Difference between revisions of "1966 AHSME Problems/Problem 33"
(→Solution) |
(→Solution 2) |
||
| Line 12: | Line 12: | ||
== Solution 2 == | == Solution 2 == | ||
| + | |||
| + | Let <math>m=\frac{x-a}{b}</math> and <math>n=\frac{x-b}{a}</math> then we have | ||
| + | <cmath>m+n=\frac{1}{m}+\frac{1}{n}</cmath> | ||
| + | <cmath>m+n=\frac{m+n}{mn}</cmath> | ||
| + | Notice that the equation is possible iff <math>m+n=0</math> or <math>mn=1</math>. | ||
| + | |||
| + | If <math>m+n=0</math> then | ||
| + | <math></math>\frac{ | ||
== See also == | == See also == | ||
Revision as of 19:34, 23 December 2019
Contents
Problem
If 
and 
, the number of distinct values of 
satisfying the equation
![\[\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b},\]](http://latex.artofproblemsolving.com/b/4/d/b4dfad7746d087a93d579bcbceadafc0c3ca8d55.png)
is:
Solution
Solution 2
Let 
and 
then we have
![\[m+n=\frac{1}{m}+\frac{1}{n}\]](http://latex.artofproblemsolving.com/2/d/9/2d9a16cf378036b0d90044e1501ce0fab7cac286.png)
![\[m+n=\frac{m+n}{mn}\]](http://latex.artofproblemsolving.com/c/5/2/c52338c48d698a51f3795fa1dbff72e32ee8b735.png)
Notice that the equation is possible iff 
or 
.
If then
$$ (Error compiling LaTeX. Unknown error_msg)\frac{
See also
| 1966 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 32 |
Followed by Problem 34 | |
| 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 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
