Difference between revisions of "1987 AIME Problems/Problem 4"
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Find the [[area]] of the region enclosed by the [[graph]] of <math>|x-60|+|y|=\left|\frac{x}{4}\right|.</math> | 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]] | [[Image:1987_AIME-4.png]] | ||
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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 shoelace, we get <math>2 \cdot \frac{1}{2}(80 - 48)(15) = \boxed{480}</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 shoelace, 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>60</math> or <math>x=60</math>: <math>x-60+y=0.25x</math>. This rearranges to <math>y=-0.75x+60</math>. | ||
+ | |||
+ | When <math>0<=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 480 as our final answer. I can't box the answer for some reason. Please let me know what's wrong. | ||
+ | |||
+ | ~hastapasta | ||
== See also == | == See also == |
Revision as of 13:18, 2 February 2022
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 shoelace, 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 or : . 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 480 as our final answer. I can't box the answer for some reason. Please let me know what's wrong.
~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.