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2021 AMC 10A Problem 15

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2021 AMC 10A Problem 15AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Counting · difficulty 3 of 5 · about 4 min · position P11-15

Values for A,B,C,A,B,C, and DD are to be selected from {1,2,3,4,5,6}\{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=Ax2+By=Ax^2+B and y=Cx2+Dy=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=1A=3, B=2, C=4, D=1 is considered the same as the choices A=4,B=1,C=3,D=2.A=4, B=1, C=3, D=2. )

  1. A)

    3030

  2. B)

    6060

  3. C)

    9090

  4. D)

    180180

  5. E)

    360360

Answer

C

Key insight

Parabolas meet iff the larger leading coefficient pairs with the larger intercept; swapping B and D exchanges good and bad configurations, so exactly half of 180 work.

Solution

Setting the curves equal gives (A−C)x2=D−B(A - C)x^2 = D - B, so x2=D−BA−Cx^2 = \frac{D - B}{A - C}. Since all four values are distinct, neither side is zero, and a real solution exists exactly when D−BD - B and A−CA - C have the same sign: the parabola with the larger coefficient must also have the larger constant term.

Count the total first. There are 6⋅5⋅4⋅3=3606 \cdot 5 \cdot 4 \cdot 3 = 360 ordered assignments, and swapping the two curves (A,B)↔(C,D)(A,B) \leftrightarrow (C,D) pairs them up without changing anything, so there are 180180 distinct pairs of curves.

Now pair each configuration with the one obtained by swapping only the constants B↔DB \leftrightarrow D. This flips the sign of D−BD - B while leaving A−CA - C alone, so it turns every intersecting pair into a non-intersecting pair and vice versa. Exactly half of the 180180 configurations intersect:
1802=90.\frac{180}{2} = 90.

The answer is (C) 90\boxed{\textbf{(C)}\ 90}.

Why this works

Two upward parabolas with vertices on the yy-axis intersect only if the "steeper" one starts lower. Rather than counting the favorable configurations, find an involution (swap the intercepts) that exchanges good and bad configurations; then good and bad are equinumerous and the answer is half the total.

Alternative approach

Direct count: choose which 44 of the 66 values are used ((64)=15\binom64 = 15), then decide which two are coefficients and which two are intercepts ((42)=6\binom42 = 6); the larger coefficient must go with the larger intercept, a single valid pairing. Total 15⋅6=9015 \cdot 6 = 90.

The trap

Forgetting that the two curves are unordered, giving 180 (or 360) instead of 90.

Common mistakes

  • Forgetting that the two curves are unordered, giving 180 (or 360) instead of 90.
  • Requiring D−BD - B and A−CA - C to have opposite signs, which describes the parabolas that miss each other.

Techniques

Map the objects to something easier to count · Exploit symmetry to reduce work or pair up objects