AMC 10 · 2022 · #17

Grade 11 algebra

Archetype Coordinate Plane Figure Computation └ Sub-archetype Figure Constraint Equations for Unknown Quantity

double-angle-formulatrigonometric-ratioszero-product-propertydomain-restrictionquadratic-equations convert-to-algebraidentify-subproblemscasework ↑ Prerequisites: double-angle-formulaquadratic-equations

📏 Medium solution 💡 3 insights

Problem

For a real number, an equation mixing sines may have one solution, several, or none in the open interval from zero to pi. Collecting every value giving more than one solution yields an interval with a single point punched out. Find the sum of its three boundary numbers.

Pick an answer.

(A)

${-}4$

(B)

${-}1$

(C)

(D)

(E)

How to solve

Strategy Introduce a Variable

Three different multiples of x appear, so nothing can be compared until they are put on one footing. The angle addition formulas rewrite both sin 2x and sin 3x as sin x times a polynomial in cos x, which is the reorganization that makes the whole problem collapse (tool #15). A common factor of sin x then peels off, and because sin x is never zero on (0,π) it can be cancelled outright (tool #7). What is left is a statement purely about cos x, and the block 2cos x + 1 shows up on both sides, so naming it u turns the equation into a factored quadratic u(u-2-a)=0 (tool #4). One factor gives a solution that exists no matter what a is, so the real question shifts to whether the second factor can also produce a legal x (tool #16). That is a pure range question about cos x on an open interval, plus one boundary case where the two solutions collide into one (tool #14).

1STEP 1

Put every angle on one footing

Multiple-angle formulas reduce it to one angle.

sin 2x = 2sin xcos x, sin 3x = sin x (4cos² x - 1)

2STEP 2

Cancel the common sine

The sine is nonzero inside, so it cancels.

asin x (2cos x + 1) = sin x (2cos x - 1)(2cos x + 1) ⟹ a(2cos x + 1) = (2cos x - 1)(2cos x + 1)

3STEP 3

Name the repeated block

The same block appears on both sides.

u = 2cos x + 1 ⟹ au = u(u-2) ⟹ u(u - (a+2)) = 0

4STEP 4

The free solution

One solution comes for free.

u = 0 ⟹ cos x = -1/2 ⟹ x = 2π/3; u = a+2 ⟹ cos x = (a+1)/2

5STEP 5

Make the second branch legal

The cosine's range fixes the interval.

-1 ＜ (a+1)/2 ＜ 1 ⟹ -2 ＜ a+1 ＜ 2 ⟹ -3 ＜ a ＜ 1

6STEP 6

Remove the collision and add

Removing the collision and adding gives negative four.

(a+1)/2 = -1/2 → a = -2; a ∈ (-3,-2)∪(-2,1); p+q+r = -3 + (-2) + 1 = -4

Answer

-4

The shape (p,q)∪(q,r) promised by the problem matches exactly what came out: one interval with one interior point removed, and the removed point q = -2 really does sit between -3 and 1. Spot-check three values of a. With a = 0 the equation is sin 3x = 0, whose solutions in (0,π) are x = π/3 and x = 2π/3 — two of them, and 0 is indeed inside (-2,1). With a = 4, the branch needs cos x = 5/2, impossible, leaving only x = 2π/3 — and 4 is outside the set. With a = -2 the branch needs cos x = -1/2, the same point as the free solution, so again only one solution — matching the hole. Finally p+q+r = -4 is one of the listed choices.

💡Key takeaway

Rewrite every angle in terms of one angle, cancel the factor that cannot be zero, and then ask not whether a solution exists but whether a second, different one does.

Put every angle on one footing
Cancel the common sine factor
Name the repeated block
One solution comes for free
Make the second branch legal
Remove the collision and add up