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9
2013 HMMT Geometry # 9
2013 HMMT Geometry # 9
Source:
March 3, 2024
geometry
Problem Statement
Pentagon
A
B
C
D
E
ABCDE
A
BC
D
E
is given with the following conditions:(a)
∠
C
B
D
+
∠
D
A
E
=
∠
B
A
D
=
4
5
o
\angle CBD + \angle DAE = \angle BAD = 45^o
∠
CB
D
+
∠
D
A
E
=
∠
B
A
D
=
4
5
o
,
∠
B
C
D
+
∠
D
E
A
=
30
0
o
\angle BCD + \angle DEA = 300^o
∠
BC
D
+
∠
D
E
A
=
30
0
o
(b)
B
A
D
A
=
2
2
3
\frac{BA}{DA} =\frac{ 2\sqrt2}{3}
D
A
B
A
=
3
2
2
,
C
D
=
7
5
3
CD =\frac{ 7\sqrt5}{3}
C
D
=
3
7
5
, and
D
E
=
15
2
4
DE = \frac{15\sqrt2}{4}
D
E
=
4
15
2
(c)
A
D
2
⋅
B
C
=
A
B
⋅
A
E
⋅
B
D
AD^2 \cdot BC = AB \cdot AE \cdot BD
A
D
2
⋅
BC
=
A
B
⋅
A
E
⋅
B
D
Compute
B
D
BD
B
D
.
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