Difference between revisions of "2013 AMC 12B Problems/Problem 17"
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Since this cubic has <math>3</math> real solutions, it must have both a maximum and a minimum (noticed by looking at the graph). Then the greatest solution is maximized when the other two solutions are the same and is minimized when the other two solutions are the same. | Since this cubic has <math>3</math> real solutions, it must have both a maximum and a minimum (noticed by looking at the graph). Then the greatest solution is maximized when the other two solutions are the same and is minimized when the other two solutions are the same. | ||
− | Thus, equating <math>a</math> and <math>b</math>, we have <cmath>2a+c=2</cmath>and< | + | Thus, equating <math>a</math> and <math>b</math>, we have <cmath>2a+c=2</cmath>and<cmath>2a^2+c^2=12</cmath>Solving this, we get the quadratic <cmath>3c^2-4c+4=24 \Rightarrow c=\frac{10}{3}, -2</cmath>implying the answer. |
== See also == | == See also == |
Revision as of 15:05, 10 February 2018
Contents
Problem
Let and
be real numbers such that
What is the difference between the maximum and minimum possible values of ?
Solution 1
. Now, by Cauchy-Schwarz, we have that
. Therefore, we have that
. We then find the roots of
that satisfy equality and find the difference of the roots. This gives the answer,
.
Solution 2
This is similar to the first solution but is far more intuitive. From the given, we have
This immediately suggests use of the Cauchy-Schwarz inequality. By Cauchy, we have
Substitution of the above results and some algebra yields
This quadratic inequality is easily solved, and it is seen that equality holds for
and
.
The difference between these two values is .
Solution 3
(no Cauchy-Schwarz)
From the first equation, we know that . We substitute this into the second equation to find that
This simplifies to
, which we can write as the quadratic
. We wish to find real values for
and
that satisfy this equation. Therefore, the discriminant is nonnegative. Hence,
or
. This factors as
. Therefore,
, and by symmetry this must be true for
and
as well.
Now and
satisfy both equations, so we see that
must be the minimum possible value of
. Also,
and
satisfy both equations, so we see that
is the maximum possible value of
. The difference between these is
, or
.
Solution 4
We take a geometrical approach.
From the given, we have and
. The first equation is a line with x and y intercepts of
and the second equation is a circle centered at the origin with radius
. Intuitively, if we want to find the minimum / maximum
such that there still exist real solutions, the two graphs of the equations should be tangent.
Thus, we have that , which simplifies to
. Solving the quadratic, we get that the values of
for which the two graphs are tangent are
and
. Thus, our answer is
.
Solution 5
Draw the sphere and the plane represented by the two equations in Cartesian space, with the -axis representing
. The intersection between the sphere and plane is a circle. We wish to find the point on the circle where
is minimized and the point where
is maximized.
Looking at the graph, it is clear by symmetry that
when
is maximized or minimized. Thus, we can set
. This gives us the following system of equations:
Solving gives , which are the maximum and minimum values of
respectively. The answer follows from here.
Solution 6
We can consider ,
, and
to be solutions to a cubic equation. Then, given our information, we have
and
, so our cubic equation looks like this:
where
can be any real number.
Since this cubic has real solutions, it must have both a maximum and a minimum (noticed by looking at the graph). Then the greatest solution is maximized when the other two solutions are the same and is minimized when the other two solutions are the same.
Thus, equating and
, we have
and
Solving this, we get the quadratic
implying the answer.
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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.