AMC 10 · 2025 · #18

Grade 7 algebra

Archetype Arithmetic Expression Decomposition └ Sub-archetype Fraction / Expression Sum Evaluation

vieta-formulasquadratic-equationsfraction-arithmetic easier-related-problem ↑ Prerequisites: vieta-formulasquadratic-equations

📏 Medium solution 💡 2 insights

Problem

The harmonic mean of a collection of numbers is the reciprocal of the arithmetic mean of the reciprocals of the numbers in the collection. For example, the harmonic mean of $4,4,$ and $5$ is

$\frac{1}{\frac{1}{3}(\frac{1}{4}+\frac{1}{4}+\frac{1}{5})}=\frac{30}{7}.$

What is the harmonic mean of all the real roots of the $4050$ th degree polynomial

$\prod_{k=1}^{2025} (kx^2-4x-3)=(x^2-4x-3)(2x^2-4x-3)(3x^2-4x-3)...(2025x^2-4x-3)?$

Pick an answer.

(A)

$~-\frac{5}{3}$

(B)

$~-\frac{3}{2}$

(C)

$~-\frac{6}{5}$

(D)

$~-\frac{5}{6}$

(E)

$~-\frac{2}{3}$

AMC 10 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

Try it yourself first — the explanation is most useful after you’ve attempted it.

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Toolkit + CCSS Solution

Understand

Restated: The harmonic mean of a list of numbers is the reciprocal of the average of their reciprocals. A giant polynomial of degree 4050 is written as a product of 2025 quadratics; the k-th factor is k*x^2 - 4x - 3, for k running from 1 to 2025. Find the harmonic mean of all of its real roots.

Givens: Harmonic mean = 1 divided by the average of the reciprocals of the numbers.; The polynomial is the product (1*x^2-4x-3)(2*x^2-4x-3)...(2025*x^2-4x-3).; Every factor is a quadratic k*x^2 - 4x - 3; only the leading coefficient k changes, while -4 and -3 stay fixed.; There are 2025 quadratic factors, so the polynomial has degree 4050.

Unknowns: The harmonic mean of all real roots of the degree-4050 polynomial.

Understand

Plan

Primary tool: #9 Solve an Easier Related Problem

Secondary: #16 Change Focus / Count the Complement, #7 Identify Subproblems, #4 Introduce a Variable, #5 Look for a Pattern

Finding all 4050 roots is hopeless, so change what we chase: the harmonic mean depends only on the sum of the reciprocals of the roots, not the roots themselves. That sum splits factor by factor, so we solve one easy quadratic instead of the monster. For a single quadratic, the sum of the reciprocals of its two roots equals -b/c, which here erases k entirely. Every one of the 2025 factors then contributes the same amount, and a short multiplication and division finish the job.

Execute — Answer: B

#16 Change Focus / Count the Complement 6.NS.A.1 Step 1

Rewrite the harmonic mean

The harmonic mean of n numbers is 1 divided by the average of their reciprocals.
The average of the reciprocals is their sum divided by n, and taking the reciprocal flips that fraction over.
So the harmonic mean equals n divided by the sum of all the reciprocals.
Here n = 4050, so the only quantity we actually need is the sum of 1/root over all 4050 roots.

$$HM=\dfrac{1}{\frac{1}{n}\sum_{i=1}^{n}\frac{1}{x_i}}=\dfrac{n}{\sum_{i=1}^{n}\frac{1}{x_i}},\qquad n=4050$$

💡 The reciprocal of an average of reciprocals is just the count divided by their total, so the roots matter only through one sum.

#9 Solve an Easier Related Problem 7.EE.A.1 Step 2

Shrink it to one quadratic

The polynomial is a product of 2025 separate quadratics, and its roots are just all the roots of those quadratics collected together.
A sum breaks apart across the factors, so the sum of every 1/root equals the sum, over the 2025 quadratics, of the reciprocal-sum of that one quadratic's two roots.
That means we only have to understand a single quadratic k*x^2 - 4x - 3 and then add up 2025 copies.

$$\sum_{i=1}^{4050}\frac{1}{x_i}=\sum_{k=1}^{2025}\left(\frac{1}{r_k}+\frac{1}{s_k}\right),\quad r_k,s_k\text{ are the roots of }kx^2-4x-3$$

💡 A product of quadratics just pools all their roots, so a sum over the roots becomes a sum over the factors.

#4 Introduce a Variable 7.NS.A.1 Step 3

Reciprocal sum for one quadratic

For the quadratic k*x^2 - 4x - 3 with roots r and s, Vieta's formulas give the sum r + s = 4/k and the product r*s = -3/k.
Add the two reciprocals over a common denominator: 1/r + 1/s = (s + r) divided by (r*s).
Substituting, that is (4/k) divided by (-3/k).
The k in the top and bottom cancel, leaving -4/3.

$$\frac{1}{r}+\frac{1}{s}=\frac{r+s}{rs}=\frac{4/k}{-3/k}=-\frac{4}{3}$$

💡 Adding two reciprocals gives (sum of roots) over (product of roots), and for these quadratics that ratio is fixed.

#5 Look for a Pattern 7.NS.A.2 Step 4

Same answer for every k

The result -4/3 has no k left in it, so every one of the 2025 quadratics contributes the exact same amount, no matter its leading coefficient.
This makes sense: for any quadratic a*x^2 + b*x + c, the reciprocal-sum of its roots is -b/c, and here b = -4 and c = -3 are identical in all factors.
Adding 2025 copies of -4/3 gives the total reciprocal sum -2700.

$$\sum_{k=1}^{2025}\left(-\frac{4}{3}\right)=2025\times\left(-\frac{4}{3}\right)=-2700$$

💡 Only b and c decide the reciprocal-sum of a quadratic's roots, and those never change, so every factor pitches in the same -4/3.

#4 Introduce a Variable 6.NS.A.1 Step 5

Divide to get the harmonic mean

Put the total reciprocal sum back into the harmonic-mean formula from Step 1.
The harmonic mean is 4050 divided by -2700.
Both numbers share a factor of 1350: 4050 = 3*1350 and 2700 = 2*1350, so the fraction reduces to -3/2.
That matches choice (B).

$$HM=\frac{4050}{-2700}=-\frac{4050}{2700}=-\frac{3}{2}$$

💡 Once the reciprocal total is known, one division finishes the harmonic mean.

[1] #16 6.NS.A.1 The harmonic mean of n numbers is 1 divided by the average of their reciprocals.

[2] #9 7.EE.A.1 The polynomial is a product of 2025 separate quadratics, and its roots are just

[3] #4 7.NS.A.1 For the quadratic k*x^2 - 4x - 3 with roots r and s, Vieta's formulas give the s

[4] #5 7.NS.A.2 The result -4/3 has no k left in it, so every one of the 2025 quadratics contrib

[5] #4 6.NS.A.1 Put the total reciprocal sum back into the harmonic-mean formula from Step 1. Th

Review

Reasonableness: Test the smallest case by hand. The k = 1 factor is x^2 - 4x - 3, whose roots are 2 + sqrt(7) and 2 - sqrt(7); their reciprocal-sum is (2+sqrt7 + 2-sqrt7) / ((2)^2 - 7) = 4 / (-3) = -4/3, and their harmonic mean is 2 / (-4/3) = -3/2. Since every quadratic contributes the identical -4/3, stacking all 2025 of them keeps the harmonic mean at -3/2. The value is negative, which fits the fact that -3 in each factor forces a negative product of roots, so it is believable and equals choice (B).

Alternative: Skip Vieta and use the reciprocal directly. If y = 1/x, then x = 1/y, and substituting into k*x^2 - 4x - 3 = 0 and multiplying through by y^2 gives -3y^2 - 4y + k = 0, i.e. 3y^2 + 4y - k = 0. The sum of the reciprocals is the sum of the y-roots, which by Vieta is -4/3 for every k. This confirms the per-factor contribution without ever computing r + s and r*s separately.

CCSS standards used (min grade 7)

6.NS.A.1 Interpret and compute quotients of fractions and solve word problems (Rewriting the reciprocal of the average of reciprocals as n divided by the reciprocal sum, and dividing 4050 by -2700 to get the harmonic mean.)
7.EE.A.1 Add, subtract, factor, and expand linear expressions to produce equivalent expressions (Splitting the sum of reciprocals across the 2025 quadratic factors so the whole polynomial reduces to one quadratic.)
7.NS.A.1 Apply and extend understanding of addition and subtraction to rational numbers (Adding the two reciprocals 1/r + 1/s over a common denominator to get (r+s)/(rs) and simplifying it to -4/3.)
7.NS.A.2 Apply and extend previous understandings of multiplication and division to rational numbers (Multiplying the fixed contribution -4/3 by the 2025 factors to obtain the total reciprocal sum -2700.)

⭐ When a mean only cares about the reciprocals of the roots, you can skip solving and use that the reciprocal-sum of a quadratic's roots is just -b/c.

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: B

Rewrite the harmonic mean

Shrink it to one quadratic

Reciprocal sum for one quadratic

Same answer for every k

Divide to get the harmonic mean

Review

CCSS standards used (min grade 7)

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