Difference between revisions of "1995 AHSME Problems/Problem 28"
(no problem 1=2 --------- solution (2) by scorpius) |
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Revision as of 22:15, 9 January 2008
Problem
Two parallel chords in a circle have lengths and
, and the distance between them is
. The chord parallel to these chords and midway between them is of length
where
is
Solution
Solution 1
We let be the center,
,
represent the chords with length
respectively (as shown below). Connecting the endpoints of the chords with the center, we have several right triangles. However, we do not yet know whether the two chords are on the same side of are two different sides of the center of the circle.
By the Pythagorean Theorem on , we get
, where
is the length of the other leg. Now the length of the leg of
is either
or
depending whether or not
are on the same side of the center of the circle:
Only the negative works here (thus the two chords are on opposite sides of the center), and solving we get . The leg formed in the right triangle with the third chord is
, and by the Pythagorean Theorem again
Solution 2
Let be the chord of length
,
be the chord of length
,
be the point on
such that
,
be the point on
such that
. Then
. Extend
intersecting the circle at
and let
. By the Power of a Point Theorem,
Since
is a right angle and
and
are on the circle, the diameter
by Pythagorean Theorem, so the radius is
.
Let
be the center of the circle,
be the midpoint of
. Then
and
is a right angle. Then
, again by Pythagorean Theorem.
The chord midway between
and
is a distance of 3 away from each, so 2 away from
. Using the Pythagorean Theorem one more time, half the length of the chord is equal to
. Then
.
See also
1995 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 27 |
Followed by Problem 29 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |