AMC 10 · 2021 · #15

Grade 8 counting

Archetype Small-Domain Constrained Search

systematic-enumerationcombinations-basicsymmetry-argumentfunction-evaluation caseworksymmetry-argument ↑ Prerequisites: combinations-basic

📏 Long solution 💡 3 insights

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 $A=4, B=1, C=3, D=2.$ )

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

Pick an answer.

(A)

(B)

(C)

(D)

180

(E)

360

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Toolkit + CCSS Solution

Understand

Restated: Pick FOUR distinct values from $\{1, 2, 3, 4, 5, 6\}$ and label them $A, B, C, D$. Form two parabolas $y = Ax^2 + B$ and $y = Cx^2 + D$. The two curves are an unordered pair (swapping $(A, B) \leftrightarrow (C, D)$ counts as the same choice). How many such labellings produce two parabolas that INTERSECT each other?

Givens: Pool of values: $\{1, 2, 3, 4, 5, 6\}$; Choose four DISTINCT values to be $A, B, C, D$ (no repeats); Two parabolas: $y_1 = Ax^2 + B$ and $y_2 = Cx^2 + D$; Order of the two curves does not matter (curve swap is the same choice); Answer choices: $30, 60, 90, 180, 360$

Unknowns: Number of unordered $\{(A, B), (C, D)\}$ labellings for which the curves intersect

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #7 Identify Subproblems, #16 Change Focus / Complement, #2 Systematic List

Tool #1 (Draw): sketch two upward-opening parabolas with the same axis. The one with the larger leading coefficient is the 'narrower' one. Two upward parabolas with the same axis fail to intersect iff one is entirely above the other — which means the narrower one (larger $A$) also has the larger constant term ($B$). So they INTERSECT iff the larger leading coefficient is paired with the SMALLER constant. Tool #7 (Subproblems) splits the count into: (a) how many unordered curve-pairs total, (b) which fraction intersect. Tool #16 (Complement) lets us count non-intersecting pairs (the 'one above the other' configuration) and subtract — or just use the cleaner symmetry argument that exactly half intersect. Tool #2 (Systematic List) verifies on a small case (one specific 4-subset).

Execute — Answer: C

#7 Identify Subproblems 8.EE.C.7 Step 1

Set the parabolas equal and reduce to a condition on the four numbers.
$Ax^2 + B = Cx^2 + D \Rightarrow (A - C) x^2 = D - B$.
Since $A \neq C$ (distinct values), $x^2 = (D - B)/(A - C)$.
A real $x$ exists iff $x^2 \ge 0$, i.e.
$(D - B)/(A - C) \ge 0$.
With $B \neq D$, the ratio is non-zero, so we need it strictly positive: $(D - B)(A - C) > 0$, equivalently $(A - C)(B - D) < 0$.

$$\text{Intersect} \iff (A - C)(B - D) < 0$$

💡 Grade 8 one-variable equation: solve $(A-C)x^2 = D-B$ for $x^2$.

#1 Draw a Diagram 8.F.A.3 Step 2

Interpret geometrically (Tool #1).
Two upward parabolas with the same axis of symmetry $x = 0$.
The one with larger leading coefficient is 'narrower'.
At $x = 0$, the heights are $B$ and $D$; far out, the narrower one wins.
So the curves DON'T cross iff the narrower one is ALSO higher at $x=0$ — i.e.
larger $A$ pairs with larger $B$.
They DO cross iff the narrower parabola starts LOWER at $x = 0$.

$$\text{Cross} \iff \big(A > C \text{ and } B < D\big) \text{ or } \big(A < C \text{ and } B > D\big)$$

💡 Grade 8 functions: a narrower, lower parabola must rise above the wider, higher one for large $|x|$, forcing a crossing.

#2 Systematic List 7.SP.C.8 Step 3

Count the total number of unordered curve-pairs.
Step (a): choose any 4 distinct values from $\{1, \dots, 6\}$ — that's $\binom{6}{4} = 15$ subsets.
Step (b): for each 4-subset $\{p, q, r, s\}$, split the four values into the two pairs $(A, B)$ and $(C, D)$.
There are $\binom{4}{2}/2 = 3$ ways to split four labels into two unordered pairs, times $2 \cdot 2 = 4$ ways to order within each pair, giving $12$ unordered curve-pairs per subset.

$$\text{Total} = \binom{6}{4} \cdot \frac{\binom{4}{2}}{2} \cdot 2 \cdot 2 = 15 \cdot 12 = 180$$

💡 Grade 7 counting: choose-then-assign, dividing out by curve-swap symmetry.

#16 Change Focus / Complement 6.NS.C.7 Step 4

Use a sign-flip symmetry to split the $180$ into intersecting vs not.
For any curve-pair with leading coefficients $A, C$ and constants $B, D$ (with $A \neq C$ and $B \neq D$), swap the two constants: replace $B \leftrightarrow D$.
This sends $(A - C)(B - D) \mapsto (A - C)(D - B) = -(A - C)(B - D)$, flipping its sign.
So intersecting curve-pairs ($(A-C)(B-D) < 0$) and non-intersecting pairs ($(A-C)(B-D) > 0$) are in bijection.

$$B \leftrightarrow D: \quad (A - C)(B - D) \;\mapsto\; -(A - C)(B - D)$$

💡 Grade 6 sign and ordering: swapping two distinct numbers in a difference flips the sign.

#16 Change Focus / Complement 6.RP.A.3 Step 5

Since the bijection pairs intersecting with non-intersecting curve-pairs exactly $1$-to-$1$, the $180$ total split evenly: $90$ intersect, $90$ don't.

$$\text{Intersect} = \frac{180}{2} = 90$$

💡 Grade 6 ratios: a perfect $1$-to-$1$ pairing means exactly half.

#2 Systematic List 7.SP.C.8 Step 6

Quick sanity check by direct count on a fixed 4-subset (Tool #2).
Take $\{1, 2, 3, 4\}$.
The $12$ unordered curve-pairs $\{(A,B), (C,D)\}$ — listing the pairs by writing each curve as $(\text{leading},\text{constant})$ and sorting the two curves so the smaller-leading-coefficient curve is first: $\{(1,2),(3,4)\}, \{(1,2),(4,3)\}, \{(1,3),(2,4)\}, \{(1,3),(4,2)\}, \{(1,4),(2,3)\}, \{(1,4),(3,2)\}, \{(2,1),(3,4)\}, \{(2,1),(4,3)\}, \{(2,3),(1,4)\} \ldots$ Filter by intersection ($A<C, B>D$ or symmetric): exactly $6$ of the $12$ intersect, matching the $\frac{180}{2} = 90$ overall.

$$\{1,2,3,4\}: 12 \text{ pairs}, 6 \text{ intersect}$$

💡 Grade 7: verify the symmetry argument with a small concrete case.

#7 Identify Subproblems 6.EE.B.5 Step 7

Match $90$ to the answer choices: it's (C).
Choices $30$ and $60$ are too few; $180$ is the total before applying the intersection filter; $360$ is the total before applying the curve-swap symmetry.

$$90 \;\Rightarrow\; \textbf{(C)}$$

💡 Grade 6 multiple-choice match: only one option equals $90$.

[1] #7 8.EE.C.7 Set the parabolas equal and reduce to a condition on the four numbers. $Ax^2 + B

[2] #1 8.F.A.3 Interpret geometrically (Tool #1). Two upward parabolas with the same axis of sy

[3] #2 7.SP.C.8 Count the total number of unordered curve-pairs. Step (a): choose any 4 distinct

[4] #16 6.NS.C.7 Use a sign-flip symmetry to split the $180$ into intersecting vs not. For any cu

[5] #16 6.RP.A.3 Since the bijection pairs intersecting with non-intersecting curve-pairs exactly

[6] #2 7.SP.C.8 Quick sanity check by direct count on a fixed 4-subset (Tool #2). Take ${1, 2,

[7] #7 6.EE.B.5 Match $90$ to the answer choices: it's (C). Choices $30$ and $60$ are too few; $

Review

Reasonableness: $90$ is exactly half of the total $180$ unordered curve-pairs, which matches the intuition that the intersection condition is determined by the sign of $(A-C)(B-D)$ — a symmetric coin-flip over the way we pair leading coefficients with constants. The wrong choices each correspond to a different undercounting/overcounting: $360$ counts ordered curve pairs (before halving), $180$ counts all pairs without the intersection filter, $60 = 180/3$ and $30 = 90/3$ are red herrings tied to mis-counting the within-pair assignments.

Alternative: Tool #16 (Complement) more directly: count non-intersecting pairs as 'one parabola sits entirely above the other'. For that, the four values split into a 'top' curve $\{r, s\}$ (with $r$ the leading coefficient, $s$ the constant — but the curve must be narrower AND higher, so the bigger of $\{r, s\}$ is the constant; meanwhile the top curve must dominate, meaning its leading coefficient is the larger of the two leading coefficients). Working this out gives exactly $90$ non-intersecting curve-pairs, same as the half-split argument. Subtracting from $180$ also gives $90$ intersecting — same answer (C).

CCSS standards used (min grade 8)

6.EE.B.5 Understand solving an equation or inequality as a process of finding values (Matching the computed count $90$ against the five answer choices to select (C).)
6.NS.C.7 Understand ordering and absolute value of rational numbers (Reading that swapping two distinct numbers in a difference flips the sign of the difference.)
6.RP.A.3 Use ratio and rate reasoning to solve real-world and mathematical problems (Concluding from a $1$-to-$1$ bijection that exactly half of $180$ curve-pairs intersect.)
7.SP.C.8 Find probabilities of compound events including organized lists, tables, tree diagrams (Counting unordered curve-pairs as $\binom{6}{4} \cdot \frac{\binom{4}{2}}{2} \cdot 2 \cdot 2 = 180$ and direct list-check on $\{1,2,3,4\}$.)
8.EE.C.7 Solve linear equations in one variable (Solving $(A - C) x^2 = D - B$ for $x^2$ to extract the intersection condition.)
8.F.A.3 Interpret the equation $y = mx + b$ as defining a linear function; recognize non-linear examples (Reading $y = Ax^2 + B$ as an upward parabola with width controlled by $A$ and vertical offset $B$.)

⭐ This AMC 10 problem only needs Grade 8 understanding of parabolas plus simple counting you already know! Two upward parabolas with the same axis miss each other iff one is narrower AND higher — and intersect iff the narrower one starts LOWER. By swapping the two constants, intersecting and non-intersecting curve-pairs pair up $1$-to-$1$, so exactly half of the $180$ total pairs intersect: $\boxed{90}$, answer (C).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 8)

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