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

Latest revision as of 20:25, 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

@@ Line 15: / Line 15: @@
 ==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}=\boxed{2}</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>
+==Solution 3==
+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>
+<math>\implies \boxed{(E)}</math>.
+<math>\textbf{Proof of the used property:}</math>
+Let <cmath>\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3} = ... = \frac{a_n}{b_n} = r.</cmath>
+Therefore, <cmath>\begin{align*} a_1 &= b_1r,\\ a_2 &= b_2r,\\ ...,\\ a_n &= b_nr \end{align*}</cmath>
+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 ==

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
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