Skip to content
🎉 DeepSeek-R1 🥳
Find the equivalent cartesian equation for the given parametric equations:begin{cases}x = sqrt{t} y = 2sqrt{1-t}end{cases}where t is a parameter, and 0 leq t leq 1.

💡Okay, so I have these parametric equations:x equals the square root of t, and y equals 2 times the square root of (1 minus t). And t is between 0 and 1. I need to find the Cartesian equation that represents the same curve without the parameter t.Hmm, parametric equations often involve a parameter like t that defines both x and y. To eliminate the parameter, I think I need to solve for t in one equation and substitute it into the other. Let me try that.Starting with the first equation: x equals the square root of t. If I square both sides, I can get rid of the square root. So, x squared equals t. That seems straightforward. So, t equals x squared.Now, moving on to the second equation: y equals 2 times the square root of (1 minus t). I can also square both sides here to eliminate the square root. So, y squared equals 4 times (1 minus t). That gives me y squared equals 4 minus 4t.But I already have t expressed in terms of x from the first equation. So, I can substitute t with x squared in this equation. Let me do that. So, y squared equals 4 minus 4 times x squared.Now, I can rearrange this equation to get it into a more familiar form. Let me bring all the terms to one side. So, y squared plus 4x squared equals 4.To make it look like the standard form of an ellipse, I can divide every term by 4. That would give me (x squared)/1 plus (y squared)/4 equals 1.Wait, that looks like the equation of an ellipse centered at the origin with semi-major axis 2 along the y-axis and semi-minor axis 1 along the x-axis. But I should double-check my steps to make sure I didn't make a mistake.Starting again, x equals sqrt(t), so x squared is t. Then y equals 2 sqrt(1 - t), so y squared is 4(1 - t). Substituting t from the first equation into the second gives y squared equals 4 minus 4x squared. Moving terms around, y squared plus 4x squared equals 4. Dividing by 4, x squared over 1 plus y squared over 4 equals 1. Yep, that seems correct.I should also consider the domain of x and y based on the parameter t. Since t is between 0 and 1, x equals sqrt(t) means x is between 0 and 1. Similarly, y equals 2 sqrt(1 - t) means y is between 0 and 2. So, the Cartesian equation represents only the portion of the ellipse where x is between 0 and 1 and y is between 0 and 2.Wait, but in the Cartesian equation, x squared over 1 plus y squared over 4 equals 1, x can technically be between -1 and 1, and y between -2 and 2. But in our case, since t is between 0 and 1, x and y are both non-negative. So, the actual graph is just the first quadrant portion of that ellipse.Is there a way to represent that in the equation? Or is it sufficient to just note the domain restrictions? I think in the context of parametric equations, it's acceptable to provide the Cartesian equation along with the domain restrictions if necessary. But the problem didn't specify, so maybe just the equation is enough.Let me recap:1. Start with x = sqrt(t), so x squared = t.2. Then y = 2 sqrt(1 - t), so y squared = 4(1 - t).3. Substitute t = x squared into the second equation: y squared = 4 - 4x squared.4. Rearrange to get y squared + 4x squared = 4.5. Divide by 4: (x squared)/1 + (y squared)/4 = 1.Yes, that seems correct. So the Cartesian equation is (x squared)/1 + (y squared)/4 = 1, which is an ellipse.I think that's the answer. But just to be thorough, let me plug in some values of t to see if both parametric and Cartesian equations give the same points.For t = 0:x = sqrt(0) = 0y = 2 sqrt(1 - 0) = 2So the point is (0, 2). Plugging into the Cartesian equation: (0)^2 + (2)^2 /4 = 0 + 4/4 = 1. Correct.For t = 1:x = sqrt(1) = 1y = 2 sqrt(1 - 1) = 0Point is (1, 0). Plugging into Cartesian: (1)^2 + (0)^2 /4 = 1 + 0 = 1. Correct.For t = 0.25:x = sqrt(0.25) = 0.5y = 2 sqrt(1 - 0.25) = 2 sqrt(0.75) ≈ 2 * 0.866 ≈ 1.732Point is (0.5, ≈1.732). Plugging into Cartesian: (0.5)^2 + (1.732)^2 /4 ≈ 0.25 + (3)/4 ≈ 0.25 + 0.75 = 1. Correct.Another test: t = 0.5x = sqrt(0.5) ≈ 0.707y = 2 sqrt(1 - 0.5) = 2 sqrt(0.5) ≈ 1.414Point is (≈0.707, ≈1.414). Plugging into Cartesian: (0.707)^2 + (1.414)^2 /4 ≈ 0.5 + (2)/4 ≈ 0.5 + 0.5 = 1. Correct.Looks like all the points satisfy the Cartesian equation. So, I'm confident that the Cartesian equation is correct.Just to visualize, this ellipse is stretched more along the y-axis because the denominator under y squared is larger. The major axis is along y, and the minor axis is along x. The curve starts at (0, 2) when t=0 and moves to (1, 0) when t=1, tracing the top-right quarter of the ellipse.I think I've covered all the steps and checked my work. So, the final Cartesian equation is x squared plus y squared over 4 equals 1.

If 25% of x is a certain amount less than 15% of 1500, then x is 780. What is the amount?

💡First, I need to determine the certain amount by which 25% of ( x ) is less than 15% of 1500.I'll start by calculating 15% of 1500. This can be done by multiplying 1500 by 0.15.Next, I'll calculate 25% of ( x ). Since ( x ) is given as 780, I'll multiply 780 by 0.25.The problem states that 25% of ( x ) is a certain amount less than 15% of 1500. To find this amount, I'll subtract 25% of ( x ) from 15% of 1500.Finally, the result of this subtraction will give me the certain amount.

Let mathbf{a} = begin{pmatrix} 1 2 1 end{pmatrix} and mathbf{b} = begin{pmatrix} 2 0 -1 end{pmatrix}. Find the vector mathbf{v} that satisfies mathbf{v} times mathbf{a} = mathbf{b} times mathbf{a} and mathbf{v} times mathbf{b} = mathbf{a} times mathbf{b}.

💡Okay, so I have this problem where I need to find a vector **v** that satisfies two cross product equations. The vectors **a** and **b** are given as:**a** = (begin{pmatrix} 1 2 1 end{pmatrix})**b** = (begin{pmatrix} 2 0 -1 end{pmatrix})The equations I need to satisfy are:1. **v** × **a** = **b** × **a**2. **v** × **b** = **a** × **b**Hmm, cross products. I remember that the cross product of two vectors is another vector that's perpendicular to both of them. Also, cross product is anti-commutative, meaning **a** × **b** = -(**b** × **a**). Maybe that property will come in handy here.Let me start by computing **b** × **a** because that's in the first equation. I'll use the determinant method for cross products.**b** × **a** = (begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 2 & 0 & -1 1 & 2 & 1 end{vmatrix})Calculating the determinant:- The i-component: (0)(1) - (-1)(2) = 0 + 2 = 2- The j-component: -[(2)(1) - (-1)(1)] = -[2 + 1] = -3- The k-component: (2)(2) - (0)(1) = 4 - 0 = 4So, **b** × **a** = 2**i** - 3**j** + 4**k** = (begin{pmatrix} 2 -3 4 end{pmatrix})Wait, but in the first equation, it's **v** × **a** = **b** × **a**, which is 2**i** - 3**j** + 4**k**. So, **v** × **a** equals that. Hmm, cross product is anti-commutative, so if **v** × **a** = **b** × **a**, that implies (**v** - **b**) × **a** = 0. Because if I subtract **b** × **a** from both sides, I get **v** × **a** - **b** × **a** = 0, which is (**v** - **b**) × **a** = 0.If the cross product of (**v** - **b**) and **a** is zero, that means (**v** - **b**) is parallel to **a**. So, (**v** - **b**) must be a scalar multiple of **a**. Let's denote that scalar as t. So, **v** - **b** = t**a**, which implies **v** = t**a** + **b**.Okay, so **v** is expressed in terms of **a**, **b**, and a scalar t. Now, let's move to the second equation: **v** × **b** = **a** × **b**.Similarly, let me compute **a** × **b**.**a** × **b** = (begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 1 & 2 & 1 2 & 0 & -1 end{vmatrix})Calculating the determinant:- The i-component: (2)(-1) - (1)(0) = -2 - 0 = -2- The j-component: -[(1)(-1) - (1)(2)] = -[-1 - 2] = -(-3) = 3- The k-component: (1)(0) - (2)(2) = 0 - 4 = -4So, **a** × **b** = -2**i** + 3**j** - 4**k** = (begin{pmatrix} -2 3 -4 end{pmatrix})Therefore, the second equation becomes **v** × **b** = -2**i** + 3**j** - 4**k**. Similar to the first equation, this implies (**v** - **a**) × **b** = 0. Because if I subtract **a** × **b** from both sides, I get **v** × **b** - **a** × **b** = 0, which is (**v** - **a**) × **b** = 0.Again, since the cross product is zero, (**v** - **a**) must be parallel to **b**. So, (**v** - **a**) is a scalar multiple of **b**. Let's denote that scalar as s. Therefore, **v** - **a** = s**b**, which implies **v** = s**b** + **a**.Now, from the first equation, I have **v** = t**a** + **b**, and from the second equation, **v** = s**b** + **a**. So, both expressions equal **v**, so they must be equal to each other:t**a** + **b** = s**b** + **a**Let me rearrange this equation:t**a** - **a** = s**b** - **b**Factor out the vectors:(t - 1)**a** = (s - 1)**b**So, (t - 1)**a** is equal to (s - 1)**b**. Now, **a** and **b** are given vectors. Are they scalar multiples of each other? Let's check.If **a** and **b** are scalar multiples, then there exists some scalar k such that **a** = k**b**. Let's see:**a** = (begin{pmatrix} 1 2 1 end{pmatrix})**b** = (begin{pmatrix} 2 0 -1 end{pmatrix})Is there a k such that 1 = 2k, 2 = 0k, 1 = -1k?From the second component: 2 = 0k implies 2 = 0, which is impossible. So, **a** and **b** are not scalar multiples. Therefore, the only way (t - 1)**a** = (s - 1)**b** is if both coefficients are zero. That is:t - 1 = 0 and s - 1 = 0Therefore, t = 1 and s = 1.So, plugging t = 1 into **v** = t**a** + **b**:**v** = 1**a** + **b** = **a** + **b**Compute **a** + **b**:**a** + **b** = (begin{pmatrix} 1 + 2 2 + 0 1 + (-1) end{pmatrix}) = (begin{pmatrix} 3 2 0 end{pmatrix})Let me verify this solution to make sure it satisfies both original equations.First equation: **v** × **a** = **b** × **a**Compute **v** × **a** where **v** = (begin{pmatrix} 3 2 0 end{pmatrix}):**v** × **a** = (begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 3 & 2 & 0 1 & 2 & 1 end{vmatrix})Calculating:- i-component: (2)(1) - (0)(2) = 2 - 0 = 2- j-component: -[(3)(1) - (0)(1)] = -[3 - 0] = -3- k-component: (3)(2) - (2)(1) = 6 - 2 = 4So, **v** × **a** = 2**i** - 3**j** + 4**k**, which matches **b** × **a** as computed earlier. Good.Second equation: **v** × **b** = **a** × **b**Compute **v** × **b**:**v** × **b** = (begin{vmatrix} mathbf{i} & mathbf{j} & mathbf{k} 3 & 2 & 0 2 & 0 & -1 end{vmatrix})Calculating:- i-component: (2)(-1) - (0)(0) = -2 - 0 = -2- j-component: -[(3)(-1) - (0)(2)] = -[-3 - 0] = 3- k-component: (3)(0) - (2)(2) = 0 - 4 = -4So, **v** × **b** = -2**i** + 3**j** - 4**k**, which matches **a** × **b** as computed earlier. Perfect.Therefore, the vector **v** that satisfies both equations is (begin{pmatrix} 3 2 0 end{pmatrix}).**Final Answer**The vector (mathbf{v}) is boxed{begin{pmatrix} 3 2 0 end{pmatrix}}.

Given that the lines (x+ay+2=0) and (2x+3y+1=0) are perpendicular to each other, find the value of a.A: - frac {2}{3}B: - frac {3}{2}C: frac {2}{3}D: frac {3}{2}

💡To determine the value of ( a ) such that the lines ( x + ay + 2 = 0 ) and ( 2x + 3y + 1 = 0 ) are perpendicular, follow these steps:1. **Find the slopes of both lines:** - For the line ( x + ay + 2 = 0 ), rewrite it in slope-intercept form: [ y = -frac{1}{a}x - frac{2}{a} ] The slope (( m_1 )) is ( -frac{1}{a} ). - For the line ( 2x + 3y + 1 = 0 ), rewrite it in slope-intercept form: [ y = -frac{2}{3}x - frac{1}{3} ] The slope (( m_2 )) is ( -frac{2}{3} ).2. **Use the condition for perpendicularity:** - If two lines are perpendicular, the product of their slopes is ( -1 ): [ m_1 times m_2 = -1 ] Substitute the slopes: [ left(-frac{1}{a}right) times left(-frac{2}{3}right) = -1 ]3. **Solve for ( a ):** - Multiply the left side: [ frac{2}{3a} = -1 ] - Multiply both sides by ( 3a ): [ 2 = -3a ] - Divide both sides by ( -3 ): [ a = -frac{2}{3} ]Thus, the value of ( a ) is ( -frac{2}{3} ).

Two perpendicular radii of a circle, O A and O B, are given. Point P is an arbitrary point on the circumference (different from A and B). The intersection point of the lines P A and O B is Q, and the intersection point of the lines P B and O A is R.a) How does the position of Q and R depend on the position of P?b) If Q is selected on the line O B, how can the point P that defines Q be constructed, and hence R? How does the position of R depend on the position of Q?c) Relate our findings to the examination of exercise 662.d) For which positions of P is the area of the quadrilateral A B Q R constant?

💡Okay, so I have this geometry problem here about a circle with two perpendicular radii, OA and OB. Point P is somewhere on the circumference, not overlapping with A or B. Then, they define Q as the intersection of PA and OB, and R as the intersection of PB and OA. The questions are about how Q and R move as P moves, constructing P from Q, relating this to another problem, and figuring out when the area of quadrilateral ABQR is constant.First, I need to visualize this setup. Let me sketch a circle with center O, and two radii OA and OB perpendicular to each other. Let's say OA is along the negative y-axis and OB is along the negative x-axis for simplicity. Point P is somewhere on the circumference, not at A or B. Then, lines PA and PB are drawn, and their intersections with OB and OA are Q and R, respectively.For part (a), I need to see how Q and R depend on P. Maybe I can parametrize P using an angle. Let's let P be at an angle θ from the positive x-axis. Then, the coordinates of P would be (cosθ, sinθ). Since OA is along the negative y-axis, A is at (0, -1), and B is at (-1, 0).Now, I need to find the equations of lines PA and PB. Let's start with PA. The line PA connects P(cosθ, sinθ) and A(0, -1). The slope of PA is (sinθ - (-1))/(cosθ - 0) = (sinθ + 1)/cosθ. So, the equation of PA is y + 1 = [(sinθ + 1)/cosθ]x.Similarly, the line PB connects P(cosθ, sinθ) and B(-1, 0). The slope of PB is (sinθ - 0)/(cosθ - (-1)) = sinθ/(cosθ + 1). So, the equation of PB is y = [sinθ/(cosθ + 1)](x + 1).Now, to find Q, which is the intersection of PA and OB. Since OB is the x-axis (y=0), plug y=0 into the equation of PA:0 + 1 = [(sinθ + 1)/cosθ]x_Q=> x_Q = cosθ / (sinθ + 1)So, Q is at (cosθ / (sinθ + 1), 0).Similarly, to find R, which is the intersection of PB and OA. OA is the y-axis (x=0), so plug x=0 into the equation of PB:y_R = [sinθ/(cosθ + 1)](0 + 1) = sinθ / (cosθ + 1)So, R is at (0, sinθ / (cosθ + 1)).So, Q and R are determined entirely by θ, which is the position of P on the circumference. As P moves around the circle, θ changes, and thus Q and R move along OB and OA respectively.For part (b), if Q is selected on OB, how do we construct P and R? Well, from part (a), we have x_Q = cosθ / (sinθ + 1). If we know x_Q, we can solve for θ. Let me denote x_Q as q for simplicity. So,q = cosθ / (sinθ + 1)I can rewrite this as:q(sinθ + 1) = cosθ=> q sinθ + q = cosθ=> q sinθ - cosθ = -qThis is a linear combination of sinθ and cosθ. I can write this as:A sinθ + B cosθ = CWhere A = q, B = -1, and C = -q.To solve for θ, I can use the identity:A sinθ + B cosθ = R sin(θ + φ)Where R = sqrt(A² + B²) and tanφ = B/A.So,R = sqrt(q² + 1)tanφ = (-1)/qThus,sqrt(q² + 1) sin(θ + φ) = -qSo,sin(θ + φ) = -q / sqrt(q² + 1)This gives θ + φ = arcsin(-q / sqrt(q² + 1))But this might be a bit complicated. Alternatively, I can square both sides of the equation:(q sinθ + q)^2 = cos²θ=> q² sin²θ + 2q² sinθ + q² = cos²θ=> q² sin²θ + 2q² sinθ + q² = 1 - sin²θ=> (q² + 1) sin²θ + 2q² sinθ + (q² - 1) = 0This is a quadratic in sinθ:Let me denote s = sinθ:(q² + 1)s² + 2q² s + (q² - 1) = 0Solving for s:s = [-2q² ± sqrt(4q⁴ - 4(q² + 1)(q² - 1))]/[2(q² + 1)]= [-2q² ± sqrt(4q⁴ - 4(q⁴ - 1))]/[2(q² + 1)]= [-2q² ± sqrt(4q⁴ - 4q⁴ + 4)]/[2(q² + 1)]= [-2q² ± 2]/[2(q² + 1)]= [-q² ± 1]/(q² + 1)So, sinθ = (-q² + 1)/(q² + 1) or sinθ = (-q² - 1)/(q² + 1)The second solution is sinθ = (-q² - 1)/(q² + 1) = -1, which would correspond to θ = 270°, but P cannot be at B, so we discard this.Thus, sinθ = (1 - q²)/(1 + q²)Then, cosθ can be found using sin²θ + cos²θ = 1:cosθ = sqrt(1 - [(1 - q²)/(1 + q²)]²) = sqrt([ (1 + q²)² - (1 - q²)² ] / (1 + q²)² )Calculating numerator:(1 + 2q² + q⁴) - (1 - 2q² + q⁴) = 4q²Thus,cosθ = sqrt(4q² / (1 + q²)²) = 2|q| / (1 + q²)Since θ is between 0 and 360°, and P is not at A or B, we can take cosθ as positive or negative depending on θ.But since Q is on OB, which is the negative x-axis, x_Q = q is negative. So, q is negative.Thus, cosθ = -2q / (1 + q²)Therefore, we can find θ from sinθ and cosθ, and then construct P.Once P is constructed, R can be found as in part (a). The position of R depends on Q through the relation we found earlier.For part (c), relating to exercise 662, which I don't have the exact details of, but likely involves similar geometric constructions or properties. Perhaps it's about harmonic division or projective geometry, given the intersections and dependencies.For part (d), we need to find when the area of quadrilateral ABQR is constant. Let's compute the area.Quadrilateral ABQR has vertices at A(0, -1), B(-1, 0), Q(q, 0), and R(0, r), where q = x_Q and r = y_R.We can compute the area using the shoelace formula.Coordinates in order: A(0, -1), B(-1, 0), Q(q, 0), R(0, r), back to A(0, -1).Compute the area:Area = 1/2 |(0*0 + (-1)*0 + q*r + 0*(-1)) - (-1*(-1) + 0*q + 0*0 + r*0)|= 1/2 |0 + 0 + q*r + 0 - (1 + 0 + 0 + 0)|= 1/2 |q*r - 1|But from part (a), we have:q = cosθ / (sinθ + 1)r = sinθ / (cosθ + 1)So,q*r = [cosθ / (sinθ + 1)] * [sinθ / (cosθ + 1)] = [cosθ sinθ] / [(sinθ + 1)(cosθ + 1)]Thus,Area = 1/2 | [cosθ sinθ] / [(sinθ + 1)(cosθ + 1)] - 1 |We need this area to be constant. Let's see if we can find θ such that this expression is constant.Alternatively, perhaps there's a specific position of P where this area is constant regardless of θ. Maybe when P is at certain symmetric positions.Alternatively, maybe the area is constant for all P, but that seems unlikely since Q and R move as P moves.Wait, let's compute the area expression more carefully.Area = 1/2 | q*r - 1 | = 1/2 | [cosθ sinθ] / [(sinθ + 1)(cosθ + 1)] - 1 |Let me compute the numerator:[cosθ sinθ] - [(sinθ + 1)(cosθ + 1)] = cosθ sinθ - (sinθ cosθ + sinθ + cosθ + 1) = - sinθ - cosθ - 1Thus,Area = 1/2 | (- sinθ - cosθ - 1) / [(sinθ + 1)(cosθ + 1)] |Simplify numerator:- (sinθ + cosθ + 1)Denominator:(sinθ + 1)(cosθ + 1) = sinθ cosθ + sinθ + cosθ + 1Thus,Area = 1/2 | - (sinθ + cosθ + 1) / (sinθ cosθ + sinθ + cosθ + 1) |Notice that the numerator is - (denominator). So,Area = 1/2 | -1 | = 1/2Wait, that's interesting. So, the area is always 1/2, regardless of θ.But that contradicts my earlier thought that it might not be constant. Let me check the calculation.Compute q*r:q = cosθ / (sinθ + 1)r = sinθ / (cosθ + 1)q*r = [cosθ sinθ] / [(sinθ + 1)(cosθ + 1)]Then, Area = 1/2 | q*r - 1 | = 1/2 | [cosθ sinθ - (sinθ + 1)(cosθ + 1)] / [(sinθ + 1)(cosθ + 1)] |Compute numerator:cosθ sinθ - (sinθ cosθ + sinθ + cosθ + 1) = - sinθ - cosθ - 1Thus,Area = 1/2 | (- sinθ - cosθ - 1) / [(sinθ + 1)(cosθ + 1)] |Factor numerator:- (sinθ + cosθ + 1)Denominator:(sinθ + 1)(cosθ + 1) = sinθ cosθ + sinθ + cosθ + 1Notice that sinθ + cosθ + 1 = (sinθ + 1) + cosθBut in the denominator, we have sinθ cosθ + sinθ + cosθ + 1 = (sinθ + 1)(cosθ + 1)Thus, the fraction becomes:- (sinθ + cosθ + 1) / [(sinθ + 1)(cosθ + 1)] = -1Because (sinθ + cosθ + 1) = (sinθ + 1) + cosθ, but in the denominator, it's (sinθ + 1)(cosθ + 1). Wait, no, that's not exactly the case.Wait, let me think differently. Let me factor the numerator and denominator.Numerator: - (sinθ + cosθ + 1)Denominator: (sinθ + 1)(cosθ + 1) = sinθ cosθ + sinθ + cosθ + 1Notice that sinθ + cosθ + 1 is part of the denominator. So, the fraction is:- (sinθ + cosθ + 1) / [ (sinθ + cosθ + 1) + sinθ cosθ ]Hmm, not immediately obvious. But let's plug in specific values to test.Let me take θ = 0°, so P is at (1, 0). But P cannot be at B, which is (-1, 0), so θ=0° is allowed.Wait, θ=0°, P is at (1,0). Then,q = cos0 / (sin0 + 1) = 1 / (0 + 1) = 1But Q is on OB, which is the negative x-axis, so q should be negative. Hmm, maybe θ=0° is not allowed since P is at (1,0), which is not A or B, but Q would be at (1,0), which is not on OB (negative x-axis). So, maybe θ=0° is not a valid position for P in this context.Let me try θ=45°, so P is at (√2/2, √2/2).Then,q = cos45 / (sin45 + 1) = (√2/2) / (√2/2 + 1) = (√2/2) / ( (√2 + 2)/2 ) = √2 / (√2 + 2)Multiply numerator and denominator by (√2 - 2):√2(√2 - 2) / ( (√2 + 2)(√2 - 2) ) = (2 - 2√2) / (2 - 4) = (2 - 2√2)/(-2) = (√2 - 1)Similarly,r = sin45 / (cos45 + 1) = (√2/2) / (√2/2 + 1) = same as q, so r = √2 - 1Thus, q*r = (√2 - 1)^2 = 3 - 2√2Then, Area = 1/2 | q*r - 1 | = 1/2 | (3 - 2√2) - 1 | = 1/2 | 2 - 2√2 | = 1/2 * 2(√2 - 1) = √2 - 1 ≈ 0.414But earlier, I thought the area was always 1/2, which is 0.5. So, that contradicts. So, my earlier conclusion was wrong.Wait, let's recalculate the area expression.Area = 1/2 | q*r - 1 |, where q = cosθ / (sinθ + 1) and r = sinθ / (cosθ + 1)Compute q*r:q*r = [cosθ / (sinθ + 1)] * [sinθ / (cosθ + 1)] = [cosθ sinθ] / [(sinθ + 1)(cosθ + 1)]Then, Area = 1/2 | [cosθ sinθ] / [(sinθ + 1)(cosθ + 1)] - 1 |Let me compute this for θ=45°:cos45 sin45 = (√2/2)(√2/2) = 1/2(sin45 + 1)(cos45 + 1) = (√2/2 + 1)^2 = ( (√2 + 2)/2 )^2 = ( (√2 + 2)^2 ) / 4 = (2 + 4√2 + 4) / 4 = (6 + 4√2)/4 = (3 + 2√2)/2Thus,q*r = (1/2) / ( (3 + 2√2)/2 ) = 1 / (3 + 2√2) = (3 - 2√2) / ( (3 + 2√2)(3 - 2√2) ) = (3 - 2√2)/ (9 - 8) = 3 - 2√2Thus,Area = 1/2 | (3 - 2√2) - 1 | = 1/2 | 2 - 2√2 | = 1/2 * 2(√2 - 1) = √2 - 1 ≈ 0.414So, the area is not constant. It depends on θ.Wait, but earlier when I thought the area was 1/2, that was incorrect because I miscalculated the numerator.So, the area is not constant. It varies with θ.But the question is asking for which positions of P the area is constant. So, we need to find θ such that the area is constant.Let me denote the area as T:T = 1/2 | q*r - 1 | = 1/2 | [cosθ sinθ] / [(sinθ + 1)(cosθ + 1)] - 1 |We need T to be constant. Let's set T = k, a constant.Thus,| [cosθ sinθ] / [(sinθ + 1)(cosθ + 1)] - 1 | = 2kBut since T is positive, we can drop the absolute value:[cosθ sinθ] / [(sinθ + 1)(cosθ + 1)] - 1 = ±2kBut this seems complicated. Maybe instead, we can express everything in terms of tan(θ/2).Let me use the substitution t = tan(θ/2). Then,sinθ = 2t/(1 + t²)cosθ = (1 - t²)/(1 + t²)Then,q = cosθ / (sinθ + 1) = [ (1 - t²)/(1 + t²) ] / [ 2t/(1 + t²) + 1 ] = [ (1 - t²) ] / [ 2t + 1 + t² ] = (1 - t²)/(1 + t² + 2t) = (1 - t²)/(1 + t)^2Similarly,r = sinθ / (cosθ + 1) = [ 2t/(1 + t²) ] / [ (1 - t²)/(1 + t²) + 1 ] = [ 2t ] / [ 1 - t² + 1 + t² ] = 2t / 2 = tThus, q = (1 - t²)/(1 + t)^2 and r = tNow, let's compute q*r:q*r = [ (1 - t²)/(1 + t)^2 ] * t = t(1 - t²)/(1 + t)^2 = t(1 - t)(1 + t)/(1 + t)^2 = t(1 - t)/(1 + t)Thus,q*r = t(1 - t)/(1 + t)Then, the area T = 1/2 | q*r - 1 | = 1/2 | [ t(1 - t)/(1 + t) ] - 1 |Simplify inside the absolute value:[ t(1 - t)/(1 + t) - 1 ] = [ t(1 - t) - (1 + t) ] / (1 + t) = [ t - t² - 1 - t ] / (1 + t) = [ -t² - 1 ] / (1 + t) = - (t² + 1)/(1 + t)Thus,T = 1/2 | - (t² + 1)/(1 + t) | = 1/2 (t² + 1)/(1 + t)Since t = tan(θ/2), and θ is between 0 and 2π, t can be any real number except where cosθ = -1, which is θ=π, but P is not at A or B.Thus, T = (t² + 1)/(2(1 + t))We need T to be constant. Let's set T = k:(t² + 1)/(2(1 + t)) = kMultiply both sides by 2(1 + t):t² + 1 = 2k(1 + t)Rearrange:t² - 2k t + (1 - 2k) = 0This is a quadratic in t:t² - 2k t + (1 - 2k) = 0For real solutions, discriminant D ≥ 0:(2k)^2 - 4*1*(1 - 2k) ≥ 04k² - 4 + 8k ≥ 04k² + 8k - 4 ≥ 0Divide by 4:k² + 2k - 1 ≥ 0Solve k² + 2k - 1 = 0:k = [-2 ± sqrt(4 + 4)]/2 = [-2 ± sqrt(8)]/2 = [-2 ± 2√2]/2 = -1 ± √2Thus, k ≥ -1 + √2 ≈ 0.414 or k ≤ -1 - √2 (but k is positive, so only k ≥ -1 + √2)But since T is positive, and k must be positive, so k ≥ √2 - 1 ≈ 0.414But we are looking for T to be constant, so for each k ≥ √2 - 1, there are solutions for t, hence θ.But the question is asking for which positions of P the area is constant. So, for any k ≥ √2 - 1, there are positions of P such that the area is k.But the problem might be asking for specific positions where the area is constant, perhaps for all P, but that's not the case as we saw.Alternatively, maybe the area is constant for certain symmetric positions of P.Wait, but from the expression T = (t² + 1)/(2(1 + t)), we can see that T is constant if t is constant. So, for each t, T is determined. Thus, the area is constant only when t is fixed, i.e., when P is at a specific position.But the problem asks for which positions of P the area is constant. So, for any specific P, the area is fixed, but it's not constant across different P's. So, perhaps the area is constant only when P is at certain specific positions, but not for all P.Wait, but the area expression T = (t² + 1)/(2(1 + t)) can be rewritten as:T = (t² + 1)/(2(t + 1)) = [ (t + 1)^2 - 2t ] / (2(t + 1)) = (t + 1)/2 - t/(t + 1)But not sure if that helps.Alternatively, let's see if T can be expressed in terms of t + 1/t or something.Let me set u = t + 1/t, but t can be any real number except -1.Wait, but t = tan(θ/2), so t can be any real number except where θ=π, which is excluded.Alternatively, maybe express T in terms of t + 1/t.But I'm not sure. Alternatively, let's see if T can be expressed as a function of t + 1/t.Wait, T = (t² + 1)/(2(t + 1)) = [ (t + 1)^2 - 2t ] / (2(t + 1)) = (t + 1)/2 - t/(t + 1)Hmm, not helpful.Alternatively, let's set s = t + 1, then t = s - 1.Then,T = ( (s - 1)^2 + 1 ) / (2s ) = (s² - 2s + 1 + 1)/2s = (s² - 2s + 2)/2s = (s² + 2)/2s - 1Hmm, not helpful.Alternatively, maybe set t = tan(φ), but not sure.Alternatively, let's consider that T = (t² + 1)/(2(t + 1)) = [ (t + 1)^2 - 2t ] / (2(t + 1)) = (t + 1)/2 - t/(t + 1)But I don't see a simplification.Alternatively, let's consider that T = (t² + 1)/(2(t + 1)) = [ (t² + 2t + 1) - 2t ] / (2(t + 1)) = ( (t + 1)^2 - 2t ) / (2(t + 1)) = (t + 1)/2 - t/(t + 1)Still not helpful.Alternatively, let's consider that T = (t² + 1)/(2(t + 1)) = [ (t + 1)^2 - 2t ] / (2(t + 1)) = (t + 1)/2 - t/(t + 1)Wait, maybe set u = t + 1, then T = u/2 - (u - 1)/u = u/2 - 1 + 1/uSo,T = u/2 - 1 + 1/uBut not sure.Alternatively, perhaps take derivative with respect to t and see if T has a minimum or maximum.Compute dT/dt:dT/dt = [ (2t)(2(t + 1)) - (t² + 1)(2) ] / [4(t + 1)^2] ?Wait, no, better to compute derivative directly.T = (t² + 1)/(2(t + 1))dT/dt = [ (2t)(2(t + 1)) - (t² + 1)(2) ] / [4(t + 1)^2] ?Wait, no, quotient rule:dT/dt = [ (2t)(2(t + 1)) - (t² + 1)(2) ] / [4(t + 1)^2] ?Wait, no, quotient rule is [num’ den - num den’ ] / den²So,dT/dt = [ (2t)(2(t + 1)) - (t² + 1)(2) ] / [4(t + 1)^2] ?Wait, no, wait:Wait, T = (t² + 1)/(2(t + 1)) = (t² + 1)/[2(t + 1)]So, dT/dt = [ (2t)(2(t + 1)) - (t² + 1)(2) ] / [4(t + 1)^2] ?Wait, no, let me compute correctly.Let me write T = (t² + 1)/(2(t + 1)) = (t² + 1) * (2(t + 1))^{-1}Thus,dT/dt = [ (2t)(2(t + 1))^{-1} ] + [ (t² + 1)(-1)(2(t + 1))^{-2}(2) ]Wait, no, better to use quotient rule:If T = N/D, where N = t² + 1, D = 2(t + 1)Then,dT/dt = (N’ D - N D’) / D²N’ = 2tD’ = 2Thus,dT/dt = [2t * 2(t + 1) - (t² + 1)*2 ] / [4(t + 1)^2]Simplify numerator:4t(t + 1) - 2(t² + 1) = 4t² + 4t - 2t² - 2 = 2t² + 4t - 2Thus,dT/dt = (2t² + 4t - 2) / [4(t + 1)^2] = [2(t² + 2t - 1)] / [4(t + 1)^2] = (t² + 2t - 1)/(2(t + 1)^2)Set dT/dt = 0:t² + 2t - 1 = 0Solutions:t = [-2 ± sqrt(4 + 4)]/2 = [-2 ± sqrt(8)]/2 = [-2 ± 2√2]/2 = -1 ± √2Thus, critical points at t = -1 + √2 ≈ 0.414 and t = -1 - √2 ≈ -2.414Since t = tan(θ/2), and θ is between 0 and 2π, t can be any real number except where cosθ = -1, which is θ=π, but P is not at A or B.Thus, the area T has a minimum at t = -1 + √2 and a maximum at t = -1 - √2, but since t can be negative, we need to consider both.But since t = tan(θ/2), and θ is between 0 and 2π, t can be any real number except where θ=π, which is excluded.Thus, the area T has a minimum value when t = -1 + √2, which is approximately 0.414, and a maximum when t approaches -1 - √2, but since t can be negative, we need to check.Wait, when t = -1 - √2, which is approximately -2.414, which is valid since t can be negative.Thus, the area T has a minimum value of T_min = (t² + 1)/(2(t + 1)) evaluated at t = -1 + √2.Compute T_min:t = -1 + √2t + 1 = √2t² = (-1 + √2)^2 = 1 - 2√2 + 2 = 3 - 2√2Thus,T_min = (3 - 2√2 + 1)/(2√2) = (4 - 2√2)/(2√2) = (2 - √2)/√2 = (2√2 - 2)/2 = √2 - 1 ≈ 0.414Similarly, at t = -1 - √2,t + 1 = -√2t² = (-1 - √2)^2 = 1 + 2√2 + 2 = 3 + 2√2Thus,T = (3 + 2√2 + 1)/(2*(-√2)) = (4 + 2√2)/(-2√2) = -(4 + 2√2)/(2√2) = -(2 + √2)/√2 = -(√2 + 1)But since T is positive, we take absolute value, so T = √2 + 1 ≈ 2.414Thus, the area T ranges from √2 - 1 to √2 + 1 as P moves around the circle.But the question is asking for which positions of P the area is constant. So, for any specific P, the area is fixed, but it's not constant across different P's. However, if we fix T to be a specific value between √2 - 1 and √2 + 1, there are two positions of P (one in each half-circle) that give that area.But the problem might be asking for specific positions where the area is constant, perhaps when P is at certain symmetric points.Alternatively, maybe when P is at 45°, 135°, etc., the area is constant.Wait, earlier when I took θ=45°, the area was √2 - 1, which is the minimum area.Similarly, when θ=225°, which is θ=180°+45°, P is at (-√2/2, -√2/2), let's compute the area.q = cos225 / (sin225 + 1) = (-√2/2) / (-√2/2 + 1) = (-√2/2) / ( ( -√2 + 2 ) / 2 ) = (-√2)/(-√2 + 2) = √2/(√2 - 2)Multiply numerator and denominator by (√2 + 2):√2(√2 + 2) / ( (√2 - 2)(√2 + 2) ) = (2 + 2√2)/ (2 - 4) = (2 + 2√2)/(-2) = - (1 + √2)But Q is on OB, which is the negative x-axis, so q should be negative, which it is.Similarly,r = sin225 / (cos225 + 1) = (-√2/2) / (-√2/2 + 1) = same as q, so r = - (1 + √2)But r is on OA, which is the negative y-axis, so r is negative, which it is.Thus, q*r = [ - (1 + √2) ] * [ - (1 + √2) ] = (1 + √2)^2 = 1 + 2√2 + 2 = 3 + 2√2Thus, Area = 1/2 | q*r - 1 | = 1/2 | (3 + 2√2) - 1 | = 1/2 | 2 + 2√2 | = 1/2 * 2(1 + √2) = 1 + √2 ≈ 2.414So, when P is at θ=45°, the area is √2 - 1, and when P is at θ=225°, the area is √2 + 1.Thus, the area is constant for P at θ=45° and θ=225°, giving the minimum and maximum areas respectively.But the problem is asking for which positions of P the area is constant. So, for each specific P, the area is fixed, but it's not constant across all P's. However, if we consider that the area can take any value between √2 - 1 and √2 + 1, then for each value in that interval, there are two positions of P that give that area.But perhaps the problem is asking for specific positions where the area is constant, such as when P is at 45°, 135°, 225°, 315°, etc., which are symmetric positions.Alternatively, maybe the area is constant when P is at certain specific angles where the expression simplifies.But from the earlier analysis, the area T = (t² + 1)/(2(t + 1)) can be rewritten as T = (t + 1)/2 - t/(t + 1). This might suggest that T is constant when t + 1 is proportional to t/(t + 1), but I'm not sure.Alternatively, perhaps when t = 1, which corresponds to θ=90°, let's check.At θ=90°, P is at (0,1).Then,q = cos90 / (sin90 + 1) = 0 / (1 + 1) = 0But Q is on OB, which is the x-axis, so Q is at (0,0), which is O. But P is at (0,1), so PA is the line from (0,1) to (0,-1), which is the y-axis, intersecting OB at O(0,0). Similarly, PB is the line from (0,1) to (-1,0), which has slope (1 - 0)/(0 - (-1)) = 1/1 = 1, so equation y = x + 1. Intersecting OA (x=0) at y=1, so R is at (0,1), which is P. But P is not allowed to be at A or B, but (0,1) is not A or B, so it's allowed.But then, quadrilateral ABQR has points A(0,-1), B(-1,0), Q(0,0), R(0,1). The area is the area of the quadrilateral from (0,-1) to (-1,0) to (0,0) to (0,1). This is a trapezoid with bases along the y-axis from (0,-1) to (0,1), length 2, and the other base from (-1,0) to (0,0), length 1. The height is the horizontal distance from x=0 to x=-1, which is 1.Area = 1/2 * (2 + 1) * 1 = 1.5But according to our earlier formula, T = (t² + 1)/(2(t + 1)). At θ=90°, t = tan(45°) = 1.Thus,T = (1 + 1)/(2(1 + 1)) = 2/4 = 0.5But the actual area is 1.5, which contradicts. So, my earlier formula might be incorrect.Wait, let's recalculate T for θ=90°.q = cos90 / (sin90 + 1) = 0 / (1 + 1) = 0r = sin90 / (cos90 + 1) = 1 / (0 + 1) = 1Thus, q*r = 0*1 = 0Thus, Area = 1/2 | 0 - 1 | = 1/2But the actual area is 1.5, so there's a discrepancy.This suggests that my earlier approach to compute the area using shoelace formula might have been incorrect because when Q or R coincide with O, the quadrilateral becomes degenerate or the shoelace formula needs to be adjusted.Alternatively, perhaps I should compute the area differently.Let me try again. Quadrilateral ABQR has points A(0,-1), B(-1,0), Q(q,0), R(0,r). Let's compute the area using the shoelace formula correctly.List the coordinates in order:A(0, -1), B(-1, 0), Q(q, 0), R(0, r), back to A(0, -1).Compute the sum of x_i y_{i+1}:0*0 + (-1)*0 + q*r + 0*(-1) = 0 + 0 + q*r + 0 = q*rCompute the sum of y_i x_{i+1}:(-1)*(-1) + 0*q + 0*0 + r*0 = 1 + 0 + 0 + 0 = 1Thus, Area = 1/2 | q*r - 1 | = 1/2 | q*r - 1 |But when θ=90°, q=0, r=1, so Area = 1/2 |0 - 1| = 1/2, but the actual area is 1.5. So, the shoelace formula is giving a different result.Wait, perhaps because when Q is at (0,0), the quadrilateral becomes a triangle ABR, but in reality, it's a quadrilateral with Q at (0,0). So, maybe the shoelace formula is correct, but the area is indeed 1/2.But visually, the area seems larger. Let me plot the points:A(0,-1), B(-1,0), Q(0,0), R(0,1). So, the quadrilateral is a trapezoid with vertices at (0,-1), (-1,0), (0,0), (0,1). This is actually a triangle from (0,-1) to (-1,0) to (0,1), with area 1.5, but the shoelace formula is giving 1/2. So, there must be a mistake in the order of the points.Wait, the correct order should be A(0,-1), B(-1,0), Q(0,0), R(0,1). But when connecting these points, the quadrilateral is actually a triangle because Q and R are on the axes, making the quadrilateral degenerate into a triangle when Q is at O.Thus, the shoelace formula might not be appropriate here because the quadrilateral becomes degenerate. So, perhaps the area expression T = 1/2 | q*r - 1 | is not valid when Q or R coincide with O.Thus, to avoid this, we should consider P not at θ=90°, where Q or R coincide with O.Thus, the area expression T = 1/2 | q*r - 1 | is valid when Q and R are distinct from O.Therefore, for positions of P where Q and R are distinct from O, the area is T = 1/2 | q*r - 1 |, which varies between √2 - 1 and √2 + 1.But the problem is asking for which positions of P the area is constant. So, for any specific P, the area is fixed, but it's not constant across different P's. However, if we fix T to be a specific value between √2 - 1 and √2 + 1, there are two positions of P (one in each half-circle) that give that area.But perhaps the problem is asking for specific positions where the area is constant, such as when P is at 45°, 135°, 225°, 315°, etc., which are symmetric positions.Alternatively, maybe the area is constant when P is at certain specific angles where the expression simplifies.But from the earlier analysis, the area T = (t² + 1)/(2(t + 1)) can be rewritten as T = (t + 1)/2 - t/(t + 1). This might suggest that T is constant when t + 1 is proportional to t/(t + 1), but I'm not sure.Alternatively, perhaps when t = 1, which corresponds to θ=90°, but as we saw, that leads to a degenerate quadrilateral.Alternatively, maybe when t = √2 - 1, which is approximately 0.414, which corresponds to θ=45°, as we saw earlier, giving T = √2 - 1.Similarly, when t = -√2 - 1, which corresponds to θ=225°, giving T = √2 + 1.Thus, the area is constant for P at θ=45° and θ=225°, giving the minimum and maximum areas respectively.Therefore, the positions of P where the area of quadrilateral ABQR is constant are when P is at 45° and 225°, i.e., when P is at (√2/2, √2/2) and (-√2/2, -√2/2).But to express this in terms of the circle, these are the points where P is at 45° and 225° from the positive x-axis.Thus, the area is constant when P is at these specific positions.

2 circles Γ and Σ, with centers O and P, respectively, are such that P lies on Γ. Let A be a point on Σ, and let M be the midpoint of AP. Let B be another point on Σ, such that AB||OM. Then prove that the midpoint of AB lies on Γ.

💡Okay, so I have this geometry problem here about two circles, Γ and Σ. Γ has center O, and Σ has center P. It says that P lies on Γ, which is interesting because that means the distance from O to P is equal to the radius of Γ. Let me note that down: OP is the radius of Γ.Now, there's a point A on Σ. So, A is somewhere on the circumference of circle Σ. Then, M is defined as the midpoint of AP. That means AM = MP. Got it. Next, there's another point B on Σ such that AB is parallel to OM. So, vector AB is parallel to vector OM. I need to visualize this. Maybe I can draw a rough sketch in my mind: two circles, one with center O, the other with center P on the first circle. Point A is on the second circle, M is halfway between A and P, and then B is another point on the second circle so that the line AB is parallel to OM.The goal is to prove that the midpoint of AB lies on Γ. So, if I can find the midpoint of AB and show that it's equidistant from O as the radius of Γ, that would do it.Let me think about properties of circles and midpoints. Since M is the midpoint of AP, and AB is parallel to OM, maybe there's some similar triangles or parallelogram properties involved here.Wait, if AB is parallel to OM, and M is the midpoint of AP, perhaps triangle ABM is similar to some other triangle? Or maybe quadrilateral OMPB has some properties because of the parallel lines.Hold on, since AB is parallel to OM, the vector AB is a scaled version of OM. Maybe I can use vectors here. Let me assign coordinates to make it easier. Let me place point O at the origin (0,0). Since P is on Γ, let's say P is at (r,0), where r is the radius of Γ. So, OP = r.Now, circle Σ has center P at (r,0). Let me denote the radius of Σ as s. So, any point A on Σ can be represented as (r + s cos θ, s sin θ) for some angle θ.Then, point M is the midpoint of AP. So, coordinates of M would be the average of coordinates of A and P. So, M_x = (r + (r + s cos θ))/2 = (2r + s cos θ)/2 = r + (s cos θ)/2. Similarly, M_y = (0 + s sin θ)/2 = (s sin θ)/2. So, M is at (r + (s cos θ)/2, (s sin θ)/2).Now, AB is parallel to OM. Vector OM is from O(0,0) to M(r + (s cos θ)/2, (s sin θ)/2). So, vector OM is (r + (s cos θ)/2, (s sin θ)/2). Vector AB is from A(r + s cos θ, s sin θ) to B. Let me denote B as (r + s cos φ, s sin φ), since it's another point on Σ.So, vector AB is (s (cos φ - cos θ), s (sin φ - sin θ)). Since AB is parallel to OM, their direction ratios must be proportional. So, the vector AB is a scalar multiple of vector OM.Therefore, (s (cos φ - cos θ), s (sin φ - sin θ)) = k*(r + (s cos θ)/2, (s sin θ)/2) for some scalar k.So, equating components:s (cos φ - cos θ) = k*(r + (s cos θ)/2)s (sin φ - sin θ) = k*(s sin θ / 2)From the second equation:s (sin φ - sin θ) = (k s sin θ)/2Divide both sides by s (assuming s ≠ 0):sin φ - sin θ = (k sin θ)/2Similarly, from the first equation:s (cos φ - cos θ) = k*(r + (s cos θ)/2)Divide both sides by s:cos φ - cos θ = (k / s)*(r + (s cos θ)/2)Hmm, this seems a bit complicated. Maybe I can express k from the second equation and substitute into the first.From the second equation:k = 2 (sin φ - sin θ) / sin θAssuming sin θ ≠ 0.Then, substitute into the first equation:cos φ - cos θ = [2 (sin φ - sin θ) / sin θ] * [r / s + (cos θ)/2]This is getting messy. Maybe there's a better approach.Alternatively, since AB is parallel to OM, the slope of AB equals the slope of OM.Slope of OM is ( (s sin θ)/2 - 0 ) / ( r + (s cos θ)/2 - 0 ) = (s sin θ / 2) / (r + (s cos θ)/2 )Slope of AB is (s sin φ - s sin θ) / (s cos φ - s cos θ) = (sin φ - sin θ) / (cos φ - cos θ)So, setting them equal:(sin φ - sin θ) / (cos φ - cos θ) = (s sin θ / 2) / (r + (s cos θ)/2 )Cross-multiplying:(sin φ - sin θ)(r + (s cos θ)/2 ) = (s sin θ / 2)(cos φ - cos θ)This still looks complicated, but maybe I can use trigonometric identities to simplify.Recall that sin φ - sin θ = 2 cos((φ+θ)/2) sin((φ-θ)/2)And cos φ - cos θ = -2 sin((φ+θ)/2) sin((φ-θ)/2)So, substituting these into the equation:[2 cos((φ+θ)/2) sin((φ-θ)/2)] [r + (s cos θ)/2 ] = [s sin θ / 2] [ -2 sin((φ+θ)/2) sin((φ-θ)/2) ]Simplify both sides:Left side: 2 cos((φ+θ)/2) sin((φ-θ)/2) [r + (s cos θ)/2 ]Right side: [s sin θ / 2] * (-2) sin((φ+θ)/2) sin((φ-θ)/2 ) = -s sin θ sin((φ+θ)/2) sin((φ-θ)/2 )So, left side: 2 [r + (s cos θ)/2 ] cos((φ+θ)/2) sin((φ-θ)/2 )Right side: -s sin θ sin((φ+θ)/2) sin((φ-θ)/2 )We can factor out sin((φ-θ)/2 ) from both sides, assuming sin((φ-θ)/2 ) ≠ 0, which would mean φ ≠ θ, which is reasonable since B is another point.So, dividing both sides by sin((φ-θ)/2 ):2 [r + (s cos θ)/2 ] cos((φ+θ)/2 ) = -s sin θ sin((φ+θ)/2 )Let me denote α = (φ+θ)/2 and β = (φ-θ)/2. Then, φ = α + β and θ = α - β.So, substituting:2 [r + (s cos(α - β))/2 ] cos α = -s sin(α - β) sin αLet me expand cos(α - β) and sin(α - β):cos(α - β) = cos α cos β + sin α sin βsin(α - β) = sin α cos β - cos α sin βSo, substituting back:2 [ r + (s/2)(cos α cos β + sin α sin β) ] cos α = -s [ sin α cos β - cos α sin β ] sin αLet me expand the left side:2r cos α + s (cos α cos β + sin α sin β) cos α= 2r cos α + s cos α (cos α cos β + sin α sin β )= 2r cos α + s cos^2 α cos β + s cos α sin α sin βRight side:-s [ sin α cos β - cos α sin β ] sin α= -s sin^2 α cos β + s sin α cos α sin βSo, equating left and right:2r cos α + s cos^2 α cos β + s cos α sin α sin β = -s sin^2 α cos β + s sin α cos α sin βLet me bring all terms to the left side:2r cos α + s cos^2 α cos β + s cos α sin α sin β + s sin^2 α cos β - s sin α cos α sin β = 0Simplify:2r cos α + s cos β (cos^2 α + sin^2 α) + s cos α sin α sin β - s cos α sin α sin β = 0Notice that cos^2 α + sin^2 α = 1, and the other terms cancel:2r cos α + s cos β = 0So, 2r cos α + s cos β = 0But α = (φ + θ)/2 and β = (φ - θ)/2.So, 2r cos[(φ + θ)/2] + s cos[(φ - θ)/2] = 0Hmm, not sure if this helps directly. Maybe I need another approach.Let me think about midpoints. The midpoint of AB is what we need to show lies on Γ. Let me denote the midpoint as N.So, coordinates of N would be the average of A and B.If A is (r + s cos θ, s sin θ) and B is (r + s cos φ, s sin φ), then N is:N_x = [ (r + s cos θ) + (r + s cos φ) ] / 2 = r + (s/2)(cos θ + cos φ)N_y = [ s sin θ + s sin φ ] / 2 = (s/2)(sin θ + sin φ)We need to show that N lies on Γ, which is centered at O(0,0) with radius OP = r. So, the distance from O to N should be r.So, distance ON^2 = (r + (s/2)(cos θ + cos φ))^2 + ( (s/2)(sin θ + sin φ) )^2 = r^2Let me compute this:= [ r + (s/2)(cos θ + cos φ) ]^2 + [ (s/2)(sin θ + sin φ) ]^2= r^2 + r s (cos θ + cos φ) + (s^2 /4)(cos θ + cos φ)^2 + (s^2 /4)(sin θ + sin φ)^2Now, let's compute the terms:First, expand (cos θ + cos φ)^2 + (sin θ + sin φ)^2:= cos^2 θ + 2 cos θ cos φ + cos^2 φ + sin^2 θ + 2 sin θ sin φ + sin^2 φ= (cos^2 θ + sin^2 θ) + (cos^2 φ + sin^2 φ) + 2(cos θ cos φ + sin θ sin φ)= 1 + 1 + 2 cos(θ - φ)= 2 + 2 cos(θ - φ)So, the last two terms in ON^2 become:(s^2 /4)(2 + 2 cos(θ - φ)) = (s^2 /4)(2(1 + cos(θ - φ))) = (s^2 /2)(1 + cos(θ - φ))So, putting it all together:ON^2 = r^2 + r s (cos θ + cos φ) + (s^2 /2)(1 + cos(θ - φ))We need this to equal r^2, so:r^2 + r s (cos θ + cos φ) + (s^2 /2)(1 + cos(θ - φ)) = r^2Subtract r^2 from both sides:r s (cos θ + cos φ) + (s^2 /2)(1 + cos(θ - φ)) = 0So, we have:r s (cos θ + cos φ) + (s^2 /2)(1 + cos(θ - φ)) = 0Let me factor out s:s [ r (cos θ + cos φ) + (s /2)(1 + cos(θ - φ)) ] = 0Since s ≠ 0 (as Σ is a circle), we have:r (cos θ + cos φ) + (s /2)(1 + cos(θ - φ)) = 0Hmm, this is a key equation. Let me see if I can relate this to the earlier equation we had.Earlier, we had:2r cos α + s cos β = 0Where α = (φ + θ)/2 and β = (φ - θ)/2.So, 2r cos α + s cos β = 0Let me express cos θ + cos φ in terms of α and β.cos θ + cos φ = cos(α - β) + cos(α + β) = 2 cos α cos βSimilarly, 1 + cos(θ - φ) = 1 + cos(2β) = 2 cos^2 βSo, substituting back into our key equation:r (2 cos α cos β) + (s /2)(2 cos^2 β) = 0Simplify:2 r cos α cos β + s cos^2 β = 0Factor out cos β:cos β (2 r cos α + s cos β) = 0So, either cos β = 0 or 2 r cos α + s cos β = 0If cos β = 0, then β = π/2 or 3π/2, meaning θ - φ = π or -π, so θ = φ ± π. But if θ and φ differ by π, then points A and B are diametrically opposite on Σ. Let's see if that satisfies the condition AB || OM.If θ = φ + π, then cos φ = cos(θ - π) = -cos θ, sin φ = sin(θ - π) = -sin θ.So, vector AB would be (-2 s cos θ, -2 s sin θ), and vector OM is (r + (s cos θ)/2, (s sin θ)/2). For AB to be parallel to OM, their direction ratios must be proportional.So, (-2 s cos θ, -2 s sin θ) = k*(r + (s cos θ)/2, (s sin θ)/2)Which would require:-2 s cos θ = k (r + (s cos θ)/2 )-2 s sin θ = k (s sin θ / 2 )From the second equation:-2 s sin θ = (k s sin θ)/2Assuming sin θ ≠ 0, divide both sides by s sin θ:-2 = k / 2 => k = -4Then, substitute into the first equation:-2 s cos θ = -4 (r + (s cos θ)/2 )Simplify:-2 s cos θ = -4 r - 2 s cos θBring all terms to one side:-2 s cos θ + 4 r + 2 s cos θ = 0 => 4 r = 0 => r = 0But r is the radius of Γ, which can't be zero. So, this leads to a contradiction. Therefore, cos β ≠ 0, so we must have:2 r cos α + s cos β = 0Which is exactly the equation we had earlier. So, this condition must hold.Therefore, our key equation reduces to 2 r cos α + s cos β = 0, which is consistent.So, going back, we have:2 r cos α + s cos β = 0But we also have from the midpoint N:We need to show that N lies on Γ, which reduces to the equation above. Since we've derived that equation under the given conditions, it must hold, meaning that N lies on Γ.Therefore, the midpoint of AB lies on Γ.**Final Answer**boxed{text{The midpoint of } AB text{ lies on } Gamma.}

Released under the MIT License.

has loaded