ a^2 + 4ad + 4d^2$.

We know that $a^2 = 36 + k$ and $(a + d)^2 = 300 + k$, and subtracting these two we get $264 = 2ad + d^2$ (1). Similarly, using $(a + d)^2 = 300 + k$ and $(a + 2d)^2 = 596 + k$, subtraction yields $296 = 2ad + 3d^2$ (2).

Subtracting the first equation from the second, we get $2d^2 = 32$, so $d = 4$. Substituting backwards yields that $a = 31$ and $k = \boxed{925}$.

Solution 2 (Straighforward, but has big numbers)

Since terms in an arithmetic progression have constant differences, \[\sqrt{300+k}-\sqrt{36+k}=\sqrt{596+k}-\sqrt{300+k}\] \[\implies 2\sqrt{300+k} = \sqrt{596+k}+\sqrt{36+k}\] \[\implies 4k+1200=596+k+36+k+2\sqrt{(596+k)(36+k)}\] \[\implies 2k+568=2\sqrt{(596+k)(36+k)}\] \[\implies k+284=\sqrt{(596+k)(36+k)}\] \[\implies k^2+568k+80656=k^2+632k+21456\] \[\implies 568k+80656=632k+21456\] \[\implies 64k = 59200\] \[\implies k = \boxed{925}\]

Solution 3

Let the arithmetic sequence be $a-d$, $a$, and $a+d$. Then $(a+d)^2-a^2 = 296$, but using the difference of squares, $d(2a+d)=296$. Also, $a^2-(a-d)^2 = 264$, and using the difference of squares we get $d(2a-d) = 264$. Subtracting both equations gives $2d^2 = 32$, $d = 4$, and $a = 35$. Since $a = 35$, $a^2 = 1225 = 300+k$ and $k = \boxed{925}$.

~~Disphenoid_lover

Video Solution by OmegaLearn

https://youtu.be/qL0OOYZiaqA?t=251

~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_7&oldid=227050"