Difference between revisions of "1959 AHSME Problems/Problem 35"
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| − | == Problem | + | == Problem == |
The symbol <math>\ge</math> means "greater than or equal to"; the symbol <math>\le</math> means "less than or equal to". | The symbol <math>\ge</math> means "greater than or equal to"; the symbol <math>\le</math> means "less than or equal to". | ||
In the equation <math>(x-m)^2-(x-n)^2=(m-n)^2; m</math> is a fixed positive number, and <math>n</math> is a fixed negative number. The set of values x satisfying the equation is: | In the equation <math>(x-m)^2-(x-n)^2=(m-n)^2; m</math> is a fixed positive number, and <math>n</math> is a fixed negative number. The set of values x satisfying the equation is: | ||
<math>\textbf{(A)}\ x\ge 0 \qquad\textbf{(B)}\ x\le n\qquad\textbf{(C)}\ x=0\qquad\textbf{(D)}\ \text{the set of all real numbers}\qquad\textbf{(E)}\ \text{none of these} </math> | <math>\textbf{(A)}\ x\ge 0 \qquad\textbf{(B)}\ x\le n\qquad\textbf{(C)}\ x=0\qquad\textbf{(D)}\ \text{the set of all real numbers}\qquad\textbf{(E)}\ \text{none of these} </math> | ||
| − | + | == Solution == | |
| − | ==See also== | + | Applying the [[difference of squares]] technique on this problem, we can see that <cmath>(x-m)^2-(x-n)^2=((x-m)+(x-n)) \cdot ((x-m)-(x-n)),</cmath> so <cmath>(x-m)^2-(x-n)^2=((x-m)+(x-n)) \cdot ((x-m)-(x-n))=(m-n)^2.</cmath> Simplifying gives us<cmath>(2x-m-n)\cdot(n-m)=(m-n)^2.</cmath>Negating <math>n-m</math> creates:<cmath>-(2x-m-n)\cdot(m-n)=(m-n)^2.</cmath>Dividing by <math>m-n</math>, <cmath>-(2x-m-n)=m-n</cmath> <cmath>-2x+m+n=m-n</cmath> <cmath>-2x+n=-n</cmath> <cmath>-2x=-2n</cmath> <cmath>x=n</cmath>Lastly, since <math>n</math> is a fixed negative number, <math>x</math> must also be a fixed negative number, so <math>x<0</math>. Since this answer is not <math>A, B, C, </math> or <math> D</math>, the solution must be <math>\fbox{\textbf{(E) }none of these}</math>. |
| + | |||
| + | == See also == | ||
{{AHSME 50p box|year=1959|num-b=34|num-a=36}} | {{AHSME 50p box|year=1959|num-b=34|num-a=36}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 14:44, 21 July 2024
Problem
The symbol 
means "greater than or equal to"; the symbol 
means "less than or equal to".
In the equation 
is a fixed positive number, and 
is a fixed negative number. The set of values x satisfying the equation is:
Solution
Applying the difference of squares technique on this problem, we can see that ![\[(x-m)^2-(x-n)^2=((x-m)+(x-n)) \cdot ((x-m)-(x-n)),\]](http://latex.artofproblemsolving.com/0/7/d/07dd4b5f5c37003cdddd485440ca0541191b67d8.png)
so ![\[(x-m)^2-(x-n)^2=((x-m)+(x-n)) \cdot ((x-m)-(x-n))=(m-n)^2.\]](http://latex.artofproblemsolving.com/b/4/5/b455f0781ca206ff3dde5ede63a42d7d82875a2e.png)
Simplifying gives us![\[(2x-m-n)\cdot(n-m)=(m-n)^2.\]](http://latex.artofproblemsolving.com/b/b/5/bb5de395006ede0be188fe58d8d2394a66386af7.png)
Negating 
creates:![\[-(2x-m-n)\cdot(m-n)=(m-n)^2.\]](http://latex.artofproblemsolving.com/1/0/5/10518cb021e9238ad92bb4bd5c0b84d52a7e2998.png)
Dividing by 
, ![\[-(2x-m-n)=m-n\]](http://latex.artofproblemsolving.com/2/f/e/2fe9a92efd89d4a2edda535b2074c0db80a998a0.png)
![\[-2x+m+n=m-n\]](http://latex.artofproblemsolving.com/d/8/7/d87ef4bc7c0b9d83284f9884ddf856155bd154d5.png)
![\[-2x+n=-n\]](http://latex.artofproblemsolving.com/d/4/f/d4f516fc1155b4c4c4d603c8201b129e00550d09.png)
![\[-2x=-2n\]](http://latex.artofproblemsolving.com/6/9/8/6981914030017c37f4012eb98053ee8abf98a1b0.png)
![\[x=n\]](http://latex.artofproblemsolving.com/e/6/1/e61228f5d6ce64239bbae8874abb7b6533a33bbb.png)
Lastly, since is a fixed negative number, 
must also be a fixed negative number, so 
. Since this answer is not 
or 
, the solution must be 
.
See also
| 1959 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 34 |
Followed by Problem 36 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
