Difference between revisions of "2021 AMC 10A Problems/Problem 15"

Revision as of 23:20, 11 February 2021

Problem

Values for $A,B,C,$ and $D$ are to be selected from $\{1, 2, 3, 4, 5, 6\}$ without replacement (i.e. no two letters have the same value). How many ways are there to make such choices so that the two curves $y=Ax^2+B$ and $y=Cx^2+D$ intersect? (The order in which the curves are listed does not matter; for example, the choices $A=3, B=2, C=4, D=1$ is considered the same as the choices $A=4, B=1, C=3, D=2.$ )

$\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360$

Solution 1 (Intuition):

Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn't matter. Then $C>A$ and $B>D$ . Therefore the number of ways to choose the four integers is $\tbinom{6}{2}\tbinom{4}{2}=90$ , and the answer is $\boxed{C}$ . ~IceWolf10

Solution 2 (Algebra):

Setting $y = Ax^2+B = Cx^2+D$ , we find that $Ax^2-Cx^2 = x^2(A-C) = D-B$ , so $x^2 = \frac {D-B}{A-C} \ge 0$ by the trivial inequality. This implies that $D-B$ and $A-C$ must both be positive or negative. If two distinct values are chosen for $(A, C)$ and $(B, D)$ respectively, there are $2$ ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). Note that we must divide by $2$ at the end, however, since the $2$ curves aren't considered distinct. Calculating, we get $\frac {1}{2} \cdot \binom {6}{2} \binom {4}{2} \cdot 2 = 90 = \boxed{C}.$ ~ ike.chen

Video Solution (Using Vieta's Formulas and clever combinatorics)

https://youtu.be/l85Qah1vGgc

~ pi_is_3.14

@@ Line 4: / Line 4: @@
 <math>\textbf{(A) }30 \qquad \textbf{(B) }60 \qquad \textbf{(C) }90 \qquad \textbf{(D) }180 \qquad \textbf{(E) }360</math>
-==Solution==
+==Solution 1 (Intuition):==
 Visualizing the two curves, we realize they are both parabolas with the same axis of symmetry. Now assume that the first equation is above the second, since order doesn't matter. Then <math>C>A</math> and <math>B>D</math>. Therefore the number of ways to choose the four integers is <math>\tbinom{6}{2}\tbinom{4}{2}=90</math>, and the answer is <math>\boxed{C}</math>. ~IceWolf10
+==Solution 2 (Algebra):==
+Setting <math>y = Ax^2+B = Cx^2+D</math>, we find that <math>Ax^2-Cx^2 = x^2(A-C) = D-B</math>, so <math>x^2 = \frac {D-B}{A-C} \ge 0</math> by the trivial inequality. This implies that <math>D-B</math> and <math>A-C</math> must both be positive or negative. If two distinct values are chosen for <math>(A, C)</math> and <math>(B, D)</math> respectively, there are <math>2</math> ways to order them so that both the numerator and denominator are positive/negative (increasing and decreasing). Note that we must divide by <math>2</math> at the end, however, since the <math>2</math> curves aren't considered distinct. Calculating, we get <cmath>\frac {1}{2} \cdot \binom {6}{2} \binom {4}{2} \cdot 2 = 90 = \boxed{C}.</cmath> ~ ike.chen
 == Video Solution (Using Vieta's Formulas and clever combinatorics) ==

2021 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Problem 16
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