1978 AHSME Problems/Problem 24

Problem

If the distinct non-zero numbers $x ( y - z),~ y(z - x),~ z(x - y )$ form a geometric progression with common ratio $r$ , then $r$ satisfies the equation

$\textbf{(A) }r^2+r+1=0\qquad \textbf{(B) }r^2-r+1=0\qquad \textbf{(C) }r^4+r^2-1=0\qquad\\ \textbf{(D) }(r+1)^4+r=0\qquad \textbf{(E) }(r-1)^4+r=0$

Solution

Let the geometric progression be $a,$ $ar,$ $ar^2,$ so $a = x(y - z)$ , $ar = y(z - x)$ , and $ar^2 = z(x - y).$ Adding these equations, we get $a + ar + ar^2 = xy - xz + yz - xy + xz - yz = 0.$ Since $a$ is nonzero, we can divide by $a$ , to get $\boxed{r^2 + r + 1 = 0}$ . The answer is (A).

Page

Toolbox

Search

1978 AHSME Problems/Problem 24

Problem

Solution