AMC 10 · 2015 · #24

Grade 11 probability

Archetype Overlap / Complement Counting └ Sub-archetype Two-Set Inclusion-Exclusion: Solve for Missing Region

complex-numbersbinomial-theoremtrigonometric-ratiosprobability-basic caseworkcomplementary-counting ↑ Prerequisites: complex-numbersprobability-basic

📏 Long solution 💡 4 insights

Problem

Two values are drawn from one list and a complex expression raised to a power must come out real. Find the probability.

Pick an answer.

(A)

$\frac{3}{50}$

(B)

$\frac{4}{25}$

(C)

$\frac{41}{200}$

(D)

$\frac{6}{25}$

(E)

$\frac{13}{50}$

How to solve

Strategy Introduce a Variable

Nothing about the fourth power is visible while cos(aπ) and sin(bπ) are still written as trig. Name them x and y, expand, and factor: the imaginary part becomes a product, and a product is zero only when a factor is zero. That converts a vague realness question into four exact equations, with no case left implicit. The hard part is then the equation |x| = |y|. Comparing decimals there is worthless, because sizes like cosπ/5 ≈ 0.809 and sin2π/5 ≈ 0.951 are close enough that eyeballing proves nothing either way. So I convert that equation into an exact statement about a and b as fractions, using the cofunction identity and the fact that cosines agree only when angles agree up to reflection and full turns. Once the condition reads 'a+b or a-b is an odd half', the reduced denominators of a and b decide everything, and the counting is pure arithmetic. Finally the zero cases and the equal-size cases overlap, so they get merged before dividing.

1STEP 1

Build the pool of 20 values

The pool holds 20 values.

|S| = 2+2+4+4+8 = 20, 20 × 20 = 400 ordered pairs

2STEP 2

Expand the fourth power

Expanding isolates the imaginary part.

(x+iy)⁴ = (x⁴ - 6x²y² + y⁴) + i (4x³y - 4xy³), x = cos(aπ), y = sin(bπ)

3STEP 3

Factor the imaginary part

It factors into three simple conditions.

4x³y - 4xy³ = 4xy(x-y)(x+y) = 0 ⇔ x = 0 or y = 0 or |x| = |y|

4STEP 4

Locate the two kinds of zero

Two of them are easy zeroes.

cos(aπ) = 0 ⇔ a ∈ {1/2, 3/2}, sin(bπ) = 0 ⇔ b ∈ {0, 1}

5STEP 5

Count the pairs with a zero

Those account for 76 pairs.

40 + 40 - 4 = 76 = 400 - 18 × 18

6STEP 6

Turn equal sizes into exact arithmetic

The third becomes exact arithmetic.

|cos(aπ)| = |sin(bπ)| ⇔ a+b ≡ 1/2 (mod 1) or a-b ≡ 1/2 (mod 1)

7STEP 7

Screen the denominators

Denominators screen it to 24 more.

(q,s) = (1,2): 4, (q,s) = (2,1): 4, (q,s) = (4,4): 16 ⟹ 24

8STEP 8

Merge the overlap and divide

Merging and dividing gives 6/25, choice (D).

(76 + 20)/400 = 96/400 = 6/25

Answer

6/25

Spot-check both verdicts. With a = 1/4 and b = 3/4, both cos(aπ) and sin(bπ) equal √(2)/2, so the base is √(2)/2(1+i) and its fourth power is 1/4(1+i)⁴ = 1/4(-4) = -1, real as predicted. With a = 1/5 and b = 2/5, neither value is zero, and a+b = 3/5 and a-b = -1/5 are not odd halves, so the test says not real; the decimals agree, since 0.809 ≠ 0.951. The size is sensible too. The zero clauses alone already force the probability to be at least 76/400 = 0.19, so 3/50 = 0.06 and 4/25 = 0.16 are impossible before any of the harder work is done, and 3/50 = 24/400 is precisely the equal-size count with the zero pairs forgotten. Adding the 20 new pairs lifts 0.19 to 0.24, and 6/25 = 0.24 matches. Choice 13/50 = 104/400 would need 8 more good pairs than exist, and 41/200 = 82/400 would need the denominator-4 block to be only 6 pairs instead of 16.

💡Key takeaway

Expand and factor first: once the imaginary part is a product, being real just means some factor is zero, and each of those small equations turns into an exact test on the fractions instead of a comparison of decimals.

Build the pool of 20 values
Expand the fourth power
Factor the imaginary part
Locate the two kinds of zero
Count the pairs with a zero
Turn equal sizes into exact arithmetic
Screen the denominators
Merge the overlap and divide