Intersecting Chords
Shown in the figure above is circle
O
with radius 6 inches. Chord
C
D
is drawn perpendicular to radius
A
O
so that its midpoint is 3 inches from the center of the circle. From point
A
, any chord
A
B
is drawn intersecting
C
D
at point
M
. Let
v
be equal to the product
(
A
B
)
(
A
M
)
, as chord
A
B
is made to rotate in the circle about the fixed point
A
. Find
v
.
The answer is 36.
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2 solutions
Simple but beautiful. Thank you.
(
A
B
)
(
A
M
)
=
(
A
M
+
M
B
)
(
A
M
)
=
(
A
M
)
2
+
(
M
B
)
(
A
M
)
=
(
A
M
)
2
+
(
C
M
)
(
M
D
)
(
1
)
let E M = x .
C E = E D = 6 2 − 3 2 = 3 6 − 9 = 2 7
( C M ) ( M D ) = ( 2 7 − x ) ( 2 7 + x ) ( 2 )
Consider △ A E M :
( A M ) 2 = 9 + x 2 ( 3 )
Substitute ( 2 ) and ( 3 ) in ( 1 ) .
( A B ) ( A M ) = 9 + x 2 + ( 2 7 − x ) ( 2 7 + x ) = 9 + x 2 + ( 2 7 + 2 7 x − 2 7 x − x 2 ) = 9 + x 2 + 2 7 − x 2 = 9 + 2 7 = 3 6