💡Alright, so I have this problem here where I need to find the number of lines that pass through the point P(1,3) and also pass through points A(a,0) and B(0,b), where a and b are positive integers. The options given are A: 1 line, B: 2 lines, C: 3 lines, and D: Infinitely many lines. I need to figure out which one is correct.First, let me visualize what's going on. There's a point P at (1,3), and I need to draw lines that go through this point and also intersect the x-axis at some point A(a,0) and the y-axis at some point B(0,b). Both a and b have to be positive integers, so they can't be fractions or negative numbers.I remember that the equation of a line can be written in intercept form as x/a + y/b = 1, where a is the x-intercept and b is the y-intercept. Since the line passes through P(1,3), I can substitute x=1 and y=3 into this equation to find a relationship between a and b.So, substituting, I get:1/a + 3/b = 1Hmm, okay. Now I need to solve this equation for positive integers a and b. Let me rearrange this equation to make it easier to handle. If I multiply both sides by ab to eliminate the denominators, I get:b + 3a = abNow, let's rearrange terms to get all terms on one side:ab - 3a - b = 0Hmm, this looks a bit tricky. Maybe I can factor this somehow. Let me try adding 3 to both sides to see if that helps:ab - 3a - b + 3 = 3Now, let's factor the left side. I notice that I can factor by grouping:a(b - 3) - 1(b - 3) = 3So, factoring out (b - 3), I get:(a - 1)(b - 3) = 3Oh, that's much better! Now, since a and b are positive integers, (a - 1) and (b - 3) must also be positive integers because a and b are greater than 0. So, I need to find pairs of positive integers whose product is 3.The positive integer pairs that multiply to 3 are (1,3) and (3,1). Let me list them out:1. (a - 1) = 1 and (b - 3) = 32. (a - 1) = 3 and (b - 3) = 1Let me solve each case separately.**Case 1: (a - 1) = 1 and (b - 3) = 3**If a - 1 = 1, then a = 2.If b - 3 = 3, then b = 6.So, one solution is a = 2 and b = 6.**Case 2: (a - 1) = 3 and (b - 3) = 1**If a - 1 = 3, then a = 4.If b - 3 = 1, then b = 4.So, another solution is a = 4 and b = 4.Are there any other pairs? Well, 3 is a prime number, so its only positive integer factors are 1 and 3. Therefore, these are the only two possible pairs.Let me check if these solutions make sense.For a = 2 and b = 6:The equation of the line is x/2 + y/6 = 1.Does this pass through (1,3)? Let's check:1/2 + 3/6 = 1/2 + 1/2 = 1. Yes, it does.For a = 4 and b = 4:The equation of the line is x/4 + y/4 = 1.Does this pass through (1,3)? Let's check:1/4 + 3/4 = 1. Yes, it does.So both solutions are valid.Wait, but the problem says "the number of such lines l that can be drawn." So, each pair (a,b) corresponds to a unique line. Since we have two pairs, (2,6) and (4,4), that means there are two such lines.Let me just make sure I didn't miss any other pairs. Since 3 is prime, the only positive integer pairs are (1,3) and (3,1). So, no, there are no other pairs. Therefore, there are exactly two lines.I think that's it. So the answer should be B: 2 lines.**Final Answer**The number of such lines ( l ) that can be drawn is boxed{B}.

💡Okay, so I need to find the equation of the normal line to the curve ( y = x - x^3 ) at the point where the abscissa (which is the x-coordinate) is ( x_0 = -1 ). Hmm, let me think about how to approach this step by step.First, I remember that to find the equation of a line, I generally need a point and a slope. Since it's the normal line, it's perpendicular to the tangent line at that point. So, I think I need to find the slope of the tangent line first and then determine the slope of the normal line from that.Alright, let's start by finding the point on the curve where ( x = -1 ). To do that, I'll substitute ( x_0 = -1 ) into the equation of the curve:( y = (-1) - (-1)^3 )Calculating that, ( (-1)^3 ) is ( -1 ), so:( y = -1 - (-1) = -1 + 1 = 0 )So, the point is ( (-1, 0) ). Got that down.Next, I need the slope of the tangent line at this point. To find that, I should compute the derivative of the function ( y = x - x^3 ). The derivative, ( y' ), will give me the slope of the tangent at any point ( x ).Let's compute the derivative:( y' = frac{d}{dx}(x) - frac{d}{dx}(x^3) = 1 - 3x^2 )So, the derivative is ( y' = 1 - 3x^2 ). Now, I need to evaluate this derivative at ( x = -1 ) to get the slope of the tangent at that specific point.Substituting ( x = -1 ) into the derivative:( y' = 1 - 3(-1)^2 )Calculating ( (-1)^2 ) gives 1, so:( y' = 1 - 3(1) = 1 - 3 = -2 )So, the slope of the tangent line at ( x = -1 ) is ( -2 ).But I need the slope of the normal line, which is perpendicular to the tangent. I recall that the slopes of two perpendicular lines are negative reciprocals of each other. That means if the slope of the tangent is ( m ), then the slope of the normal is ( -1/m ).Given that the slope of the tangent is ( -2 ), the slope of the normal should be:( m_{text{normal}} = -1/(-2) = 1/2 )Okay, so the slope of the normal line is ( 1/2 ).Now, I have the slope of the normal line and a point through which it passes, which is ( (-1, 0) ). I can use the point-slope form of a line to write the equation of the normal line.The point-slope form is:( y - y_1 = m(x - x_1) )Where ( (x_1, y_1) ) is the point and ( m ) is the slope.Substituting the values I have:( y - 0 = (1/2)(x - (-1)) )Simplifying that:( y = (1/2)(x + 1) )So, the equation of the normal line is ( y = (1/2)x + 1/2 ).Wait, let me double-check my calculations to make sure I didn't make any mistakes.First, the point: ( x = -1 ), ( y = (-1) - (-1)^3 = -1 - (-1) = 0 ). That seems correct.Derivative: ( y' = 1 - 3x^2 ). At ( x = -1 ), ( y' = 1 - 3(1) = -2 ). That looks right.Slope of normal: ( -1/(-2) = 1/2 ). Correct.Point-slope form: ( y - 0 = (1/2)(x + 1) ), which simplifies to ( y = (1/2)x + 1/2 ). Yep, that seems right.I think I've got it. The equation of the normal line is ( y = frac{1}{2}x + frac{1}{2} ).**Final Answer**The equation of the normal line is boxed{y = frac{1}{2}x + frac{1}{2}}.

💡Alright, so I have this geometry problem here about an acute triangle ABC where AB is less than BC, and BH is the altitude from B to AC. Then, there's a point P which is the reflection of H over the line connecting the midpoints of AC and BC. I need to prove that the line BP passes through the circumcenter of triangle ABC. Hmm, okay, let me try to unpack this step by step.First, let me visualize the triangle. Since it's acute, all altitudes are inside the triangle. AB is less than BC, so maybe triangle ABC is such that point A is closer to B than point C is. BH is the altitude, so H is the foot of the perpendicular from B to AC. Got it.Now, the midpoints of AC and BC. Let me denote the midpoint of AC as M and the midpoint of BC as N. So, the line MN connects these two midpoints. I remember that in a triangle, the line connecting the midpoints of two sides is parallel to the third side and half its length. So, MN should be parallel to AB and half its length. Interesting.Point P is the reflection of H over the line MN. Reflection over a line means that P is such that MN is the perpendicular bisector of segment HP. So, if I were to draw MN, then H and P would be symmetric with respect to MN. That means the distance from H to MN is equal to the distance from P to MN, and the line HP is perpendicular to MN.Okay, so I need to show that BP passes through the circumcenter of triangle ABC. The circumcenter is the intersection of the perpendicular bisectors of the sides of the triangle. In an acute triangle, it lies inside the triangle. So, maybe if I can show that BP is a perpendicular bisector or relates to one in some way, that could help.Let me recall some properties of reflections in geometry. Reflecting a point over a line often relates to symmetries or midpoints. Since MN is the midline, it has some nice properties. Also, since H is the orthocenter-related point, reflecting it might have some connection to other significant points in the triangle.Wait, actually, in triangle ABC, the orthocenter is the intersection of the altitudes, so BH is one of them. But since we're reflecting H over MN, which is a midline, maybe P has some relation to the circumcircle or the circumcenter.Let me try to draw this out mentally. If I have triangle ABC, with midpoints M and N, then MN is parallel to AB. Reflecting H over MN gives me P. So, P is somewhere on the other side of MN from H. Since MN is parallel to AB, reflecting H over MN might place P in a position relative to AB.I wonder if there's a homothety or some transformation that maps H to P. Since reflection is a type of transformation, maybe properties under reflection can help.Alternatively, maybe I can use coordinate geometry. Assign coordinates to the triangle and compute everything algebraically. Let me try that approach.Let's place triangle ABC in a coordinate system. Let me set point B at the origin (0,0) for simplicity. Let me let point C be at (c,0) and point A at (a,b), where a, b, c are positive real numbers since the triangle is acute and AB < BC.Since AB < BC, the length AB is sqrt(a² + b²) and BC is c, so sqrt(a² + b²) < c. That might be useful later.Now, the midpoint M of AC is ((a + c)/2, b/2), and the midpoint N of BC is (c/2, 0). So, the line MN connects ((a + c)/2, b/2) to (c/2, 0). Let me find the equation of line MN.The slope of MN is (0 - b/2) / (c/2 - (a + c)/2) = (-b/2) / (-a/2) = (b/2) / (a/2) = b/a. So, the slope is b/a. Therefore, the equation of MN is y - 0 = (b/a)(x - c/2). So, y = (b/a)x - (b c)/(2a).Now, point H is the foot of the altitude from B to AC. Let me find the coordinates of H.The line AC connects points A(a,b) and C(c,0). The slope of AC is (0 - b)/(c - a) = -b/(c - a). Therefore, the equation of AC is y - b = (-b/(c - a))(x - a).The altitude BH is perpendicular to AC, so its slope is the negative reciprocal of the slope of AC. So, slope of BH is (c - a)/b.Since BH passes through B(0,0), its equation is y = [(c - a)/b]x.To find H, we need the intersection of BH and AC.So, set the equations equal:[(c - a)/b]x = (-b/(c - a))(x - a) + bMultiply both sides by b(c - a) to eliminate denominators:(c - a)^2 x = -b^2 (x - a) + b^2 (c - a)Expand:(c² - 2ac + a²)x = -b²x + a b² + b² c - a b²Simplify right side:- b²x + a b² + b² c - a b² = -b²x + b² cSo, left side: (c² - 2ac + a²)xRight side: -b²x + b² cBring all terms to left:(c² - 2ac + a²)x + b²x - b² c = 0Factor x:[(c² - 2ac + a²) + b²]x - b² c = 0So,x = [b² c] / [c² - 2ac + a² + b²]Similarly, y = [(c - a)/b]x = [(c - a)/b] * [b² c] / [c² - 2ac + a² + b²] = [b(c - a)c] / [c² - 2ac + a² + b²]So, H has coordinates:H_x = [b² c] / [c² - 2ac + a² + b²]H_y = [b c (c - a)] / [c² - 2ac + a² + b²]Okay, that's H.Now, I need to find the reflection of H over line MN. Let's denote this reflection as P.To find the reflection of a point over a line, I can use the formula for reflection over a line ax + by + c = 0. The formula is:If you have a point (x0, y0), its reflection over the line ax + by + c = 0 is:(x', y') = (x0 - 2a(ax0 + by0 + c)/(a² + b²), y0 - 2b(ax0 + by0 + c)/(a² + b²))First, let me write the equation of MN in standard form. Earlier, I had y = (b/a)x - (b c)/(2a). Let's rearrange this:(b/a)x - y - (b c)/(2a) = 0Multiply both sides by 2a to eliminate denominators:2b x - 2a y - b c = 0So, the standard form is 2b x - 2a y - b c = 0. Therefore, a_line = 2b, b_line = -2a, c_line = -b c.So, using the reflection formula, for point H(H_x, H_y), its reflection P(P_x, P_y) is:P_x = H_x - 2*(2b)*(2b*H_x + (-2a)*H_y + (-b c))/( (2b)^2 + (-2a)^2 )Similarly,P_y = H_y - 2*(-2a)*(2b*H_x + (-2a)*H_y + (-b c))/( (2b)^2 + (-2a)^2 )Let me compute the denominator first:(2b)^2 + (-2a)^2 = 4b² + 4a² = 4(a² + b²)Now, compute the numerator for the fraction:2b*H_x - 2a*H_y - b cLet me compute this:2b*H_x = 2b * [b² c / (c² - 2ac + a² + b²)] = 2b³ c / D, where D = c² - 2ac + a² + b²-2a*H_y = -2a * [b c (c - a) / D] = -2a b c (c - a) / D- b c = -b cSo, altogether:2b³ c / D - 2a b c (c - a)/D - b cCombine terms over D:[2b³ c - 2a b c (c - a)] / D - b cFactor out 2b c from the first two terms:2b c [b² - a(c - a)] / D - b cSimplify inside the brackets:b² - a c + a²So,2b c (a² + b² - a c) / D - b cBut D = c² - 2ac + a² + b² = (a² + b²) + c² - 2acSo, 2b c (D - c²) / D - b cWait, let me see:Wait, a² + b² - a c = D - c² + a c - a c? Hmm, maybe not. Let me just proceed step by step.So, 2b c (a² + b² - a c) / D - b cLet me factor out b c:b c [2(a² + b² - a c)/D - 1]So, that's the numerator.Therefore, the fraction is [2(a² + b² - a c)/D - 1] * b cSo, putting it back into P_x:P_x = H_x - 2*(2b)*[numerator]/[4(a² + b²)]Simplify:P_x = H_x - (4b)*[numerator]/[4(a² + b²)] = H_x - (b * numerator)/(a² + b²)Similarly for P_y:P_y = H_y - 2*(-2a)*[numerator]/[4(a² + b²)] = H_y + (4a * numerator)/(4(a² + b²)) = H_y + (a * numerator)/(a² + b²)So, let me compute numerator:numerator = 2b³ c / D - 2a b c (c - a)/D - b cWait, perhaps instead of going through all this algebra, maybe there's a smarter way.Alternatively, since MN is the midline, reflecting H over MN might have some geometric interpretation.Wait, MN is the midline, so it's parallel to AB and half its length. Reflecting H over MN would place P such that MN is the perpendicular bisector of HP.Alternatively, since MN is the midline, maybe there's a homothety or translation that maps H to P.Wait, another thought: in triangle ABC, the nine-point circle passes through the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments from each vertex to the orthocenter.Since M and N are midpoints, and H is the foot of the altitude, perhaps P is another point on the nine-point circle?But I'm not sure if that's directly helpful.Alternatively, maybe I can consider vectors.Let me denote vectors with origin at B(0,0). So, vector BA is (a, b), vector BC is (c, 0).Midpoint M of AC is ((a + c)/2, b/2), midpoint N of BC is (c/2, 0).So, vector MN is from M to N: (c/2 - (a + c)/2, 0 - b/2) = (-a/2, -b/2)So, direction vector of MN is (-a, -b). So, the line MN has direction (-a, -b). Therefore, the reflection over MN would involve some transformation.But maybe this is getting too abstract.Wait, perhaps instead of computing coordinates, I can use properties of midlines and reflections.Since MN is the midline, it's parallel to AB and half its length. So, reflecting H over MN would produce a point P such that HP is perpendicular to MN and twice the distance from H to MN.But since MN is parallel to AB, HP is perpendicular to AB as well.Wait, but BH is already perpendicular to AC, not necessarily to AB.Hmm, maybe not directly.Alternatively, since MN is the midline, reflecting H over MN would place P such that P is the midpoint of some segment related to H and another point.Wait, actually, reflecting H over MN would mean that MN is the perpendicular bisector of HP. So, P is such that HP is perpendicular to MN, and MN bisects HP.Given that MN is parallel to AB, and HP is perpendicular to MN, then HP is perpendicular to AB as well.But BH is the altitude, so BH is perpendicular to AC, not necessarily to AB.Hmm, perhaps not directly.Wait, maybe I can consider the midpoint of HP. Since MN is the perpendicular bisector, the midpoint of HP lies on MN.So, if I can find the midpoint of HP, it should lie on MN.Given that, maybe I can express the midpoint in terms of coordinates and see if it satisfies the equation of MN.But this might get too involved.Alternatively, let me think about the circumcenter.In triangle ABC, the circumcenter O is the intersection of the perpendicular bisectors of the sides.Since ABC is acute, O lies inside the triangle.If I can show that BP passes through O, then I'm done.Alternatively, maybe I can show that O lies on BP by showing that O satisfies the equation of BP.But perhaps that's too computational.Wait, another idea: in triangle ABC, the reflection of the orthocenter over the midline MN lies on the circumcircle. Is that a known property?Wait, reflecting the orthocenter over the midline MN might give a point on the circumcircle. If that's the case, then P would lie on the circumcircle, and BP would pass through O.But I'm not sure about this property. Maybe I need to verify.Alternatively, perhaps I can use the fact that the reflection of H over MN is P, and then show that BP is the Euler line or something similar.Wait, the Euler line passes through the orthocenter, centroid, and circumcenter. But I don't know if BP is the Euler line.Alternatively, maybe I can consider the midpoint of BH. Since MN is the midline, reflecting H over MN might relate to the midpoint.Wait, let me think differently.Since MN is the midline, it's the line connecting midpoints of AC and BC. So, it's parallel to AB and half its length.Reflecting H over MN gives P. So, P is such that MN is the perpendicular bisector of HP.Therefore, HP is perpendicular to MN, and the midpoint of HP lies on MN.Given that MN is parallel to AB, then HP is perpendicular to AB as well.Wait, but BH is the altitude, so BH is perpendicular to AC, not AB.Hmm, maybe not directly.Alternatively, since MN is parallel to AB, reflecting H over MN would place P such that BP is symmetric with respect to MN.Wait, maybe I can consider triangle BHP. Since P is the reflection of H over MN, and MN is the midline, perhaps BP is related to some median or symmedian.Alternatively, perhaps I can use vectors.Let me denote the position vectors of points with respect to B as the origin.So, vector BA = A - B = (a, b), vector BC = C - B = (c, 0).Midpoint M of AC: (A + C)/2 = ((a + c)/2, b/2)Midpoint N of BC: (B + C)/2 = (c/2, 0)So, vector MN = N - M = (c/2 - (a + c)/2, 0 - b/2) = (-a/2, -b/2)So, the line MN has direction vector (-a, -b). Therefore, the reflection over MN can be represented as a linear transformation.But reflections are a bit involved. Maybe instead, I can use the formula for reflection over a line in vector form.Alternatively, since MN is the midline, reflecting H over MN would place P such that P is the midpoint of some segment.Wait, another approach: since MN is the midline, reflecting H over MN would map H to P such that P is the midpoint of some other segment related to H.Wait, actually, in triangle ABC, the nine-point circle passes through the midpoints of the sides, the feet of the altitudes, and the midpoints of the segments from each vertex to the orthocenter.Since H is the foot of the altitude, and M and N are midpoints, perhaps P is another point on the nine-point circle.But I'm not sure how that helps with BP passing through the circumcenter.Alternatively, maybe I can consider that reflecting H over MN gives P, and then BP is the reflection of BH over MN.But BH is the altitude, so reflecting it over MN might give a line that passes through the circumcenter.Alternatively, perhaps I can use the fact that the circumcenter lies on the perpendicular bisector of AB and BC.Wait, let me think about the circumcenter O.In triangle ABC, the circumcenter O is the intersection of the perpendicular bisectors of AB, BC, and AC.Since MN is the midline parallel to AB, maybe reflecting H over MN relates to the perpendicular bisector.Alternatively, perhaps I can use coordinate geometry to find O and then show that it lies on BP.Let me try that.Given triangle ABC with coordinates:B(0,0), A(a,b), C(c,0)The circumcenter O is the intersection of the perpendicular bisectors of AB and BC.First, find the perpendicular bisector of AB.Midpoint of AB: (a/2, b/2)Slope of AB: (b - 0)/(a - 0) = b/aTherefore, the perpendicular bisector has slope -a/b.Equation: y - b/2 = (-a/b)(x - a/2)Similarly, find the perpendicular bisector of BC.Midpoint of BC: (c/2, 0)Slope of BC: (0 - 0)/(c - 0) = 0, so it's a horizontal line.Therefore, the perpendicular bisector is vertical, with undefined slope, so it's the line x = c/2.So, the perpendicular bisector of BC is x = c/2.Now, find the intersection point O of the two perpendicular bisectors.From the perpendicular bisector of BC: x = c/2Plug into the perpendicular bisector of AB:y - b/2 = (-a/b)(c/2 - a/2) = (-a/b)((c - a)/2) = (-a(c - a))/(2b)Therefore,y = b/2 + (-a(c - a))/(2b) = [b² - a(c - a)]/(2b) = [b² - a c + a²]/(2b)So, circumcenter O is at (c/2, [a² + b² - a c]/(2b))Now, I need to find the equation of BP and see if O lies on it.Point B is at (0,0), and point P is the reflection of H over MN.I already have coordinates for H:H_x = [b² c] / D, where D = c² - 2ac + a² + b²H_y = [b c (c - a)] / DNow, I need to find P, the reflection of H over MN.As before, the equation of MN is 2b x - 2a y - b c = 0Using the reflection formula:P_x = H_x - 2*(2b)*(2b H_x - 2a H_y - b c)/(4a² + 4b²)Similarly,P_y = H_y - 2*(-2a)*(2b H_x - 2a H_y - b c)/(4a² + 4b²)Simplify denominator: 4(a² + b²)Compute numerator: 2b H_x - 2a H_y - b cLet me compute this:2b H_x = 2b * [b² c / D] = 2b³ c / D-2a H_y = -2a * [b c (c - a) / D] = -2a b c (c - a) / D- b c = -b cSo, numerator = 2b³ c / D - 2a b c (c - a)/D - b cCombine terms over D:[2b³ c - 2a b c (c - a)] / D - b cFactor out 2b c:2b c [b² - a(c - a)] / D - b cSimplify inside the brackets:b² - a c + a²So,2b c (a² + b² - a c) / D - b cFactor out b c:b c [2(a² + b² - a c)/D - 1]So, numerator = b c [2(a² + b² - a c)/D - 1]Now, plug back into P_x and P_y:P_x = H_x - (4b * numerator)/(4(a² + b²)) = H_x - (b * numerator)/(a² + b²)Similarly,P_y = H_y + (4a * numerator)/(4(a² + b²)) = H_y + (a * numerator)/(a² + b²)So, compute P_x:P_x = [b² c / D] - [b * b c (2(a² + b² - a c)/D - 1)] / (a² + b²)Simplify:P_x = [b² c / D] - [b² c (2(a² + b² - a c)/D - 1)] / (a² + b²)Similarly for P_y:P_y = [b c (c - a) / D] + [a * b c (2(a² + b² - a c)/D - 1)] / (a² + b²)This is getting really messy. Maybe there's a simplification.Wait, let me note that D = c² - 2ac + a² + b² = (a² + b²) + c² - 2acSo, D = (a² + b²) + (c² - 2ac) = (a² + b²) + (c - a)^2 - a² = Hmm, not sure.Alternatively, let me denote S = a² + b², so D = S + c² - 2acSo, numerator becomes b c [2(S - a c)/D - 1] = b c [2(S - a c - D/2)]Wait, not sure.Alternatively, let me compute the term inside the brackets:2(a² + b² - a c)/D - 1 = 2(S - a c)/D - 1But S = a² + b², so:= 2(S - a c)/D - 1 = 2(S - a c)/D - D/D = [2(S - a c) - D]/DBut D = S + c² - 2acSo,2(S - a c) - D = 2S - 2a c - S - c² + 2a c = S - c²Therefore,2(a² + b² - a c)/D - 1 = (S - c²)/DSo, numerator = b c (S - c²)/DTherefore,P_x = [b² c / D] - [b * b c (S - c²)/D] / (a² + b²)= [b² c / D] - [b² c (S - c²)] / [D (a² + b²)]Similarly,P_y = [b c (c - a) / D] + [a * b c (S - c²)] / [D (a² + b²)]Now, let's compute P_x:Factor out b² c / D:P_x = [b² c / D] [1 - (S - c²)/(a² + b²)]But S = a² + b², so:P_x = [b² c / D] [1 - (S - c²)/S] = [b² c / D] [ (S - (S - c²)) / S ] = [b² c / D] [c² / S] = [b² c * c²] / [D S] = [b² c³] / [D S]Similarly, compute P_y:P_y = [b c (c - a) / D] + [a b c (S - c²)] / [D S]Factor out b c / D:P_y = [b c / D] [ (c - a) + a (S - c²)/S ]= [b c / D] [ (c - a) + a (1 - c²/S) ]= [b c / D] [c - a + a - a c²/S ]= [b c / D] [c - a c²/S ]= [b c / D] [c (1 - a c/S ) ]= [b c² / D] [1 - a c/S ]But S = a² + b², so:= [b c² / D] [1 - a c/(a² + b²) ]= [b c² / D] [ (a² + b² - a c)/ (a² + b²) ]= [b c² (a² + b² - a c) ] / [ D (a² + b²) ]But D = a² + b² + c² - 2ac, so:= [b c² (S - a c) ] / [ (S + c² - 2ac) S ]Hmm, not sure if this helps.Alternatively, maybe I can find the equation of BP and see if O lies on it.Point B is (0,0), and point P is (P_x, P_y). So, the parametric equation of BP is t*(P_x, P_y), where t is a parameter.So, any point on BP can be written as (t P_x, t P_y).Now, the circumcenter O is at (c/2, [S - a c]/(2b)).So, to check if O lies on BP, we need to see if there exists a t such that:t P_x = c/2t P_y = [S - a c]/(2b)So, from the first equation, t = c/(2 P_x)Plug into the second equation:(c/(2 P_x)) P_y = [S - a c]/(2b)Simplify:(c P_y)/(2 P_x) = [S - a c]/(2b)Multiply both sides by 2:(c P_y)/P_x = (S - a c)/bSo, we need to check if (c P_y)/P_x = (S - a c)/bFrom earlier, we have expressions for P_x and P_y.From P_x:P_x = [b² c³] / [D S]From P_y:P_y = [b c² (S - a c) ] / [ D S ]So, compute (c P_y)/P_x:(c * [b c² (S - a c) / (D S)]) / [b² c³ / (D S) ] =(c * b c² (S - a c) / (D S)) * (D S / b² c³) =Simplify:(c * b c² (S - a c) / (D S)) * (D S / b² c³) =The D S cancels out, and c * c² / c³ = 1, b / b² = 1/bSo, we have:(S - a c)/bWhich is exactly the right-hand side.Therefore, (c P_y)/P_x = (S - a c)/bWhich means that O lies on BP.Therefore, BP passes through the circumcenter O.So, after a lot of coordinate computations, I've shown that O lies on BP, hence BP passes through the circumcenter.**Final Answer**boxed{BP text{ passes through the circumcenter of } triangle ABC}

💡Okay, so I have this problem about three friends who went mushroom picking and found a total of 55 mushrooms. They made a dish and divided it into four equal portions, inviting their friend Pepa to join them. Líba contributed six of her mushrooms, Maruška eight, and Sárka five. After that, each of the three friends had the same number of mushrooms left. Then, Pepa gave them a box of chocolates with 38 chocolates and told them to divide it fairly according to their contributions to his meal.Alright, let's break this down. First, I need to figure out how many mushrooms each friend found. Then, I need to determine how they should divide the chocolates based on their contributions.Starting with the first part: How many mushrooms did each friend find?They found a total of 55 mushrooms. Líba contributed 6, Maruška 8, and Sárka 5 to the dish. So, the total number of mushrooms used in the dish is 6 + 8 + 5 = 19 mushrooms.That means the remaining mushrooms are 55 - 19 = 36 mushrooms. These 36 mushrooms are equally divided among the three friends, so each friend has 36 / 3 = 12 mushrooms left.Now, to find out how many mushrooms each friend originally found, we need to add the mushrooms they contributed to the dish to the mushrooms they have left.- Líba contributed 6 mushrooms and has 12 left, so she found 6 + 12 = 18 mushrooms.- Maruška contributed 8 mushrooms and has 12 left, so she found 8 + 12 = 20 mushrooms.- Sárka contributed 5 mushrooms and has 12 left, so she found 5 + 12 = 17 mushrooms.Okay, that seems straightforward. Now, moving on to the second part: How should they divide the chocolates according to Pepa's suggestion?Pepa gifted them 38 chocolates and told them to divide it fairly according to their contributions to his meal. So, we need to figure out how much each friend contributed to Pepa's portion.First, let's understand how the dish was divided. They made a mushroom dish and divided it into four equal portions. So, each portion is 1/4 of the total dish.The total number of mushrooms used in the dish is 19, as we calculated earlier. So, each portion is 19 / 4 = 4.75 mushrooms.Now, each friend contributed some mushrooms to the dish:- Líba contributed 6 mushrooms.- Maruška contributed 8 mushrooms.- Sárka contributed 5 mushrooms.But each of them also ate one portion of the dish, which is 4.75 mushrooms. So, the amount of mushrooms that went into Pepa's portion can be calculated by subtracting the mushrooms they ate from the total they contributed.- Líba contributed 6 mushrooms and ate 4.75, so her contribution to Pepa's portion is 6 - 4.75 = 1.25 mushrooms.- Maruška contributed 8 mushrooms and ate 4.75, so her contribution to Pepa's portion is 8 - 4.75 = 3.25 mushrooms.- Sárka contributed 5 mushrooms and ate 4.75, so her contribution to Pepa's portion is 5 - 4.75 = 0.25 mushrooms.Now, we have the contributions to Pepa's portion:- Líba: 1.25 mushrooms- Maruška: 3.25 mushrooms- Sárka: 0.25 mushroomsTo divide the chocolates fairly, we need to find the ratio of their contributions. Let's express these contributions as fractions to make it easier:- Líba: 1.25 = 5/4- Maruška: 3.25 = 13/4- Sárka: 0.25 = 1/4So, the ratio of their contributions is 5/4 : 13/4 : 1/4. To simplify this ratio, we can multiply each part by 4 to eliminate the denominators:5 : 13 : 1Now, the total number of chocolates is 38. We need to divide these chocolates according to the ratio 5 : 13 : 1.First, let's find the total number of parts in the ratio:5 + 13 + 1 = 19 partsEach part is equal to 38 chocolates / 19 parts = 2 chocolates per part.Now, we can calculate how many chocolates each friend gets:- Líba: 5 parts * 2 = 10 chocolates- Maruška: 13 parts * 2 = 26 chocolates- Sárka: 1 part * 2 = 2 chocolatesSo, Líba gets 10 chocolates, Maruška gets 26 chocolates, and Sárka gets 2 chocolates.Wait a minute, does this make sense? Sárka contributed the least to Pepa's portion, so she gets the least chocolates, which seems fair. Maruška contributed the most, so she gets the most chocolates. Líba is in the middle. Yeah, that seems fair based on their contributions.I think that's it. So, summarizing:1. Each friend found: - Líba: 18 mushrooms - Maruška: 20 mushrooms - Sárka: 17 mushrooms2. The chocolates should be divided as: - Líba: 10 chocolates - Maruška: 26 chocolates - Sárka: 2 chocolatesI don't see any mistakes in my calculations, but let me double-check.For the first part:Total mushrooms found: 55Mushrooms used in dish: 6 + 8 + 5 = 19Mushrooms left: 55 - 19 = 36Mushrooms left per friend: 36 / 3 = 12Mushrooms found by each friend:- Líba: 12 + 6 = 18- Maruška: 12 + 8 = 20- Sárka: 12 + 5 = 17Yes, that adds up to 55.For the second part:Mushrooms per portion: 19 / 4 = 4.75Contributions to Pepa's portion:- Líba: 6 - 4.75 = 1.25- Maruška: 8 - 4.75 = 3.25- Sárka: 5 - 4.75 = 0.25Ratio: 1.25 : 3.25 : 0.25 = 5 : 13 : 1Total parts: 5 + 13 + 1 = 19Chocolates per part: 38 / 19 = 2Chocolates per friend:- Líba: 5 * 2 = 10- Maruška: 13 * 2 = 26- Sárka: 1 * 2 = 2Yes, that adds up to 38.So, I think my answers are correct.

💡Okay, so I've got this function f(x) = x² - 4x + 5 - a/eˣ, where a is a real number. There are two parts to this problem. Let me tackle them one by one.Starting with part (I): If f(x) is monotonically increasing on (-∞, +∞), find the range of values for a.Alright, so if a function is monotonically increasing, its derivative should be non-negative everywhere. So, I need to find the derivative of f(x) and ensure it's always greater than or equal to zero.Let me compute f'(x). The derivative of x² is 2x, the derivative of -4x is -4, the derivative of 5 is 0, and the derivative of -a/eˣ is a/eˣ because the derivative of eˣ is eˣ, and with the negative exponent, it becomes positive. So, putting it all together:f'(x) = 2x - 4 + a/eˣNow, for f(x) to be monotonically increasing, f'(x) ≥ 0 for all x ∈ ℝ. So, we have:2x - 4 + a/eˣ ≥ 0 for all xI need to find the range of a such that this inequality holds for all real numbers x.Let me rearrange the inequality:a/eˣ ≥ -2x + 4Multiply both sides by eˣ (which is always positive, so the inequality sign doesn't change):a ≥ (-2x + 4)eˣSo, a must be greater than or equal to (-2x + 4)eˣ for all x. To find the range of a, I need to find the maximum value of the function h(x) = (-2x + 4)eˣ because a has to be at least as big as the maximum of h(x).Let me define h(x) = (-2x + 4)eˣ. To find its maximum, I can take the derivative and set it equal to zero.Compute h'(x):h'(x) = derivative of (-2x + 4)eˣUsing the product rule: derivative of (-2x + 4) is -2, times eˣ, plus (-2x + 4) times derivative of eˣ, which is eˣ.So,h'(x) = (-2)eˣ + (-2x + 4)eˣ = (-2 - 2x + 4)eˣ = (-2x + 2)eˣSet h'(x) = 0:(-2x + 2)eˣ = 0Since eˣ is never zero, we can divide both sides by eˣ:-2x + 2 = 0Solving for x:-2x + 2 = 0 ⇒ -2x = -2 ⇒ x = 1So, the critical point is at x = 1. Now, let's check if this is a maximum or a minimum.Looking at the sign of h'(x):- For x < 1, say x = 0: h'(0) = (-0 + 2)e⁰ = 2 > 0. So, h(x) is increasing on (-∞, 1).- For x > 1, say x = 2: h'(2) = (-4 + 2)e² = (-2)e² < 0. So, h(x) is decreasing on (1, ∞).Therefore, x = 1 is a maximum point.Compute h(1):h(1) = (-2*1 + 4)e¹ = (2)e = 2eSo, the maximum value of h(x) is 2e. Therefore, for a to satisfy a ≥ h(x) for all x, a must be at least 2e.Hence, the range of a is [2e, ∞).Alright, that was part (I). Now, moving on to part (II):Let g(x) = eˣ f(x). When m ≥ 1, if g(x₁) + g(x₂) = 2g(m), and x₁ ≠ x₂, prove that x₁ + x₂ < 2m.Hmm, okay. Let's first write down what g(x) is.Given f(x) = x² - 4x + 5 - a/eˣ, so:g(x) = eˣ f(x) = eˣ (x² - 4x + 5 - a/eˣ) = (x² - 4x + 5)eˣ - aSo, g(x) = (x² - 4x + 5)eˣ - aWe are given that g(x₁) + g(x₂) = 2g(m), with x₁ ≠ x₂, and m ≥ 1. We need to prove that x₁ + x₂ < 2m.Let me think about this. It seems like we might need to use some properties of the function g(x). Maybe its convexity or monotonicity.First, let me compute the derivative of g(x) to see its behavior.g(x) = (x² - 4x + 5)eˣ - aCompute g'(x):g'(x) = derivative of (x² - 4x + 5)eˣ - derivative of aThe derivative of a is zero. For the first term, use the product rule:Derivative of (x² - 4x + 5) is (2x - 4), times eˣ, plus (x² - 4x + 5) times derivative of eˣ, which is eˣ.So,g'(x) = (2x - 4)eˣ + (x² - 4x + 5)eˣ = [2x - 4 + x² - 4x + 5]eˣ = (x² - 2x + 1)eˣSimplify x² - 2x + 1: that's (x - 1)².So, g'(x) = (x - 1)² eˣSince (x - 1)² is always non-negative and eˣ is always positive, g'(x) is always non-negative. Therefore, g(x) is monotonically increasing on ℝ.Wait, but (x - 1)² eˣ is zero only at x = 1, right? So, g(x) is increasing everywhere, but with a stationary point at x = 1.So, g(x) is increasing on (-∞, ∞), with a critical point at x = 1 where the derivative is zero.Given that, let's think about the equation g(x₁) + g(x₂) = 2g(m). Since g is increasing, if x₁ < x₂, then g(x₁) < g(x₂). But since x₁ ≠ x₂, we can assume without loss of generality that x₁ < x₂.But wait, the problem says x₁ ≠ x₂, but doesn't specify which is larger. Hmm.But given that g is increasing, if x₁ < m < x₂, then g(x₁) < g(m) < g(x₂). So, if we have g(x₁) + g(x₂) = 2g(m), it's like the average of g(x₁) and g(x₂) is g(m). So, maybe x₁ and x₂ are symmetric around m? But we need to show that x₁ + x₂ < 2m.Wait, so if x₁ and x₂ were symmetric around m, then x₁ + x₂ = 2m, but since the function is increasing, maybe the sum is less? Hmm.Alternatively, perhaps using convexity. Let me check if g(x) is convex.Compute the second derivative g''(x):We have g'(x) = (x - 1)² eˣSo, g''(x) = derivative of (x - 1)² eˣAgain, use the product rule:Derivative of (x - 1)² is 2(x - 1), times eˣ, plus (x - 1)² times derivative of eˣ, which is eˣ.So,g''(x) = 2(x - 1)eˣ + (x - 1)² eˣ = (x - 1)eˣ [2 + (x - 1)] = (x - 1)(x + 1)eˣWait, let me compute that again:Wait, 2(x - 1) + (x - 1)² = (x - 1)(2 + x - 1) = (x - 1)(x + 1). So, yes, g''(x) = (x - 1)(x + 1)eˣ.So, g''(x) = (x² - 1)eˣ.Therefore, for x > 1, g''(x) > 0 because x² - 1 > 0 and eˣ > 0. For x < -1, g''(x) > 0 as well because x² - 1 > 0. For x between -1 and 1, g''(x) < 0.So, g(x) is convex on (-∞, -1) and (1, ∞), and concave on (-1, 1).But since m ≥ 1, we might be dealing with x in the convex region.Wait, but the problem says m ≥ 1, so perhaps x₁ and x₂ are greater than or equal to 1? Not necessarily, because x₁ and x₂ could be on either side of m.But given that g is increasing, if m ≥ 1, and if x₁ < m < x₂, then x₁ could be less than 1, and x₂ greater than 1.But since m is at least 1, maybe x₂ is greater than m, which is at least 1.Hmm, not sure yet.Let me think about the equation g(x₁) + g(x₂) = 2g(m). Since g is increasing, if x₁ < m < x₂, then g(x₁) < g(m) < g(x₂). So, the average of g(x₁) and g(x₂) is g(m). So, maybe x₁ and x₂ are symmetric around m in some way, but not necessarily.But we need to show that x₁ + x₂ < 2m.Wait, maybe we can use the concept of convexity or concavity.Alternatively, maybe consider the function φ(x) = g(x) + g(2m - x). Let me see.Wait, let me try to define a function that relates x₁ and x₂.Alternatively, let me consider that since g is increasing, if x₁ < m < x₂, then 2m - x₂ < m - x₁. Hmm, not sure.Wait, another approach: Let me define x₂ = 2m - x₁ + t, where t is some positive number. Then, if I can show that t > 0, then x₁ + x₂ = 2m + t > 2m, but we need to show the opposite. Wait, maybe not.Alternatively, perhaps consider the function h(x) = g(x) + g(2m - x). Let's compute h(m):h(m) = g(m) + g(2m - m) = g(m) + g(m) = 2g(m)So, h(m) = 2g(m). Now, if x₁ and x₂ are such that x₂ = 2m - x₁, then h(x₁) = g(x₁) + g(2m - x₁) = 2g(m). So, if x₂ = 2m - x₁, then h(x₁) = 2g(m). But in our case, we have g(x₁) + g(x₂) = 2g(m), so if x₂ ≠ 2m - x₁, then perhaps h(x₁) ≠ 2g(m). Hmm, not sure.Wait, maybe I can consider the function h(x) = g(x) + g(2m - x). Let's compute its derivative:h'(x) = g'(x) - g'(2m - x)Since g'(x) = (x - 1)² eˣ, which is always non-negative, and it's symmetric in a way.Wait, let me compute h'(x):h'(x) = g'(x) - g'(2m - x) = (x - 1)² eˣ - (2m - x - 1)² e^{2m - x}Simplify the second term:(2m - x - 1)² = (2m - 1 - x)² = (x - (2m - 1))²So,h'(x) = (x - 1)² eˣ - (x - (2m - 1))² e^{2m - x}Now, let's analyze the sign of h'(x):Since m ≥ 1, and x is somewhere around m, maybe.But this seems complicated. Maybe another approach.Wait, let's think about the function φ(x) = g(x) - g(2m - x). If we can show that φ(x) is increasing or something.Alternatively, maybe consider the function F(x) = g(x) + g(2m - x). We know that F(m) = 2g(m). If we can show that F(x) is increasing for x > m, then since x₂ > m, F(x₂) > F(m) = 2g(m), which would imply that g(x₂) + g(2m - x₂) > 2g(m). But we have g(x₁) + g(x₂) = 2g(m). Hmm, maybe not directly.Wait, let me try to consider that since g is increasing, if x₁ < m, then g(x₁) < g(m). Similarly, if x₂ > m, then g(x₂) > g(m). So, the sum g(x₁) + g(x₂) would be less than g(m) + g(x₂). But we have g(x₁) + g(x₂) = 2g(m), so g(x₂) must be greater than g(m), and g(x₁) must be less than g(m). So, x₂ > m and x₁ < m.Now, let me define x₁ = m - t and x₂ = m + s, where t, s > 0. Then, x₁ + x₂ = 2m + (s - t). We need to show that s - t < 0, i.e., s < t.But how?Alternatively, maybe consider the function φ(x) = g(x) + g(2m - x). We know that φ(m) = 2g(m). If we can show that φ(x) is increasing for x > m, then for x > m, φ(x) > φ(m) = 2g(m). But in our case, g(x₁) + g(x₂) = 2g(m). If x₂ > m, then x₁ = 2m - x₂ < m. So, φ(x₂) = g(x₂) + g(2m - x₂) = g(x₂) + g(x₁) = 2g(m). But if φ(x) is increasing for x > m, then x₂ must be less than m? Wait, that contradicts.Wait, maybe I need to think differently.Let me consider the function h(x) = g(m + x) + g(m - x). Let's compute h(0):h(0) = g(m) + g(m) = 2g(m)Now, compute h'(x):h'(x) = g'(m + x) - g'(m - x)Since g'(x) = (x - 1)² eˣ, so:h'(x) = (m + x - 1)² e^{m + x} - (m - x - 1)² e^{m - x}Simplify the terms:(m + x - 1)² = (m - 1 + x)²(m - x - 1)² = (m - 1 - x)²So,h'(x) = (m - 1 + x)² e^{m + x} - (m - 1 - x)² e^{m - x}Let me factor out e^{m}:h'(x) = e^{m} [ (m - 1 + x)² e^{x} - (m - 1 - x)² e^{-x} ]Now, let me denote y = x, so:h'(x) = e^{m} [ (m - 1 + y)² e^{y} - (m - 1 - y)² e^{-y} ]Let me compute this expression:Let me expand (m - 1 + y)² and (m - 1 - y)²:(m - 1 + y)² = (m - 1)² + 2(m - 1)y + y²(m - 1 - y)² = (m - 1)² - 2(m - 1)y + y²So,(m - 1 + y)² e^{y} - (m - 1 - y)² e^{-y} = [ (m - 1)² + 2(m - 1)y + y² ] e^{y} - [ (m - 1)² - 2(m - 1)y + y² ] e^{-y}Let me group terms:= (m - 1)² (e^{y} - e^{-y}) + 2(m - 1)y (e^{y} + e^{-y}) + y² (e^{y} - e^{-y})Hmm, interesting. Notice that e^{y} - e^{-y} = 2 sinh y, and e^{y} + e^{-y} = 2 cosh y.So,= (m - 1)² * 2 sinh y + 2(m - 1)y * 2 cosh y + y² * 2 sinh y= 2 sinh y [ (m - 1)² + y² ] + 4(m - 1)y cosh ySo, h'(x) = e^{m} [ 2 sinh y ( (m - 1)² + y² ) + 4(m - 1)y cosh y ]Since y = x, and x is in (0, ∞) because we're considering x > 0 (since x₂ > m, x = x₂ - m > 0).Now, sinh y and cosh y are both positive for y > 0. Also, (m - 1)² + y² is positive, and (m - 1)y is non-negative since m ≥ 1 and y > 0.Therefore, each term in the expression is positive, so h'(x) > 0 for x > 0.This means that h(x) is increasing for x > 0.Given that h(0) = 2g(m), and h(x) is increasing, for x > 0, h(x) > 2g(m). So, if we have g(x₁) + g(x₂) = 2g(m), and x₂ > m, then x₁ must be less than m, and the sum x₁ + x₂ must be less than 2m.Wait, let me see:Suppose x₂ > m, then x₁ = 2m - x₂ < m. But we have g(x₁) + g(x₂) = 2g(m). If we consider h(x) = g(m + x) + g(m - x), which is increasing, then for x > 0, h(x) > h(0) = 2g(m). So, if we have g(x₁) + g(x₂) = 2g(m), and x₂ > m, then x₁ must be such that x₂ - m < m - x₁, because otherwise, h(x) would be greater than 2g(m). Therefore, x₂ - m < m - x₁ ⇒ x₂ + x₁ < 2m.Yes, that makes sense.Alternatively, since h(x) is increasing, and h(x) = g(m + x) + g(m - x) > 2g(m) for x > 0, then if we have g(x₁) + g(x₂) = 2g(m), and x₂ > m, then x₁ must be such that x₂ - m < m - x₁, leading to x₁ + x₂ < 2m.Therefore, we've shown that x₁ + x₂ < 2m.So, to summarize:For part (I), we found that a must be at least 2e, so a ∈ [2e, ∞).For part (II), by analyzing the function g(x) and its properties, we showed that x₁ + x₂ < 2m.**Final Answer**(I) The range of values for ( a ) is (boxed{[2e, +infty)}).(II) It is proven that ( x_{1} + x_{2} < 2m ), so the final result is (boxed{x_{1} + x_{2} < 2m}).

💡First, I identify the length of each CD based on the given information.The first two CDs are each 1.5 hours long.The third CD is twice as long as the first CD, so it is 1.5 hours multiplied by 2, which equals 3 hours.The fourth CD is half the length of the second CD, so it is 1.5 hours divided by 2, resulting in 0.75 hours.The fifth CD is equal to the combined length of the first and second CDs, which is 1.5 hours plus 1.5 hours, totaling 3 hours.Finally, I add up the lengths of all five CDs: 1.5 hours + 1.5 hours + 3 hours + 0.75 hours + 3 hours, which equals 9.75 hours.