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Difference between revisions of "1978 AHSME Problems/Problem 24"

(Created page with "Let the geometric progression be <math>a,</math> <math>ar,</math> <math>ar^2,</math> so <math>a = x(y - z)</math>, <math>ar = y(z - x)</math>, and <math>ar^2 = z(x - y).</math...")
 
 
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==Problem==
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If the distinct non-zero numbers <math>x ( y - z),~ y(z - x),~ z(x - y )</math> form a geometric progression with common ratio <math>r</math>, then <math>r</math> satisfies the equation
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<math>\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</math>
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==Solution==
 
Let the geometric progression be <math>a,</math> <math>ar,</math> <math>ar^2,</math> so <math>a = x(y - z)</math>, <math>ar = y(z - x)</math>, and <math>ar^2 = z(x - y).</math> Adding these equations, we get
 
Let the geometric progression be <math>a,</math> <math>ar,</math> <math>ar^2,</math> so <math>a = x(y - z)</math>, <math>ar = y(z - x)</math>, and <math>ar^2 = z(x - y).</math> Adding these equations, we get
 
<cmath>a + ar + ar^2 = xy - xz + yz - xy + xz - yz = 0.</cmath>
 
<cmath>a + ar + ar^2 = xy - xz + yz - xy + xz - yz = 0.</cmath>
 
Since <math>a</math> is nonzero, we can divide by <math>a</math>, to get <math>\boxed{r^2 + r + 1 = 0}</math>. The answer is (A).
 
Since <math>a</math> is nonzero, we can divide by <math>a</math>, to get <math>\boxed{r^2 + r + 1 = 0}</math>. The answer is (A).

Latest revision as of 14:06, 20 June 2021

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).

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