Difference between revisions of "2007 AMC 12A Problems/Problem 23"
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== Problem == | == Problem == | ||
− | + | Square <math>ABCD</math> has area <math>36,</math> and <math>\overline{AB}</math> is [[parallel]] to the [[x-axis]]. Vertices <math>A,</math> <math>B</math>, and <math>C</math> are on the graphs of <math>y = \log_{a}x,</math> <math>y = 2\log_{a}x,</math> and <math>y = 3\log_{a}x,</math> respectively. What is <math>a?</math> | |
<math>\mathrm{(A)}\ \sqrt [6]{3}\qquad \mathrm{(B)}\ \sqrt {3}\qquad \mathrm{(C)}\ \sqrt [3]{6}\qquad \mathrm{(D)}\ \sqrt {6}\qquad \mathrm{(E)}\ 6</math> | <math>\mathrm{(A)}\ \sqrt [6]{3}\qquad \mathrm{(B)}\ \sqrt {3}\qquad \mathrm{(C)}\ \sqrt [3]{6}\qquad \mathrm{(D)}\ \sqrt {6}\qquad \mathrm{(E)}\ 6</math> | ||
− | == Solution == | + | == Solution 1== |
− | Let <math>x</math> be the x-coordinate of <math>B</math> and <math>C</math>, and <math>x_2</math> be the x-coordinate of <math>A</math> and <math>y</math> be the y-coordinate of <math>A</math> and <math>B</math>. Then <math>2\log_ax | + | Let <math>x</math> be the x-coordinate of <math>B</math> and <math>C</math>, and <math>x_2</math> be the x-coordinate of <math>A</math> and <math>y</math> be the y-coordinate of <math>A</math> and <math>B</math>. Then <math>2\log_ax= y \Longrightarrow a^{y/2} = x</math> and <math>\log_ax_2 = y \Longrightarrow x_2 = a^y = \left(a^{y/2}\right)^2 = x^2</math>. Since the distance between <math>A</math> and <math>B</math> is <math>6</math>, we have <math>x^2 - x - 6 = 0</math>, yielding <math>x = -2, 3</math>. |
− | < | + | However, we can discard the negative root (all three [[logarithm]]ic equations are underneath the line <math>y = 3</math> and above <math>y = 0</math> when <math>x</math> is negative, hence we can't squeeze in a square of side 6). Thus <math>x = 3</math>. |
+ | |||
+ | Substituting back, <math>3\log_{a}x - 2\log_{a}x = 6 \Longrightarrow a^6 = x</math>, so <math>a = \sqrt[6]{3}\ \ \mathrm{(A)}</math>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | Notice that all of the graphs <math>y = \log_{a}x,</math> <math>y = 2\log_{a}x,</math> and <math>y = 3\log_{a}x</math> have a domain of <math>(0, \infty]</math>. Also notice that <math>y = \log_{a}x</math> is the furthest to the right, as adding coefficients in front of the <math>\log</math> part only makes the graph steeper. Since <math>A</math> is on the graph of <math>y = \log_{a}x</math> and <math>B</math> is on the graph of <math>y = 2\log_{a}x</math>, <math>\,</math> <math>A</math> must be to the right of <math>B</math>. We are told that <math>\overline{AB}</math> is [[parallel]] to the [[x-axis]]. Let <math>A</math> be the point <math>(x, y)</math>. Then the points <math>B</math> and <math>C</math> are | ||
+ | |||
+ | <math>(x-6, y)</math> | ||
+ | |||
+ | and | ||
+ | |||
+ | <math>(x-6, y+6)</math> | ||
+ | |||
+ | respectively. | ||
+ | |||
+ | Substituting these coordinates into the equations given yields <math>y = \log_{a}x,</math> <math>y = 2\log_{a}x-6,</math> and <math>y+6 = 3\log_{a}x-6</math>. Rearranging a bit, we get the following equations: | ||
+ | |||
+ | <math>1) a^y = x</math> | ||
+ | |||
+ | <math>2) a^y = (x-6)^2</math> | ||
− | + | <math>3) a^{y+6} = (x-6)^3</math> | |
+ | |||
+ | Using equations 1 and 2, we get <math>x=(x-6)^2</math>. Solving yields <math>x=9, x=-4</math>. However, <math>-4</math> is extraneous so <math>x=9</math> (also because we know <math>A</math> has to have a positive <math>x</math>-coordinate). Using the second and third equations, we get | ||
− | + | <math>\frac{a^{y+6}}{a^y} = \frac{(x-6)^3}{(x-6)^2}</math> <math>\Rightarrow</math> <math>a^6 = x-6</math>. | |
− | + | Plugging in <math>9</math> for <math>x</math> yields <math>a^6 = 3</math>, or <math>a= \sqrt[6]{3} \Rightarrow \boxed{\text{A}}</math>. | |
== See also == | == See also == | ||
Line 20: | Line 41: | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 18:09, 21 July 2020
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.