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AMC 10 · 2019 · #24

Grade 8 algebra
Archetype Arithmetic Expression Decomposition
vieta-formulaspolynomial-rootssymmetric-polynomialspolynomial-factoringpolynomial-substitution identify-subproblemsconvert-to-algebra ↑ Prerequisites: vieta-formulaspolynomial-roots
📏 Medium solution 💡 3 insights

Problem

Let pp, qq, and rr be the distinct roots of the polynomial x3−22x2+80x−67x^3 - 22x^2 + 80x - 67. It is given that there exist real numbers AA, BB, and CC such that 1s3−22s2+80s−67=As−p+Bs−q+Cs−r\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \frac{C}{s-r}for all s∉{p,q,r}s\not\in\{p,q,r\}. What is 1A+1B+1C\tfrac1A+\tfrac1B+\tfrac1C?

(A) 243(B) 244(C) 245(D) 246(E) 247\textbf{(A) }243\qquad\textbf{(B) }244\qquad\textbf{(C) }245\qquad\textbf{(D) }246\qquad\textbf{(E) } 247

Pick an answer.

(A)
243
(B)
244
(C)
245
(D)
246
(E)
247
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Toolkit + CCSS Solution

Understand

Restated: $p, q, r$ are the three distinct roots of $x^3 - 22x^2 + 80x - 67$. There exist constants $A, B, C$ with $\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \dfrac{C}{s-r}$ for every valid $s$. Find $\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C}$.

Givens: Cubic $x^3 - 22x^2 + 80x - 67$ with three distinct real roots $p, q, r$; Partial-fraction decomposition $\dfrac{1}{(s-p)(s-q)(s-r)} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \dfrac{C}{s-r}$; Vieta sums: $p + q + r = 22$, $pq + qr + rp = 80$, $pqr = 67$; Answer choices: (A) $243$, (B) $244$, (C) $245$, (D) $246$, (E) $247$

Unknowns: $\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C}$

Understand

Restated: $p, q, r$ are the three distinct roots of $x^3 - 22x^2 + 80x - 67$. There exist constants $A, B, C$ with $\dfrac{1}{s^3 - 22s^2 + 80s - 67} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \dfrac{C}{s-r}$ for every valid $s$. Find $\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C}$.

Givens: Cubic $x^3 - 22x^2 + 80x - 67$ with three distinct real roots $p, q, r$; Partial-fraction decomposition $\dfrac{1}{(s-p)(s-q)(s-r)} = \dfrac{A}{s-p} + \dfrac{B}{s-q} + \dfrac{C}{s-r}$; Vieta sums: $p + q + r = 22$, $pq + qr + rp = 80$, $pqr = 67$; Answer choices: (A) $243$, (B) $244$, (C) $245$, (D) $246$, (E) $247$

Plan

Primary tool: #7 Identify Subproblems

Secondary: #11 Work Backwards, #13 Convert to Algebra, #9 Solve an Easier Related Problem

Tool #7 (Subproblems): split into (a) find a closed form for each of $A, B, C$, (b) write $\tfrac{1}{A} + \tfrac{1}{B} + \tfrac{1}{C}$ 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 $p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq+qr+rp)$ is a smaller fact we can invoke.

Execute — Answer: B

#11 Work Backwards 8.EE.C.7 Step 1
  • Multiply both sides of the identity by $(s-p)(s-q)(s-r) = s^3 - 22s^2 + 80s - 67$ to clear denominators: $1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)$.
  • This is a polynomial identity that holds for every $s$.
$$1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q)$$

💡 Multiplying clears the fractions and gives a single polynomial identity to manipulate.

#7 Identify Subproblems 8.EE.C.7 Step 2
  • Plug $s = p$: the last two terms vanish ($s - p = 0$), giving $1 = A(p - q)(p - r)$, so $A = \dfrac{1}{(p - q)(p - r)}$.
  • The same trick with $s = q$ and $s = r$ gives the symmetric formulas for $B$ and $C$.
$$A = \dfrac{1}{(p - q)(p - r)}, \;\; B = \dfrac{1}{(q - p)(q - r)}, \;\; C = \dfrac{1}{(r - p)(r - q)}$$

💡 Substituting $s = p$ isolates $A$ because the other two terms have factor $(s - p)$.

#13 Convert to Algebra 6.EE.A.3 Step 3
  • Take reciprocals: $\dfrac{1}{A} = (p-q)(p-r)$, $\dfrac{1}{B} = (q-p)(q-r)$, $\dfrac{1}{C} = (r-p)(r-q)$.
  • Sum these.
$$\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C} = (p-q)(p-r) + (q-p)(q-r) + (r-p)(r-q)$$

💡 Reciprocal of a product equals product of reciprocals undone.

#13 Convert to Algebra 7.EE.A.1 Step 4
  • Expand each product: $(p-q)(p-r) = p^2 - p(q+r) + qr$.
  • Adding the three cyclic versions: \\ $(p^2 + q^2 + r^2) - [p(q+r) + q(p+r) + r(p+q)] + (qr + pr + pq) = (p^2 + q^2 + r^2) - 2(pq + qr + rp) + (pq + qr + rp)$.
$$\sum = (p^2 + q^2 + r^2) - (pq + qr + rp)$$

💡 Cross terms $p(q+r) + q(p+r) + r(p+q) = 2(pq + qr + rp)$ collapse the middle to $-(pq+qr+rp)$.

#9 Solve an Easier Related Problem 7.EE.A.1 Step 5
  • Use the identity $p^2 + q^2 + r^2 = (p + q + r)^2 - 2(pq + qr + rp)$ from $(p+q+r)^2$ expansion.
  • With Vieta: $p + q + r = 22$ and $pq + qr + rp = 80$.
  • So $p^2 + q^2 + r^2 = 22^2 - 2 \cdot 80 = 484 - 160 = 324$.
$$p^2 + q^2 + r^2 = 484 - 160 = 324$$

💡 Square of a sum minus twice the pairwise products gives the sum of squares.

#7 Identify Subproblems 4.NBT.B.4 Step 6
  • Combine: $\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C} = 324 - 80 = 244$.
  • Answer (B).
$$324 - 80 = 244$$

💡 Sum of squares minus pairwise-product sum gives the target value.

[1] #11 8.EE.C.7 Multiply both sides of the identity by $(s-p)(s-q)(s-r) = s^3 - 22s^2 + 80s - 67
[2] #7 8.EE.C.7 Plug $s = p$: the last two terms vanish ($s - p = 0$), giving $1 = A(p - q)(p -
[3] #13 6.EE.A.3 Take reciprocals: $\dfrac{1}{A} = (p-q)(p-r)$, $\dfrac{1}{B} = (q-p)(q-r)$, $\df
[4] #13 7.EE.A.1 Expand each product: $(p-q)(p-r) = p^2 - p(q+r) + qr$. Adding the three cyclic v
[5] #9 7.EE.A.1 Use the identity $p^2 + q^2 + r^2 = (p + q + r)^2 - 2(pq + qr + rp)$ from $(p+q+
[6] #7 4.NBT.B.4 Combine: $\dfrac{1}{A} + \dfrac{1}{B} + \dfrac{1}{C} = 324 - 80 = 244$. Answer (

Review

Reasonableness: Vieta's formulas for $x^3 + bx^2 + cx + d = 0$ (here $b = -22, c = 80, d = -67$) give $\sum p = -b = 22$, $\sum pq = c = 80$, $\prod p = -d = 67$ — 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 $p^2$ appears once and each cross $pq$ appears with net coefficient $-1$, confirming the formula $\sum p_i^2 - \sum_{i<j} p_ip_j$.

Alternative: Tool #6 (Guess and Check) with a concrete cubic. Pick any cubic with known roots — say roots $1, 2, 3$: $(x-1)(x-2)(x-3)$ gives Vieta sums $6, 11, 6$. By the same partial-fraction formula, $\tfrac{1}{A} + \tfrac{1}{B} + \tfrac{1}{C} = (1-2)(1-3) + (2-1)(2-3) + (3-1)(3-2) = 2 - 1 + 2 = 3$. And $\sum p^2 - \sum pq = (1 + 4 + 9) - 11 = 14 - 11 = 3$. ✓ The formula checks. Plugging Vieta values $22, 80$ then gives $244$ instantly.

CCSS standards used (min grade 8)

  • 4.NBT.B.4 Fluently add and subtract multi-digit whole numbers (Final arithmetic $324 - 80 = 244$.)
  • 6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Taking reciprocals of $A, B, C$ to flip $\tfrac{1}{(p-q)(p-r)}$ into $(p-q)(p-r)$.)
  • 7.EE.A.1 Apply properties of operations to add, subtract, factor, and expand linear expressions (Expanding each $(p-q)(p-r)$, collecting like terms, and using $(p+q+r)^2 = p^2+q^2+r^2 + 2(pq+qr+rp)$.)
  • 8.EE.C.7 Solve linear equations in one variable (Clearing the partial-fraction denominators and substituting $s = p$ (a linear-equation move on $s$) to isolate $A$.)

⭐ 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 $A = \tfrac{1}{(p-q)(p-r)}$, then sum the three reciprocals to get $(p^2 + q^2 + r^2) - (pq + qr + rp) = 324 - 80 = 244$.

⭐ 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 $A = \tfrac{1}{(p-q)(p-r)}$, then sum the three reciprocals to get $(p^2 + q^2 + r^2) - (pq + qr + rp) = 324 - 80 = 244$.

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