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 4,4,5 is
31(41+41+51)1=730.
What is the harmonic mean of all the real roots of the 4050th degree polynomial
k=1∏2025(kx2−4x−3)?
Answer Choices:
A. −35
B. −23
C. −56
D. −65
E. −32
Solution:
Let
P(x)=k=1∏2025(kx2−4x−3)
In order to compute the harmonic mean of all the roots of the polynomial P(x), we need to find the sum of the reciprocals of all the roots of P(x).
We observe that each quadratic factor in the polynomial P(x) is kx2−4x−3. Let its roots be rk,1 and rk,2.
For any quadratic ax2+bx+c=0, we know:
r1+r2=−abandr1r2=ac.
Thus,
r11+r21=r1r2r1+r2=c/a−b/a=c−b.
For the quadratic kx2−4x−3.here a=k,b=−4,c=−3, therefore
rk,11+rk,21=−3−(−4)=−34.
Since there are 2025 such quadratics, the total number of roots is 4050, and the sum of the reciprocals of all roots is
∑r1=2025(−34)=−2700.
The harmonic mean H of all 4050 roots is given by
H=∑r14050=−27004050=(B) -23.
The problems on this page are the property of the MAA's American Mathematics Competitions