2007 AMC 12A Problems/Problem 23
Contents
Problem
Square has area and is parallel to the x-axis. Vertices , and are on the graphs of and respectively. What is
Solution 1
Let be the x-coordinate of and , and be the x-coordinate of and be the y-coordinate of and . Then and . Since the distance between and is , we have , yielding .
However, we can discard the negative root (all three logarithmic equations are underneath the line and above when is negative, hence we can't squeeze in a square of side 6). Thus .
Substituting back, , so .
Solution 2
Notice that all of the graphs and have a domain of . Also notice that is the furthest to the right, as adding coefficients in front of the part only makes the graph steeper. Since is on the graph of and is on the graph of , must be to the right of . We are told that is parallel to the x-axis. Let be the point . Then the points and are
and
respectively.
Substituting these coordinates into the equations given yields and . Rearranging a bit, we get the following equations:
Using equations 1 and 2, we get . Solving yields . However, is extraneous so (also because we know has to have a positive -coordinate). Using the second and third equations, we get
.
Plugging in for yields , or .
See also
2007 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 |
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