Difference between revisions of "1965 AHSME Problems/Problem 12"
(added solution page) |
(→Solution: with diagram) |
||
(One intermediate revision by the same user not shown) | |||
Line 11: | Line 11: | ||
== Solution == | == Solution == | ||
− | <math>\ | + | |
+ | <asy> | ||
+ | |||
+ | draw((0,0)--(6,0)); | ||
+ | dot((0,0)); | ||
+ | label("A", (-0.75,-0.5)); | ||
+ | dot((6,0)); | ||
+ | label("C", (6.75,-0.5)); | ||
+ | label("$6$", (3,-1.75)); | ||
+ | |||
+ | |||
+ | draw((0,0)--(12*29/36, 12*sqrt(455)/36)--(6,0)); | ||
+ | dot((12*29/36, 12*sqrt(455)/36)); | ||
+ | label("B", (12*29/36+0.4, 12*sqrt(455)/36+0.75)); | ||
+ | label("$12$", (6*29/36-1.5, 6*sqrt(455)/36+1.5)); | ||
+ | label("$8$", (6*29/36+4.5, 6*sqrt(455)/36-0.5)); | ||
+ | |||
+ | dot((4*29/36, 4*sqrt(455)/36)); | ||
+ | label("F", (4*29/36-0.75, 4*sqrt(455)/36)); | ||
+ | dot((4*29/36+4, 4*sqrt(455)/36)); | ||
+ | label("E", (4*29/36+4+0.65, 4*sqrt(455)/36-0.25)); | ||
+ | dot((4,0)); | ||
+ | label("D", (4,-0.75)); | ||
+ | |||
+ | draw((0,0)--(4,0)--(4*29/36+4, 4*sqrt(455)/36)--(4*29/36, 4*sqrt(455)/36)); | ||
+ | |||
+ | </asy> | ||
+ | |||
+ | As in the diagram, suppose the rhombus <math>ADEF</math> is inscribed in <math>\triangle ABC</math> with <math>D</math> on <math>\overline{AC}</math>, <math>E</math> on <math>\overline{BC}</math>, and <math>F</math> on <math>\overline{AB}</math>. Let the side length of the rhombus be <math>x</math>. Because a rhombus is a parallelogram, its opposite sides are parallel. Thus, by [[AA similarity]], <math>\triangle BFE \sim \triangle EDC</math>, and so <math>\frac{FE}{FB}=\frac{DC}{DE}</math>. Because <math>\overline{FE}</math> and <math>\overline{DE}</math> are sides of the rhombus, they both have length <math>x</math>. Furthermore, <math>FB=12-x</math>, and <math>DC=6-x</math>, because, combined with sides of the rhombus, they form sides of the triangle. Thus, by substituting into the proportion derived above, we see that: | ||
+ | \begin{align*} | ||
+ | \frac{FE}{FB}&=\frac{DC}{DE} \\ | ||
+ | \frac{x}{12-x}&=\frac{6-x}{x} \\ | ||
+ | x^2&=72-12x-6x+x^2 \\ | ||
+ | 18x&=72 \\ | ||
+ | x&=4 \\ | ||
+ | \end{align*} | ||
+ | |||
+ | Thus, the side of the rhombus has length <math>\boxed{\text{(D) }4}</math>. | ||
==See Also== | ==See Also== | ||
− | {{AHSME 40p box|year=1965|num-b= | + | {{AHSME 40p box|year=1965|num-b=11|num-a=13}} |
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] |
Revision as of 12:15, 18 July 2024
Problem
A rhombus is inscribed in in such a way that one of its vertices is
and two of its sides lie along
and
.
If
inches,
inches, and
inches, the side of the rhombus, in inches, is:
Solution
As in the diagram, suppose the rhombus is inscribed in
with
on
,
on
, and
on
. Let the side length of the rhombus be
. Because a rhombus is a parallelogram, its opposite sides are parallel. Thus, by AA similarity,
, and so
. Because
and
are sides of the rhombus, they both have length
. Furthermore,
, and
, because, combined with sides of the rhombus, they form sides of the triangle. Thus, by substituting into the proportion derived above, we see that:
\begin{align*}
\frac{FE}{FB}&=\frac{DC}{DE} \\
\frac{x}{12-x}&=\frac{6-x}{x} \\
x^2&=72-12x-6x+x^2 \\
18x&=72 \\
x&=4 \\
\end{align*}
Thus, the side of the rhombus has length .
See Also
1965 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 | ||
All AHSME Problems and Solutions |