2007 AMC 12A Problems/Problem 23
Problem
Square has area
and
is parallel to the x-axis. Vertices
, and
are on the graphs of
and
respectively. What is
Solution
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 6:
![$x^2 - x - 6 = 0$](http://latex.artofproblemsolving.com/1/e/b/1eb5e101e858d10029d9241f2f0a9ebb2a46ae79.png)
![$x = -2, 3$](http://latex.artofproblemsolving.com/a/e/f/aef0a7c8d9e420dae286aa07ceea76e5383c6adb.png)
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
.
![$\displaystyle 3\log_{a}x - 2\log_{a}x = 6$](http://latex.artofproblemsolving.com/6/7/8/6787687e854abe23ba832846fb89ecb5c9429e8b.png)
![$\log_{a}\frac{x^3}{x^2} = 6$](http://latex.artofproblemsolving.com/d/d/7/dd7370c96c5748cbb6e191075277303f08800d17.png)
![$a^6 = x$](http://latex.artofproblemsolving.com/d/b/f/dbfc39a74201f0faae00510fbd383cffaacf3beb.png)
So .
See also
2007 AMC 12 (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 |