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JHMT problems
2005 JHMT
5
5
Part of
2005 JHMT
Problems
(1)
2005 JHMT Geometry #5
Source:
8/31/2023
Equilateral triangle
A
B
C
ABC
A
BC
has
A
D
=
D
B
=
F
G
=
A
E
=
E
C
=
4
AD = DB = FG = AE = EC = 4
A
D
=
D
B
=
FG
=
A
E
=
EC
=
4
and
B
F
=
G
C
=
2
BF = GC = 2
BF
=
GC
=
2
. From
D
D
D
and
G
G
G
are drawn perpendiculars to
E
F
EF
EF
intersecting at
H
H
H
and
I
I
I
, respectively. The three polygons
E
C
G
I
ECGI
ECG
I
,
F
G
I
FGI
FG
I
, and
B
F
H
D
BFHD
BF
HD
are rearranged to
E
A
N
L
EANL
E
A
N
L
,
M
N
K
MNK
MN
K
, and
A
M
J
D
AMJD
A
M
J
D
so that the rectangle
H
L
K
J
HLKJ
H
L
K
J
is formed. Find its area. https://cdn.artofproblemsolving.com/attachments/d/4/7e6667f0f0544b6fbc860f8d86c8ceaaf85cc1.png
geometry