AMC 10 · 2019 · #24

Grade 8 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Successive Multiplicative Steps

vieta-formulaspolynomial-rootssymmetric-polynomialspolynomial-factoringpolynomial-substitution identify-subproblemsconvert-to-algebra ↑ Prerequisites: vieta-formulaspolynomial-roots

📏 Medium solution 💡 3 insights

Problem

p, q, r are the three distinct roots of x³ - 22x² + 80x - 67. There exist constants A, B, C with

\frac{1}{s³ - 22s² + 80s - 67}

\frac{A}{s-p}

\frac{B}{s-q}

\frac{C}{s-r}

for every valid s. Find

\frac{1}{A}

\frac{1}{B}

\frac{1}{C}

Pick an answer.

(A)

243

(B)

244

(C)

245

(D)

246

(E)

247

AMC 10 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Tool #7 (Subproblems): split into (a) find a closed form for each of A, B, C, (b) write $\frac{1}{A}$ + $\frac{1}{B}$ + $\frac{1}{C}$ as a symmetric expression in p, q, r, (c) evaluate that symmetric expression using Vieta. Tool #11 (Work Backwards): undo partial fractions by multiplying by (s-p)(s-q)(s-r) then plug s = p to isolate A. Tool #13 (Algebra): symbolic manipulation is unavoidable. Tool #9 (Easier Problem): the identity p² + q² + r² = (p+q+r)² - 2(pq+qr+rp) is a smaller fact we can invoke.

1STEP 1

Clear denominators by multiplying by (s-p)(s-q)(s-r): 1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q), true for every s.

1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)

2STEP 2

Substitute s = p so the other two terms vanish: 1 = A(p-q)(p-r), giving A = $\frac{1}{(p-q)(p-r)}$ ; by symmetry B and C follow.

A =

\frac{1}{(p - q)(p - r)}

, B =

\frac{1}{(q - p)(q - r)}

, C =

\frac{1}{(r - p)(r - q)}

3STEP 3

Take reciprocals and add: $\frac{1}{A}$ + $\frac{1}{B}$ + $\frac{1}{C}$ = (p-q)(p-r) + (q-p)(q-r) + (r-p)(r-q).

\frac{1}{A}

\frac{1}{B}

\frac{1}{C}

= (p-q)(p-r) + (q-p)(q-r) + (r-p)(r-q)

4STEP 4

Expanding each product and adding cancels the cross terms, leaving (p²+q²+r²) - (pq+qr+rp).

Σ = (p² + q² + r²) - (pq + qr + rp)

5STEP 5

Rewrite p²+q²+r² = (p+q+r)² - 2(pq+qr+rp); with Vieta's 22 and 80 this is 484 - 160 = 324.

p² + q² + r² = 484 - 160 = 324

6STEP 6

Subtract: $\frac{1}{A}$ + $\frac{1}{B}$ + $\frac{1}{C}$ = 324 - 80 = 244, choice (B).

324 - 80 = 244

Answer

244

Vieta's formulas for x³ + bx² + cx + d = 0 (here b = -22, c = 80, d = -67) give Σ p = -b = 22, Σ pq = c = 80, Π p = -d = 67 — all positive, consistent with three real distinct roots. The constant C = -67 in the original polynomial is not used here, but feeds into a check: by Vieta pqr = 67, and we never needed it (the answer only depends on the first two elementary symmetric sums). The expression 324 - 80 = 244 falls right inside the tight range 243 – 247, and only choice (B) matches. Sanity on cancellation: in the expansion (p-q)(p-r) + (q-p)(q-r) + (r-p)(r-q) each diagonal p² appears once and each cross pq appears with net coefficient -1, confirming the formula Σ p_i² - Σ_i ＜ j p_ip_j.

💡Key takeaway

This AMC 10 problem only needs Grade 8 equation-clearing tricks and Grade 7 expression expansion you already know — multiply out, substitute s = p to get A = $\frac{1}{(p-q)(p-r)}$ , then sum the three reciprocals to get (p² + q² + r²) - (pq + qr + rp) = 324 - 80 = 244.

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