Difference between revisions of "1966 AHSME Problems/Problem 33"

Revision as of 18:38, 23 December 2019

Problem

If $ab \ne 0$ and $|a| \ne |b|$ , the number of distinct values of $x$ satisfying the equation

$\frac{x-a}{b}+\frac{x-b}{a}=\frac{b}{x-a}+\frac{a}{x-b},$

is:

$\text{(A) zero} \quad \text{(B) one} \quad \text{(C) two} \quad \text{(D) three} \quad \text{(E) four}$

Solution

$\fbox{D}$

Solution 2

Let $m=\frac{x-a}{b}$ and $n=\frac{x-b}{a}$ then we have $m+n=\frac{1}{m}+\frac{1}{n}$ $m+n=\frac{m+n}{mn}$ Notice that the equation is possible iff $m+n=0$ or $mn=1$ .

If $m+n=0$ then $\frac{x-a}{b}+\frac{x-b}{a}=0$ $\frac{x-a}{b}=\frac{b-x}{a}$ $x=\frac{a^2+b^2}{a+b}$ Which yields $1$ solution for $x$ .

If $mn=0$ then $(\frac{x-a}{b})(\frac{x-b}{a})=1$ $x^2-(a+b)x=0$ Solving the quadratic gets another $2$ solutions for $x$ .

Thus there are $3 \boxed{D}$ solutions in total.

~ Nafer

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.

@@ Line 30: / Line 30: @@
 Thus there are <math>3 \boxed{D}</math> solutions in total.
+~ Nafer
 == See also ==

Page

Toolbox

Search

Difference between revisions of "1966 AHSME Problems/Problem 33"

Revision as of 18:38, 23 December 2019

Contents

Problem

Solution

Solution 2

See also