Difference between revisions of "1957 AHSME Problems/Problem 49"
(diagram) |
(→Solution) |
||
Line 72: | Line 72: | ||
</asy> | </asy> | ||
− | <math>\boxed{\textbf{(C) }4:1}</math>. | + | 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*} | ||
+ | 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 <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} \\ | ||
+ | &= \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 15:19, 27 July 2024
Problem
The parallel sides of a trapezoid are and
. The non-parallel sides are
and
.
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:
Solution
Let the points be labeled as in the new diagram above, with and
(from the problem). Because
and
,
and
. Solving these equations for
and
, respectively yields
and
. Let
. Thus, because
,
. Solving this equation for
yields
. Similarly,
, and solving this equation for
yields
.
Now, we can set the perimeters of and
equal to each other to solve for
:
\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
, we substitute
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
.
See Also
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.