AMC 10 · 2019 · #9

Grade 11 algebra

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

logarithm-propertiespolygon-inequalityinterval-arithmeticconsecutive-integer-count convert-to-algebrabound-inequality-then-enumeratesystematic-enumeration ↑ Prerequisites: logarithm-propertiespolygon-inequality

📏 Medium solution 💡 3 insights

Problem

Three lengths are proposed: the base-two logarithm of x, the base-four logarithm of x, and 3. Count how many integer values of x let those three lengths be the sides of a triangle with positive area.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

How to solve

Strategy Convert to Algebra

The word 'triangle' is geometric, but the only geometry involved is the triangle inequality, which is three algebraic statements. So the plan is to write those three inequalities, then clean up the logarithms. The two log sides look different but are not: base 4 is base 2 squared, so log₄ x is exactly half of log₂ x. Naming t = log₂ x (Introduce a Variable) collapses the whole problem into linear inequalities in one variable. After solving for t I convert back to x, test the two boundary values where the triangle goes flat (Extreme Principle), and count the surviving integers (Make a Systematic List).

1STEP 1

Turn positive area into inequalities

Positive area becomes three inequalities.

log₂ x + log₄ x ＞ 3, log₂ x + 3 ＞ log₄ x, log₄ x + 3 ＞ log₂ x

2STEP 2

Write both logs in base 2

Put both logarithms in one base.

log₄ x = (log₂ x)/(log₂ 4) = 1/2log₂ x, t = log₂ x → sides t, t/2, 3

3STEP 3

Solve the three inequalities for t

Solving gives a range.

t + t/2 ＞ 3 → t ＞ 2; t/2 + 3 ＞ t → t ＜ 6; t + 3 ＞ t/2 (always true) → 2 ＜ t ＜ 6

4STEP 4

Undo the logarithm

Undo the logarithm.

2 ＜ t ＜ 6 ⟺ 2² ＜ 2^t ＜ 2⁶ ⟺ 4 ＜ x ＜ 64

5STEP 5

Check the two endpoints

The endpoints go flat and fail.

x = 4: (2, 1, 3), 1 + 2 = 3; x = 64: (6, 3, 3), 3 + 3 = 6

6STEP 6

Count the integers in range

Counting the integers gives 59.

{5, 6, 7, …, 63} → 63 - 5 + 1 = 59

Answer

Spot-check one value inside the range and one outside. At x = 16 the sides are 4, 2, 3, and the two shorter ones give 2 + 3 = 5 ＞ 4, so a real triangle exists. At x = 100 the sides are about 6.64, 3.32, 3, and 3.32 + 3 = 6.32 ＜ 6.64, so it fails — which matches the cutoff at 64. The count is also worth checking against the obvious trap: keeping 4 and 64 would give 64 - 4 + 1 = 61, which is choice (C), and step 5 showed those two are exactly the flat cases. Dropping both leaves 59, choice (B).

💡Key takeaway

A base-4 log is exactly half a base-2 log, so name that one log t and the triangle rule turns into a plain range for t.

Turn positive area into inequalities
Write both logs in base 2
Solve the three inequalities for t
Undo the logarithm
Check the two endpoints
Count the integers in range