AMC 10 · 2005 · #23

Grade 11 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Symmetric Function of Roots via Vieta

logarithm-propertiessymmetric-polynomialsexponents convert-to-algebraidentify-subproblems ↑ Prerequisites: logarithm-propertiessymmetric-polynomialsexponents

📏 Long solution 💡 3 insights

Problem

Two logarithm conditions tie three real numbers together. A cubic combination of them always equals a fixed pair of coefficients times two powers of ten. Find the sum of those coefficients.

Pick an answer.

(A)

$\frac {15}{2}$

(B)

$\frac {29}{2}$

(C)

(D)

$\frac {39}{2}$

(E)

How to solve

Strategy Introduce a Variable

The two logarithms are just a disguised statement about x+y and x²+y², so tool #4 strips the disguise and names s = 10^z, putting all the z-dependence in one letter. The target x³+y³ is symmetric in x and y, so tool #15 rewrites the givens as the sum and the product of x and y, and tool #7 splits the cube into the two standard identities (x+y)² and (x+y)³. Tool #6 closes the loop with one explicit legal triple, which both checks the algebra and shows that a+b could not have been anything else.

1STEP 1

Undo both logarithms

Undoing the logarithms turns both conditions into algebra.

x+y = 10^z, x²+y² = 10^z+1 = 10 · 10^z

2STEP 2

Rename the power of 10

Naming the power of ten makes the third letter disappear.

s = 10^z ＞ 0, x+y = s, x²+y² = 10s

3STEP 3

Squeeze out the product xy

Squaring the sum is the only bridge to the product.

s² = 10s + 2xy ⟹ xy = (s²-10s)/2

4STEP 4

Apply the cube-of-a-sum identity

The cube identity then leaves an expression in one letter.

x³+y³ = s³ - 3s·(s²-10s)/2 = s³ - 3/2s³ + 15s² = -1/2s³ + 15s²

5STEP 5

Translate s back into powers of 10

Translating back reads off both coefficients, summing to 29/2.

x³+y³ = -1/2 · 10³z + 15 · 10²z, a = -1/2, b = 15, a+b = 29/2

6STEP 6

Confirm with one real triple

One real triple confirms it, so the answer is 29/2, choice (B).

z = 0: x+y = 1, xy = -9/2, x³+y³ = 1³ - 3(-9/2)(1) = 29/2 = a+b

Answer

29/2

Real x and y with a given sum s and product (s²-10s)/2 exist exactly when the discriminant s²-4·(s²-10s)/2 = 20s-s² is non-negative, that is when 0 ＜ s ≤ 20. So S contains triples for infinitely many values of z, and across those values 10³z and 10²z are not proportional to each other, so the pair (a,b) that fits them all is unique — 29/2 is the only possible value of a+b, not merely one that happens to work. The value also sits sensibly between the choices 15 and 39/2, and the independent z = 0 computation reproduces it exactly.

💡Key takeaway

Anything symmetric in x and y can be rebuilt from just their sum and their product, so two facts about x and y are enough to pin down x³+y³.

Undo both logarithms
Rename the power of 10
Squeeze out the product xy
Apply the cube-of-a-sum identity
Translate s back into powers of 10
Confirm with one real triple