Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1958 AHSME Problems/Problem 9"

(wrote solution)
 
m
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
A value of <math> x</math> satisfying the equation <math> x^2 \plus{} b^2 \equal{} (a \minus{} x)^2</math> is:
+
A value of <math> x</math> satisfying the equation <math> x^2 + b^2 = (a - x)^2</math> is:
  
−
<math> \textbf{(A)}\ \frac{b^2 \plus{} a^2}{2a}\qquad  
+
<math> \textbf{(A)}\ \frac{b^2 + a^2}{2a}\qquad  
−
\textbf{(B)}\ \frac{b^2 \minus{} a^2}{2a}\qquad  
+
\textbf{(B)}\ \frac{b^2 - a^2}{2a}\qquad  
−
\textbf{(C)}\ \frac{a^2 \minus{} b^2}{2a}\qquad  
+
\textbf{(C)}\ \frac{a^2 - b^2}{2a}\qquad  
−
\textbf{(D)}\ \frac{a \minus{} b}{2}\qquad  
+
\textbf{(D)}\ \frac{a - b}{2}\qquad  
−
\textbf{(E)}\ \frac{a^2 \minus{} b^2}{2}</math>
+
\textbf{(E)}\ \frac{a^2 - b^2}{2}</math>
  
 
== Solution ==
 
== Solution ==
Line 23: Line 23:
  
  
−
{{AHSME box|year=1958|num-b=8|num-a=10}}
+
{{AHSME 50p box|year=1958|num-b=8|num-a=10}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:06, 13 March 2015

Problem

A value of $x$ satisfying the equation $x^2 + b^2 = (a - x)^2$ is:

$\textbf{(A)}\ \frac{b^2 + a^2}{2a}\qquad \textbf{(B)}\ \frac{b^2 - a^2}{2a}\qquad \textbf{(C)}\ \frac{a^2 - b^2}{2a}\qquad \textbf{(D)}\ \frac{a - b}{2}\qquad \textbf{(E)}\ \frac{a^2 - b^2}{2}$

Solution

Solve for x:

\[x^2+b^2=(a-x)^2\] \[x^2+b^2=x^2-2ax+a^2\] \[b^2=-2ax+a^2\] \[2ax=a^2-b^2\] \[x=\frac{a^2-b^2}{2a} \to \boxed{\text{(C)}}\]


See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1958_AHSME_Problems/Problem_9&oldid=68942"