Difference between revisions of "2013 AMC 12A Problems/Problem 23"
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The area also includes <math>4</math> circular segments. Two of them are quarter-segments of circles centered at <math>P</math> of radii <math>\sqrt{2}</math> (the segment bounded by <math> \overline{PA} </math> and <math> \overline{PA'}</math>) and <math>\sqrt{6}</math> (that bounded by <math>\overline{PC}</math> and <math>\overline{PC'}</math>). Assuming <math>A</math> is the bottom-left vertex and <math>B</math> is the bottom-right one, it is clear that the third segment is formed as <math>B</math> swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when <math>D</math> overshoots the final square's left edge. To find the areas of these segments, consider the perpendicular from <math>P</math> to <math>\overline{BC}</math>. Call the point of intersection <math>E</math>. From the previous paragraph, it is clear that <math>PE = \sqrt{3}</math> and <math>BE = 1</math>. This means <math>PB = 2</math>, and <math>B</math> swings back inside edge <math>\overline{BC}</math> at a point <math>1</math> unit above <math>E</math> (since it left the edge <math>1</math> unit below). The triangle of the circular sector is therefore an equilateral triangle of side length <math>2</math>, and so the angle of the segment is <math>60^{\circ}</math>. Imagining the process in reverse, it is clear that the situation is the same with point <math>D</math>. | The area also includes <math>4</math> circular segments. Two of them are quarter-segments of circles centered at <math>P</math> of radii <math>\sqrt{2}</math> (the segment bounded by <math> \overline{PA} </math> and <math> \overline{PA'}</math>) and <math>\sqrt{6}</math> (that bounded by <math>\overline{PC}</math> and <math>\overline{PC'}</math>). Assuming <math>A</math> is the bottom-left vertex and <math>B</math> is the bottom-right one, it is clear that the third segment is formed as <math>B</math> swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when <math>D</math> overshoots the final square's left edge. To find the areas of these segments, consider the perpendicular from <math>P</math> to <math>\overline{BC}</math>. Call the point of intersection <math>E</math>. From the previous paragraph, it is clear that <math>PE = \sqrt{3}</math> and <math>BE = 1</math>. This means <math>PB = 2</math>, and <math>B</math> swings back inside edge <math>\overline{BC}</math> at a point <math>1</math> unit above <math>E</math> (since it left the edge <math>1</math> unit below). The triangle of the circular sector is therefore an equilateral triangle of side length <math>2</math>, and so the angle of the segment is <math>60^{\circ}</math>. Imagining the process in reverse, it is clear that the situation is the same with point <math>D</math>. | ||
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Revision as of 21:35, 2 September 2013
Problem
is a square of side length
. Point
is on
such that
. The square region bounded by
is rotated
counterclockwise with center
, sweeping out a region whose area is
, where
,
, and
are positive integers and
. What is
?
Solution
We first note that diagonal is of length
. It must be that
divides the diagonal into two segments in the ratio
to
. It is not difficult to visualize that when the square is rotated, the initial and final squares overlap in a rectangular region of dimensions
by
. The area of the overall region (of the initial and final squares) is therefore twice the area of the original square minus the overlap, or
.
The area also includes circular segments. Two of them are quarter-segments of circles centered at
of radii
(the segment bounded by
and
) and
(that bounded by
and
). Assuming
is the bottom-left vertex and
is the bottom-right one, it is clear that the third segment is formed as
swings out to the right of the original square [recall that the square is rotated counterclockwise], while the fourth is formed when
overshoots the final square's left edge. To find the areas of these segments, consider the perpendicular from
to
. Call the point of intersection
. From the previous paragraph, it is clear that
and
. This means
, and
swings back inside edge
at a point
unit above
(since it left the edge
unit below). The triangle of the circular sector is therefore an equilateral triangle of side length
, and so the angle of the segment is
. Imagining the process in reverse, it is clear that the situation is the same with point
.
The area of the segments can be found by subtracting the area of the triangle from that of the sector; it follows that the two quarter-segments have areas and
. The other two segments both have area
.
The total area is therefore
This means ,
, and
. So
; answer choice
is correct.
See also
2013 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.