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Difference between revisions of "1992 AHSME Problems/Problem 16"

(→‎Solution 2)
(New alternate solution using proportions (+ proof of the property I used in the solution). Delete if you want.)
 
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==Solution 2==
 
==Solution 2==
 
We cross multiply the first and third fractions and the second and third fractions, respectively, for <cmath>x(x-z)=y^2</cmath> <cmath>y(x+y)=xz</cmath> Notice how the first equation can be expanded and rearranged to contain an <math>(x+y)</math> term. <cmath>x^2-xz=y^2</cmath> <cmath>x^2-y^2=xz</cmath> <cmath>(x+y)(x-y)=xz</cmath> We can divide this by the second equation to get <cmath>\frac{(x+y)(x-y)}{y(x+y)}=\frac{xz}{xz}</cmath> <cmath>\frac{x-y}{y}=1</cmath> <cmath>\frac{x}{y}-1=1</cmath> <cmath>\frac{x}{y}=2 \rightarrow \boxed{E}</cmath>
 
We cross multiply the first and third fractions and the second and third fractions, respectively, for <cmath>x(x-z)=y^2</cmath> <cmath>y(x+y)=xz</cmath> Notice how the first equation can be expanded and rearranged to contain an <math>(x+y)</math> term. <cmath>x^2-xz=y^2</cmath> <cmath>x^2-y^2=xz</cmath> <cmath>(x+y)(x-y)=xz</cmath> We can divide this by the second equation to get <cmath>\frac{(x+y)(x-y)}{y(x+y)}=\frac{xz}{xz}</cmath> <cmath>\frac{x-y}{y}=1</cmath> <cmath>\frac{x}{y}-1=1</cmath> <cmath>\frac{x}{y}=2 \rightarrow \boxed{E}</cmath>
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==Solution 3==
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Since <cmath>\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = \frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n},</cmath> we can say that <cmath>\frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} = \frac {y+(x+y)+x}{(x-z)+z+y} = \boxed{2}.</cmath>
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<math>\implies \boxed{(E)}</math>.
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<math>\textbf{Proof of the used property:}</math>
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Let <cmath>\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = ... = \frac{a_n}{b_n} = r.</cmath>
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Therefore, <cmath>\begin{align*} a_1 &= b_1r,\\ a_2 &= b_2r,\\ ...,\\ a_n &= b_nr \end{align*}</cmath>
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so <cmath>\frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n} = \frac{b_1r + b_2r + ... + b_nr}{b_1 + b_2 + ... + b_n}</cmath> <cmath>= \frac{(b_1 + b_2 + ... + b_n)(nr)}{(b_1 + b_2 + ... + b_n)(n)} = r </cmath> --- NamelyOrange
  
 
== See also ==
 
== See also ==

Latest revision as of 21:25, 2 May 2023

Problem

If \[\frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}\] for three positive numbers $x,y$ and $z$, all different, then $\frac{x}{y}=$

$\text{(A) } \frac{1}{2}\quad \text{(B) } \frac{3}{5}\quad \text{(C) } \frac{2}{3}\quad \text{(D) } \frac{5}{3}\quad \text{(E) } 2$

Solution 1

$\fbox{E}$ We have $\frac{x+y}{z} = \frac{x}{y} \implies xy+y^2=xz$ and $\frac{y}{x-z} = \frac{x}{y} \implies y^2=x^2-xz \implies x^2-y^2=xz$. Equating the two expressions for $xz$ gives $xy+y^2=x^2-y^2 \implies x^2-xy-2y^2=0 \implies (x+y)(x-2y)=0$, so as $x+y$ cannot be $0$ for positive $x$ and $y$, we must have $x-2y=0 \implies x=2y \implies \frac{x}{y}=2$.

Solution 2

We cross multiply the first and third fractions and the second and third fractions, respectively, for \[x(x-z)=y^2\] \[y(x+y)=xz\] Notice how the first equation can be expanded and rearranged to contain an $(x+y)$ term. \[x^2-xz=y^2\] \[x^2-y^2=xz\] \[(x+y)(x-y)=xz\] We can divide this by the second equation to get \[\frac{(x+y)(x-y)}{y(x+y)}=\frac{xz}{xz}\] \[\frac{x-y}{y}=1\] \[\frac{x}{y}-1=1\] \[\frac{x}{y}=2 \rightarrow \boxed{E}\]

Solution 3

Since \[\frac{a_1}{b_1} = \frac{a_2}{b_2} = ... = \frac{a_n}{b_n} = \frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n},\] we can say that \[\frac{y}{x-z} = \frac{x+y}{z} = \frac{x}{y} = \frac {y+(x+y)+x}{(x-z)+z+y} = \boxed{2}.\] $\implies \boxed{(E)}$.

$\textbf{Proof of the used property:}$

Let \[\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = ... = \frac{a_n}{b_n} = r.\]

Therefore, \begin{align*} a_1 &= b_1r,\\ a_2 &= b_2r,\\ ...,\\ a_n &= b_nr \end{align*} so \[\frac{a_1 + a_2+ ... + a_n}{b_1 + b_2 + ... + b_n} = \frac{b_1r + b_2r + ... + b_nr}{b_1 + b_2 + ... + b_n}\] \[= \frac{(b_1 + b_2 + ... + b_n)(nr)}{(b_1 + b_2 + ... + b_n)(n)} = r\] --- NamelyOrange

See also

1992 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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

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