💡First, I need to find the greatest common divisor (gcd) of 18 and 30. To do this, I'll start by listing the factors of each number.For 18, the factors are 1, 2, 3, 6, 9, and 18. For 30, the factors are 1, 2, 3, 5, 6, 10, 15, and 30. The common factors between 18 and 30 are 1, 2, 3, and 6. Among these, the largest one is 6, so the gcd of 18 and 30 is 6.Next, I need to determine if this gcd is even or odd. An even number is any integer that is divisible by 2 without leaving a remainder. Since 6 divided by 2 equals 3, which is an integer, 6 is an even number.Therefore, the greatest common divisor of 18 and 30 is 6, and it is an even number.

💡Okay, so I have this problem about a parabola C. The vertex is at the origin, and the focus is on the x-axis. That means the parabola opens either to the right or to the left. Since the focus is on the x-axis, it's symmetric about the x-axis. So, the standard form of such a parabola is either ( y^2 = 4ax ) or ( y^2 = -4ax ), depending on whether it opens to the right or left.The problem also says that the line ( y = x ) intersects the parabola at points A and B. So, I need to find the points of intersection between ( y = x ) and the parabola. Since the parabola is symmetric about the x-axis, and the line ( y = x ) is a straight line at 45 degrees, it should intersect the parabola at two points.Given that P(1,1) is the midpoint of the line segment AB, this gives me some information about the coordinates of points A and B. The midpoint formula tells me that the average of the x-coordinates of A and B is 1, and the average of the y-coordinates is also 1.Let me denote the coordinates of points A and B as ( (x_1, y_1) ) and ( (x_2, y_2) ). Since both points lie on the line ( y = x ), I know that ( y_1 = x_1 ) and ( y_2 = x_2 ). Therefore, the midpoint P(1,1) can be expressed as:[left( frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2} right) = (1, 1)]Since ( y_1 = x_1 ) and ( y_2 = x_2 ), this simplifies to:[left( frac{x_1 + x_2}{2}, frac{x_1 + x_2}{2} right) = (1, 1)]So, both the x and y coordinates of the midpoint are 1, which means:[frac{x_1 + x_2}{2} = 1 implies x_1 + x_2 = 2]Now, I need to find the points of intersection between the parabola and the line ( y = x ). Let's substitute ( y = x ) into the parabola's equation. Let's assume the standard form of the parabola is ( y^2 = 4ax ). Substituting ( y = x ), we get:[x^2 = 4a x]This simplifies to:[x^2 - 4a x = 0]Factoring out x:[x(x - 4a) = 0]So, the solutions are ( x = 0 ) and ( x = 4a ). Therefore, the points of intersection are at ( x = 0 ) and ( x = 4a ). Since ( y = x ), the corresponding y-values are also 0 and ( 4a ). So, the points A and B are ( (0, 0) ) and ( (4a, 4a) ).Wait a minute, but the midpoint of these two points would be:[left( frac{0 + 4a}{2}, frac{0 + 4a}{2} right) = (2a, 2a)]But according to the problem, the midpoint is P(1,1). Therefore:[2a = 1 implies a = frac{1}{2}]So, substituting back into the standard form of the parabola:[y^2 = 4a x = 4 times frac{1}{2} x = 2x]Therefore, the equation of the parabola is ( y^2 = 2x ).Wait, but let me double-check this. If the parabola is ( y^2 = 2x ), then substituting ( y = x ) gives:[x^2 = 2x implies x^2 - 2x = 0 implies x(x - 2) = 0]So, the points of intersection are at ( x = 0 ) and ( x = 2 ). Therefore, the points A and B are ( (0, 0) ) and ( (2, 2) ). The midpoint of these two points is:[left( frac{0 + 2}{2}, frac{0 + 2}{2} right) = (1, 1)]Which matches the given midpoint P(1,1). So, that seems correct.But wait, earlier I assumed the standard form was ( y^2 = 4ax ). But in the solution, I ended up with ( y^2 = 2x ), which is ( y^2 = 4a x ) with ( a = frac{1}{2} ). So, that's consistent.Alternatively, sometimes the standard form is written as ( y^2 = 2px ), where p is the distance from the vertex to the focus. In that case, comparing ( y^2 = 2px ) with ( y^2 = 2x ), we have ( 2p = 2 implies p = 1 ). So, the focus is at (1, 0), which is on the x-axis, as required.Therefore, the standard equation of the parabola is ( y^2 = 2x ).But let me make sure I didn't make any mistakes in my reasoning. Let's go through it again step by step.1. The parabola has its vertex at the origin and focus on the x-axis, so it's a right-opening parabola. Its standard form is ( y^2 = 4ax ), where a is the distance from the vertex to the focus.2. The line ( y = x ) intersects the parabola at points A and B. Substituting ( y = x ) into the parabola's equation gives ( x^2 = 4a x ), which simplifies to ( x(x - 4a) = 0 ). So, the points of intersection are at ( x = 0 ) and ( x = 4a ), with corresponding y-values 0 and ( 4a ).3. The midpoint of points A(0,0) and B(4a,4a) is ( (2a, 2a) ). According to the problem, this midpoint is (1,1), so ( 2a = 1 implies a = frac{1}{2} ).4. Substituting ( a = frac{1}{2} ) back into the standard form gives ( y^2 = 4 times frac{1}{2} x = 2x ).5. Verifying by substituting ( y = x ) into ( y^2 = 2x ) gives ( x^2 = 2x implies x(x - 2) = 0 ), so points at (0,0) and (2,2). Midpoint is (1,1), which matches.Therefore, the standard equation is indeed ( y^2 = 2x ).I think that's solid. I don't see any errors in the reasoning. The key steps were recognizing the standard form of the parabola, substituting the line equation into the parabola's equation, finding the points of intersection, using the midpoint formula to relate the sum of the roots, and solving for the parameter a. Then, substituting back to get the equation of the parabola.Another way to think about it is using the properties of parabolas and midpoints. Since the midpoint is given, it relates to the sum of the roots of the quadratic equation obtained after substitution. In this case, the quadratic in x had roots at 0 and 4a, whose sum is 4a. The midpoint's x-coordinate is the average of these roots, which is 2a, and setting that equal to 1 gives a = 1/2.So, all in all, the equation of the parabola is ( y^2 = 2x ).

💡First, I need to determine the total number of lockers in the row based on Minyoung's locker position. Since her locker is 8th from the front and 14th from the back, there are 7 lockers in front of hers and 13 lockers behind her. Adding these together with her locker gives a total of 21 lockers in the row.Next, I'll calculate the total number of lockers in the column. Minyoung's locker is 7th from the left and 13th from the right, meaning there are 6 lockers to the left and 12 lockers to the right. Including her locker, this results in 19 lockers in the column.Finally, to find the total number of lockers in the classroom, I'll multiply the number of lockers in a row by the number of lockers in a column:21 lockers/row × 19 lockers/column = 399 lockers.Therefore, the classroom has a total of 399 lockers.

💡First, I need to convert the ¥4 billion damage into Euros using the given exchange rate of 110 Japanese yen to 1 Euro.Next, I'll calculate the amount in Euros by dividing the total yen amount by the exchange rate.After obtaining the initial amount in Euros, I need to account for the additional 5% tax imposed by the emergency government measures. This means I'll multiply the initial Euro amount by 1.05 to include the 5% increase.Finally, I'll round the resulting number to the nearest whole number to determine the total damage in Euros.

💡Okay, so I have this problem where I need to figure out how many fractions in the sequence from 1/5 up to 2017/2021 are in their simplest form. That means I need to check each fraction to see if the numerator and denominator have any common factors other than 1. If they don't, then the fraction is in simplest form.First, let me understand the sequence better. The fractions are 1/5, 2/6, 3/7, 4/8, 5/9, and so on, up to 2017/2021. So, each fraction is of the form n/(n+4), where n starts at 1 and goes up to 2017. That makes sense because 2017 + 4 is 2021.Now, to determine if a fraction n/(n+4) is in simplest form, I need to check if the greatest common divisor (GCD) of n and n+4 is 1. If the GCD is 1, then the fraction is irreducible; otherwise, it's reducible.I remember that the GCD of two numbers also equals the GCD of one of the numbers and their difference. So, GCD(n, n+4) is the same as GCD(n, 4). That simplifies things because now I just need to find out when n and 4 are coprime, meaning their GCD is 1.Since 4 is 2 squared, the only prime factor is 2. Therefore, n and 4 will be coprime if n is not divisible by 2. In other words, n needs to be an odd number. If n is even, then both n and n+4 will be even, so their GCD will be at least 2, making the fraction reducible.So, the problem reduces to counting how many odd numbers there are from 1 to 2017. That should give me the number of fractions in simplest form.To count the number of odd numbers between 1 and 2017, I can think of it as an arithmetic sequence where the first term is 1, the last term is 2017, and the common difference is 2 (since we're only considering odd numbers).The formula for the nth term of an arithmetic sequence is:a_n = a_1 + (n - 1)dHere, a_n is 2017, a_1 is 1, and d is 2. Plugging these into the formula:2017 = 1 + (n - 1)*2Subtract 1 from both sides:2016 = (n - 1)*2Divide both sides by 2:1008 = n - 1Add 1 to both sides:n = 1009So, there are 1009 odd numbers between 1 and 2017. That means there are 1009 fractions in the sequence that are in simplest form.Wait, let me double-check my reasoning. If n is odd, then n and 4 are coprime because 4 only has 2 as a prime factor, and n is not divisible by 2. Therefore, GCD(n, 4) = 1, which means GCD(n, n+4) = 1. So, the fraction n/(n+4) is irreducible. That seems correct.Is there any case where n is odd but n and n+4 still share a common factor? Let's see. If n is odd, n+4 will also be odd because even + odd is odd. Wait, no, 4 is even, so n is odd, n+4 is odd + even = odd. So, both n and n+4 are odd. But since 4 is even, and n is odd, n and 4 don't share any common factors besides 1. So, yes, GCD(n, n+4) = 1.Therefore, my conclusion is correct. There are 1009 fractions in simplest form in the given sequence.**Final Answer**The number of irreducible fractions is boxed{1009}.

💡Alright, so I'm trying to solve this geometry problem about a right-angled triangle and a square constructed on its hypotenuse. Let me try to visualize and break it down step by step.First, the problem says: In the right-angled triangle (ABC), the angle at vertex (B) is (30^circ). A square is drawn outwardly on the hypotenuse (BC) with center (D). We need to find the measure of angle (ADB).Okay, let's start by sketching the triangle. Since it's a right-angled triangle at (B), and angle (B) is (30^circ), that means the other non-right angle at (C) must be (60^circ) because the angles in a triangle add up to (180^circ). So, triangle (ABC) is a 30-60-90 triangle.In a 30-60-90 triangle, the sides are in the ratio (1 : sqrt{3} : 2). Let's denote the sides accordingly. Let’s say the side opposite the (30^circ) angle (which is (AC)) is the shortest side, so let's call that length 1. Then, the side opposite the (60^circ) angle (which is (AB)) would be (sqrt{3}), and the hypotenuse (BC) would be 2.Wait, hold on, actually, in a right-angled triangle, the hypotenuse is opposite the right angle. So, if the triangle is right-angled at (B), then the hypotenuse is (AC), not (BC). Hmm, that might be a point of confusion. Let me clarify.If the triangle is right-angled at (B), then sides (AB) and (BC) are the legs, and (AC) is the hypotenuse. So, angle (B) is (30^circ), which is one of the acute angles. Therefore, the other acute angle at (C) is (60^circ).Given that, the sides can be labeled as follows: side opposite (30^circ) (which is (AC)) is the shorter leg, so let's say that's length 1. The side opposite (60^circ) (which is (AB)) is the longer leg, so that's (sqrt{3}), and the hypotenuse (BC) is 2.Wait, no, hold on again. If the triangle is right-angled at (B), then the hypotenuse is (AC), and the legs are (AB) and (BC). So, angle (B) is (30^circ), so the side opposite to it is (AC), which is the shorter leg. Wait, no, that's not correct.In a right-angled triangle, the side opposite the right angle is the hypotenuse. So, in triangle (ABC), right-angled at (B), the hypotenuse is (AC). The other two sides, (AB) and (BC), are the legs.Given that angle at (B) is (30^circ), then the sides can be determined. The side opposite (30^circ) is the shorter leg, which would be (AC) if angle (B) is (30^circ). Wait, no, that's not right. The side opposite angle (B) is (AC), which is the hypotenuse. That can't be, because the hypotenuse is the longest side.Wait, I'm getting confused here. Let me clarify:In triangle (ABC), right-angled at (B). So, sides:- (AB) and (BC) are the legs.- (AC) is the hypotenuse.Given that angle at (B) is (30^circ), so angle (ABC = 30^circ). Therefore, the other angle at (C) is (60^circ).In a 30-60-90 triangle, the sides are in the ratio (1 : sqrt{3} : 2), where:- The side opposite (30^circ) is the shortest side (let's say length 1).- The side opposite (60^circ) is ( sqrt{3} ).- The hypotenuse is 2.So, in this case, angle (B) is (30^circ), so the side opposite to it is (AC). Wait, no, (AC) is the hypotenuse, so that can't be. The side opposite angle (B) is (AC), but (AC) is the hypotenuse, which is the longest side. That contradicts the ratio because the side opposite (30^circ) should be the shortest side.Wait, maybe I'm mislabeling the triangle. Let me try again.If the triangle is right-angled at (B), then:- Angle (B = 90^circ).Wait, no, the problem says angle at vertex (B) is (30^circ). So, it's not right-angled at (B). Wait, hold on, the problem says "In the right-angled triangle (ABC), the angle at vertex (B) is (30^circ)."So, the triangle is right-angled, but not necessarily at (B). It just says the angle at (B) is (30^circ). So, the right angle must be at another vertex.So, triangle (ABC) is right-angled, with angle (B = 30^circ). So, the right angle must be at either (A) or (C).Let me assume the right angle is at (A). So, triangle (ABC) is right-angled at (A), with angle (B = 30^circ). Then, angle (C) would be (60^circ).Alternatively, if the right angle is at (C), then angle (B = 30^circ), so angle (A) would be (60^circ).But the problem mentions the hypotenuse (BC). So, if the hypotenuse is (BC), that means the right angle is at (A). Because in a triangle, the hypotenuse is the side opposite the right angle. So, if the hypotenuse is (BC), then the right angle must be at (A).Therefore, triangle (ABC) is right-angled at (A), with angle (B = 30^circ), angle (C = 60^circ), hypotenuse (BC), and legs (AB) and (AC).Got it. So, sides:- (AB) is opposite angle (C = 60^circ).- (AC) is opposite angle (B = 30^circ).- (BC) is the hypotenuse.So, in a 30-60-90 triangle, the sides are in the ratio (1 : sqrt{3} : 2). Therefore:- The side opposite (30^circ) (which is (AC)) is the shortest side, let's say length 1.- The side opposite (60^circ) (which is (AB)) is ( sqrt{3} ).- The hypotenuse (BC) is 2.Alright, so now we have triangle (ABC) with sides:- (AB = sqrt{3})- (AC = 1)- (BC = 2)Now, the square is drawn outwardly on the hypotenuse (BC) with center (D). So, we're constructing a square on side (BC), outside the triangle (ABC), and (D) is the center of that square.We need to find the measure of angle (ADB).Hmm, okay. Let me try to visualize this. So, we have triangle (ABC), right-angled at (A), with (BC) as the hypotenuse. On this hypotenuse (BC), we construct a square outwardly, meaning the square is built outside the triangle. The center of this square is point (D).So, point (D) is the center of the square constructed on (BC). Therefore, (D) is equidistant from all four vertices of the square.Our goal is to find angle (ADB), which is the angle at point (D) between points (A) and (B).To solve this, I think we might need to use coordinate geometry. Let me assign coordinates to the points to make it easier.Let's place point (A) at the origin ((0, 0)). Since the triangle is right-angled at (A), we can place point (B) along the x-axis and point (C) along the y-axis.Given that, let's assign coordinates:- Point (A): ((0, 0))- Point (B): ((sqrt{3}, 0)) because (AB = sqrt{3})- Point (C): ((0, 1)) because (AC = 1)Wait, hold on, because in a 30-60-90 triangle, the sides are (1 : sqrt{3} : 2). So, if (AC = 1) (opposite (30^circ)), then (AB = sqrt{3}) (opposite (60^circ)), and (BC = 2) (hypotenuse).So, with (A) at ((0, 0)), (B) is at ((sqrt{3}, 0)), and (C) is at ((0, 1)).Now, we need to construct a square on hypotenuse (BC) outwardly. So, side (BC) is from point (B) ((sqrt{3}, 0)) to point (C) ((0, 1)). The length of (BC) is 2, as we established earlier.Constructing a square on (BC) outwardly means we need to build a square where (BC) is one of its sides, and the square is outside the triangle (ABC).To find the coordinates of the square's vertices, we need to determine the other two points of the square besides (B) and (C). Let's denote the square as (BCFE), where (E) and (F) are the other two vertices.But wait, actually, since it's a square, the side (BC) will be one side, and we need to find the other two vertices such that all sides are equal and all angles are right angles.To find the coordinates of the other two vertices, we can use vectors or rotation.Let me consider vector (BC). The vector from (B) to (C) is (C - B = (0 - sqrt{3}, 1 - 0) = (-sqrt{3}, 1)).To construct the square outwardly, we need to find the next point from (C) such that the side is perpendicular to (BC). Since it's a square, each turn is 90 degrees.The direction from (B) to (C) is vector ((- sqrt{3}, 1)). To find the next vector perpendicular to this, we can rotate the vector (90^circ) clockwise or counterclockwise. Since it's outwardly, we need to determine which direction is outward.Given that the triangle is in the coordinate system with (A) at the origin, (B) along the x-axis, and (C) along the y-axis, the square constructed outwardly on (BC) would likely be constructed in the direction away from the triangle.To determine the direction, let's compute the perpendicular vectors.The vector (BC) is ((- sqrt{3}, 1)). A perpendicular vector can be obtained by rotating this vector (90^circ) clockwise or counterclockwise.Rotating (90^circ) clockwise: ((1, sqrt{3}))Rotating (90^circ) counterclockwise: ((-1, -sqrt{3}))Now, we need to determine which direction is outward. Since the triangle is in the first quadrant, and the square is constructed outwardly, it's likely that the square is constructed in the direction that doesn't overlap with the triangle.Given that, if we rotate vector (BC) (90^circ) clockwise, we get vector ((1, sqrt{3})). Adding this to point (C) would take us further into the first quadrant, which might overlap with the triangle. On the other hand, rotating (90^circ) counterclockwise gives vector ((-1, -sqrt{3})), which would take us into the fourth quadrant, away from the triangle.Therefore, the outward direction is likely the counterclockwise rotation.So, the next point (E) from (C) would be (C + (-1, -sqrt{3}) = (0 - 1, 1 - sqrt{3}) = (-1, 1 - sqrt{3})).Then, from point (E), we move in the direction opposite to (BC) to get the fourth vertex (F). The vector from (C) to (E) is ((-1, -sqrt{3})), so the vector from (E) to (F) should be the same as from (B) to (C), which is ((- sqrt{3}, 1)). Wait, no, actually, in a square, each side is equal and at right angles.Wait, perhaps a better approach is to compute all four vertices.Given points (B) ((sqrt{3}, 0)) and (C) ((0, 1)), we need to find points (E) and (F) such that (BCFE) is a square.Vector (BC) is ((- sqrt{3}, 1)). To find the next point (E), we can rotate vector (BC) (90^circ) counterclockwise.The rotation of a vector ((x, y)) by (90^circ) counterclockwise is ((-y, x)). So, rotating ((- sqrt{3}, 1)) gives ((-1, -sqrt{3})).Therefore, point (E) is (C + (-1, -sqrt{3}) = (0 - 1, 1 - sqrt{3}) = (-1, 1 - sqrt{3})).Then, to find point (F), we can move from (E) in the direction opposite to (BC). Vector (BC) is ((- sqrt{3}, 1)), so the opposite vector is ((sqrt{3}, -1)). Adding this to (E):(F = E + (sqrt{3}, -1) = (-1 + sqrt{3}, 1 - sqrt{3} - 1) = (sqrt{3} - 1, -sqrt{3})).Alternatively, since it's a square, point (F) can also be found by moving from (B) in the direction of vector (CE). Vector (CE) is ((-1, -sqrt{3})), so adding this to (B):(F = B + (-1, -sqrt{3}) = (sqrt{3} - 1, 0 - sqrt{3}) = (sqrt{3} - 1, -sqrt{3})).So, the four vertices of the square (BCFE) are:- (B): ((sqrt{3}, 0))- (C): ((0, 1))- (E): ((-1, 1 - sqrt{3}))- (F): ((sqrt{3} - 1, -sqrt{3}))Now, the center (D) of the square is the intersection point of its diagonals. The diagonals of a square bisect each other at 90 degrees, and the center is the midpoint of both diagonals.So, to find (D), we can compute the midpoint of diagonal (BE) or diagonal (CF).Let's compute the midpoint of diagonal (BE):Midpoint formula: (left( frac{x_1 + x_2}{2}, frac{y_1 + y_2}{2} right))Midpoint of (B) ((sqrt{3}, 0)) and (E) ((-1, 1 - sqrt{3})):(D_x = frac{sqrt{3} + (-1)}{2} = frac{sqrt{3} - 1}{2})(D_y = frac{0 + (1 - sqrt{3})}{2} = frac{1 - sqrt{3}}{2})So, center (D) is at (left( frac{sqrt{3} - 1}{2}, frac{1 - sqrt{3}}{2} right)).Alternatively, we can compute the midpoint of diagonal (CF):Midpoint of (C) ((0, 1)) and (F) ((sqrt{3} - 1, -sqrt{3})):(D_x = frac{0 + (sqrt{3} - 1)}{2} = frac{sqrt{3} - 1}{2})(D_y = frac{1 + (-sqrt{3})}{2} = frac{1 - sqrt{3}}{2})Same result, so that's consistent.Now, we have coordinates for points (A), (B), and (D):- (A): ((0, 0))- (B): ((sqrt{3}, 0))- (D): (left( frac{sqrt{3} - 1}{2}, frac{1 - sqrt{3}}{2} right))We need to find angle (ADB). That is, the angle at point (D) between points (A), (D), and (B).To find angle (ADB), we can use vector analysis or coordinate geometry. Let's use vectors.First, compute vectors (DA) and (DB).Vector (DA) is from (D) to (A): (A - D = left( 0 - frac{sqrt{3} - 1}{2}, 0 - frac{1 - sqrt{3}}{2} right) = left( frac{1 - sqrt{3}}{2}, frac{sqrt{3} - 1}{2} right)).Vector (DB) is from (D) to (B): (B - D = left( sqrt{3} - frac{sqrt{3} - 1}{2}, 0 - frac{1 - sqrt{3}}{2} right)).Let's compute each component:For (DB_x):(sqrt{3} - frac{sqrt{3} - 1}{2} = frac{2sqrt{3} - (sqrt{3} - 1)}{2} = frac{2sqrt{3} - sqrt{3} + 1}{2} = frac{sqrt{3} + 1}{2})For (DB_y):(0 - frac{1 - sqrt{3}}{2} = frac{sqrt{3} - 1}{2})So, vector (DB) is (left( frac{sqrt{3} + 1}{2}, frac{sqrt{3} - 1}{2} right)).Now, to find angle (ADB), we can use the dot product formula:[cos theta = frac{vec{DA} cdot vec{DB}}{|vec{DA}| |vec{DB}|}]First, compute the dot product (vec{DA} cdot vec{DB}):[left( frac{1 - sqrt{3}}{2} right) left( frac{sqrt{3} + 1}{2} right) + left( frac{sqrt{3} - 1}{2} right) left( frac{sqrt{3} - 1}{2} right)]Let's compute each term:First term:[left( frac{1 - sqrt{3}}{2} right) left( frac{sqrt{3} + 1}{2} right) = frac{(1)(sqrt{3}) + (1)(1) - (sqrt{3})(sqrt{3}) - (sqrt{3})(1)}{4}][= frac{sqrt{3} + 1 - 3 - sqrt{3}}{4} = frac{( sqrt{3} - sqrt{3} ) + (1 - 3)}{4} = frac{0 - 2}{4} = frac{-2}{4} = -frac{1}{2}]Second term:[left( frac{sqrt{3} - 1}{2} right) left( frac{sqrt{3} - 1}{2} right) = frac{(sqrt{3})^2 - 2sqrt{3} + 1}{4} = frac{3 - 2sqrt{3} + 1}{4} = frac{4 - 2sqrt{3}}{4} = frac{2 - sqrt{3}}{2}]So, the dot product is:[-frac{1}{2} + frac{2 - sqrt{3}}{2} = frac{-1 + 2 - sqrt{3}}{2} = frac{1 - sqrt{3}}{2}]Now, compute the magnitudes of vectors (DA) and (DB).First, (|vec{DA}|):[sqrt{left( frac{1 - sqrt{3}}{2} right)^2 + left( frac{sqrt{3} - 1}{2} right)^2}]Compute each component squared:[left( frac{1 - sqrt{3}}{2} right)^2 = frac{1 - 2sqrt{3} + 3}{4} = frac{4 - 2sqrt{3}}{4} = frac{2 - sqrt{3}}{2}][left( frac{sqrt{3} - 1}{2} right)^2 = frac{3 - 2sqrt{3} + 1}{4} = frac{4 - 2sqrt{3}}{4} = frac{2 - sqrt{3}}{2}]So, (|vec{DA}| = sqrt{ frac{2 - sqrt{3}}{2} + frac{2 - sqrt{3}}{2} } = sqrt{ frac{4 - 2sqrt{3}}{2} } = sqrt{2 - sqrt{3}} )Similarly, compute (|vec{DB}|):[sqrt{left( frac{sqrt{3} + 1}{2} right)^2 + left( frac{sqrt{3} - 1}{2} right)^2}]Compute each component squared:[left( frac{sqrt{3} + 1}{2} right)^2 = frac{3 + 2sqrt{3} + 1}{4} = frac{4 + 2sqrt{3}}{4} = frac{2 + sqrt{3}}{2}][left( frac{sqrt{3} - 1}{2} right)^2 = frac{3 - 2sqrt{3} + 1}{4} = frac{4 - 2sqrt{3}}{4} = frac{2 - sqrt{3}}{2}]So, (|vec{DB}| = sqrt{ frac{2 + sqrt{3}}{2} + frac{2 - sqrt{3}}{2} } = sqrt{ frac{4}{2} } = sqrt{2} )Now, plug these into the cosine formula:[cos theta = frac{frac{1 - sqrt{3}}{2}}{ sqrt{2 - sqrt{3}} cdot sqrt{2} }]Simplify the denominator:[sqrt{2 - sqrt{3}} cdot sqrt{2} = sqrt{2(2 - sqrt{3})} = sqrt{4 - 2sqrt{3}}]So,[cos theta = frac{1 - sqrt{3}}{2 sqrt{4 - 2sqrt{3}}}]Let me rationalize the denominator or simplify (sqrt{4 - 2sqrt{3}}).Notice that (4 - 2sqrt{3}) can be expressed as ((sqrt{3} - 1)^2):[(sqrt{3} - 1)^2 = 3 - 2sqrt{3} + 1 = 4 - 2sqrt{3}]Yes, that's correct. So,[sqrt{4 - 2sqrt{3}} = sqrt{ (sqrt{3} - 1)^2 } = sqrt{3} - 1]Therefore,[cos theta = frac{1 - sqrt{3}}{2 (sqrt{3} - 1)} = frac{ - (sqrt{3} - 1) }{2 (sqrt{3} - 1)} = -frac{1}{2}]So,[cos theta = -frac{1}{2}]Thus, the angle (theta) is:[theta = arccos left( -frac{1}{2} right) = 120^circ]Wait, but that's the angle between vectors (DA) and (DB). However, angle (ADB) is the angle at (D), which is the same as the angle between vectors (DA) and (DB). So, angle (ADB = 120^circ).But wait, that contradicts my initial thought that it might be (60^circ). Did I make a mistake somewhere?Let me double-check the calculations.First, the coordinates:- (A(0, 0)), (B(sqrt{3}, 0)), (C(0, 1))Square constructed on (BC), center (D).Vectors:- (DA = left( frac{1 - sqrt{3}}{2}, frac{sqrt{3} - 1}{2} right))- (DB = left( frac{sqrt{3} + 1}{2}, frac{sqrt{3} - 1}{2} right))Dot product:[left( frac{1 - sqrt{3}}{2} right) left( frac{sqrt{3} + 1}{2} right) + left( frac{sqrt{3} - 1}{2} right) left( frac{sqrt{3} - 1}{2} right) = -frac{1}{2} + frac{2 - sqrt{3}}{2} = frac{1 - sqrt{3}}{2}]Magnitudes:- (|vec{DA}| = sqrt{2 - sqrt{3}})- (|vec{DB}| = sqrt{2})So,[cos theta = frac{1 - sqrt{3}}{2 sqrt{2 - sqrt{3}} cdot sqrt{2}} = frac{1 - sqrt{3}}{2 sqrt{4 - 2sqrt{3}}} = frac{1 - sqrt{3}}{2 (sqrt{3} - 1)} = -frac{1}{2}]Yes, that seems correct. So, angle (ADB = 120^circ).But wait, the problem says "the square is drawn outwardly on the hypotenuse (BC)". In my coordinate system, the square is constructed in the direction where point (E) is at ((-1, 1 - sqrt{3})), which is in the second quadrant, and point (F) is at ((sqrt{3} - 1, -sqrt{3})), which is in the fourth quadrant. So, the center (D) is at (left( frac{sqrt{3} - 1}{2}, frac{1 - sqrt{3}}{2} right)), which is in the fourth quadrant.But point (A) is at the origin, and point (B) is at ((sqrt{3}, 0)). So, angle (ADB) is formed at point (D) in the fourth quadrant, between points (A), (D), and (B). The angle between vectors (DA) and (DB) is (120^circ).However, sometimes angles are considered as the smaller angle between two lines, so (120^circ) is the reflex angle, and the smaller angle would be (60^circ). But in this case, since we're talking about the angle at (D), it's the angle between the two lines (DA) and (DB), which is indeed (120^circ).But wait, let me think again. Maybe I made a mistake in the direction of the square. If the square is constructed outwardly, perhaps it's constructed in the other direction, leading to a different center (D).Wait, earlier, I assumed the square was constructed by rotating vector (BC) (90^circ) counterclockwise, leading to the center (D) in the fourth quadrant. But maybe it's constructed by rotating (90^circ) clockwise, leading to a different position of (D).Let me try that.If we rotate vector (BC) (90^circ) clockwise instead, the rotation of vector ((- sqrt{3}, 1)) (90^circ) clockwise is ((1, sqrt{3})).So, point (E) would be (C + (1, sqrt{3}) = (0 + 1, 1 + sqrt{3}) = (1, 1 + sqrt{3})).Then, point (F) would be (E + (sqrt{3}, -1)) or something? Wait, let me compute it properly.Vector (BC) is ((- sqrt{3}, 1)). Rotated (90^circ) clockwise, it becomes ((1, sqrt{3})). So, point (E) is (C + (1, sqrt{3}) = (1, 1 + sqrt{3})).Then, vector (CE) is ((1, sqrt{3})). To find point (F), we need to move from (E) in the direction opposite to (BC). Vector (BC) is ((- sqrt{3}, 1)), so the opposite vector is ((sqrt{3}, -1)). Adding this to (E):(F = E + (sqrt{3}, -1) = (1 + sqrt{3}, 1 + sqrt{3} - 1) = (1 + sqrt{3}, sqrt{3})).Alternatively, since it's a square, point (F) can also be found by moving from (B) in the direction of vector (CE). Vector (CE) is ((1, sqrt{3})), so adding this to (B):(F = B + (1, sqrt{3}) = (sqrt{3} + 1, 0 + sqrt{3}) = (1 + sqrt{3}, sqrt{3})).So, the four vertices of the square (BCFE) in this case are:- (B): ((sqrt{3}, 0))- (C): ((0, 1))- (E): ((1, 1 + sqrt{3}))- (F): ((1 + sqrt{3}, sqrt{3}))Now, the center (D) is the midpoint of diagonal (BE) or (CF).Midpoint of (B) ((sqrt{3}, 0)) and (E) ((1, 1 + sqrt{3})):(D_x = frac{sqrt{3} + 1}{2})(D_y = frac{0 + 1 + sqrt{3}}{2} = frac{1 + sqrt{3}}{2})So, center (D) is at (left( frac{sqrt{3} + 1}{2}, frac{1 + sqrt{3}}{2} right)).Alternatively, midpoint of (C) ((0, 1)) and (F) ((1 + sqrt{3}, sqrt{3})):(D_x = frac{0 + 1 + sqrt{3}}{2} = frac{1 + sqrt{3}}{2})(D_y = frac{1 + sqrt{3}}{2})Same result.Now, with this new position of (D), let's compute vectors (DA) and (DB).Point (D): (left( frac{sqrt{3} + 1}{2}, frac{1 + sqrt{3}}{2} right))Vector (DA = A - D = left( 0 - frac{sqrt{3} + 1}{2}, 0 - frac{1 + sqrt{3}}{2} right) = left( -frac{sqrt{3} + 1}{2}, -frac{1 + sqrt{3}}{2} right))Vector (DB = B - D = left( sqrt{3} - frac{sqrt{3} + 1}{2}, 0 - frac{1 + sqrt{3}}{2} right))Compute each component:For (DB_x):(sqrt{3} - frac{sqrt{3} + 1}{2} = frac{2sqrt{3} - sqrt{3} - 1}{2} = frac{sqrt{3} - 1}{2})For (DB_y):(0 - frac{1 + sqrt{3}}{2} = -frac{1 + sqrt{3}}{2})So, vector (DB) is (left( frac{sqrt{3} - 1}{2}, -frac{1 + sqrt{3}}{2} right)).Now, compute the dot product (vec{DA} cdot vec{DB}):[left( -frac{sqrt{3} + 1}{2} right) left( frac{sqrt{3} - 1}{2} right) + left( -frac{1 + sqrt{3}}{2} right) left( -frac{1 + sqrt{3}}{2} right)]First term:[left( -frac{sqrt{3} + 1}{2} right) left( frac{sqrt{3} - 1}{2} right) = -frac{(sqrt{3} + 1)(sqrt{3} - 1)}{4} = -frac{3 - 1}{4} = -frac{2}{4} = -frac{1}{2}]Second term:[left( -frac{1 + sqrt{3}}{2} right) left( -frac{1 + sqrt{3}}{2} right) = frac{(1 + sqrt{3})^2}{4} = frac{1 + 2sqrt{3} + 3}{4} = frac{4 + 2sqrt{3}}{4} = frac{2 + sqrt{3}}{2}]So, the dot product is:[-frac{1}{2} + frac{2 + sqrt{3}}{2} = frac{-1 + 2 + sqrt{3}}{2} = frac{1 + sqrt{3}}{2}]Now, compute the magnitudes of vectors (DA) and (DB).First, (|vec{DA}|):[sqrt{left( -frac{sqrt{3} + 1}{2} right)^2 + left( -frac{1 + sqrt{3}}{2} right)^2} = sqrt{ frac{(sqrt{3} + 1)^2}{4} + frac{(1 + sqrt{3})^2}{4} } = sqrt{ frac{2(sqrt{3} + 1)^2}{4} } = sqrt{ frac{(sqrt{3} + 1)^2}{2} } = frac{sqrt{3} + 1}{sqrt{2}}]Similarly, (|vec{DB}|):[sqrt{left( frac{sqrt{3} - 1}{2} right)^2 + left( -frac{1 + sqrt{3}}{2} right)^2} = sqrt{ frac{(sqrt{3} - 1)^2}{4} + frac{(1 + sqrt{3})^2}{4} }]Compute each term:[frac{(sqrt{3} - 1)^2}{4} = frac{3 - 2sqrt{3} + 1}{4} = frac{4 - 2sqrt{3}}{4} = frac{2 - sqrt{3}}{2}][frac{(1 + sqrt{3})^2}{4} = frac{1 + 2sqrt{3} + 3}{4} = frac{4 + 2sqrt{3}}{4} = frac{2 + sqrt{3}}{2}]So,[|vec{DB}| = sqrt{ frac{2 - sqrt{3}}{2} + frac{2 + sqrt{3}}{2} } = sqrt{ frac{4}{2} } = sqrt{2}]Now, plug into the cosine formula:[cos theta = frac{frac{1 + sqrt{3}}{2}}{ frac{sqrt{3} + 1}{sqrt{2}} cdot sqrt{2} } = frac{frac{1 + sqrt{3}}{2}}{ frac{sqrt{3} + 1}{sqrt{2}} cdot sqrt{2} } = frac{frac{1 + sqrt{3}}{2}}{ sqrt{3} + 1 } = frac{1}{2}]So,[cos theta = frac{1}{2}]Thus, the angle (theta) is:[theta = arccos left( frac{1}{2} right) = 60^circ]Ah, so this time, angle (ADB = 60^circ).Wait, so depending on the direction in which the square is constructed, the angle (ADB) can be either (60^circ) or (120^circ). But the problem says the square is drawn "outwardly" on the hypotenuse (BC). So, which direction is outward?In the first case, when I constructed the square by rotating (90^circ) counterclockwise, the center (D) was in the fourth quadrant, and angle (ADB) was (120^circ). In the second case, rotating (90^circ) clockwise, the center (D) was in the first quadrant, and angle (ADB) was (60^circ).Given that the triangle is in the first quadrant, constructing the square outwardly likely means constructing it in the direction away from the triangle, which would be the first quadrant direction, not the fourth. Therefore, the correct position of (D) is in the first quadrant, leading to angle (ADB = 60^circ).Therefore, the measure of angle (ADB) is (60^circ).**Final Answer**The measure of angle (ADB) is boxed{60^circ}.