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2019 AMC 10A Problem 24

P21-25~7 minPrint
PolynomialsAlgebraic ManipulationApply an identity: SFFT, sum of squares, difference of cubes, VietaTest small/specific values or special cases to find or verify the answer

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 ?

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Key insight

Clear the fractions and plug in each root: 1/A is the product of p minus each other root. Adding the three needs only the sum and pair-products of the roots.

Solution

Step 1: clear the fractions. The roots are p,q,rp, q, r, so the cubic factors as (s−p)(s−q)(s−r)(s - p)(s - q)(s - r). Multiply both sides of the given equation by it:
1=A(s−q)(s−r)+B(s−p)(s−r)+C(s−p)(s−q).1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q).
This is true for every ss (both sides are polynomials that agree at infinitely many values, so they are the same polynomial).

Step 2: plug in a root to find AA. Choose s=ps = p because it makes two of the three terms vanish: the BB and CC terms both contain (s−p)(s - p), which is now 00. What is left is
1=A(p−q)(p−r),so1A=(p−q)(p−r).1 = A(p - q)(p - r), \quad\text{so}\quad \frac1A = (p - q)(p - r).
In the same way, s=qs = q and s=rs = r give
1B=(q−p)(q−r),1C=(r−p)(r−q).\frac1B = (q - p)(q - r), \qquad \frac1C = (r - p)(r - q).

Step 3: add the three products. Expand each one:

  • (p−q)(p−r)=p2−pq−pr+qr(p-q)(p-r) = p^2 - pq - pr + qr
  • (q−p)(q−r)=q2−pq−qr+pr(q-p)(q-r) = q^2 - pq - qr + pr
  • (r−p)(r−q)=r2−pr−qr+pq(r-p)(r-q) = r^2 - pr - qr + pq

Adding, each of pqpq, qrqr, prpr appears twice with a minus sign and once with a plus sign, so
1A+1B+1C=p2+q2+r2−(pq+qr+pr).\frac1A + \frac1B + \frac1C = p^2 + q^2 + r^2 - (pq + qr + pr).

Step 4: use what the cubic tells us about its roots. Multiplying out (s−p)(s−q)(s−r)(s-p)(s-q)(s-r) gives
s3−(p+q+r)s2+(pq+qr+pr)s−pqr.s^3 - (p+q+r)s^2 + (pq + qr + pr)s - pqr.
Comparing with s3−22s2+80s−67s^3 - 22s^2 + 80s - 67:
p+q+r=22,pq+qr+pr=80.p + q + r = 22, \qquad pq + qr + pr = 80.

Step 5: get the sum of squares. Squaring p+q+rp + q + r gives p2+q2+r2+2(pq+qr+pr)p^2 + q^2 + r^2 + 2(pq + qr + pr). So
p2+q2+r2=222−2⋅80=484−160=324.p^2 + q^2 + r^2 = 22^2 - 2 \cdot 80 = 484 - 160 = 324.

Step 6: finish.
1A+1B+1C=324−80=244.\frac1A + \frac1B + \frac1C = 324 - 80 = 244.

The answer is (B) 244\boxed{\textbf{(B)}\ 244}.

Why this works

To find the numbers on top of fractions like As−p\frac{A}{s-p}, clear the bottoms and then plug in the value that makes most terms zero. The answer came out as an expression that treats p,q,rp, q, r all alike, and any such expression can be written using p+q+rp + q + r and pq+qr+prpq + qr + pr, which you read straight off the cubic's coefficients. You never need the roots themselves, and the 6767 is never used.

Alternative approach

If you know derivatives: for f(x)=(x−p)(x−q)(x−r)f(x) = (x-p)(x-q)(x-r), the value f′(p)f'(p) equals (p−q)(p−r)(p-q)(p-r). So the sum is f′(p)+f′(q)+f′(r)f'(p) + f'(q) + f'(r) with f′(x)=3x2−44x+80f'(x) = 3x^2 - 44x + 80. That is 3(p2+q2+r2)−44(p+q+r)+3⋅80=3⋅324−44⋅22+240=972−968+240=2443(p^2 + q^2 + r^2) - 44(p + q + r) + 3 \cdot 80 = 3 \cdot 324 - 44 \cdot 22 + 240 = 972 - 968 + 240 = 244.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Since s3−22s2+80s−67=(s−p)(s−q)(s−r)s^3 - 22s^2 + 80s - 67 = (s-p)(s-q)(s-r), multiply the identity by this product:
1=A(s−q)(s−r)+B(s−p)(s−r)+C(s−p)(s−q).1 = A(s-q)(s-r) + B(s-p)(s-r) + C(s-p)(s-q).
This holds for all ss (both sides are polynomials that agree at infinitely many points). Substituting s=ps = p kills the BB and CC terms and leaves 1=A(p−q)(p−r)1 = A(p-q)(p-r), so
1A=(p−q)(p−r),1B=(q−p)(q−r),1C=(r−p)(r−q).\frac1A = (p-q)(p-r),\qquad \frac1B = (q-p)(q-r),\qquad \frac1C = (r-p)(r-q).

Add them. Expanding each product,
(p−q)(p−r)=p2−p(q+r)+qr,(p-q)(p-r) = p^2 - p(q+r) + qr,
and summing the three cyclic versions gives
p2+q2+r2−2(pq+qr+rp)+(pq+qr+rp)=p2+q2+r2−(pq+qr+rp).p^2 + q^2 + r^2 - 2(pq + qr + rp) + (pq + qr + rp) = p^2 + q^2 + r^2 - (pq + qr + rp).
Using p2+q2+r2=(p+q+r)2−2(pq+qr+rp)p^2 + q^2 + r^2 = (p+q+r)^2 - 2(pq+qr+rp), the total is
(p+q+r)2−3(pq+qr+rp).(p+q+r)^2 - 3(pq + qr + rp).
By Vieta, p+q+r=22p + q + r = 22 and pq+qr+rp=80pq + qr + rp = 80, so the sum is 222−3⋅80=484−240=24422^2 - 3 \cdot 80 = 484 - 240 = 244.

The answer is (B) 244\boxed{\textbf{(B)}\ 244}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.