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Difference between revisions of "1957 AHSME Problems/Problem 49"

(Created page with "== Problem == The parallel sides of a trapezoid are <math>3</math> and <math>9</math>. The non-parallel sides are <math>4</math> and <math>6</math>. A line parallel to the ba...")
 
(→‎Solution)
 
(One intermediate revision by the same user not shown)
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== Solution ==
 
== Solution ==
−
<math>\boxed{\textbf{(C) }4:1}</math>.
+
 
 +
<asy>
 +
 
 +
import geometry;
 +
 
 +
defaultpen(linewidth(.8pt));
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unitsize(2cm);
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point F = origin;
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point G = (2.25,0);
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point C = (2,1);
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point B = (1,1);
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point D = 3/4*F + 1/4*B;
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point E = 3/4 * G + 1/4 * C;
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point A;
 +
 
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// Defining A
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pair[] a = intersectionpoints(D--B*2, E--2*C-G);
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A = a[0];
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// Trapezoid, parallel segment
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draw(F--G--C--B--cycle);
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draw(D--E);
 +
 
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// Segments AB and AC
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draw(A--B);
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draw(A--C);
 +
 
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// Points w/ labels
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dot(A);
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label("A",A,N);
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dot(B);
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label("B",B,NW);
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dot(C);
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label("C",C,NE);
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dot(D);
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label("D",D,NW);
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dot(E);
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label("E",E,NE);
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dot(F);
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label("F",F,SW);
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dot(G);
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label("G",G,SE);
 +
 
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// Length Labels
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label("3",midpoint(C--B),S);
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label("9",midpoint(F--G),S);
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</asy>
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 +
Let the points be labeled as in the new diagram above, with <math>BF=6</math> and <math>CG=4</math> (from the problem). Because <math>\overline{BC} \parallel \overline{FG}</math> and <math>\tfrac{BC}{FG}=\tfrac 1 3</math>, <math>AB=\tfrac{AF}3=\tfrac{AB+6}3</math> and <math>AC=\tfrac{AG}3=\tfrac{AC+4}3</math>. Solving these equations for <math>AB</math> and <math>AC</math>, respectively yields <math>AB=3</math> and <math>AC=2</math>. Let <math>DE=x</math>. Thus, because <math>\overline{DE} \parallel \overline{BC}</math>, <math>\tfrac{BC}{DE}=\tfrac 3 x=\tfrac{AB}{AD}=\tfrac3{3+BD}</math>. Solving this equation for <math>BD</math> yields <math>BD=x-3</math>. Similarly, <math>\tfrac 3 x=\tfrac{AC}{AE}=\tfrac2{2+CE}</math>, and solving this equation for <math>CE</math> yields <math>CE=\tfrac{2x}3-2</math>.
 +
 
 +
Now, we can set the perimeters of <math>BCED</math> and <math>DEGF</math> equal to each other to solve for <math>x</math>:
 +
\begin{align*}
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3+(x-3)+x+(\frac{2x}3-2) &= x+[6-(x-3)]+9+[4-(\frac{2x}3-2)] \\
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\frac{8x}3-2 &= 9+9+4-\frac{2x}3+2 \\
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\frac{10x}3 &= 26 \\
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\frac{5x}3 &= 13 \\
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5x &= 39 \\
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x &= \frac{39}5
 +
\end{align*}
 +
To find the ratio <math>\tfrac{BD}{DF}=\tfrac{x-3}{6-(x-3)}</math>, we substitute <math>x=\tfrac{39}5</math> into this expression to find our answer:
 +
\begin{align*}
 +
\frac{x-3}{6-(x-3)} &= \frac{x-3}{9-x} \\
 +
&= \frac{\tfrac{39-15}5}{\tfrac{45-39}5} \\
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&= \frac{24}6 \\
 +
&= \frac 4 1
 +
\end{align*}
 +
Thus, our answer is <math>\boxed{\textbf{(C) }4:1}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 14:19, 27 July 2024

Problem

The parallel sides of a trapezoid are $3$ and $9$. The non-parallel sides are $4$ and $6$. A line parallel to the bases divides the trapezoid into two trapezoids of equal perimeters. The ratio in which each of the non-parallel sides is divided is:

[asy] defaultpen(linewidth(.8pt)); unitsize(2cm); pair A = origin; pair B = (2.25,0); pair C = (2,1); pair D = (1,1); pair E = waypoint(A--D,0.25); pair F = waypoint(B--C,0.25); draw(A--B--C--D--cycle); draw(E--F); label("6",midpoint(A--D),NW); label("3",midpoint(C--D),N); label("4",midpoint(C--B),NE); label("9",midpoint(A--B),S);[/asy]

$\textbf{(A)}\ 4: 3\qquad\textbf{(B)}\ 3: 2\qquad\textbf{(C)}\ 4: 1\qquad\textbf{(D)}\ 3: 1\qquad\textbf{(E)}\ 6: 1$

Solution

[asy] import geometry; defaultpen(linewidth(.8pt)); unitsize(2cm); point F = origin; point G = (2.25,0); point C = (2,1); point B = (1,1); point D = 3/4*F + 1/4*B; point E = 3/4 * G + 1/4 * C; point A; // Defining A pair[] a = intersectionpoints(D--B*2, E--2*C-G); A = a[0]; // Trapezoid, parallel segment draw(F--G--C--B--cycle); draw(D--E); // Segments AB and AC draw(A--B); draw(A--C); // Points w/ labels dot(A); label("A",A,N); dot(B); label("B",B,NW); dot(C); label("C",C,NE); dot(D); label("D",D,NW); dot(E); label("E",E,NE); dot(F); label("F",F,SW); dot(G); label("G",G,SE); // Length Labels label("3",midpoint(C--B),S); label("9",midpoint(F--G),S); [/asy]

Let the points be labeled as in the new diagram above, with $BF=6$ and $CG=4$ (from the problem). Because $\overline{BC} \parallel \overline{FG}$ and $\tfrac{BC}{FG}=\tfrac 1 3$, $AB=\tfrac{AF}3=\tfrac{AB+6}3$ and $AC=\tfrac{AG}3=\tfrac{AC+4}3$. Solving these equations for $AB$ and $AC$, respectively yields $AB=3$ and $AC=2$. Let $DE=x$. Thus, because $\overline{DE} \parallel \overline{BC}$, $\tfrac{BC}{DE}=\tfrac 3 x=\tfrac{AB}{AD}=\tfrac3{3+BD}$. Solving this equation for $BD$ yields $BD=x-3$. Similarly, $\tfrac 3 x=\tfrac{AC}{AE}=\tfrac2{2+CE}$, and solving this equation for $CE$ yields $CE=\tfrac{2x}3-2$.

Now, we can set the perimeters of $BCED$ and $DEGF$ equal to each other to solve for $x$: \begin{align*} 3+(x-3)+x+(\frac{2x}3-2) &= x+[6-(x-3)]+9+[4-(\frac{2x}3-2)] \\ \frac{8x}3-2 &= 9+9+4-\frac{2x}3+2 \\ \frac{10x}3 &= 26 \\ \frac{5x}3 &= 13 \\ 5x &= 39 \\ x &= \frac{39}5 \end{align*} To find the ratio $\tfrac{BD}{DF}=\tfrac{x-3}{6-(x-3)}$, we substitute $x=\tfrac{39}5$ into this expression to find our answer: \begin{align*} \frac{x-3}{6-(x-3)} &= \frac{x-3}{9-x} \\ &= \frac{\tfrac{39-15}5}{\tfrac{45-39}5} \\ &= \frac{24}6 \\ &= \frac 4 1 \end{align*} Thus, our answer is $\boxed{\textbf{(C) }4:1}$.

See Also

1957 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 48 		Followed by
Problem 50
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