1984 AHSME Problems/Problem 4
Problem
A rectangle intersects a circle as shown: ,
, and
. Then
equals:
Solution
Draw and
, forming a trapezoid. Since it's cyclic, this trapezoid must be isosceles. Also, drop altitudes from
to
,
to
, and
to
, and let the feet of these altitudes be
,
, and
respectively.
is a rectangle since it has
right angles. Therefore,
, and
. By the same logic,
is also a rectangle, and
.
since they're both altitudes to a trapezoid, and
since the trapezoid is isosceles. Therefore,
by HL congruence, so
. Also,
is a rectangle from
right angles, and
. Therefore,
.
See Also
1984 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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