💡Okay, so I have this problem about an acute triangle ABC where AB is not equal to AC. I need to find the geometric locus of the centers of all possible rectangles that have two vertices on BC, one on AB, and one on AC. The center is defined as the intersection of the diagonals. First, they mentioned using analytic geometry, so I should probably set up a coordinate system. They suggested taking O, the foot of the altitude from A to BC, as the origin. So, O is (0,0). Then, the x-axis is along BC, and the y-axis is along OA. So, point A is (0,a), B is (b,0), and C is (c,0). Since AB ≠ AC, b ≠ -c, meaning b + c ≠ 0.Starting with part (a): Determine the coordinates of the midpoints M and N of segments OA and BC, respectively, and then find the equation of the line MN.Okay, midpoint M of OA. OA is from (0,0) to (0,a). The midpoint would be halfway in the y-direction. So, M should be (0, a/2).Midpoint N of BC. BC is from (b,0) to (c,0). The midpoint would be the average of the x-coordinates and the y-coordinates. Since both are on the x-axis, the y-coordinate is 0. So, N is ((b + c)/2, 0).Now, equation of line MN. We have two points: M(0, a/2) and N((b + c)/2, 0). Let me find the slope first. Slope m = (0 - a/2) / ((b + c)/2 - 0) = (-a/2) / ((b + c)/2) = -a/(b + c). So, the equation of the line is y = m x + n. We know it passes through M(0, a/2), so n = a/2. Therefore, the equation is y = (-a/(b + c))x + a/2.Alright, that seems straightforward. So, part (a) is done.Moving on to part (b): If we consider the coordinates of point D = (z, w), what are the coordinates of points E, F, and G?Hmm, so D is a point on AB. Since AB is from A(0,a) to B(b,0), any point on AB can be parameterized. But here, D is given as (z, w). So, E, F, G are the other vertices of the rectangle. Since it's a rectangle, opposite sides are equal and parallel. Two vertices are on BC, so E and F must be on BC. Wait, but BC is the base from B(b,0) to C(c,0). So, E and F are on BC, which is the x-axis. Wait, but the rectangle has two vertices on BC, one on AB, and one on AC. So, D is on AB, and G is on AC. So, DEFG is the rectangle, with D on AB, E and F on BC, and G on AC.So, if D is (z, w), then E is the projection of D onto BC, which is (z, 0). Similarly, G is the projection of D onto AC. But AC is from A(0,a) to C(c,0). So, the line AC has the equation y = (-a/c)x + a.So, to find G, which is the projection of D onto AC, we can use the projection formula. Alternatively, since DEFG is a rectangle, the sides DE and DF must be perpendicular. So, the vector DE is (0, -w), and the vector DF should be horizontal or something? Wait, maybe not.Alternatively, since DEFG is a rectangle, the sides DE and DF must be perpendicular. So, if DE is vertical from D(z, w) to E(z, 0), then DF must be horizontal from D(z, w) to F(f, w). But wait, F is on BC, which is the x-axis, so F must be (f, 0). Hmm, that doesn't make sense because DF would have to go from D(z, w) to F(f, 0), which is not necessarily horizontal.Wait, maybe I need to think differently. Since DEFG is a rectangle, DE and DF are adjacent sides. So, DE is from D(z, w) to E(z, 0), which is vertical. Then DF should be from D(z, w) to F(f, w), but F is on BC, which is the x-axis. So, that can't be because F would have to be (f, 0). So, maybe DF is not horizontal.Wait, perhaps I need to consider that DEFG is a rectangle, so DE and DF are perpendicular. So, DE is vertical, so DF must be horizontal? But F is on BC, which is the x-axis. So, if DE is vertical, then DF must be horizontal, but F is on BC, so DF would have to go from D(z, w) to F(z, 0). But that would make DF vertical, which contradicts DE being vertical.Wait, maybe I'm getting confused. Let me try to visualize. So, D is on AB, E is on BC, F is on BC, and G is on AC. So, DEFG is a rectangle with DE and DF as sides. Since DE is from D(z, w) to E(z, 0), which is vertical, then DF must be from D(z, w) to F(f, w), which is horizontal. But F is on BC, which is the x-axis, so F must be (f, 0). So, DF cannot be horizontal because F is on BC. Hmm, this is confusing.Wait, maybe DEFG is such that DE is one side, EF is another side, FG is another, and GD is the last. So, DE is from D(z, w) to E(z, 0), which is vertical. Then EF is from E(z, 0) to F(f, 0), which is horizontal. Then FG is from F(f, 0) to G(f, w'), which is vertical. Then GD is from G(f, w') to D(z, w), which is horizontal. But since it's a rectangle, GD must be equal and parallel to EF, and FG must be equal and parallel to DE.So, DE is vertical, so FG must also be vertical. So, if DE is from (z, w) to (z, 0), then FG must be from (f, 0) to (f, w'). Since DE has length w, FG must also have length w. So, G is (f, w). Then, GD is from G(f, w) to D(z, w), which is horizontal, so GD must be equal to EF. EF is from E(z, 0) to F(f, 0), which is length |f - z|. So, GD is from (f, w) to (z, w), which is also length |f - z|. So, that works.But G is on AC, so G(f, w) must satisfy the equation of AC. The equation of AC is y = (-a/c)x + a. So, substituting G(f, w):w = (-a/c)f + aSo, f = (a - w)c / aTherefore, F is (f, 0) = ((a - w)c / a, 0)Similarly, E is (z, 0), and G is ((a - w)c / a, w)So, to summarize:- D = (z, w)- E = (z, 0)- F = ((a - w)c / a, 0)- G = ((a - w)c / a, w)Wait, but E and F are both on BC, which is from (b,0) to (c,0). So, z must be between b and c? Or is it possible for z to be outside? Hmm, since ABC is acute, and the rectangle is inside the triangle, probably z is between b and c.But in any case, the coordinates are as above.Now, part (c): The center of the rectangle is the intersection of the diagonals, which is the midpoint of DF. So, find the coordinates of this midpoint and show it lies on line MN.So, midpoint of DF. D is (z, w), F is ((a - w)c / a, 0). So, midpoint P is:x-coordinate: (z + (a - w)c / a) / 2y-coordinate: (w + 0) / 2 = w / 2So, P = ((z + (a - w)c / a)/2, w / 2)Simplify the x-coordinate:(z + (a - w)c / a)/2 = (z a + (a - w)c) / (2a)So, P = ((z a + c a - w c) / (2a), w / 2)Now, we need to show that this point lies on line MN, whose equation is y = (-a/(b + c))x + a/2.So, substitute x and y into the equation:w / 2 = (-a/(b + c)) * ((z a + c a - w c) / (2a)) + a/2Simplify the right-hand side:= (-a/(b + c)) * ((z a + c a - w c) / (2a)) + a/2= (-1/(b + c)) * ((z a + c a - w c) / 2) + a/2= (- (z a + c a - w c) ) / (2(b + c)) + a/2Now, let's compute the left-hand side minus the right-hand side:w / 2 - [ (- (z a + c a - w c) ) / (2(b + c)) + a/2 ]= w / 2 + (z a + c a - w c) / (2(b + c)) - a / 2Factor out 1/2:= (1/2)[ w + (z a + c a - w c)/(b + c) - a ]Let me compute the term inside the brackets:w + (z a + c a - w c)/(b + c) - a= w - a + (z a + c a - w c)/(b + c)Now, let's see if this is zero.Wait, but I need to relate z and w. Since D(z, w) is on AB. The equation of AB is from A(0,a) to B(b,0). So, the slope of AB is (0 - a)/(b - 0) = -a/b. So, the equation is y = (-a/b)x + a.Therefore, since D(z, w) is on AB, we have:w = (-a/b)z + aSo, w = a - (a/b)zTherefore, we can express z in terms of w:z = (a - w) b / aSo, z = (b/a)(a - w)Now, substitute z into the expression:w - a + (z a + c a - w c)/(b + c)= w - a + ( (b/a)(a - w) * a + c a - w c ) / (b + c)Simplify:= w - a + ( b(a - w) + c a - w c ) / (b + c )= w - a + ( a b - b w + a c - c w ) / (b + c )Factor numerator:= w - a + ( a(b + c) - w(b + c) ) / (b + c )= w - a + ( (a - w)(b + c) ) / (b + c )= w - a + (a - w)= w - a + a - w= 0So, the expression inside the brackets is zero, meaning the entire expression is zero. Therefore, the left-hand side equals the right-hand side, so P lies on line MN.Therefore, the center of the rectangle lies on line MN. Since this is true for any rectangle constructed as described, the locus of all such centers is the line segment MN.So, the answer is that the locus is the line segment MN.**Final Answer**The geometric locus formed by the centers of these rectangles is the line segment ( MN ). Therefore, the final answer is boxed{MN}.

💡Alright, so I have this problem here: "If the function ( y = (k^2 - k - 5)x^3 ) is a power function, then the value of the real number ( k ) is ( ) A: 3, B: -2, C: 3 or -2, D: ( k neq 3 ) and ( k neq -2 )." Hmm, okay. Let me try to figure this out step by step.First, I need to recall what a power function is. From what I remember, a power function is of the form ( y = ax^n ), where ( a ) is a constant and ( n ) is an exponent. So, in this case, the function given is ( y = (k^2 - k - 5)x^3 ). Comparing this to the general form, ( a ) here is ( (k^2 - k - 5) ) and ( n ) is 3.Wait, but the question is asking about the value of ( k ). So, I think the key here is to figure out what ( k ) must be for this function to be a power function. But hold on, isn't any function of the form ( y = ax^n ) already a power function? So, is there something more specific here?Let me think. Maybe the problem is implying that the coefficient ( a ) must be 1? Because sometimes power functions are defined with a leading coefficient of 1, like ( y = x^n ). If that's the case, then ( (k^2 - k - 5) ) must equal 1. Let me check that.So, if ( (k^2 - k - 5) = 1 ), then solving for ( k ) would give me the possible values. Let's write that equation down:( k^2 - k - 5 = 1 )Subtracting 1 from both sides:( k^2 - k - 6 = 0 )Now, I need to solve this quadratic equation. I can try factoring it. Let me see if I can find two numbers that multiply to -6 and add up to -1. Hmm, 2 and -3? Because 2 * (-3) = -6 and 2 + (-3) = -1. Yes, that works.So, factoring the quadratic:( (k + 2)(k - 3) = 0 )Setting each factor equal to zero:1. ( k + 2 = 0 ) => ( k = -2 )2. ( k - 3 = 0 ) => ( k = 3 )So, the solutions are ( k = -2 ) and ( k = 3 ). That means the value of ( k ) can be either 3 or -2 for the function to be a power function. Looking back at the options, option C says "3 or -2," which matches my solutions.But wait, let me make sure I didn't make a mistake. Is the coefficient required to be 1 for it to be a power function? I think sometimes power functions are defined as monomials, which can have any coefficient, not necessarily 1. So, maybe my initial assumption is wrong.Let me double-check the definition of a power function. A power function is a function of the form ( y = ax^n ), where ( a ) is a constant and ( n ) is an exponent. So, actually, the coefficient ( a ) can be any real number, not necessarily 1. That means the function ( y = (k^2 - k - 5)x^3 ) is already a power function regardless of the value of ( k ), as long as ( k^2 - k - 5 ) is a real number.Wait, so does that mean that ( k ) can be any real number except those that make the coefficient undefined or something? But in this case, ( k^2 - k - 5 ) is defined for all real numbers ( k ). So, does that mean that the function is always a power function, regardless of ( k )?But then why would the question ask for specific values of ( k )? Maybe I'm misunderstanding the question. Perhaps it's not about the function being a power function in general, but about it being a specific type of power function, like a monomial with a certain exponent or something else.Wait, the exponent here is already fixed at 3. So, maybe the question is testing something else. Let me read the problem again: "If the function ( y = (k^2 - k - 5)x^3 ) is a power function, then the value of the real number ( k ) is ( )..."Hmm, maybe the question is implying that the function is a power function only if the coefficient is non-zero? Because if the coefficient were zero, then the function would just be ( y = 0 ), which is a constant function, not a power function. So, perhaps we need to ensure that ( k^2 - k - 5 neq 0 ).Let me test that idea. If ( k^2 - k - 5 = 0 ), then the function becomes ( y = 0 cdot x^3 = 0 ), which is a constant function, not a power function. So, to ensure that the function is a power function, we need ( k^2 - k - 5 neq 0 ). Therefore, ( k ) cannot be the roots of the equation ( k^2 - k - 5 = 0 ).Let me solve ( k^2 - k - 5 = 0 ) to find those values. Using the quadratic formula:( k = frac{1 pm sqrt{1 + 20}}{2} = frac{1 pm sqrt{21}}{2} )So, ( k ) cannot be ( frac{1 + sqrt{21}}{2} ) or ( frac{1 - sqrt{21}}{2} ). But looking at the options given, none of them mention these values. The options are 3, -2, 3 or -2, or ( k neq 3 ) and ( k neq -2 ).Wait, so maybe my initial approach was correct after all. If the problem is expecting ( k ) to be 3 or -2, perhaps it's assuming that the coefficient should be 1, making it a monic power function. So, in that case, setting ( k^2 - k - 5 = 1 ) gives ( k = 3 ) or ( k = -2 ).But I'm confused because, according to the general definition, a power function doesn't require the coefficient to be 1. It can be any constant. So, why would the problem specify that ( k ) must be 3 or -2? Maybe the problem is from a specific context where power functions are defined with a leading coefficient of 1. Or perhaps it's a translation issue, and the term "power function" is being used differently.Alternatively, maybe the problem is testing whether the function is a monomial, which is a type of power function with a single term. In that case, as long as ( k^2 - k - 5 ) is not zero, it's a monomial, hence a power function. But then, the answer would be that ( k ) can be any real number except the roots of ( k^2 - k - 5 = 0 ), which are approximately 2.791 and -1.791. But again, the options don't include these.Wait, looking back at the options: A is 3, B is -2, C is 3 or -2, D is ( k neq 3 ) and ( k neq -2 ). So, if I go back to my initial thought, that the coefficient must be 1, then ( k ) must be 3 or -2, which is option C. But if the coefficient can be any real number except zero, then ( k ) cannot be the roots of ( k^2 - k - 5 = 0 ), which are approximately 2.791 and -1.791, but these aren't among the options.Therefore, it's more likely that the problem expects the coefficient to be 1, making the function a monic power function. Hence, solving ( k^2 - k - 5 = 1 ) gives ( k = 3 ) or ( k = -2 ), which is option C.But just to be thorough, let me consider another angle. Maybe the problem is referring to the function being a power function in the sense that it's a polynomial of degree 3. But in that case, any function of the form ( y = ax^3 ) is a cubic polynomial, which is a power function. So, again, unless there's a specific condition on the coefficient, ( k ) can be any real number.However, since the options include specific values, it's likely that the problem is expecting the coefficient to be 1, hence leading to ( k = 3 ) or ( k = -2 ). Therefore, I think the correct answer is C: 3 or -2.But just to make sure, let me plug in ( k = 3 ) and ( k = -2 ) into the coefficient:For ( k = 3 ):( 3^2 - 3 - 5 = 9 - 3 - 5 = 1 ). So, the function becomes ( y = 1x^3 = x^3 ), which is a power function.For ( k = -2 ):( (-2)^2 - (-2) - 5 = 4 + 2 - 5 = 1 ). So, the function becomes ( y = 1x^3 = x^3 ), which is also a power function.Therefore, both ( k = 3 ) and ( k = -2 ) make the coefficient equal to 1, resulting in a power function. Hence, option C is correct.I think I've considered the different angles here. The key was recognizing whether the coefficient needs to be 1 or not. Given the options, it makes sense that the problem expects the coefficient to be 1, leading to ( k = 3 ) or ( k = -2 ).

💡First, I need to determine how many unbroken seashells Tom found. Tom found a total of 23 seashells, and out of these, 11 were broken. To find the number of unbroken seashells, I subtract the number of broken seashells from the total number of seashells.So, 23 total seashells minus 11 broken seashells equals 12 unbroken seashells.

💡Okay, so I have this function f(x) defined piecewise. It's equal to 1/x minus 1 when x is greater than 1, and it's equal to -2x plus a when x is less than or equal to 1. The problem says that for any two different x values, x1 and x2, the function values f(x1) and f(x2) are also different. So, that means the function is injective, or one-to-one, right? I remember that for a function to be injective, it needs to be strictly increasing or strictly decreasing throughout its entire domain. So, I need to check if this piecewise function is either always increasing or always decreasing. If it is, then it will satisfy the condition that f(x1) ≠ f(x2) whenever x1 ≠ x2.Let me look at each piece separately first. For x > 1, the function is 1/x - 1. The derivative of this part would be the derivative of 1/x, which is -1/x², so the derivative is negative. That means this part is decreasing for x > 1. Okay, so the right side of the function is decreasing.Now, for x ≤ 1, the function is -2x + a. The derivative here is just -2, which is also negative. So, this part is also decreasing for x ≤ 1. Hmm, so both pieces are decreasing. That’s good because if both parts are decreasing, the function could potentially be injective if the pieces connect properly at x = 1.But wait, just because both pieces are decreasing doesn't automatically mean the whole function is injective. I need to make sure that the function doesn't have any jumps or overlaps where the left side and the right side could produce the same value. So, I should check the value of the function at x = 1 from both sides.From the right side (x > 1), as x approaches 1 from above, the function approaches 1/1 - 1, which is 0. From the left side (x ≤ 1), at x = 1, the function is -2(1) + a, which is -2 + a. For the function to be injective, the left side at x = 1 should be greater than or equal to the right side's limit as x approaches 1 from above. Otherwise, there might be a point where the left side is less than the right side, causing the function to not be strictly decreasing overall.So, setting up the inequality: -2 + a ≥ 0. Solving for a, I get a ≥ 2. Wait, let me think again. If a is greater than or equal to 2, then at x = 1, the left side is -2 + a, which is at least 0. The right side as x approaches 1 is 0. So, if a is exactly 2, then at x = 1, the left side is 0, which is equal to the right side's limit. But since the function is defined as -2x + a for x ≤ 1, at x = 1, it's exactly 0, and for x > 1, it's approaching 0 from below. So, the function is continuous at x = 1 when a = 2.But wait, the problem doesn't say the function is continuous, just that it's injective. So, maybe even if there's a jump, as long as the function is strictly decreasing, it can still be injective. Let me consider that.If a is greater than 2, then at x = 1, the left side is greater than 0, and the right side approaches 0. So, the function would have a jump discontinuity at x = 1, but since the left side is higher than the right side, the function is still strictly decreasing overall. Because for x ≤ 1, the function is decreasing, and for x > 1, it's also decreasing, and the left side at x = 1 is higher than the right side near x = 1. So, no two different x's would map to the same f(x).But if a is less than 2, then at x = 1, the left side would be less than 0. The right side approaches 0 from below as x approaches 1 from above. So, the left side at x = 1 is less than 0, and the right side is approaching 0. That means there's a point where the left side is less than the right side, which could cause the function to not be injective because the left side might dip below the right side, creating an overlap in function values.Wait, let me visualize this. If a is less than 2, say a = 1, then at x = 1, the left side is -2 + 1 = -1. The right side as x approaches 1 from above is approaching 0. So, the function jumps from -1 at x = 1 to approaching 0 as x approaches 1 from the right. But since the right side is decreasing towards 0, and the left side is also decreasing, but starting at -1, which is below 0, does that cause any issues?Actually, no, because the left side is decreasing to the left of 1, and the right side is decreasing to the right of 1. But the left side at x = 1 is -1, and the right side is approaching 0. So, the function is decreasing on both sides, but there's a jump from -1 to 0. Since the function is decreasing on both intervals, and the left side at x =1 is less than the right side near x =1, the function remains injective because it's always decreasing.Wait, that contradicts my earlier thought. Maybe I need to think more carefully.If a is less than 2, say a = 0, then at x =1, the left side is -2(1) + 0 = -2. The right side approaches 0 as x approaches 1 from above. So, the function jumps from -2 to near 0. But the left side is decreasing for x ≤1, so as x decreases from 1, f(x) becomes more negative. The right side is decreasing for x >1, so as x increases beyond 1, f(x) becomes more negative. But wait, 1/x -1 for x >1 is actually decreasing towards 0, not negative infinity.Wait, no. For x >1, 1/x is between 0 and 1, so 1/x -1 is between -1 and 0. So, as x increases, 1/x approaches 0, so 1/x -1 approaches -1. So, the right side is decreasing from 0 towards -1 as x increases. The left side is -2x +a, which is a linear function with slope -2, so it's decreasing as x increases.Wait, but for x ≤1, as x decreases, f(x) increases because it's a linear function with a negative slope. So, as x approaches negative infinity, f(x) approaches positive infinity. So, the left side goes from positive infinity down to -2 + a at x =1, and the right side goes from 0 down to -1 as x approaches infinity.So, if a is less than 2, then at x =1, the left side is less than 0. The right side is approaching 0 from below. So, the function is decreasing on both sides, but the left side at x =1 is less than the right side near x =1. So, the function is continuous in its decreasing nature, but there's a jump from the left side to the right side.But does that affect injectivity? Because injectivity requires that each y-value corresponds to exactly one x-value. If the left side at x =1 is less than the right side near x =1, then there might be a y-value that is achieved both by the left side and the right side.Wait, let me think. Suppose a is less than 2, say a =1. Then at x =1, f(x) = -1. For x just above 1, f(x) is just below 0. So, the function jumps from -1 to near 0. So, the function is decreasing on both sides, but there's a jump. So, the left side goes from infinity down to -1, and the right side goes from near 0 down to -1 as x approaches infinity.Wait, no, the right side as x approaches infinity, 1/x -1 approaches -1. So, the right side approaches -1 from above, since 1/x is positive, so 1/x -1 is greater than -1. So, the right side is decreasing from 0 towards -1, but never actually reaching -1. The left side at x =1 is -1, and as x decreases, it goes to infinity.So, in this case, the function f(x) on the left side reaches -1 at x =1, and the right side approaches -1 as x approaches infinity. So, does that mean that the function f(x) = -1 is achieved both at x =1 and as x approaches infinity? But wait, as x approaches infinity, f(x) approaches -1, but never actually reaches -1 except at x =1. So, is -1 achieved only once at x =1?Wait, no, because as x approaches infinity, f(x) approaches -1, but doesn't reach it. So, f(x) = -1 is only achieved at x =1. So, in that case, the function is still injective because each y-value is achieved at most once.Wait, but if a is less than 2, say a =1, then at x =1, f(x) = -1, and for x >1, f(x) approaches -1 as x approaches infinity. So, f(x) = -1 is only achieved at x =1, and for x >1, f(x) is always greater than -1, approaching it asymptotically. So, in that case, the function is still injective because no two different x's map to the same y.But wait, if a is less than 2, say a =0, then at x =1, f(x) = -2, and for x >1, f(x) approaches -1. So, f(x) on the left side goes from infinity down to -2, and on the right side, it goes from 0 down to -1. So, the function is decreasing on both sides, but the left side ends at -2, and the right side starts near 0. So, there's a gap between -2 and 0 where the function doesn't take any values. But does that affect injectivity?No, because injectivity is about each y-value being achieved at most once, not about covering all y-values. So, even if there are gaps, as long as each y is achieved at most once, the function is injective.Wait, but in this case, if a is less than 2, the left side at x =1 is less than 0, and the right side is approaching 0 from below. So, the function is decreasing on both sides, but the left side ends at a lower value than the right side starts. So, the function is injective because it's strictly decreasing on both intervals, and there's no overlap in the y-values.Wait, but earlier I thought that if a is less than 2, the function might not be injective because of overlapping y-values. But now, I'm thinking that it is injective regardless of a, as long as both pieces are decreasing.But that can't be right because the answer choices suggest that a has to be at least 2. So, maybe I'm missing something.Let me think again. The function is injective if it's strictly monotonic. Since both pieces are decreasing, the function is injective if the left side at x =1 is greater than or equal to the right side's limit as x approaches 1 from above.Wait, that makes sense. Because if the left side at x =1 is greater than the right side near x =1, then the function is strictly decreasing across the entire domain. If the left side is less than the right side near x =1, then there might be a point where the function increases, which would violate injectivity.Wait, no. Because both sides are decreasing, but if the left side at x =1 is less than the right side near x =1, then as x increases through 1, the function jumps from a lower value to a higher value, which would mean that the function is not strictly decreasing overall. It would have a jump upwards, which could cause the function to not be injective because it might not be strictly decreasing.Wait, but injectivity doesn't require the function to be continuous, just that each y is achieved at most once. So, even if there's a jump, as long as the function doesn't repeat any y-values, it's injective.But in reality, if the left side at x =1 is less than the right side near x =1, then there's a possibility that some y-values between the left side's value at x =1 and the right side's value near x =1 are achieved both by the left side and the right side. Wait, no, because the left side is decreasing to the left of 1, and the right side is decreasing to the right of 1. So, the left side goes from infinity down to, say, -2 + a, and the right side goes from 0 down to -1. So, if a is less than 2, the left side ends at a lower value than the right side starts. So, the function is decreasing on both sides, but there's a gap between the left side's end and the right side's start. So, the function is injective because each y is achieved at most once.Wait, but if a is less than 2, say a =1, then the left side at x =1 is -1, and the right side near x =1 is approaching 0. So, the function jumps from -1 to near 0. So, the function is decreasing on both sides, but there's a jump upwards. So, does that affect injectivity?No, because injectivity is about each y being achieved at most once, not about the function being continuous or having a certain behavior. So, even if there's a jump, as long as no y is achieved twice, the function is injective.But wait, if the left side at x =1 is less than the right side near x =1, then the function is decreasing on both sides, but the left side ends lower than the right side starts. So, the function is injective because it's strictly decreasing on both intervals, and there's no overlap in the y-values.Wait, but the answer choices suggest that a has to be at least 2. So, maybe I'm misunderstanding something.Let me try plugging in a =2. At x =1, the left side is -2 +2 =0, and the right side approaches 0 as x approaches 1 from above. So, the function is continuous at x =1 when a =2, and it's strictly decreasing everywhere. So, it's injective.If a is greater than 2, say a =3, then at x =1, the left side is -2 +3 =1, and the right side approaches 0. So, the function jumps from 1 to near 0. So, the left side is decreasing from infinity down to 1, and the right side is decreasing from 0 down to -1. So, the function is injective because each y is achieved at most once.But if a is less than 2, say a =1, then at x =1, the left side is -1, and the right side approaches 0. So, the function is decreasing on both sides, but the left side ends at -1, and the right side starts near 0. So, the function is injective because each y is achieved at most once.Wait, but the answer choices suggest that a has to be at least 2. So, maybe I'm missing something about the function's behavior.Wait, perhaps if a is less than 2, the function might not be strictly decreasing overall because the left side at x =1 is less than the right side near x =1, causing the function to have a local maximum or something. But no, because both sides are decreasing, just with a jump.Wait, no, because the function is decreasing on both sides, but the left side ends lower than the right side starts. So, the function is still injective because it's strictly decreasing on each interval, and there's no overlap in the y-values.Wait, but maybe the function is not strictly decreasing overall because of the jump. For example, if a is less than 2, then as x increases through 1, the function jumps from a lower value to a higher value, which would mean that the function is not strictly decreasing across the entire domain. So, even though each piece is decreasing, the overall function is not strictly decreasing because of the jump upwards.But injectivity doesn't require the function to be strictly decreasing; it just requires that each y is achieved at most once. So, even if the function has a jump, as long as no y is achieved twice, it's injective.Wait, but if the function jumps from a lower value to a higher value, then there's a possibility that some y-values are achieved both by the left side and the right side. For example, suppose a =1, so the left side at x =1 is -1, and the right side near x =1 is approaching 0. So, the function jumps from -1 to near 0. Now, consider y = -0.5. Is there an x ≤1 such that f(x) = -0.5? Yes, because the left side is a linear function decreasing from infinity to -1, so it must cross y = -0.5 somewhere. Also, on the right side, as x increases from 1 onwards, f(x) decreases from 0 to -1, so it also crosses y = -0.5 somewhere. So, in this case, y = -0.5 is achieved by both the left side and the right side, which means the function is not injective.Ah, that's the key! If a is less than 2, then the left side at x =1 is less than the right side near x =1, which means that there's an overlap in the y-values between the two pieces. So, some y-values are achieved by both the left side and the right side, making the function not injective.Therefore, to ensure that the function is injective, the left side at x =1 must be greater than or equal to the right side's limit as x approaches 1 from above. That way, there's no overlap in the y-values between the two pieces.So, setting up the inequality: -2 + a ≥ 0. Solving for a, we get a ≥ 2. Therefore, the range of values for a is [2, ∞).So, the correct answer is C: [2, +∞).

💡First, I recognize that one dozen equals 12 eggs.Dan bought 9 dozen eggs.To find the total number of eggs, I multiply the number of dozens by the number of eggs in a dozen.So, 9 dozen multiplied by 12 eggs per dozen equals 108 eggs.Therefore, Dan bought 108 eggs.

💡Alright, so I have this problem where a square is divided into several convex polygons, and each of these polygons has a distinct number of sides. I need to prove that among these polygons, there must be at least one triangle. Hmm, okay, let me think about how to approach this.First, let me visualize the square. It's a four-sided figure, and it's being divided into smaller convex polygons. Each of these polygons has a different number of sides. So, for example, one could be a triangle (3 sides), another a quadrilateral (4 sides), another a pentagon (5 sides), and so on. But the problem states that all these polygons have pairwise distinct numbers of sides, meaning no two polygons share the same number of sides.Now, since the square is being divided into these polygons, the total number of sides contributed by all these polygons should relate to the original square's sides. But I need to think carefully about how the sides interact when the square is divided.Each time we divide the square, we're essentially adding internal edges. These internal edges are shared between two polygons. So, if I consider the total number of edges in all the polygons combined, it's equal to the number of edges on the square plus twice the number of internal edges. This is because each internal edge is shared by two polygons.Let me denote the number of polygons as ( n ). Each polygon has a certain number of sides, say ( k_1, k_2, ldots, k_n ), where each ( k_i ) is distinct. The total number of edges, counting each internal edge twice, would be ( k_1 + k_2 + ldots + k_n ). On the other hand, the square has 4 edges, and if there are ( m ) internal edges, the total number of edges is also ( 4 + 2m ).So, we have:[k_1 + k_2 + ldots + k_n = 4 + 2m]Since each ( k_i ) is at least 3 (because a polygon must have at least 3 sides), the sum ( k_1 + k_2 + ldots + k_n ) is at least ( 3n ). Therefore:[3n leq 4 + 2m]But I don't know ( m ) yet. Maybe I can find another relationship.Each internal edge is shared by two polygons, so the number of internal edges ( m ) is related to the number of polygons ( n ). Specifically, each internal edge can be thought of as a connection between two polygons. If I have ( n ) polygons, the maximum number of internal edges would be something like ( binom{n}{2} ), but that's probably too high. Actually, in a planar graph, the number of edges is related to the number of faces and vertices, but maybe that's complicating things.Wait, maybe I can think about the minimum number of edges required. If all polygons are triangles, then each triangle has 3 edges, but each internal edge is shared by two triangles. So, the total number of edges would be ( 4 + 2m = 3n ). But in our case, the polygons have distinct numbers of sides, so they can't all be triangles.Hmm, perhaps I should consider the minimum total number of edges given that all polygons have distinct side counts. The smallest possible sum of sides would be if the polygons have 3, 4, 5, ..., up to some number of sides. So, the sum would be ( 3 + 4 + 5 + ldots + (n+2) ) because starting from 3, each subsequent polygon has one more side.Wait, is that right? If there are ( n ) polygons, starting from 3 sides, the number of sides would be 3, 4, 5, ..., ( n+2 ). So, the sum would be:[sum_{k=3}^{n+2} k = frac{(n+2)(n+3)}{2} - 3]Wait, no, that's not correct. The sum from 1 to ( n+2 ) is ( frac{(n+2)(n+3)}{2} ), so subtracting the sum from 1 to 2, which is 3, gives:[frac{(n+2)(n+3)}{2} - 3 = frac{(n+2)(n+3) - 6}{2} = frac{n^2 + 5n + 6 - 6}{2} = frac{n^2 + 5n}{2}]So, the total number of edges is ( frac{n^2 + 5n}{2} ). But we also have that the total number of edges is ( 4 + 2m ). Therefore:[frac{n^2 + 5n}{2} = 4 + 2m]But this seems too large because ( m ) can't be that big. Maybe my assumption about the number of sides is wrong.Wait, actually, the polygons don't necessarily have to start from 3 sides. They just have to have distinct numbers of sides, but they could start from higher numbers. However, since we're trying to find the minimum total number of edges, starting from the smallest possible sides makes sense.But maybe I'm overcomplicating this. Let's think differently. Each polygon must have at least 3 sides, and all have distinct numbers of sides. So, the minimum total number of sides is ( 3 + 4 + 5 + ldots + (n+2) ), as before. But we also know that the total number of sides is ( 4 + 2m ), which is less than or equal to ( 4 + 2 times ) (some function of ( n )).Wait, perhaps I should consider Euler's formula for planar graphs. Euler's formula states that ( V - E + F = 2 ), where ( V ) is the number of vertices, ( E ) is the number of edges, and ( F ) is the number of faces. In our case, the square is divided into ( n ) polygons, so ( F = n + 1 ) (including the outer face). So, ( V - E + (n + 1) = 2 ), which simplifies to ( V - E = 1 - n ).But I'm not sure if this helps directly. Maybe I need another approach.Let me think about the number of edges. Each polygon contributes its number of sides to the total edge count, but each internal edge is shared by two polygons. So, if I denote ( E ) as the total number of edges, then:[E = frac{k_1 + k_2 + ldots + k_n + 4}{2}]Wait, no. The total number of edges, counting each internal edge twice, is ( k_1 + k_2 + ldots + k_n ). But the actual number of edges is ( 4 + m ), where ( m ) is the number of internal edges. So:[k_1 + k_2 + ldots + k_n = 2(4 + m) = 8 + 2m]But I don't know ( m ). However, I can relate ( m ) to ( n ) somehow.Each internal edge connects two polygons, so the number of internal edges ( m ) is at most ( binom{n}{2} ), but in reality, it's less because each internal edge can only connect adjacent polygons.Wait, maybe I can use the fact that each polygon has a certain number of sides and relate that to the number of edges.If all polygons have distinct numbers of sides, then the number of sides must be at least 3, 4, 5, ..., up to some maximum. So, the total number of sides is at least ( 3 + 4 + 5 + ldots + (n+2) ), which is ( frac{n(n+5)}{2} ).But we also have that the total number of sides is ( 8 + 2m ). So:[frac{n(n+5)}{2} leq 8 + 2m]But I don't know ( m ), so maybe this isn't helpful.Wait, perhaps I can find a contradiction. Suppose there are no triangles. Then all polygons have at least 4 sides. So, the total number of sides would be at least ( 4n ). Therefore:[4n leq 8 + 2m]Which simplifies to:[2n leq 4 + m]So, ( m geq 2n - 4 ).But the number of internal edges ( m ) is also related to the number of polygons. Each internal edge can be thought of as a connection between two polygons, so the maximum number of internal edges is ( binom{n}{2} ), but in reality, it's less because each polygon can only have a certain number of edges.Wait, maybe I can think about the number of edges per polygon. Each polygon has ( k_i ) edges, and each internal edge is shared by two polygons. So, the total number of edges is ( frac{sum k_i + 4}{2} ).If there are no triangles, then ( k_i geq 4 ) for all ( i ), so ( sum k_i geq 4n ). Therefore:[frac{4n + 4}{2} leq E]Which simplifies to:[2n + 2 leq E]But ( E = 4 + m ), so:[2n + 2 leq 4 + m]Which gives:[m geq 2n - 2]But earlier, I had ( m geq 2n - 4 ). So, combining these, ( m geq 2n - 2 ).But the number of internal edges ( m ) cannot exceed a certain limit. For example, each polygon can only have a certain number of edges. If each polygon has at least 4 sides, then the number of edges per polygon is at least 4, but the total number of edges is ( 4 + m ).Wait, maybe I can use the fact that each polygon must have at least 3 sides, but if there are no triangles, then each has at least 4. So, the total number of edges is at least ( 4n ), but also equals ( 8 + 2m ). So:[4n leq 8 + 2m implies 2n leq 4 + m implies m geq 2n - 4]But earlier, I had ( m geq 2n - 2 ). So, combining these, ( m geq 2n - 2 ).But how does this help me? Maybe I need to find a relationship between ( n ) and ( m ) that leads to a contradiction if there are no triangles.Alternatively, perhaps I can consider the maximum number of sides a polygon can have. Since the square is divided into ( n ) polygons, the maximum number of sides any polygon can have is ( n + 2 ). Wait, why?Because each polygon can border at most one side of the square, and each internal edge can border at most one other polygon. So, the maximum number of sides a polygon can have is 4 (from the square) plus ( n - 1 ) internal edges, but that would be ( n + 3 ). Wait, no, that's not quite right.Actually, a polygon inside the square can border multiple sides of the square, but in reality, it can only border up to four sides, but since it's convex, it can't wrap around the square. So, the maximum number of sides a polygon can have is 4 (from the square) plus the number of internal edges it has. But each internal edge is shared with another polygon, so the maximum number of sides a polygon can have is 4 + (n - 1), because it can share an edge with each of the other ( n - 1 ) polygons. So, the maximum number of sides is ( n + 3 ).But if all polygons have distinct numbers of sides, then the largest polygon would have ( n + 3 ) sides, the next largest ( n + 2 ), and so on, down to the smallest, which would have 3 sides if there are no triangles. Wait, but if there are no triangles, the smallest would have 4 sides.Wait, let's think carefully. If there are no triangles, then the smallest polygon has 4 sides, the next has 5, and so on, up to ( n + 3 ) sides. So, the total number of sides would be:[4 + 5 + 6 + ldots + (n + 3)]This is an arithmetic series with first term 4, last term ( n + 3 ), and number of terms ( n ). The sum is:[frac{n}{2} times (4 + n + 3) = frac{n(n + 7)}{2}]But we also have that the total number of sides is ( 8 + 2m ). So:[frac{n(n + 7)}{2} = 8 + 2m]But earlier, we had ( m geq 2n - 2 ). Substituting:[frac{n(n + 7)}{2} = 8 + 2(2n - 2) = 8 + 4n - 4 = 4n + 4]So:[frac{n(n + 7)}{2} = 4n + 4]Multiply both sides by 2:[n(n + 7) = 8n + 8]Simplify:[n^2 + 7n = 8n + 8][n^2 - n - 8 = 0]Solving this quadratic equation:[n = frac{1 pm sqrt{1 + 32}}{2} = frac{1 pm sqrt{33}}{2}]Since ( n ) must be positive, ( n = frac{1 + sqrt{33}}{2} approx 3.37 ). But ( n ) must be an integer greater than 1, so ( n = 4 ) is the smallest possible.But if ( n = 4 ), then the total number of sides would be:[4 + 5 + 6 + 7 = 22]But according to the earlier equation, ( frac{4(4 + 7)}{2} = frac{44}{2} = 22 ), and ( 4n + 4 = 16 + 4 = 20 ). Wait, that doesn't match. There's a discrepancy here.Wait, I think I made a mistake in the substitution. Let me check.We had:[frac{n(n + 7)}{2} = 4n + 4]For ( n = 4 ):[frac{4(11)}{2} = 22]And ( 4n + 4 = 20 ). So, 22 ≠ 20. That's a problem.This suggests that our assumption that there are no triangles leads to a contradiction because the total number of sides doesn't match up. Therefore, there must be at least one triangle among the polygons.Wait, but let me make sure. Maybe I messed up the arithmetic somewhere.Alternatively, perhaps I should consider that if there are no triangles, then the smallest polygon has 4 sides, and the total number of sides is ( 4 + 5 + ldots + (n + 3) ), which is ( frac{n(n + 7)}{2} ). But this must equal ( 8 + 2m ). Also, from Euler's formula, we have ( V - E + F = 2 ), where ( F = n + 1 ). So, ( V - E = 1 - n ).But I don't know ( V ) or ( E ). Maybe I can express ( V ) in terms of ( E ). Each polygon has ( k_i ) edges and ( k_i ) vertices, but each vertex is shared by multiple polygons. So, the total number of vertices is less than the sum of all ( k_i ).Wait, maybe this is getting too complicated. Let me try a different approach.Suppose there are no triangles. Then all polygons have at least 4 sides. The total number of sides is ( sum k_i geq 4n ). But the total number of sides is also ( 8 + 2m ). So:[4n leq 8 + 2m implies 2n leq 4 + m implies m geq 2n - 4]Now, consider the number of internal edges ( m ). Each internal edge is shared by two polygons, so the number of internal edges is at most ( binom{n}{2} ), but in reality, it's much less because each polygon can only have a certain number of edges.But if ( m geq 2n - 4 ), and ( m leq ) something like ( 2n - 4 ), then maybe equality holds. But I'm not sure.Wait, let's think about the maximum number of sides a polygon can have. If there are ( n ) polygons, the maximum number of sides any one polygon can have is ( n + 3 ), as I thought earlier. But if all polygons have distinct numbers of sides, then the largest polygon would have ( n + 3 ) sides, the next ( n + 2 ), and so on, down to the smallest, which would have 4 sides if there are no triangles.So, the total number of sides would be:[4 + 5 + 6 + ldots + (n + 3) = frac{n(n + 7)}{2}]But this must equal ( 8 + 2m ). So:[frac{n(n + 7)}{2} = 8 + 2m]But we also have ( m geq 2n - 4 ). Substituting:[frac{n(n + 7)}{2} = 8 + 2(2n - 4) = 8 + 4n - 8 = 4n]So:[frac{n(n + 7)}{2} = 4n]Multiply both sides by 2:[n(n + 7) = 8n][n^2 + 7n = 8n][n^2 - n = 0][n(n - 1) = 0]So, ( n = 0 ) or ( n = 1 ). But ( n ) must be greater than 1, as the square is divided into more than one polygon. Therefore, this is a contradiction.This means our assumption that there are no triangles is false. Therefore, there must be at least one triangle among the polygons.I think this makes sense. By assuming there are no triangles, we arrived at a contradiction, which means our assumption was wrong. Therefore, there must be at least one triangle.