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MAA AMC
AMC 12/AHSME
1950 AMC 12/AHSME
48
48
Part of
1950 AMC 12/AHSME
Problems
(1)
AHSME 1950- part 3
Source:
7/20/2008
A point is selected at random inside an equilateral triangle. From this point perpendiculars are dropped to each side. The sum of these perpendiculars is:
<
s
p
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c
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(
A
)
<
/
s
p
a
n
>
Least when the point is the center of gravity of the triangle
<
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x
−
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>
(
B
)
<
/
s
p
a
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>
Greater than the altitude of the triangle
<
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c
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a
s
s
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a
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x
−
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>
(
C
)
<
/
s
p
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Equal to the altitude of the triangle
<
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a
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>
(
D
)
<
/
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One-half the sum of the sides of the triangle
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a
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x
−
b
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(
E
)
<
/
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Greatest when the point is the center of gravity
<span class='latex-bold'>(A)</span>\ \text{Least when the point is the center of gravity of the triangle}\qquad\\ <span class='latex-bold'>(B)</span>\ \text{Greater than the altitude of the triangle} \qquad\\ <span class='latex-bold'>(C)</span>\ \text{Equal to the altitude of the triangle}\qquad\\ <span class='latex-bold'>(D)</span>\ \text{One-half the sum of the sides of the triangle} \qquad\\ <span class='latex-bold'>(E)</span>\ \text{Greatest when the point is the center of gravity}
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
A
)
<
/
s
p
an
>
Least when the point is the center of gravity of the triangle
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
B
)
<
/
s
p
an
>
Greater than the altitude of the triangle
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
C
)
<
/
s
p
an
>
Equal to the altitude of the triangle
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
D
)
<
/
s
p
an
>
One-half the sum of the sides of the triangle
<
s
p
an
c
l
a
ss
=
′
l
a
t
e
x
−
b
o
l
d
′
>
(
E
)
<
/
s
p
an
>
Greatest when the point is the center of gravity
geometry