Difference between revisions of "1987 AIME Problems/Problem 4"
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== Problem == | == Problem == | ||
+ | Find the [[area]] of the region enclosed by the [[graph]] of <math>|x-60|+|y|=\left|\frac{x}{4}\right|.</math> | ||
− | == Solution == | + | == Solution 1 == |
+ | [[Image:1987_AIME-4.png]] | ||
+ | |||
+ | Since <math>|y|</math> is [[nonnegative]], <math>\left|\frac{x}{4}\right| \ge |x - 60|</math>. Solving this gives us two equations: <math>\frac{x}{4} \ge x - 60\ \mathrm{and} \ -\frac{x}{4} \le x - 60</math>. Thus, <math>48 \le x \le 80</math>. The [[maximum]] and [[minimum]] y value is when <math>|x - 60| = 0</math>, which is when <math>x = 60</math> and <math>y = \pm 15</math>. Since the graph is [[symmetry|symmetric]] about the y-axis, we just need [[casework]] upon <math>x</math>. <math>\frac{x}{4} > 0</math>, so we break up the condition <math>|x-60|</math>: | ||
+ | |||
+ | *<math>x - 60 > 0</math>. Then <math>y = -\frac{3}{4}x+60</math>. | ||
+ | *<math>x - 60 < 0</math>. Then <math>y = \frac{5}{4}x-60</math>. | ||
+ | |||
+ | The area of the region enclosed by the graph is that of the quadrilateral defined by the points <math>(48,0),\ (60,15),\ (80,0), \ (60,-15)</math>. Breaking it up into triangles and solving or using the [[Shoelace Theorem]], we get <math>2 \cdot \frac{1}{2}(80 - 48)(15) = \boxed{480}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | Since <math>|y|</math> is the only present <math>y</math> "term" in this equation, we know that the area must be symmetrical about the x-axis. | ||
+ | |||
+ | We'll consider the area when <math>y>0</math> and we only consider the portion enclosed with <math>y=0</math>. Then, we'll double that area since the graph is symmetrical. | ||
+ | |||
+ | Now, let us remove the absolute values: | ||
+ | |||
+ | When <math>x\ge 60</math>: <math>x-60+y=0.25x</math>. This rearranges to <math>y=-0.75x+60</math>. | ||
+ | |||
+ | When <math>0\le x<60</math>: <math>60-x+y=0.25</math>. So <math>y=1.25x-60</math>. | ||
+ | |||
+ | When <math>x<0</math>: <math>60-x+y=-0.25x</math>. So <math>y=0.75x-60</math>. | ||
+ | |||
+ | By simple sketching, we see the shape that looks like the image in Solution 1 (graph it out and you'll see). We see that the partial area we seek in this part is the triangle with the vertices <math>(60,15)</math>, <math>(48,0)</math>, <math>(80,0)</math>. This triangle has an area of <math>(80-48)*15*0.5=240</math>. | ||
+ | |||
+ | Simply double the area and we get <math>\boxed{480}</math> as our final answer. | ||
+ | ~hastapasta | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1987|num-b=3|num-a=5}} | |
− | {{ | + | [[Category:Intermediate Algebra Problems]] |
+ | {{MAA Notice}} |
Latest revision as of 17:25, 13 February 2024
Contents
Problem
Find the area of the region enclosed by the graph of
Solution 1
Since is nonnegative,
. Solving this gives us two equations:
. Thus,
. The maximum and minimum y value is when
, which is when
and
. Since the graph is symmetric about the y-axis, we just need casework upon
.
, so we break up the condition
:
. Then
.
. Then
.
The area of the region enclosed by the graph is that of the quadrilateral defined by the points . Breaking it up into triangles and solving or using the Shoelace Theorem, we get
.
Solution 2
Since is the only present
"term" in this equation, we know that the area must be symmetrical about the x-axis.
We'll consider the area when and we only consider the portion enclosed with
. Then, we'll double that area since the graph is symmetrical.
Now, let us remove the absolute values:
When :
. This rearranges to
.
When :
. So
.
When :
. So
.
By simple sketching, we see the shape that looks like the image in Solution 1 (graph it out and you'll see). We see that the partial area we seek in this part is the triangle with the vertices ,
,
. This triangle has an area of
.
Simply double the area and we get as our final answer.
~hastapasta
See also
1987 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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.