💡Alright, so I have this geometry problem here. Let me try to visualize it first. There's a sector OAB of a circle with center O and radius 8 cm. The angle at O, which is angle AOB, is 120 degrees. Now, another circle is supposed to pass through the points O, A, and B, and I need to find the radius r of this new circle.Hmm, okay. So, sector OAB is part of a circle with radius 8 cm, and the angle between OA and OB is 120 degrees. That means points A and B are on the circumference of this circle, each 8 cm away from O, and the angle between them at O is 120 degrees. Now, another circle is to be drawn through these three points: O, A, and B. So, this new circle will pass through the center of the original circle and the two endpoints of the sector.I think the key here is to find the circumradius of triangle OAB because the new circle passing through O, A, and B is essentially the circumcircle of triangle OAB. So, if I can find the circumradius of triangle OAB, that should give me the radius r of the new circle.Okay, so let's recall some geometry. The circumradius R of a triangle with sides a, b, and c is given by the formula:[ R = frac{a b c}{4 Delta} ]where Δ is the area of the triangle.Alternatively, if I know two sides and the included angle, I can use the formula:[ R = frac{a}{2 sin A} ]where a is the length of a side, and A is the angle opposite that side.In triangle OAB, sides OA and OB are both 8 cm because they are radii of the original circle. The angle between them, angle AOB, is 120 degrees. So, triangle OAB is an isosceles triangle with two sides of 8 cm and an included angle of 120 degrees.Wait, actually, triangle OAB is not just any triangle; it's a specific triangle with two sides known and the included angle known. So, maybe I can use the Law of Cosines to find the third side, which is AB.Law of Cosines states that:[ c^2 = a^2 + b^2 - 2ab cos C ]In this case, side AB is opposite the angle at O, which is 120 degrees. So, let's denote AB as c, OA as a = 8 cm, OB as b = 8 cm, and angle C = 120 degrees.Plugging into the formula:[ c^2 = 8^2 + 8^2 - 2 times 8 times 8 times cos 120^circ ]Calculating that:First, 8 squared is 64, so:[ c^2 = 64 + 64 - 2 times 8 times 8 times cos 120^circ ][ c^2 = 128 - 128 cos 120^circ ]Now, cos 120 degrees is equal to cos(180 - 60) degrees, which is -cos 60 degrees, and cos 60 degrees is 0.5. So, cos 120 degrees is -0.5.Therefore:[ c^2 = 128 - 128 times (-0.5) ][ c^2 = 128 + 64 ][ c^2 = 192 ]So, c is the square root of 192. Simplifying that:[ c = sqrt{192} ][ c = sqrt{64 times 3} ][ c = 8 sqrt{3} ]So, the length of AB is 8√3 cm.Now, going back to the formula for the circumradius R:[ R = frac{a}{2 sin A} ]But in this case, I need to figure out which angle to use. Since I have sides OA, OB, and AB, and I know all the angles in triangle OAB.Wait, actually, in triangle OAB, we have sides OA = 8 cm, OB = 8 cm, and AB = 8√3 cm. The angles at A and B can be found since the triangle is isosceles.Let me calculate the angles at A and B. Since the triangle is isosceles with OA = OB, the angles at A and B are equal.The sum of angles in a triangle is 180 degrees. We already know angle at O is 120 degrees, so the remaining 60 degrees are split equally between angles at A and B.So, angles at A and B are each 30 degrees.Now, using the formula for circumradius:[ R = frac{a}{2 sin A} ]Here, side a can be AB, which is 8√3 cm, and angle A is 30 degrees.So,[ R = frac{8 sqrt{3}}{2 sin 30^circ} ]We know that sin 30 degrees is 0.5, so:[ R = frac{8 sqrt{3}}{2 times 0.5} ][ R = frac{8 sqrt{3}}{1} ][ R = 8 sqrt{3} ]Wait, that can't be right because the original circle has a radius of 8 cm, and the new circle is passing through O, A, and B, which are points on the original circle. So, the radius of the new circle should be larger than 8 cm, right? Because it's passing through the center and two points on the circumference.But according to my calculation, R is 8√3, which is approximately 13.85 cm, which is larger than 8 cm, so that makes sense.Wait, but let me double-check my formula. I used the formula:[ R = frac{a}{2 sin A} ]where a is the side opposite angle A. In this case, I took a as AB, which is opposite angle O, which is 120 degrees. Wait, maybe I made a mistake here.Let me clarify. In the formula, R is the circumradius, and for any triangle, it can be calculated using any side and its opposite angle. So, if I take side AB, which is opposite angle O (120 degrees), then:[ R = frac{AB}{2 sin 120^circ} ]So, AB is 8√3 cm, and sin 120 degrees is sin(60 degrees) which is √3/2.So,[ R = frac{8 sqrt{3}}{2 times (sqrt{3}/2)} ]Simplify the denominator:2 times √3/2 is √3.So,[ R = frac{8 sqrt{3}}{sqrt{3}} ]The √3 cancels out:[ R = 8 ]Wait, that's different. So, now I'm getting R = 8 cm, which is the same as the original circle's radius. But that seems contradictory because the new circle is supposed to pass through O, A, and B, which are points on the original circle. If the original circle has radius 8 cm, and the new circle also has radius 8 cm, then the new circle would coincide with the original circle, but the original circle doesn't pass through O because O is the center, not on the circumference.Wait, that doesn't make sense. The original circle has center O, so O is not on the circumference of the original circle. Therefore, the new circle passing through O, A, and B cannot be the same as the original circle.So, where did I go wrong? Let's go back.I think I confused the sides and angles in the formula. Let me try a different approach.Another formula for the circumradius is:[ R = frac{abc}{4 Delta} ]where a, b, c are the sides of the triangle, and Δ is the area.We have triangle OAB with sides OA = 8 cm, OB = 8 cm, AB = 8√3 cm.First, let's compute the area Δ of triangle OAB.Since it's an isosceles triangle with two sides of 8 cm and included angle of 120 degrees, the area can be calculated using the formula:[ Delta = frac{1}{2} ab sin C ]where a and b are the sides, and C is the included angle.So,[ Delta = frac{1}{2} times 8 times 8 times sin 120^circ ][ Delta = frac{1}{2} times 64 times (sqrt{3}/2) ][ Delta = 32 times (sqrt{3}/2) ][ Delta = 16 sqrt{3} ]So, the area Δ is 16√3 cm².Now, using the formula for circumradius:[ R = frac{abc}{4 Delta} ]Plugging in the values:a = 8 cm, b = 8 cm, c = 8√3 cm, Δ = 16√3 cm².So,[ R = frac{8 times 8 times 8 sqrt{3}}{4 times 16 sqrt{3}} ]Simplify numerator and denominator:Numerator: 8 × 8 × 8√3 = 512√3Denominator: 4 × 16√3 = 64√3So,[ R = frac{512 sqrt{3}}{64 sqrt{3}} ]The √3 terms cancel out:[ R = frac{512}{64} ][ R = 8 ]Again, I get R = 8 cm. But this seems contradictory because the new circle passes through O, which is the center of the original circle, and A and B, which are on the circumference. So, the new circle should have a larger radius than the original circle.Wait, maybe I'm misunderstanding the problem. Let me read it again."The diagram shows a sector OAB of a circle, center O and radius 8 cm, in which angle AOB = 120 degrees. Another circle of radius r cm is to be drawn through the points O, A, and B. Find the value of r."So, the original circle has center O and radius 8 cm. The sector OAB is part of this circle. Now, another circle is to be drawn through O, A, and B. So, this new circle is different from the original circle because it passes through O, which is the center of the original circle, and A and B, which are on the original circle.Therefore, the new circle is the circumcircle of triangle OAB, which we calculated to have a radius of 8 cm. But wait, that would mean the new circle coincides with the original circle, but O is the center of the original circle, not on its circumference. Therefore, the new circle cannot coincide with the original circle.This suggests that my calculation is wrong. Let me think again.Wait, maybe the triangle OAB is not a triangle with sides OA, OB, and AB, but rather, O is the center, and A and B are points on the circumference. So, OA and OB are radii, each 8 cm, and AB is a chord of the original circle.Now, the new circle passes through O, A, and B. So, O is a point inside the original circle, and A and B are on the original circle.Therefore, the new circle is not the circumcircle of triangle OAB in the original circle, but rather, it's a different circle passing through three points: O, A, and B.Wait, but O is the center of the original circle, so OA and OB are radii, each 8 cm, and AB is a chord subtending 120 degrees at the center.So, perhaps I need to consider the circumradius of triangle OAB, but O is not on the circumference of the original circle, it's the center. So, triangle OAB has sides OA = 8 cm, OB = 8 cm, and AB = 8√3 cm, as calculated before.But when I calculated the circumradius R of triangle OAB, I got R = 8 cm, which suggests that the circumcircle of triangle OAB has the same radius as the original circle, but that can't be because O is the center of the original circle, not on its circumference.Wait, maybe I'm confusing the roles. Let me clarify.In the original circle, O is the center, and A and B are points on the circumference, each 8 cm from O. The angle AOB is 120 degrees. Now, we need to find a new circle that passes through O, A, and B.So, this new circle is different from the original circle because it passes through O, which is the center of the original circle, and A and B, which are on the original circle.Therefore, the new circle is the circumcircle of triangle OAB, but O is not on the original circle, it's the center.Wait, but in triangle OAB, O is a vertex, and A and B are points on the original circle. So, the circumradius of triangle OAB is the radius of the new circle passing through O, A, and B.But when I calculated the circumradius R of triangle OAB, I got R = 8 cm, which is the same as the original circle's radius. But that would mean that the new circle coincides with the original circle, but O is the center, not on the circumference, so that can't be.Therefore, there must be a mistake in my calculation.Wait, let's go back to the formula:[ R = frac{abc}{4 Delta} ]We have a = 8 cm, b = 8 cm, c = 8√3 cm, and Δ = 16√3 cm².So,[ R = frac{8 times 8 times 8 sqrt{3}}{4 times 16 sqrt{3}} ]Simplify numerator: 8 × 8 × 8√3 = 512√3Denominator: 4 × 16√3 = 64√3So,[ R = frac{512√3}{64√3} = frac{512}{64} = 8 ]So, R = 8 cm.But this suggests that the circumradius of triangle OAB is 8 cm, which is the same as the original circle's radius. But that would mean that the new circle coincides with the original circle, which is impossible because O is the center of the original circle, not on its circumference.Wait, perhaps I'm misunderstanding the problem. Maybe the new circle is not the circumcircle of triangle OAB, but rather, it's another circle passing through O, A, and B, but not necessarily the circumcircle.Wait, but any three non-collinear points define a unique circle, so the new circle must be the circumcircle of triangle OAB. Therefore, my calculation must be correct, and the radius is 8 cm.But that contradicts the intuition because O is the center of the original circle, and the new circle passes through O, which is inside the original circle.Wait, perhaps the original circle and the new circle are the same? But O is the center of the original circle, so it cannot be on the circumference of the original circle. Therefore, the new circle must be different.Wait, maybe I made a mistake in calculating the circumradius. Let me try another approach.Let me consider the coordinates. Let's place point O at the origin (0,0). Let me place point A at (8,0). Since angle AOB is 120 degrees, point B will be at an angle of 120 degrees from point A.So, the coordinates of point B can be calculated using polar coordinates:x = 8 cos 120°, y = 8 sin 120°cos 120° = -0.5, sin 120° = √3/2So, point B is at (8 × -0.5, 8 × √3/2) = (-4, 4√3)So, points O(0,0), A(8,0), and B(-4,4√3)Now, to find the circle passing through these three points, we can find the circumradius.The general equation of a circle is:[ x^2 + y^2 + Dx + Ey + F = 0 ]Since the circle passes through O(0,0), plugging in:0 + 0 + 0 + 0 + F = 0 ⇒ F = 0So, the equation simplifies to:[ x^2 + y^2 + Dx + Ey = 0 ]Now, plug in point A(8,0):8^2 + 0^2 + D×8 + E×0 = 064 + 8D = 0 ⇒ 8D = -64 ⇒ D = -8So, the equation becomes:[ x^2 + y^2 - 8x + Ey = 0 ]Now, plug in point B(-4,4√3):(-4)^2 + (4√3)^2 - 8×(-4) + E×(4√3) = 016 + 16×3 + 32 + 4√3 E = 016 + 48 + 32 + 4√3 E = 096 + 4√3 E = 0 ⇒ 4√3 E = -96 ⇒ E = -96 / (4√3) = -24 / √3 = -8√3So, E = -8√3Therefore, the equation of the circle is:[ x^2 + y^2 - 8x - 8√3 y = 0 ]To find the radius, we can rewrite the equation in standard form by completing the squares.Group x terms and y terms:[ x^2 - 8x + y^2 - 8√3 y = 0 ]Complete the square for x:x^2 - 8x = (x - 4)^2 - 16Complete the square for y:y^2 - 8√3 y = (y - 4√3)^2 - (4√3)^2 = (y - 4√3)^2 - 48So, substituting back:[ (x - 4)^2 - 16 + (y - 4√3)^2 - 48 = 0 ]Combine constants:[ (x - 4)^2 + (y - 4√3)^2 - 64 = 0 ][ (x - 4)^2 + (y - 4√3)^2 = 64 ]So, the center of the new circle is at (4, 4√3), and the radius is √64 = 8 cm.Wait, so the radius is 8 cm, same as the original circle. But this seems contradictory because the new circle passes through O, which is the center of the original circle, and A and B, which are on the original circle.But according to this calculation, the new circle has the same radius as the original circle, which suggests that it's the same circle, but that can't be because O is the center, not on the circumference.Wait, but in the calculation, the center of the new circle is at (4, 4√3), which is different from O(0,0). So, the new circle is a different circle with the same radius as the original circle.But how is that possible? Let me check the distance from the new center (4, 4√3) to point O(0,0):Distance = √[(4 - 0)^2 + (4√3 - 0)^2] = √[16 + 48] = √64 = 8 cmSo, the distance from the new center to O is 8 cm, which is equal to the radius of the new circle. Therefore, point O lies on the circumference of the new circle, which makes sense because the new circle passes through O.Similarly, the distance from the new center (4, 4√3) to point A(8,0):Distance = √[(8 - 4)^2 + (0 - 4√3)^2] = √[16 + 48] = √64 = 8 cmAnd the distance from the new center to point B(-4,4√3):Distance = √[(-4 - 4)^2 + (4√3 - 4√3)^2] = √[(-8)^2 + 0] = √64 = 8 cmSo, all three points O, A, and B lie on the circumference of the new circle, which has center at (4, 4√3) and radius 8 cm.Therefore, the radius r of the new circle is 8 cm.But wait, that seems to contradict the initial intuition that the new circle should have a larger radius because it passes through the center of the original circle and two points on its circumference. However, the calculation shows that the radius is the same as the original circle.But let's think about it geometrically. The original circle has center O and radius 8 cm. The new circle passes through O, A, and B, which are points such that OA and OB are radii of the original circle, and AB is a chord subtending 120 degrees at O.When we construct the circumcircle of triangle OAB, it turns out that the radius is the same as the original circle because the distances from the new center to O, A, and B are all equal to 8 cm.Therefore, despite O being the center of the original circle, the new circle passing through O, A, and B has the same radius as the original circle.So, the value of r is 8 cm.

💡First, I need to determine the number of women on the train. Since 70% of the passengers are women, I can calculate this by multiplying the total number of passengers by 70%.Next, I need to find out how many of these women are in business class. Given that 15% of the women are in business class, I can calculate this by multiplying the number of women by 15%.Finally, since the number of passengers must be a whole number, I will round the result to the nearest whole number to determine the exact number of women in business class.

💡Okay, so I have this problem here where I need to compare three values: a, b, and c. They're defined as follows:- ( a = frac{31}{32} )- ( b = cosleft(frac{1}{4}right) )- ( c = 4sinleft(frac{1}{4}right) )And I need to figure out which of the options A, B, C, or D is correct based on their relationships. The options are:A: ( c > b > a )B: ( b > a > c )C: ( a > b > c )D: ( a > c > b )Alright, let's break this down step by step. I'm a bit nervous because I'm not super confident with trigonometric functions, but I'll try my best.First, let's understand what each of these values represents.Starting with ( a = frac{31}{32} ). That's a fraction, so I can convert it to a decimal to get a better sense of its value. Dividing 31 by 32:( 31 ÷ 32 = 0.96875 )So, ( a = 0.96875 ). Got that.Next, ( b = cosleft(frac{1}{4}right) ). Cosine of what? The argument is ( frac{1}{4} ), but is that in radians or degrees? Hmm, in higher mathematics, especially in calculus, angles are typically measured in radians unless specified otherwise. So, I think it's safe to assume that ( frac{1}{4} ) is in radians.Similarly, ( c = 4sinleft(frac{1}{4}right) ). Again, the argument is ( frac{1}{4} ), which I believe is in radians.Alright, so I need to find the approximate values of ( cosleft(frac{1}{4}right) ) and ( sinleft(frac{1}{4}right) ) in radians.I remember that for small angles (in radians), the cosine and sine functions can be approximated using their Taylor series expansions. The Taylor series for cosine around 0 is:( cos(x) = 1 - frac{x^2}{2!} + frac{x^4}{4!} - frac{x^6}{6!} + dots )Similarly, the Taylor series for sine around 0 is:( sin(x) = x - frac{x^3}{3!} + frac{x^5}{5!} - frac{x^7}{7!} + dots )Since ( frac{1}{4} ) is a relatively small angle (about 0.25 radians, which is roughly 14.33 degrees), these approximations should be reasonably accurate.Let's compute ( cosleft(frac{1}{4}right) ) first.Using the Taylor series for cosine up to the ( x^4 ) term:( cosleft(frac{1}{4}right) ≈ 1 - frac{left(frac{1}{4}right)^2}{2} + frac{left(frac{1}{4}right)^4}{24} )Calculating each term:1. ( 1 ) is just 1.2. ( frac{left(frac{1}{4}right)^2}{2} = frac{frac{1}{16}}{2} = frac{1}{32} ≈ 0.03125 )3. ( frac{left(frac{1}{4}right)^4}{24} = frac{frac{1}{256}}{24} = frac{1}{6144} ≈ 0.00016276 )Adding these up:( 1 - 0.03125 + 0.00016276 ≈ 0.96891276 )So, ( cosleft(frac{1}{4}right) ≈ 0.9689 ). Hmm, interesting. That's very close to ( a = 0.96875 ). So, ( b ≈ 0.9689 ) and ( a = 0.96875 ). So, ( b ) is slightly larger than ( a ).Wait, but let me check if adding more terms in the Taylor series would change this. Let's include the next term, which is ( -frac{x^6}{720} ).Calculating ( frac{left(frac{1}{4}right)^6}{720} ):( left(frac{1}{4}right)^6 = frac{1}{4096} )So, ( frac{1}{4096} ÷ 720 ≈ frac{1}{2949120} ≈ 0.000000339 )So, subtracting that:( 0.96891276 - 0.000000339 ≈ 0.96891242 )So, it's still approximately 0.968912. So, even with more terms, it's still about 0.9689, which is just a tiny bit larger than ( a = 0.96875 ). So, ( b > a ).Okay, so that's one relationship: ( b > a ).Now, let's compute ( c = 4sinleft(frac{1}{4}right) ).Again, using the Taylor series for sine:( sinleft(frac{1}{4}right) ≈ frac{1}{4} - frac{left(frac{1}{4}right)^3}{6} + frac{left(frac{1}{4}right)^5}{120} - dots )Calculating each term:1. ( frac{1}{4} = 0.25 )2. ( frac{left(frac{1}{4}right)^3}{6} = frac{frac{1}{64}}{6} = frac{1}{384} ≈ 0.002604167 )3. ( frac{left(frac{1}{4}right)^5}{120} = frac{frac{1}{1024}}{120} = frac{1}{122880} ≈ 0.00000813 )Adding these up:( 0.25 - 0.002604167 + 0.00000813 ≈ 0.24740396 )So, ( sinleft(frac{1}{4}right) ≈ 0.2474 )Therefore, ( c = 4 times 0.2474 ≈ 0.9896 )So, ( c ≈ 0.9896 )Comparing this to ( b ≈ 0.9689 ) and ( a = 0.96875 ), we can see that ( c > b > a )Wait, hold on. So, ( c ≈ 0.9896 ), which is larger than both ( b ) and ( a ). ( b ≈ 0.9689 ) is slightly larger than ( a = 0.96875 ). So, the order is ( c > b > a ), which is option A.But wait, let me double-check my calculations because sometimes approximations can be misleading.First, let's verify ( cosleft(frac{1}{4}right) ). Maybe I can use a calculator for a more precise value.Using a calculator, ( cos(0.25) ) radians is approximately 0.96891242. So, my approximation was accurate.Similarly, ( sin(0.25) ) radians is approximately 0.247403959. So, multiplying by 4 gives approximately 0.989615836.So, ( c ≈ 0.9896 ), which is indeed larger than ( b ≈ 0.9689 ), which is just slightly larger than ( a = 0.96875 ).Therefore, the correct order is ( c > b > a ), which is option A.But wait, just to be thorough, let me consider the exact values without approximation.Is there a way to compare ( cosleft(frac{1}{4}right) ) and ( frac{31}{32} ) without approximating?I recall that for ( 0 < x < frac{pi}{2} ), ( cos(x) ) is decreasing, and ( cos(0) = 1 ), ( cosleft(frac{pi}{2}right) = 0 ). Since ( frac{1}{4} ) is less than ( frac{pi}{2} ) (which is approximately 1.5708), ( cosleft(frac{1}{4}right) ) is positive and less than 1.But how does it compare to ( frac{31}{32} )?Alternatively, perhaps using inequalities involving trigonometric functions.I remember that for ( 0 < x < frac{pi}{2} ), ( cos(x) > 1 - frac{x^2}{2} ). Let's check that.So, ( cos(x) > 1 - frac{x^2}{2} ). Plugging in ( x = frac{1}{4} ):( cosleft(frac{1}{4}right) > 1 - frac{left(frac{1}{4}right)^2}{2} = 1 - frac{1}{32} = frac{31}{32} )So, ( cosleft(frac{1}{4}right) > frac{31}{32} ), which means ( b > a ). That's a more precise way to see it without approximating.Similarly, for ( sin(x) ), I remember that ( sin(x) < x ) for ( x > 0 ). Wait, but we have ( c = 4sinleft(frac{1}{4}right) ). So, ( sinleft(frac{1}{4}right) < frac{1}{4} ), which would imply ( c < 1 ). But we already saw that ( c ≈ 0.9896 ), which is less than 1 but still larger than both ( a ) and ( b ).But wait, perhaps there's a better inequality to compare ( c ) and ( b ).I recall that ( tan(x) = frac{sin(x)}{cos(x)} ), and for ( 0 < x < frac{pi}{2} ), ( tan(x) > x ). So, ( tanleft(frac{1}{4}right) > frac{1}{4} ).Which implies ( frac{sinleft(frac{1}{4}right)}{cosleft(frac{1}{4}right)} > frac{1}{4} )Multiplying both sides by ( cosleft(frac{1}{4}right) ):( sinleft(frac{1}{4}right) > frac{1}{4}cosleft(frac{1}{4}right) )Multiplying both sides by 4:( 4sinleft(frac{1}{4}right) > cosleft(frac{1}{4}right) )So, ( c > b )Therefore, combining both results:( c > b > a )Which is option A.So, even without approximating, using these inequalities, we can establish the correct order.I think that's solid. So, the answer should be A.**Final Answer**boxed{A}

💡First, I recognize that the speeds of A and B are in the ratio of 3:4. Since speed is inversely proportional to time when the distance is constant, the ratio of their times taken to cover the same distance will be the inverse of the ratio of their speeds, which is 4:3.Next, I denote the time taken by A as 2 hours and set up the proportion based on the time ratio:2 hours / t hours = 4 / 3Solving for t, I find that t = 1.5 hours.Finally, I calculate the difference in time between A and B:2 hours - 1.5 hours = 0.5 hoursConverting this time difference into minutes:0.5 hours * 60 minutes/hour = 30 minutesTherefore, A takes 30 minutes more than B to reach the destination.

💡Okay, so I need to find the point on the line y = 3x - 1 that is closest to the point (4, -2). Hmm, I remember that the closest point on a line to a given point is where the perpendicular from the given point meets the line. So, I think I need to find the equation of the line perpendicular to y = 3x - 1 that passes through (4, -2), and then find the intersection point of these two lines. That intersection should be the closest point.First, let me recall that the slope of a line perpendicular to another line is the negative reciprocal of the original line's slope. The given line has a slope of 3, so the perpendicular slope should be -1/3. That makes sense because 3 * (-1/3) = -1, which confirms they are perpendicular.So, the equation of the perpendicular line passing through (4, -2) would have the slope -1/3. Using the point-slope form of a line, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point, I can plug in the values.Let me write that out:y - (-2) = (-1/3)(x - 4)Simplifying that:y + 2 = (-1/3)(x - 4)Now, distribute the -1/3 on the right side:y + 2 = (-1/3)x + 4/3Subtract 2 from both sides to solve for y:y = (-1/3)x + 4/3 - 2Wait, 4/3 - 2 is the same as 4/3 - 6/3, which equals -2/3. So the equation of the perpendicular line is:y = (-1/3)x - 2/3Now, I have two lines:1. The original line: y = 3x - 12. The perpendicular line: y = (-1/3)x - 2/3I need to find their intersection point. To do that, I can set the two equations equal to each other since at the intersection point, the y-values will be the same.So, set 3x - 1 equal to (-1/3)x - 2/3:3x - 1 = (-1/3)x - 2/3Hmm, let me solve for x. First, I can eliminate the fractions by multiplying every term by 3 to make it easier.3*(3x) - 3*1 = 3*(-1/3)x - 3*(2/3)Calculating each term:9x - 3 = -x - 2Now, let's get all the x terms on one side and constants on the other. I'll add x to both sides:9x + x - 3 = -2Which simplifies to:10x - 3 = -2Now, add 3 to both sides:10x = 1So, x = 1/10Now that I have x, I can plug it back into either of the two equations to find y. I'll choose the original line y = 3x - 1 because it looks simpler.Plugging x = 1/10 into y = 3x - 1:y = 3*(1/10) - 1 = 3/10 - 1Convert 1 to 10/10 to subtract:y = 3/10 - 10/10 = -7/10So, the point of intersection is (1/10, -7/10). This should be the closest point on the line y = 3x - 1 to the point (4, -2).Wait, let me double-check my calculations to make sure I didn't make any mistakes. Starting from the beginning, the slope of the original line is 3, so the perpendicular slope is -1/3. The equation of the perpendicular line through (4, -2) is y = (-1/3)x - 2/3. Setting that equal to the original line:3x - 1 = (-1/3)x - 2/3Multiply both sides by 3:9x - 3 = -x - 2Adding x to both sides:10x - 3 = -2Adding 3:10x = 1x = 1/10Plugging back in:y = 3*(1/10) - 1 = 3/10 - 10/10 = -7/10Yes, that seems correct. So, the closest point is (1/10, -7/10).Alternatively, I remember there's a formula for the distance from a point to a line, but since we're asked for the point on the line closest to (4, -2), finding the intersection of the perpendicular is the way to go.Just to be thorough, maybe I can use another method, like minimizing the distance function. The distance between a point (x, y) on the line and (4, -2) is sqrt[(x - 4)^2 + (y + 2)^2]. Since the square root is a monotonic function, minimizing the distance is equivalent to minimizing the square of the distance, which is easier.So, let's define the distance squared as D = (x - 4)^2 + (y + 2)^2. But since y = 3x - 1, substitute that into D:D = (x - 4)^2 + (3x - 1 + 2)^2 = (x - 4)^2 + (3x + 1)^2Now, expand both terms:(x - 4)^2 = x^2 - 8x + 16(3x + 1)^2 = 9x^2 + 6x + 1So, D = x^2 - 8x + 16 + 9x^2 + 6x + 1 = 10x^2 - 2x + 17Now, to find the minimum, take the derivative of D with respect to x and set it to zero.dD/dx = 20x - 2Set equal to zero:20x - 2 = 020x = 2x = 2/20 = 1/10So, x = 1/10. Plugging back into y = 3x - 1:y = 3*(1/10) - 1 = 3/10 - 10/10 = -7/10Same result. So, this confirms that the closest point is indeed (1/10, -7/10).I think that's solid. Both methods give the same answer, so I'm confident that's correct.**Final Answer**The point on the line closest to (4, -2) is boxed{left( dfrac{1}{10}, -dfrac{7}{10} right)}.

💡Okay, so I need to find the set of positive integers ( n ) such that the sum of the floor of log base 2 of each integer from 1 to ( n ) equals 2006. That is, I need to solve:[leftlfloor log_2 1 rightrfloor + leftlfloor log_2 2 rightrfloor + leftlfloor log_2 3 rightrfloor + cdots + leftlfloor log_2 n rightrfloor = 2006]First, let me recall what the floor function does. The floor of a number is the greatest integer less than or equal to that number. So, for example, ( leftlfloor 2.3 rightrfloor = 2 ) and ( leftlfloor 3 rightrfloor = 3 ).Now, ( log_2 k ) is the logarithm of ( k ) with base 2. So, ( log_2 1 = 0 ), ( log_2 2 = 1 ), ( log_2 4 = 2 ), and so on. The floor of these logarithms will give me integers that represent the exponent needed to reach the nearest lower power of 2.Let me think about how ( leftlfloor log_2 k rightrfloor ) behaves as ( k ) increases. For ( k ) in the range ( 2^m leq k < 2^{m+1} ), ( leftlfloor log_2 k rightrfloor = m ). So, for example:- When ( k = 1 ), ( leftlfloor log_2 1 rightrfloor = 0 )- When ( k = 2 ) to ( k = 3 ), ( leftlfloor log_2 k rightrfloor = 1 )- When ( k = 4 ) to ( k = 7 ), ( leftlfloor log_2 k rightrfloor = 2 )- When ( k = 8 ) to ( k = 15 ), ( leftlfloor log_2 k rightrfloor = 3 )- And so on.So, for each ( m ), the value ( m ) is added ( 2^m ) times. For example, ( m = 1 ) is added twice (for ( k = 2 ) and ( k = 3 )), ( m = 2 ) is added four times (for ( k = 4 ) to ( k = 7 )), etc.Therefore, the sum ( S(n) ) can be expressed as:[S(n) = sum_{k=1}^{n} leftlfloor log_2 k rightrfloor]To compute this sum, I can break it down into intervals where ( leftlfloor log_2 k rightrfloor ) is constant. Let's denote ( m ) as the integer such that ( 2^m leq k < 2^{m+1} ). Then, for each ( m ), the number of terms where ( leftlfloor log_2 k rightrfloor = m ) is ( 2^m ), except possibly for the last interval if ( n ) doesn't reach ( 2^{m+1} - 1 ).So, the sum ( S(n) ) can be written as:[S(n) = sum_{m=0}^{M} m cdot 2^m + text{remaining terms}]where ( M ) is the largest integer such that ( 2^{M+1} - 1 leq n ). The remaining terms would be for ( k ) from ( 2^{M+1} ) to ( n ), each contributing ( M+1 ) to the sum.Now, I need to compute this sum and set it equal to 2006. Let's start by computing the sum for each ( m ) and see where it gets us.First, let's compute the sum ( sum_{m=0}^{M} m cdot 2^m ). There's a formula for this sum:[sum_{m=0}^{M} m cdot 2^m = (M - 1) cdot 2^{M+1} + 2]Wait, let me verify that formula. Let me compute it manually for small ( M ):- For ( M = 0 ): ( 0 cdot 2^0 = 0 ). The formula gives ( (0 - 1) cdot 2^{1} + 2 = (-1) cdot 2 + 2 = -2 + 2 = 0 ). Correct.- For ( M = 1 ): ( 0 cdot 2^0 + 1 cdot 2^1 = 0 + 2 = 2 ). The formula gives ( (1 - 1) cdot 2^{2} + 2 = 0 cdot 4 + 2 = 2 ). Correct.- For ( M = 2 ): ( 0 cdot 2^0 + 1 cdot 2^1 + 2 cdot 2^2 = 0 + 2 + 8 = 10 ). The formula gives ( (2 - 1) cdot 2^{3} + 2 = 1 cdot 8 + 2 = 10 ). Correct.- For ( M = 3 ): ( 0 + 2 + 8 + 24 = 34 ). The formula gives ( (3 - 1) cdot 2^{4} + 2 = 2 cdot 16 + 2 = 34 ). Correct.Okay, so the formula seems to hold. Therefore, the sum up to ( M ) is ( (M - 1) cdot 2^{M+1} + 2 ).Now, let's compute this for increasing ( M ) until we get close to 2006.Let's start:- ( M = 0 ): Sum = 0- ( M = 1 ): Sum = 2- ( M = 2 ): Sum = 10- ( M = 3 ): Sum = 34- ( M = 4 ): Sum = 106- ( M = 5 ): Sum = 322- ( M = 6 ): Sum = 994- ( M = 7 ): Sum = 2986Wait, at ( M = 7 ), the sum is 2986, which is greater than 2006. So, the sum up to ( M = 6 ) is 994, and up to ( M = 7 ) is 2986. Therefore, the value of ( M ) we need is 7, but we have to adjust for the remaining terms.Wait, actually, let me compute the sum step by step without using the formula to make sure.Compute the sum manually:- ( m = 0 ): 1 term, value 0. Sum = 0- ( m = 1 ): 2 terms, each 1. Sum += 2*1 = 2. Total sum = 2- ( m = 2 ): 4 terms, each 2. Sum += 4*2 = 8. Total sum = 10- ( m = 3 ): 8 terms, each 3. Sum += 8*3 = 24. Total sum = 34- ( m = 4 ): 16 terms, each 4. Sum += 16*4 = 64. Total sum = 98- ( m = 5 ): 32 terms, each 5. Sum += 32*5 = 160. Total sum = 258- ( m = 6 ): 64 terms, each 6. Sum += 64*6 = 384. Total sum = 642- ( m = 7 ): 128 terms, each 7. Sum += 128*7 = 896. Total sum = 1538- ( m = 8 ): 256 terms, each 8. Sum += 256*8 = 2048. Total sum = 3586Wait, now I'm confused because earlier I thought the formula gave 2986 for ( M = 7 ), but manually adding up to ( m = 7 ) gives 1538, and up to ( m = 8 ) is 3586.Wait, perhaps I made a mistake in the formula. Let me check the formula again.Wait, the formula I used earlier was ( (M - 1) cdot 2^{M+1} + 2 ). Let's test it for ( M = 7 ):( (7 - 1) cdot 2^{8} + 2 = 6 * 256 + 2 = 1536 + 2 = 1538 ). Okay, that matches the manual sum up to ( m = 7 ). So, the formula is correct.Similarly, for ( M = 8 ):( (8 - 1) cdot 2^{9} + 2 = 7 * 512 + 2 = 3584 + 2 = 3586 ). Correct.So, up to ( m = 7 ), the sum is 1538, and up to ( m = 8 ), it's 3586. But we need the sum to be 2006, which is between 1538 and 3586.Therefore, the value of ( M ) is 7, and we need to find how many additional terms beyond ( 2^8 - 1 = 255 ) we need to add to reach the sum of 2006.Wait, let me clarify:The sum up to ( m = 7 ) is 1538, which corresponds to ( k ) up to ( 2^8 - 1 = 255 ). So, ( S(255) = 1538 ).Now, we need to find ( n ) such that ( S(n) = 2006 ). Since 2006 > 1538, ( n ) must be greater than 255. Let's denote ( n = 255 + t ), where ( t ) is the number of terms beyond 255. Each of these terms contributes ( m = 8 ) to the sum because ( leftlfloor log_2 k rightrfloor = 8 ) for ( 256 leq k < 512 ).So, the additional sum contributed by these ( t ) terms is ( 8t ). Therefore, the total sum ( S(n) = 1538 + 8t ).We need:[1538 + 8t = 2006]Solving for ( t ):[8t = 2006 - 1538 = 468 t = 468 / 8 = 58.5]Wait, ( t ) must be an integer because we're counting the number of terms. 58.5 is not an integer, which suggests that there's no integer ( t ) such that ( 1538 + 8t = 2006 ). Therefore, there is no integer ( n ) such that ( S(n) = 2006 ).Wait, that can't be right. Let me double-check my calculations.Sum up to ( m = 7 ): 1538.We need 2006 - 1538 = 468 more.Each term beyond 255 contributes 8, so 468 / 8 = 58.5.But since ( t ) must be an integer, 58.5 is not possible. Therefore, the sum cannot reach exactly 2006 because the next possible sum after 1538 would be 1538 + 8*58 = 1538 + 464 = 2002, and then 2002 + 8 = 2010. So, the sum jumps from 2002 to 2010, skipping 2006.Therefore, there is no integer ( n ) such that ( S(n) = 2006 ).Wait, but let me make sure I didn't make a mistake in the sum up to ( m = 7 ). Let me recalculate:- ( m = 0 ): 1 term, sum = 0- ( m = 1 ): 2 terms, sum += 2*1 = 2, total = 2- ( m = 2 ): 4 terms, sum += 4*2 = 8, total = 10- ( m = 3 ): 8 terms, sum += 8*3 = 24, total = 34- ( m = 4 ): 16 terms, sum += 16*4 = 64, total = 98- ( m = 5 ): 32 terms, sum += 32*5 = 160, total = 258- ( m = 6 ): 64 terms, sum += 64*6 = 384, total = 642- ( m = 7 ): 128 terms, sum += 128*7 = 896, total = 1538Yes, that's correct. So, up to ( k = 255 ), the sum is 1538. Then, each additional ( k ) from 256 onwards adds 8 to the sum. So, the next sums are:- ( k = 256 ): sum = 1538 + 8 = 1546- ( k = 257 ): sum = 1546 + 8 = 1554- ...- Each time, adding 8.So, the sum increases by 8 each time ( k ) increases by 1 beyond 255.We need to reach 2006. The difference is 2006 - 1538 = 468.468 divided by 8 is 58.5, which is not an integer. Therefore, we cannot reach exactly 2006 by adding 8s. The closest we can get is 58 terms beyond 255, which would give us 1538 + 58*8 = 1538 + 464 = 2002. Then, the next term would be 2002 + 8 = 2010, which is beyond 2006.Therefore, there is no integer ( n ) such that the sum equals 2006.Wait, but the problem asks for the set of positive integers ( n ) such that the sum equals 2006. If no such ( n ) exists, then the set is empty.Alternatively, maybe I made a mistake in the initial sum. Let me check the formula again.The formula for the sum ( sum_{m=0}^{M} m cdot 2^m = (M - 1) cdot 2^{M+1} + 2 ).For ( M = 7 ):( (7 - 1) cdot 2^{8} + 2 = 6 * 256 + 2 = 1536 + 2 = 1538 ). Correct.So, up to ( k = 255 ), sum is 1538.Then, each ( k ) from 256 to ( n ) adds 8 to the sum.So, the sum is 1538 + 8*(n - 255).Set this equal to 2006:1538 + 8*(n - 255) = 2006Solve for ( n ):8*(n - 255) = 2006 - 1538 = 468n - 255 = 468 / 8 = 58.5n = 255 + 58.5 = 313.5But ( n ) must be an integer, so there's no solution.Therefore, the set of positive integers ( n ) satisfying the equation is empty.