Difference between revisions of "2021 Fall AMC 10A Problems/Problem 20"
Line 62: | Line 62: | ||
~Arcticturn | ~Arcticturn | ||
+ | |||
+ | == Solution 4 == | ||
+ | We need to solve the following system of inequalities: | ||
+ | <cmath> | ||
+ | \[ | ||
+ | \left\{ | ||
+ | \begin{array}{ll} | ||
+ | b^2 - 4 c \leq 0 \ | ||
+ | c^2 - 4 b \leq 0 | ||
+ | \end{array} | ||
+ | \right.. | ||
+ | \] | ||
+ | </cmath> | ||
+ | |||
+ | Feasible solutions are in the region formed between two parabolas <math>b^2 - 4 c = 0</math> and <math>c^2 - 4 b = 0</math>. | ||
+ | |||
+ | Define <math>f \left( b \right) = \frac{b^2}{4}</math> and <math>g \left( b \right) = 2 \sqrt{b}</math>. | ||
+ | Therefore, all feasible solutions are in the region formed between the graphs of these two functions. | ||
+ | |||
+ | For <math>b = 1</math>, <math>f \left( b \right) = \frac{1}{4}</math> and <math>g \left( b \right) = 2</math>. | ||
+ | Hence, the feasible <math>c</math> are 1, 2. | ||
+ | |||
+ | For <math>b = 2</math>, <math>f \left( b \right) = 1</math> and <math>g \left( b \right) = 2 \sqrt{2}</math>. | ||
+ | Hence, the feasible <math>c</math> are 1, 2. | ||
+ | |||
+ | For <math>b = 3</math>, <math>f \left( b \right) = \frac{9}{4}</math> and <math>g \left( b \right) = 2 \sqrt{3}</math>. | ||
+ | Hence, the feasible <math>c</math> is 3. | ||
+ | |||
+ | For <math>b = 4</math>, <math>f \left( b \right) = 4</math> and <math>g \left( b \right) = 4</math>. | ||
+ | Hence, the feasible <math>c</math> is 4. | ||
+ | |||
+ | For <math>b > 4</math>, <math>f \left( b \right) > g \left( b \right)</math>. Hence, there is no feasible <math>c</math>. | ||
+ | |||
+ | Putting all cases together, the correct answer is <math>\boxed{\textbf{(B) }6}</math>. | ||
+ | |||
+ | ~Steven Chen (www.professorchenedu.com) | ||
Revision as of 20:25, 25 November 2021
Contents
[hide]Problem
How many ordered pairs of positive integers exist where both
and
do not have distinct, real solutions?
Solution 1 (Casework)
A quadratic equation does not have real solutions if and only if the discriminant is nonpositive. We conclude that:
- Since
does not have real solutions, we have
- Since
does not have real solutions, we have
Squaring the first inequality, we get Multiplying the second inequality by
we get
Combining these results, we get
We apply casework to the value of
- If
then
from which
- If
then
from which
- If
then
from which
- If
then
from which
Together, there are ordered pairs
namely
and
~MRENTHUSIASM
Solution 2 (Graphing)
Similar to Solution 1, use the discriminant to get and
. These can be rearranged to
and
. Now, we can roughly graph these two inequalities, letting one of them be the
axis and the other be
.
The graph of solutions should be above the parabola and under its inverse, meaning we want points on the graph or in the first area enclosed by the two graphs:
We are looking for lattice points (since
and
are positive integers), of which we can count
.
~aop2014
Solution 3 (Oversimplified but Risky)
A quadratic equation has one real solution if and only if
Similarly, it has imaginary solutions if and only if
We proceed as following:
We want both to be
value or imaginary and
to be
value or imaginary.
is one such case since
is
Also,
are always imaginary for both
and
We also have
along with
since the latter has one solution, while the first one is imaginary. Therefore, we have
total ordered pairs of integers, which is
~Arcticturn
Solution 4
We need to solve the following system of inequalities:
Feasible solutions are in the region formed between two parabolas and
.
Define and
.
Therefore, all feasible solutions are in the region formed between the graphs of these two functions.
For ,
and
.
Hence, the feasible
are 1, 2.
For ,
and
.
Hence, the feasible
are 1, 2.
For ,
and
.
Hence, the feasible
is 3.
For ,
and
.
Hence, the feasible
is 4.
For ,
. Hence, there is no feasible
.
Putting all cases together, the correct answer is .
~Steven Chen (www.professorchenedu.com)
Video Solution by Mathematical Dexterity
https://www.youtube.com/watch?v=EkaKfkQgFbI
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 19 |
Followed by Problem 21 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.