Difference between revisions of "2005 AIME I Problems/Problem 11"
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== Problem == | == Problem == | ||
| − | A semicircle with diameter <math> d </math> is contained in a square whose sides have length 8. Given the maximum value of <math> d </math> is <math> m - \sqrt{n},</math> find <math> m+n. </math> | + | A [[semicircle]] with [[diameter]] <math> d </math> is contained in a [[square (geometry) | square]] whose sides have length 8. Given the maximum value of <math> d </math> is <math> m - \sqrt{n},</math> find <math> m+n. </math> |
== Solution == | == 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. | ||
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| + | Now, let the square be <math>ABCD</math>, and let <math>E \in AB</math> and <math>F \in DA</math> be the points at which the "corners" of the semicircle touch the square. Then by the comments above, <math>AE = AF = a</math>. By the [[Pythagorean Theorem]], <math>d^2 = 2a^2</math>. | ||
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| + | Now, if we draw a line through the center, <math>O</math>, of the semicircle and its point of tangency with <math>BC</math>, we see that this line is perpendicular to <math>BC</math> and so parallel to <math>AB</math>. Thus, by [[triangle similarity]] it cuts <math>AF</math> in half, and so by symmetry the distance from <math>O</math> to <math>AD</math> is <math>\frac{a}{2}</math> and so the distance from <math>O</math> to <math>BC</math> is <math>8 - \frac a2</math>. But this latter quantity is also the radius of the semicircle, so <math>d = 16 - a</math>. | ||
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| + | Our two previous paragraphs give <math>2a^2 = (16 - a)^2</math> so <math>a^2 + 32a - 256 = 0</math> and <math>a = 16\sqrt{2} - 16</math> (where we discard the negative root of that quadratic) and so <math>d = a\sqrt{2} = 32 - 16\sqrt{2} = 32 - \sqrt{512}</math>, so the answer is <math>32 + 512 = 544</math>. | ||
== See also == | == See also == | ||
| + | * [[2005 AIME I Problems/Problem 10 | Previous Problem]] | ||
| + | * [[2005 AIME I Problems/Problem 12 | Next Problem]] | ||
* [[2005 AIME I Problems]] | * [[2005 AIME I Problems]] | ||
Revision as of 12:39, 17 January 2007
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.
An image is supposed to go here. You can help us out by creating one and editing it in. Thanks.
Now, let the square be 
, and let 
and 
be the points at which the "corners" of the semicircle touch the square. Then 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 
.
