Difference between revisions of "1966 AHSME Problems/Problem 33"
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== 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 | ||
+ | <cmath>\frac{x-a}{b}+\frac{x-b}{a}=0</cmath> | ||
+ | <cmath>\frac{x-a}{b}=\frac{b-x}{a}</cmath> | ||
+ | <cmath>x=\frac{a^2+b^2}{a+b}</cmath> | ||
+ | Which yields <math>1</math> solution for <math>x</math>. | ||
+ | |||
+ | If <math>mn=0</math> then | ||
+ | <cmath>(\frac{x-a}{b})(\frac{x-b}{a})=1</cmath> | ||
+ | <cmath>x^2-(a+b)x=0</cmath> | ||
+ | Solving the quadratic gets another <math>2</math> solutions for <math>x</math>. | ||
+ | |||
+ | Thus there are <math>\boxed{\text{(D) three}}</math> solutions in total. | ||
+ | |||
+ | ~ Nafer | ||
== See also == | == See also == |
Latest revision as of 13:04, 4 July 2020
Contents
Problem
If and
, the number of distinct values of
satisfying the equation
is:
Solution
Solution 2
Let and
then we have
Notice that the equation is possible iff
or
.
If then
Which yields
solution for
.
If then
Solving the quadratic gets another
solutions for
.
Thus there are solutions in total.
~ Nafer
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.