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indu1954
Joined: 02 Nov 2017
Last visit: 23 Feb 2020
Posts: 29
Given Kudos: 22
Location: India
Schools: UVA Darden Tepper Babson '21 Johnson '21 Rotman '21 Booth '22
GPA: 3.87
11 Jun 2019, 11:08
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
58% (03:16) correct
42%
(02:36)
wrong
based on 38
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
Equilateral Triangle.jpg [ 76.58 KiB | Viewed 7419 times ]
11 Jun 2019, 11:57
Kudos
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If we find the area of the big circle, and subtract the area of the triangle, we'll get the total area of the three equal sections outside of the triangle and inside the big circle. We only want the area of one of those sections, so if we do that subtraction, we'll need to divide by 3. So we want to find:
(Area of big circle - Area of triangle)/3
Now if you look at the small triangle at the top right of the diagram, the one with the dotted lines as two of its edges, notice that the line of length m in that triangle must meet the big equilateral triangle at 90 degrees, because the edge of the equilateral is tangent to the circle (and tangent lines are always perpendicular to radii at the tangent point). So that dotted-line triangle is a right triangle, and since we're bisecting the 60 degree angle at the top of the diagram, it must be a 30-60-90 triangle, and its sides are in a 1 to √3 to 2 ratio. Its shortest side is m, so its longest side, the vertical dotted line, is 2m. But that's the radius of the big circle, so the area of the big circle is π (2m)^2 = 4πm^2
Notice also that the third side of that small triangle, the downwards sloping side on the right, has length √3m. But that's half the length of a side of the equilateral triangle, so each side of the equilateral has length 2√3m. We need the area of that triangle, so we need to find a height (unless you know an equilateral triangle formula). If we take the horizontal side at the bottom as our base, then the vertical height is just made up of one radius of the small circle, of length m, plus the radius of the big circle which we found above was 2m. So the height of the triangle is 3m. Since our base is just the length of a side in the equilateral triangle, our base is 2√3m, and the area of the equilateral triangle is bh/2 = (2√3m)(3m)/2 = 3√3m^2.
Finally,
(Area of big circle - Area of triangle)/3 = (4πm^2 - 3√3m^2)/3 = (4π - 3√3)m^2/3
(Area of big circle - Area of triangle)/3
Now if you look at the small triangle at the top right of the diagram, the one with the dotted lines as two of its edges, notice that the line of length m in that triangle must meet the big equilateral triangle at 90 degrees, because the edge of the equilateral is tangent to the circle (and tangent lines are always perpendicular to radii at the tangent point). So that dotted-line triangle is a right triangle, and since we're bisecting the 60 degree angle at the top of the diagram, it must be a 30-60-90 triangle, and its sides are in a 1 to √3 to 2 ratio. Its shortest side is m, so its longest side, the vertical dotted line, is 2m. But that's the radius of the big circle, so the area of the big circle is π (2m)^2 = 4πm^2
Notice also that the third side of that small triangle, the downwards sloping side on the right, has length √3m. But that's half the length of a side of the equilateral triangle, so each side of the equilateral has length 2√3m. We need the area of that triangle, so we need to find a height (unless you know an equilateral triangle formula). If we take the horizontal side at the bottom as our base, then the vertical height is just made up of one radius of the small circle, of length m, plus the radius of the big circle which we found above was 2m. So the height of the triangle is 3m. Since our base is just the length of a side in the equilateral triangle, our base is 2√3m, and the area of the equilateral triangle is bh/2 = (2√3m)(3m)/2 = 3√3m^2.
Finally,
(Area of big circle - Area of triangle)/3 = (4πm^2 - 3√3m^2)/3 = (4π - 3√3)m^2/3
General Discussion
11 Jun 2019, 11:46
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indu1954
Let ABC are the vertices with vertices A on top.
Property of equilateral triangle is centroid = center of incircle = Center of circumcircle
Centroid divides Median in ratio 2:1
So, if radius of incircle = m
Radius of circumcircle = 2m
—> OA = OB = OC = 2m
Required shaded area = area of sector OBC - area of triangle OBC.
OBC is isosceles triangle with 120-30-30 angles (sides ratio 3^1/2:1:1) . If OB = OC = 2m
—> BC = 2*3^1/2*m
Area of sector = 120/360*pi*(2m)^2
= 4/3*pi*m^2
Area of triangle = 1/2*m*2*3^1/2*m
= sqrt(3)*m^2
Required Area = (4/3*pi - sqrt(3))*m^2
= (4*pi - 3*sqrt(3))*m^2/3
IMO Option A
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