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MAA AMC
AMC 12/AHSME
1978 AMC 12/AHSME
28
28
Part of
1978 AMC 12/AHSME
Problems
(1)
Find <A_44 A_45 A_43
Source: 1978 AHSME Problem 28
6/18/2014
[asy] import cse5; size(180); pathpen=black; pair A1=(0,0), A2=(1,0), A3=(0.5,sqrt(3)/2); D(MP("A_1",A1)--MP("A_2",A2)--MP("A_3",A3,N)--cycle); pair A4=(A1+A2)/2, A5 = (A3+A2)/2, A6 = (A4+A3)/2; D(MP("A_4",A4,S)--MP("A_6",A6,W)--A3); D(A6--MP("A_5",A5,NE)--A4); //Credit to chezbgone2 for the diagram[/asy]If
△
A
1
A
2
A
3
\triangle A_1A_2A_3
△
A
1
A
2
A
3
is equilateral and
A
n
+
3
A_{n+3}
A
n
+
3
is the midpoint of line segment
A
n
A
n
+
1
A_nA_{n+1}
A
n
A
n
+
1
for all positive integers
n
n
n
, then the measure of
∡
A
44
A
45
A
43
\measuredangle A_{44}A_{45}A_{43}
∡
A
44
A
45
A
43
equals
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(
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5
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6
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(
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9
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(
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)
<
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12
0
∘
<span class='latex-bold'>(A) </span>30^\circ\qquad<span class='latex-bold'>(B) </span>45^\circ\qquad<span class='latex-bold'>(C) </span>60^\circ\qquad<span class='latex-bold'>(D) </span>90^\circ\qquad <span class='latex-bold'>(E) </span>120^\circ
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3
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(
B
)
<
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>
4
5
∘
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(
C
)
<
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>
6
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∘
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=
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−
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>
(
D
)
<
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>
9
0
∘
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=
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−
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′
>
(
E
)
<
/
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p
an
>
12
0
∘
Asymptote
AMC