Difference between revisions of "2021 Fall AMC 12A Problems/Problem 17"
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<math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad</math> | <math>\textbf{(A) } 4 \qquad \textbf{(B) } 6 \qquad \textbf{(C) } 8 \qquad \textbf{(D) } 12 \qquad \textbf{(E) } 16 \qquad</math> | ||
− | ==Solution== | + | ==Solution 1== |
If a [[quadratic equation]] does not have two distinct real solutions, then its [[discriminant]] must be <math>\le0</math>. So, <math>b^2-4c\le0</math> and <math>c^2-4b\le0</math>. By inspection, there are <math>\boxed{\textbf{(B) } 6}</math> ordered pairs of positive integers that fulfill these criteria: <math>(1,1)</math>, <math>(1,2)</math>, <math>(2,1)</math>, <math>(2,2)</math>, <math>(3,3)</math>, and <math>(4,4)</math>. | If a [[quadratic equation]] does not have two distinct real solutions, then its [[discriminant]] must be <math>\le0</math>. So, <math>b^2-4c\le0</math> and <math>c^2-4b\le0</math>. By inspection, there are <math>\boxed{\textbf{(B) } 6}</math> ordered pairs of positive integers that fulfill these criteria: <math>(1,1)</math>, <math>(1,2)</math>, <math>(2,1)</math>, <math>(2,2)</math>, <math>(3,3)</math>, and <math>(4,4)</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | 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) | ||
+ | |||
{{AMC12 box|year=2021 Fall|ab=A|num-a=18|num-b=16}} | {{AMC12 box|year=2021 Fall|ab=A|num-a=18|num-b=16}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:26, 25 November 2021
Problem
For how many ordered pairs of positive integers does neither
nor
have two distinct real solutions?
Solution 1
If a quadratic equation does not have two distinct real solutions, then its discriminant must be . So,
and
. By inspection, there are
ordered pairs of positive integers that fulfill these criteria:
,
,
,
,
, and
.
Solution 2
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)
2021 Fall AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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