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06 Jan 2022, 07:00
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B
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Difficulty:
95%
(hard)
Question Stats:
19% (02:56) correct
81%
(02:52)
wrong
based on 175
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A square is drawn in the coordinate plane with its vertices at the points (–2, –2), (–2, 6), (6, 6), and (6, –2), and a non-vertical line is drawn that passes through the point (0, 4). The portion of the coordinate plane that lies within the square, but above the line, is then shaded as shown above. If A is the area of the shaded region in square units, which of the following specifies all the possible values of A ?
(A) 8 ≤ A ≤ 16
(B) 8 ≤ A < 48
(C) 16 ≤ A< 48
(D) 8 < A ≤ 32
(E) 16 < A ≤ 32
Attachment:
1.jpg [ 9.95 KiB | Viewed 8865 times ]
06 Jan 2022, 09:41
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The only information given regarding the line is that it passes through the point (0, 4). Hence there are different positions that the line takes.
Before we begin the analysis, let's quickly calculate the area of the square.
Area of the square = 8 * 8 = 64 units
We know that area of the shaded region (in grey) will be minimum when the area of unshaded area is maximum. Conversely, the area of the shaded region will be maximum when the area of the unshaded region is minimum.
Thus problem becomes a bit easy to manage if we can visualize the change in area at different positions of the line.
geogebra-export.png [ 77.54 KiB | Viewed 8599 times ]
Let's assume that the line segment is at a position denoted by JK, now if we move the segment in clockwise direction, we observe that the the area of the shaded region increases. When the line is at position AB, the area enclosed by the shaded region is greater than the area enclosed by the shaded region when the line was at JK. The area increases because we are decreasing the slop of the line, thereby increasing the area enclosed by the line with the top surface of the square.
Let's further move the line in the clockwise direction. At the position denoted by MN, the area of the shaded region continues to increase and we see that the area enclosed is greater than the area enclosed when the line was at AB.
The area will continue to increase as we rotate the line in the clockwise direction. The area of the shaded region becomes maximum when the line almost overlaps the Y-axis. (Why 'almost' overlaps ? Because it is given that the line is a non-vertical line, hence the line cannot overlap with Y axis)
Therefore when the line is at a point denoted by UT, the area of the shaded region will be maximum. At this point the area of the unshaded region is minimum and is the area bound by the left side of the square and the Y-axis.
Hence area of the unshaded region is 'almost' equal to 2 * 8 i.e. 16 units (of course it will not be equal to 16 units but slightly greater than 16 units)
Maximum Area of shaded region therefore is less than (64 - 16) units
Max area < 48 units
The area of the shaded region becomes minimum when the the line makes an \( \angle 45\) with the top surface of the square.
At this point the slope of the line is 1.
We can find the point of intersection of the top and left sides of the square with the line by formulating the equation of the line.
When slope =1, the equation of the line is
y = x + 4
Intersection with the left side -
x = -2
y = -2 + 4 = 2
Point of intersection = (-2,2)
Intersection with the top -
y = 6
6 = x + 4 = 2
Point of intersection = (2,6)
Area of the shaded region = Area of the triangle formed = \(\frac{1}{2} * 4 * 4\) = 8
Minimum area = 8
Combining the obtained values
8<= A < 48
IMO- B
P.S - Hope to see someone share a better approach !
Before we begin the analysis, let's quickly calculate the area of the square.
Area of the square = 8 * 8 = 64 units
We know that area of the shaded region (in grey) will be minimum when the area of unshaded area is maximum. Conversely, the area of the shaded region will be maximum when the area of the unshaded region is minimum.
Thus problem becomes a bit easy to manage if we can visualize the change in area at different positions of the line.
Attachment:
geogebra-export.png [ 77.54 KiB | Viewed 8599 times ]
Let's assume that the line segment is at a position denoted by JK, now if we move the segment in clockwise direction, we observe that the the area of the shaded region increases. When the line is at position AB, the area enclosed by the shaded region is greater than the area enclosed by the shaded region when the line was at JK. The area increases because we are decreasing the slop of the line, thereby increasing the area enclosed by the line with the top surface of the square.
Let's further move the line in the clockwise direction. At the position denoted by MN, the area of the shaded region continues to increase and we see that the area enclosed is greater than the area enclosed when the line was at AB.
The area will continue to increase as we rotate the line in the clockwise direction. The area of the shaded region becomes maximum when the line almost overlaps the Y-axis. (Why 'almost' overlaps ? Because it is given that the line is a non-vertical line, hence the line cannot overlap with Y axis)
Therefore when the line is at a point denoted by UT, the area of the shaded region will be maximum. At this point the area of the unshaded region is minimum and is the area bound by the left side of the square and the Y-axis.
Hence area of the unshaded region is 'almost' equal to 2 * 8 i.e. 16 units (of course it will not be equal to 16 units but slightly greater than 16 units)
Maximum Area of shaded region therefore is less than (64 - 16) units
Max area < 48 units
The area of the shaded region becomes minimum when the the line makes an \( \angle 45\) with the top surface of the square.
At this point the slope of the line is 1.
We can find the point of intersection of the top and left sides of the square with the line by formulating the equation of the line.
When slope =1, the equation of the line is
y = x + 4
Intersection with the left side -
x = -2
y = -2 + 4 = 2
Point of intersection = (-2,2)
Intersection with the top -
y = 6
6 = x + 4 = 2
Point of intersection = (2,6)
Area of the shaded region = Area of the triangle formed = \(\frac{1}{2} * 4 * 4\) = 8
Minimum area = 8
Combining the obtained values
8<= A < 48
IMO- B
P.S - Hope to see someone share a better approach !
General Discussion
10 Feb 2022, 08:13
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AbhiroopGhosh
Thanks for the solution.
Can you please share more on how can we know that the min area will be obtained when the line creates 2 45-degree angles with the sides of the square?
I'm really struggling to understand base on what theorem we can infer that... Thanksss
)
, however please free to comment if you see any concern or for any further questions. 
