AMC 10 · 2013 · #25
Grade 11 algebracountingPick an answer.
Counting polynomials directly is hopeless because the degree is unbounded. The way in is to re-encode each polynomial as its set of roots, then chop that set into pieces that behave like ordinary positive integers: a lone real root, or a conjugate pair a+bi and a-bi whose product is a²+b². Once every piece carries a whole-number size and the sizes must multiply to 50, the question becomes a finite factoring question. Then a systematic list does the rest: write down the few ways to split 50 into factors, and for each factor write down every real root and every conjugate pair that produces it.
A polynomial is just its root set
A polynomial here is just its root set.
A monic polynomial and the list of its roots carry exactly the same information, so counting one counts the other.
11.A-APR.B.3Organize Information In More WaysConjugate symmetry, and why it is exactly the right condition
Integer coefficients mean conjugate symmetry.
Conjugation reshuffles the roots without moving the set, so it cannot move the coefficients either, which forces them to be real.
Conjugation reshuffles the roots without moving the set, so it cannot move the coefficients either.
▸ Why?
A polynomial with real coefficients has its nonreal roots in matched conjugate pairs.
▸ Why?
Each root is sent to exactly one root and back again, so the set is carried onto itself with none left over.
Negate every root to kill the sign
Negating every root removes the sign problem.
Flipping the sign of every root converts Vieta's alternating sign into one clean product that equals 50.
7.NS.A.2Change Focus Count The ComplementCut R into blocks
The roots group into conjugate blocks.
A conjugate pair acts like a single positive whole number, so the whole root set behaves like a product of integers.
11.N-CN.A.3Identify SubproblemsEvery block value divides 50, so list all blocks
Every block's value must divide the constant.
Only divisors of 50 can show up, and for each divisor you can simply write down every real root and every conjugate pair that makes it.
4.OA.B.4Make A Systematic ListThe size-1 blocks: two free, one forced
Two single-root blocks are free.
The value +1 blocks change nothing, and the lone -1 block is a switch whose position is forced by the sign you already have.
6.NS.C.7Change Focus Count The ComplementSplit 50 into sizes and count each split
Splitting the constant gives 132 choices.
Each way of splitting 50 into factors is its own shopping list, and each factor has a known number of Gaussian integers behind it.
11.S-CP.B.9Make A Systematic ListMultiply the free units back in
Multiplying back gives 528, choice (D).
The two harmless roots each double the count, turning 132 into 528.
4.NBT.B.5Make A Systematic ListRoots with whole-number parts come in conjugate pairs that multiply to the plain whole number a² + b², so counting these polynomials turns into counting the ways to break 50 into factors and then counting the Gaussian integers behind each factor.
- A polynomial is just its root set
- Conjugate symmetry, and why it is exactly the right condition
- Negate every root to kill the sign
- Cut R into blocks
- Every block value divides 50, so list all blocks
- The size-1 blocks: two free, one forced
- Split 50 into sizes and count each split
- Multiply the free units back in