AMC 10 · 2020 · #22

Grade 11 algebra

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

exponentspolynomial-substitutioncompleting-the-squareoptimization convert-to-algebraidentify-subproblemsextreme-principle ↑ Prerequisites: exponentsquadratic-equations

📏 Long solution 💡 3 insights

Problem

Over all real numbers, consider two to the power of the number minus three times the number, all multiplied by the number and divided by four to that power. Find the largest possible value.

Pick an answer.

(A)

$\frac{1}{16}$

(B)

$\frac{1}{15}$

(C)

$\frac{1}{12}$

(D)

$\frac{1}{10}$

(E)

$\frac{1}{9}$

How to solve

Strategy Introduce a Variable

Tool #15 (Reorganize): split the single fraction into two pieces so the repeated block t/2^t becomes visible. Tool #4 (Introduce a Variable): name that block u — the whole messy expression collapses to u - 3u². Tool #9 (Easier Problem): maximizing a quadratic in u is a Grade 9 task, far easier than the original. Tool #14 (Extreme Principle): complete the square to get the vertex, which is the upper bound. Tool #6 (Guess and Check): test two easy inputs, t = 0 and t = 1, to confirm the required u value is actually reachable — an upper bound is worthless if nothing hits it.

1STEP 1

Write the denominator as a square

Write it as a square.

((2^t - 3t) t)/4^t = (t · 2^t - 3t²)/(2^t)²

2STEP 2

Split the fraction

Split it into two pieces.

(t · 2^t - 3t²)/(2^t)² = t/2^t - 3(t/2^t)²

3STEP 3

Name the repeated block

The same block appears twice.

u = t/2^t ⟹ f(t) = u - 3u²

4STEP 4

Complete the square

Completing the square gives a ceiling.

u - 3u² = 1/12 - 3(u - 1/6)² ≤ 1/12

5STEP 5

Check the peak is reachable

The intermediate value theorem confirms it.

g(0) = 0 ＜ 1/6 ＜ 1/2 = g(1), so g(t₀) = 1/6 for some t₀ ∈ (0, 1)

6STEP 6

Read the maximum

The maximum is one twelfth.

max_t ∈ R ((2^t - 3t) t)/4^t = 1/12 → (C)

Answer

1/12

Spot-check some values. At t = 1: (2 - 3)(1)/4 = -1/4. At t = 2: (4 - 6)(2)/16 = -1/4. At t = 1/2: u ≈ 0.3536, giving u - 3u² ≈ -0.021. All below 1/12 ≈ 0.0833, as required. At t = 0.2: u ≈ 0.1741, giving u - 3u² ≈ 0.0832 — just under 1/12 and very close, which is exactly what should happen near t₀ ≈ 0.190. Also note g(t) = t/2^t climbs from 0 at t = 0 to about 0.531 near t ≈ 1.443, so it actually hits u = 1/6 twice; either crossing gives the same value 1/12. The bound is tight and the answer 1/12 matches choice (C).

💡Key takeaway

Name the repeated chunk: with u = t/2^t, the monster ((2^t-3t)t)/4^t is just u - 3u² = 1/12 - 3(u - 1/6)², so the ceiling is 1/12 — and since t/2^t runs continuously from 0 to 1/2 as t goes from 0 to 1, that ceiling really is reached.

Write the denominator as a square
Split the fraction in two
Name the repeated block
Complete the square for the ceiling
Check the peak is reachable
Read off the maximum