2008 AMC 12A Problems/Problem 14: Difference between revisions

Revision as of 13:53, 22 January 2012

Problem

What is the area of the region defined by the inequality $|3x-18|+|2y+7|\le3$ ?

$\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ \frac {7}{2}\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 5$

Solution

Area is invariant under translation, so after translating left $6$ and up $7/2$ units, we have the inequality

$|3x| + |2y|\leq 3$

which forms a diamond centered at the origin and vertices at $(\pm 1, 0), (0, \pm 1.5)$ . Thus the diagonals are of length $2$ and $3$ . Using the formula $A = \frac 12 d_1 d_2$ , the answer is $\frac{1}{2}(2)(3) = 3 \Rightarrow \mathrm{(A)}$ .

2008 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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
All AMC 12 Problems and Solutions

@@ Line 5: / Line 5: @@
 == Solution ==
-Area is invariant under translation, so after translating left <math>6</math> and up <math>3/2</math> units, we have the inequality
+Area is invariant under translation, so after translating left <math>6</math> and up <math>7/2</math> units, we have the inequality
 <cmath>|3x| + |2y|\leq 3</cmath>

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2008 AMC 12A Problems/Problem 14: Difference between revisions

Revision as of 13:53, 22 January 2012

Problem

Solution

See also