1957 AHSME Problems/Problem 49

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 $\boxed{\textbf{(C) }4:1}$ .

1957 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 48	Followed by Problem 50
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
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1957 AHSME Problems/Problem 49

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