2005 AIME I Problems/Problem 11
Problem
A semicircle with diameter is contained in a square whose sides have length 8. Given the maximum value of
is
find
Solution
We note that aligning the base of the semicircle with a side of the square is certainly non-optimal. If the semicircle is tangent to only one side of the square, we will have "wiggle-room" to increase its size. Once it is tangent to two adjacent sides of the square, we will maximize its size when it touches both other sides of the square. This can happen only when it is arranged so that the center of the semicircle lies on one diagonal of the square.
Now, let the square be , and let
and
be the points at which the "corners" of the semicircle touch the square. Let
be the center of the semicircle.
Solution 1
Define the radius of the semicircle as . Draw the perpendicular from
to
, which forms a
triangle. The length of the perpendicular is
. Note also that
is equal to the length of that perpendicular plus the radius to the point of tangency on
. Thus,
, and
. The diameter is then
, and the solution is
.
Solution 2
By the comments above, . By the Pythagorean Theorem,
.
Now, if we draw a line through the center, , of the semicircle and its point of tangency with
, we see that this line is perpendicular to
and so parallel to
. Thus, by triangle similarity it cuts
in half, and so by symmetry the distance from
to
is
and so the distance from
to
is
. But this latter quantity is also the radius of the semicircle, so
.
Our two previous paragraphs give so
and
(where we discard the negative root of that quadratic) and so
, so the answer is
.
See also
2005 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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