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MAA AMC
AMC 12/AHSME
1966 AMC 12/AHSME
40
40
Part of
1966 AMC 12/AHSME
Problems
(1)
Which Relation is Deduced Correctly?
Source: 1966 AHSME #40
8/31/2011
[asy]draw(Circle((0,0), 1)); dot((0,0)); label("
O
O
O
", (0,0), S); label("
A
A
A
", (-1,0), W); label("
B
B
B
", (1,0), E); label("
a
a
a
", (-0.5,0), S); draw((-1,-1.25)--(-1,1.25)); draw((1,-1.25)--(1,1.25)); draw((-1,0)--(1,0)); draw((-1,0)--(-1,0)+2.3*dir(30)); label("
C
C
C
", (-1,0)+2.3*dir(30), E); label("
D
D
D
", (-1,0)+1.8*dir(30), N); dot((-1,0)+.4*dir(30)); label("
E
E
E
", (-1,0)+.4*dir(30), N); [/asy] In this figure
A
B
AB
A
B
is a diameter of a circle, centered at
O
O
O
, with radius
a
a
a
. A chord
A
D
AD
A
D
is drawn and extended to meet the tangent to the circle at
B
B
B
in point
C
C
C
. Point
E
E
E
is taken on
A
C
AC
A
C
so that
A
E
=
D
C
AE=DC
A
E
=
D
C
. Denoting the distances of
E
E
E
from the tangent through
A
A
A
and from the diameter
A
B
AB
A
B
by
x
x
x
and
y
y
y
, respectively, we can deduce the relation:
(A)
y
2
=
x
3
2
a
−
x
(B)
y
2
=
x
3
2
a
+
x
(C)
y
4
=
x
2
2
−
x
(D)
x
2
=
y
2
2
a
−
x
(E)
x
2
=
y
2
2
a
+
x
\text{(A)}\ y^2=\dfrac{x^3}{2a-x} \qquad \text{(B)}\ y^2=\frac{x^3}{2a+x}\qquad \text{(C)}\ y^4=\frac{x^2}{2-x}\qquad\\ \text{(D)}\ x^2=\dfrac{y^2}{2a-x}\qquad \text{(E)}\ x^2=\frac{y^2}{2a+x}
(A)
y
2
=
2
a
−
x
x
3
(B)
y
2
=
2
a
+
x
x
3
(C)
y
4
=
2
−
x
x
2
(D)
x
2
=
2
a
−
x
y
2
(E)
x
2
=
2
a
+
x
y
2
AMC