Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 16"

Latest revision as of 09:28, 27 April 2008

Problem

If $x_1,x_2$ are the roots of the equation $x^2-2kx+2m=0$ , then $\frac{1}{x_1},\frac{1}{x_2}$ are the roots of the equation

$\mathrm{(A)}\ x^2-2k^2x+2m^2=0\qquad\mathrm{(B)}\ x^2-\frac{k}{m}x+\frac{1}{2m}=0\qquad\mathrm{(C)}\ x^2-\frac{m}{k}x+\frac{1}{2m}=0\\ \mathrm{(D)}\ 2mx^2-kx+1=0\qquad\mathrm{(E)}\ 2kx^2-2mx+1=0$

Solution

By Vieta’s, $x_1 + x_2 = 2k,\ x_1 \cdot x_2 = 2m$ .

The equation with roots $x = \frac{1}{x_1}, \frac{1}{x_2}$ is $0 = \left(x - \frac{1}{x_1}\right)\left(x - \frac{1}{x_2}\right)$ $= x^2 - \left(\frac{1}{x_1} + \frac{1}{x_2}\right) + \frac{1}{x_1x_2} = x^2 - \left(\frac{x_1 + x_2}{x_1x_2}\right) + \frac{1}{x_1x_2}$ . Substituting from above, we get $x^2 - \frac{k}{m}x + \frac{1}{2m}= 0 \Longrightarrow \mathrm{(B)}$ .

@@ Line 1: / Line 1: @@
 ==Problem==
-If <math>x_1,x_2</math> are the roots of the equation <math>x^2-2kx+2m=0</math>, then <math>\frac{1}{x_1},\frac{1}{x_2}</math> are the roots of the equation
+If <math>x_1,x_2</math> are the [[root]]s of the [[equation]] <math>x^2-2kx+2m=0</math>, then <math>\frac{1}{x_1},\frac{1}{x_2}</math> are the roots of the equation
-A. <math>x^2-2k^2x+2m^2=0</math>
+<math>\mathrm{(A)}\ x^2-2k^2x+2m^2=0\qquad\mathrm{(B)}\ x^2-\frac{k}{m}x+\frac{1}{2m}=0\qquad\mathrm{(C)}\ x^2-\frac{m}{k}x+\frac{1}{2m}=0\\ \mathrm{(D)}\ 2mx^2-kx+1=0\qquad\mathrm{(E)}\ 2kx^2-2mx+1=0</math>
-B. <math>x^2-\frac{k}{m}x+\frac{1}{2m}=0</math>
-C. <math>x^2-\frac{m}{k}x+\frac{1}{2m}=0</math>
-D. <math>2mx^2-kx+1=0</math>
-E. <math>2kx^2-2mx+1=0</math>
 ==Solution==

2006 Cyprus MO, Lyceum (Problems)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30

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Difference between revisions of "2006 Cyprus MO/Lyceum/Problem 16"

Latest revision as of 09:28, 27 April 2008

Problem

Solution

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