Difference between revisions of "1964 AHSME Problems/Problem 32"

(Solution)
 
Line 31: Line 31:
 
<math>(a-c)(a+b+c+d) = 0</math>
 
<math>(a-c)(a+b+c+d) = 0</math>
  
This is equivalent to <math>\boxed{\textbf{(C)}</math>
+
This is equivalent to <math>\boxed{\textbf{(C)}}</math>
+
 
 
==See Also==
 
==See Also==
 
{{AHSME 40p box|year=1964|num-b=31|num-a=33}}
 
{{AHSME 40p box|year=1964|num-b=31|num-a=33}}

Latest revision as of 01:27, 25 July 2019

Problem

If $\dfrac{a+b}{b+c}=\dfrac{c+d}{d+a}$, then:

$\textbf{(A) }a \text{ must equal }c\qquad\textbf{(B) }a+b+c+d\text{ must equal zero}\qquad$

$\textbf{(C) }\text{either }a=c\text{ or }a+b+c+d=0\text{, or both}\qquad$

$\textbf{(D) }a+b+c+d\ne 0\text{ if }a=c\qquad$

$\textbf{(E) }a(b+c+d)=c(a+b+d)$

Solution

Cross-multiplying gives:

$(a+b)(a+d) = (b+c)(c+d)$

$a^2 + ad + ab + bd = bc + bd + c^2 + cd$

$a^2 + ad + ab   - bc - c^2 - cd = 0$

$a(a + b + d) - c(b+c+d)= 0$

This looks close to turning into option C, but we don't have a $c$ term in the first parentheses, and we don't have an $a$ term in the second parentheses to allow us to complete the factorization. However, if we both add $ac$ and subtract $ac$ on the LHS, we get:

$a(a + b + d) + ac - c(b+c+d) - ca= 0$

$a(a+b+d +c) - c(b+c+d+a) = 0$

$(a-c)(a+b+c+d) = 0$

This is equivalent to $\boxed{\textbf{(C)}}$

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 31
Followed by
Problem 33
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
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

Invalid username
Login to AoPS