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Contests
National and Regional Contests
USA Contests
MAA AMC
AMC 12/AHSME
1990 AMC 12/AHSME
4
4
Part of
1990 AMC 12/AHSME
Problems
(1)
1990 AHSME #4: Parallelogram
Source:
1/1/2013
Let
A
B
C
D
ABCD
A
BC
D
be a parallelogram with
∠
A
B
C
=
12
0
∘
\angle ABC=120^\circ
∠
A
BC
=
12
0
∘
,
A
B
=
16
AB=16
A
B
=
16
and
B
C
=
10
BC=10
BC
=
10
. Extend
C
D
‾
\overline{CD}
C
D
through
D
D
D
to
E
E
E
so that
D
E
=
4
DE=4
D
E
=
4
. If
B
E
‾
\overline{BE}
BE
intersects
A
D
‾
\overline{AD}
A
D
at
F
F
F
, then
F
D
FD
F
D
is closest to
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
a
n
>
1
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
a
n
>
2
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
a
n
>
3
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
a
n
>
4
<
s
p
a
n
c
l
a
s
s
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
a
n
>
5
<span class='latex-bold'>(A) </span>1\qquad <span class='latex-bold'>(B) </span>2\qquad <span class='latex-bold'>(C) </span>3\qquad <span class='latex-bold'>(D) </span>4\qquad <span class='latex-bold'>(E) </span>5
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
1
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
2
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
3
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
4
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
5
[asy] size(200); defaultpen(linewidth(0.8)); pair A=origin,B=(16,0),C=(26,10*sqrt(3)),D=(10,10*sqrt(3)),E=(0,10*sqrt(3)); draw(A--B--C--E--B--A--D); label("
A
A
A
",A,S); label("
B
B
B
",B,S); label("
C
C
C
",C,N); label("
D
D
D
",D,N); label("
E
E
E
",E,N); label("
F
F
F
",extension(A,D,B,E),W); label("
4
4
4
",(D+E)/2,N); label("
16
16
16
",(8,0),S); label("
10
10
10
",(B+C)/2,SE); [/asy]
geometry
parallelogram
AMC