🧠:Okay, let's see. I need to find the minimum value of the ratio of the area of triangle ABC to the area of the inscribed square PQRS, given that AB is length a (a constant) and angle B is theta, which can vary. Hmm, let's start by understanding the problem step by step.First, triangle ABC is a right triangle with angle A at 90 degrees. So sides AB and AC are the legs, and BC is the hypotenuse. The square PQRS is inscribed in this triangle such that one of its sides, RS, lies on BC. The points P and Q are on AB and AC respectively. So the square is touching all three sides of the triangle: AB, AC, and BC.Given that AB = a, and angle B is theta. Since angle A is 90 degrees, angle C must be 90 - theta degrees. The hypotenuse BC will then have a length that depends on theta. Wait, but AB is fixed at a. So if angle B is theta, then the sides can be expressed in terms of theta and a. Let me recall the trigonometric ratios in a right triangle.In triangle ABC, angle B is theta, so:- AB = adjacent side to angle B = a- BC = hypotenuse = AB / cos(theta) = a / cos(theta)- AC = opposite side to angle B = AB * tan(theta) = a * tan(theta)So the sides are AB = a, AC = a tan(theta), BC = a / cos(theta). Therefore, the area of triangle ABC is (AB * AC)/2 = (a * a tan(theta))/2 = (a² tan(theta))/2.Now, the square PQRS is inscribed in triangle ABC with RS on BC, P on AB, and Q on AC. I need to find the area of the square PQRS in terms of a and theta, and then compute the ratio S_ABC / S_PQRS and find its minimum value as theta varies.First, let's visualize the square inside the triangle. The square has one side on the hypotenuse BC. Points P and Q are on the legs AB and AC, respectively. So the square touches all three sides: AB, AC, and BC.Let me denote the side length of the square as s. Then RS = s, and since RS is on BC, the square extends from BC towards the right angle A. Points P and Q are on AB and AC, respectively. So the square divides the triangle ABC into two smaller right triangles and the square itself.Let me try to model this. Let's set up a coordinate system with point A at the origin (0,0), AB along the x-axis, and AC along the y-axis. Therefore:- Point A is (0,0)- Point B is (a,0)- Point C is (0, a tan(theta)) because AC = a tan(theta)Hypotenuse BC goes from (a,0) to (0, a tan(theta)). The square PQRS has side RS on BC. Let's denote the coordinates of the square's vertices:- R and S are on BC. Let's say S is closer to B and R is closer to C.- P is on AB, so somewhere between A and B.- Q is on AC, so somewhere between A and C.Since the square has side length s, the coordinates of the points can be determined based on the side length.Alternatively, maybe it's easier to use similar triangles here. When a square is inscribed in a right triangle with one side on the hypotenuse, the two smaller triangles formed are similar to the original triangle.Let me recall that when a square is inscribed in a right triangle with one side on the hypotenuse, the side length of the square can be found using the formula:s = (a b c) / (a^2 + b^2 + a b)But wait, that might not be the case here. Maybe I need to derive it.Alternatively, consider the coordinates.Let me parameterize BC. The equation of BC can be found. Since it connects (a, 0) to (0, a tan(theta)), the slope is (a tan(theta) - 0)/(0 - a) = -tan(theta). So the equation of BC is y = -tan(theta) x + a tan(theta).Now, the square has its side RS along BC. Let's suppose that the square has side length s. Let's find the coordinates of points R and S.Since RS is a horizontal or vertical line? Wait, no. RS is a side of the square lying on BC. But BC is the hypotenuse, which is not aligned with the axes. Wait, but the square has sides perpendicular to each other. Since RS is on BC, then the square must have sides perpendicular and parallel to BC. Wait, but in the problem statement, it's a square with side RS on BC, and points P on AB and Q on AC. So the square is oriented such that RS is along BC, and the other sides PQ, QR, and SP are perpendicular to BC? Wait, no. Because in a square, adjacent sides are perpendicular. So if RS is on BC, then the adjacent sides (say SP and RQ) must be perpendicular to BC.But BC has a slope of -tan(theta). Therefore, the sides SP and RQ must have a slope of 1/tan(theta), since they are perpendicular to BC. Wait, actually, the slope of BC is -tan(theta), so the perpendicular slope would be cot(theta). Because the product of the slopes of two perpendicular lines is -1. So if the slope of BC is -tan(theta), then the slope of the perpendicular line is 1/tan(theta) = cot(theta).So, the sides SP and RQ of the square are perpendicular to BC, hence have a slope of cot(theta). But since it's a square, those sides must have length s. Therefore, moving from points S and R along the direction perpendicular to BC (slope cot(theta)) for length s would reach points P and Q.But this might be getting complicated. Maybe using coordinate geometry here would help. Let's assign coordinates as follows:- Let’s place point A at (0,0), B at (a,0), and C at (0, a tan(theta)) as before.The hypotenuse BC has endpoints at (a,0) and (0, a tan(theta)). Let’s parameterize BC. Let’s let S be a point on BC. Let’s parameterize BC from B to C. Let’s let t be the parameter such that when t=0, we are at B (a,0), and when t=1, we are at C (0, a tan(theta)). So the coordinates of a general point on BC can be written as:x = a - a ty = 0 + a tan(theta) tSo for t in [0,1], (x,y) = (a(1 - t), a tan(theta) t)Therefore, the point S can be represented as (a(1 - t), a tan(theta) t) for some t, and point R would be another point on BC further along towards C. Since RS is a side of the square, the distance between S and R should be s. However, since BC has length sqrt(a² + (a tan(theta))²) = a sqrt(1 + tan²(theta)) = a sec(theta). Therefore, the length BC is a sec(theta), which matches earlier calculation.But the square's side RS is along BC, so the distance from S to R along BC is s. However, since BC is the hypotenuse, the actual Euclidean distance between S and R is s, but along BC, the length would be s. Wait, but BC is a straight line. Wait, but the side RS is a side of the square lying on BC. Therefore, RS is a segment of BC with length s. Therefore, the Euclidean distance between S and R is s. However, the coordinates of S and R need to be s apart along BC.But since BC is length a sec(theta), the parameter t corresponds to length along BC. The total length is a sec(theta), so a small change dt in the parameter t corresponds to a length of a sec(theta) dt. Wait, maybe not. Let's see.Alternatively, the parametric equations for BC are (x(t), y(t)) = (a(1 - t), a tan(theta) t) for t in [0,1]. The derivative dx/dt = -a, dy/dt = a tan(theta). Therefore, the speed along the parameter t is sqrt( (dx/dt)^2 + (dy/dt)^2 ) = sqrt(a² + a² tan²(theta)) = a sqrt(1 + tan²(theta)) = a sec(theta). Therefore, the length from t1 to t2 is integral from t1 to t2 of speed dt = a sec(theta) (t2 - t1). Therefore, to have a length s along BC, we need t2 - t1 = s / (a sec(theta)) = (s cos(theta))/a.Therefore, if S is at parameter t, then R is at parameter t + (s cos(theta))/a. But since R must lie on BC, t + (s cos(theta))/a ≤ 1. So t ≤ 1 - (s cos(theta))/a.But perhaps this is getting too involved. Maybe there's a better way.Alternatively, let's denote the side length of the square as s. The square touches AB at P and AC at Q. Let's consider the coordinates:- Point P is on AB. Since AB is from (0,0) to (a,0), so P is at (p, 0) where 0 ≤ p ≤ a.- Point Q is on AC. Since AC is from (0,0) to (0, a tan(theta)), so Q is at (0, q) where 0 ≤ q ≤ a tan(theta).Since PQ is a side of the square, which should be perpendicular to QR and RS. Wait, PQ is one side of the square, QR is another side. Wait, actually, the square is PQRS, so the order is P connected to Q connected to R connected to S connected to P. Wait, but according to the problem statement: "inscribed square PQRS with one side RS on hypotenuse BC, P on AB, and Q on AC". So the square has vertices P on AB, Q on AC, R and S on BC. So the order is P-Q-R-S-P?Wait, that would make PQ one side, QR another side, RS the third side (on BC), and SP the fourth side. So PQ is from AB to AC, QR is from AC to BC, RS is along BC, and SP is from BC back to AB. Wait, but how does SP go back to AB? If S is on BC and P is on AB, then SP would be another side of the square. So in this configuration, the square has sides PQ, QR, RS, and SP, each of length s. So PQ is vertical or horizontal? Wait, no. PQ connects P on AB to Q on AC. Since AB is horizontal and AC is vertical, PQ would be a diagonal line, but in a square, adjacent sides are perpendicular. Therefore, PQ and QR must be perpendicular. So PQ is from P on AB to Q on AC, then QR is from Q to R on BC, which should be perpendicular to PQ. Similarly, RS is along BC, and SP is from S to P, which should be perpendicular to RS.This seems complicated. Maybe using coordinate geometry is the way to go.Let me denote the coordinates as follows:- Let’s assume point P is at (p, 0) on AB.- Point Q is at (0, q) on AC.Since PQ is a side of the square PQRS, the vector from P to Q should have length s and direction such that PQ is one side of the square.But in a square, adjacent sides are perpendicular. So if PQ is a side, then QR must be a side perpendicular to PQ. QR goes from Q to R on BC. Similarly, RS is along BC, and SP goes back to P.Alternatively, since RS is along BC, and the square has side length s, the distance from R to S along BC is s. But BC has slope -tan(theta), so the direction vector of BC is (-a, a tan(theta)), as from B to C.Therefore, a unit vector along BC is ( -a / (a sec(theta)) , a tan(theta) / (a sec(theta)) ) = ( -cos(theta), sin(theta) ). Because BC has length a sec(theta), so dividing by that.Therefore, moving from point S towards C along BC by distance s would reach point R. So the coordinates of R can be written as S plus s times the unit vector along BC. But since BC is from B to C, the unit vector is (-cos(theta), sin(theta)).Wait, if S is a point on BC, then R is obtained by moving from S towards C by distance s along BC. Therefore, if S has coordinates (x_S, y_S), then R has coordinates:x_R = x_S + s*(-cos(theta))y_R = y_S + s*sin(theta)But also, since PQRS is a square, the vector from S to R should be the same as the vector from Q to R minus the vector from Q to S. Wait, this might not be straightforward.Alternatively, since PQRS is a square, the sides PQ, QR, RS, and SP are all equal in length and each pair of adjacent sides is perpendicular.So PQ is from P(p,0) to Q(0,q). The vector PQ is (-p, q). Then QR is from Q(0,q) to R(x_R, y_R). The vector QR should be perpendicular to PQ and have the same length s.Similarly, RS is from R to S, which is along BC, so vector RS should be a scalar multiple of the direction vector of BC, which is (-a, a tan(theta)). But since RS is a side of the square, its length is s, and direction is along BC.Wait, maybe we can set up equations based on the properties of the square.First, PQ is a side of the square, so PQ has length s, and QR is another side, perpendicular to PQ and also of length s.Let’s compute vector PQ: from P(p,0) to Q(0,q) is (-p, q). The length of PQ is sqrt(p² + q²) = s.Vector QR is from Q(0,q) to R(x_R, y_R). Since QR must be perpendicular to PQ, their dot product should be zero. The vector QR is (x_R - 0, y_R - q) = (x_R, y_R - q). The dot product of PQ and QR is (-p)(x_R) + q(y_R - q) = 0.Also, the length of QR must be s: sqrt(x_R² + (y_R - q)^2) = s.Moreover, since R lies on BC, its coordinates (x_R, y_R) must satisfy the equation of BC. The equation of BC, as previously established, is y = -tan(theta) x + a tan(theta).Therefore, y_R = -tan(theta) x_R + a tan(theta).Additionally, after R, the next vertex is S, which is along BC from R by a distance s. So vector RS is along BC with length s. The direction from R to S is the same as BC, which is from B to C, so direction vector (-a, a tan(theta)). Therefore, the unit vector in the direction of BC is (-cos(theta), sin(theta)), as previously calculated. Therefore, moving from R along BC by distance s gives S:x_S = x_R + s*(-cos(theta))y_S = y_R + s*sin(theta)But since S is on BC, it must also satisfy the equation of BC: y_S = -tan(theta) x_S + a tan(theta)So substituting:y_R + s sin(theta) = -tan(theta)(x_R - s cos(theta)) + a tan(theta)But we already have y_R = -tan(theta) x_R + a tan(theta) from R being on BC. So substituting:(-tan(theta) x_R + a tan(theta)) + s sin(theta) = -tan(theta)(x_R - s cos(theta)) + a tan(theta)Simplify the right-hand side:= -tan(theta) x_R + tan(theta) s cos(theta) + a tan(theta)Left-hand side:= -tan(theta) x_R + a tan(theta) + s sin(theta)Setting them equal:- tan(theta) x_R + a tan(theta) + s sin(theta) = -tan(theta) x_R + tan(theta) s cos(theta) + a tan(theta)Cancel out terms:Left: ... + s sin(theta) = Right: ... + tan(theta) s cos(theta)So:s sin(theta) = s tan(theta) cos(theta)Simplify RHS: tan(theta) cos(theta) = sin(theta)Thus:s sin(theta) = s sin(theta)Which is an identity. So this doesn't give new information. Therefore, we need to use other equations.Let me recap the equations we have:1. PQ vector: (-p, q), length sqrt(p² + q²) = s2. QR vector: (x_R, y_R - q), which is perpendicular to PQ, so (-p)x_R + q(y_R - q) = 03. QR length: sqrt(x_R² + (y_R - q)^2) = s4. R lies on BC: y_R = -tan(theta) x_R + a tan(theta)5. S lies on BC: y_S = -tan(theta) x_S + a tan(theta)6. RS vector: (-s cos(theta), s sin(theta)) [since moving along BC by s]But since S is obtained by moving from R along BC by s, coordinates of S are (x_R - s cos(theta), y_R + s sin(theta))But S must lie on BC, so substituting into BC equation:y_R + s sin(theta) = -tan(theta)(x_R - s cos(theta)) + a tan(theta)But we already know from R's coordinates that y_R = -tan(theta) x_R + a tan(theta). Therefore, substituting:(-tan(theta) x_R + a tan(theta)) + s sin(theta) = -tan(theta)(x_R - s cos(theta)) + a tan(theta)Which simplifies as before to an identity, so no new info.Therefore, we need to use the other equations.From equations 1, 2, 3, 4.We have four equations:1. sqrt(p² + q²) = s2. -p x_R + q(y_R - q) = 03. sqrt(x_R² + (y_R - q)^2) = s4. y_R = -tan(theta) x_R + a tan(theta)Let me attempt to solve these equations.From equation 4, we can express y_R in terms of x_R: y_R = -tan(theta) x_R + a tan(theta)Let me substitute y_R into equation 2:-p x_R + q(-tan(theta) x_R + a tan(theta) - q) = 0Simplify:-p x_R - q tan(theta) x_R + q a tan(theta) - q² = 0Factor x_R:x_R (-p - q tan(theta)) + q a tan(theta) - q² = 0Let me solve for x_R:x_R = [ q² - q a tan(theta) ] / [ -p - q tan(theta) ]Hmm, this might get messy. Let's see if we can relate p and q.From equation 1: sqrt(p² + q²) = s => p² + q² = s²From equation 3: sqrt(x_R² + (y_R - q)^2) = s. Let's square both sides:x_R² + (y_R - q)^2 = s²But y_R - q = (-tan(theta) x_R + a tan(theta) - q)Thus, substitute:x_R² + (-tan(theta) x_R + a tan(theta) - q)^2 = s²Let me expand the second term:[ -tan(theta) x_R + a tan(theta) - q ]^2= [ -tan(theta) x_R + (a tan(theta) - q) ]^2= tan²(theta) x_R² - 2 tan(theta) x_R (a tan(theta) - q) + (a tan(theta) - q)^2Therefore, equation 3 becomes:x_R² + tan²(theta) x_R² - 2 tan(theta) x_R (a tan(theta) - q) + (a tan(theta) - q)^2 = s²Combine like terms:x_R² (1 + tan²(theta)) - 2 tan(theta) x_R (a tan(theta) - q) + (a tan(theta) - q)^2 = s²But 1 + tan²(theta) = sec²(theta), so:x_R² sec²(theta) - 2 tan(theta) x_R (a tan(theta) - q) + (a tan(theta) - q)^2 = s²Hmm, this seems complicated, but maybe we can substitute x_R from equation 2.From earlier, we had:x_R = [ q² - q a tan(theta) ] / [ -p - q tan(theta) ]Let me denote numerator N = q² - q a tan(theta)Denominator D = -p - q tan(theta)Thus, x_R = N / DLet me plug x_R into equation 3's expanded form:(N/D)² sec²(theta) - 2 tan(theta) (N/D) (a tan(theta) - q) + (a tan(theta) - q)^2 = s²Multiply both sides by D² to eliminate denominators:N² sec²(theta) - 2 tan(theta) N (a tan(theta) - q) D + (a tan(theta) - q)^2 D² = s² D²But this is getting very complicated. Perhaps there's a better approach.Alternatively, notice that triangles APQ and ABC are similar? Wait, maybe not. Let me think.When the square is inscribed in the triangle, it creates two smaller right triangles and the square. The triangle APQ is not necessarily similar to ABC, but perhaps the triangles BPQ and QRC are similar to ABC?Wait, perhaps another approach. Let's consider the areas.The area of the square is s². The area of triangle ABC is (a * a tan(theta))/2 = (a² tan(theta))/2. So the ratio we need is [(a² tan(theta))/2] / s². We need to minimize this ratio, which is equivalent to maximizing s² for a given theta, or minimizing (a² tan(theta))/ (2 s²). Wait, but actually, the ratio is (Area ABC) / (Area PQRS) = [ (a² tan(theta)/2) ] / s². So to minimize this ratio, we need to maximize s² / (a² tan(theta)/2 ), i.e., maximize s² / tan(theta).Therefore, the problem reduces to finding the maximum value of s² / tan(theta) over theta, which would give the minimum of the original ratio.Alternatively, since the ratio is (a² tan(theta)/2) / s², so the ratio is (a² / (2 s²)) tan(theta). To minimize this, since a is constant, we need to minimize tan(theta) / s². Wait, no: the ratio is proportional to tan(theta) / s². Therefore, to minimize the ratio, we need to minimize tan(theta)/s².But perhaps this line of thought isn't directly helpful. Maybe I need to express s in terms of a and theta, then compute the ratio and find its minimum.Let me try to find s in terms of a and theta.Let me consider the similar triangles approach.In the original triangle ABC, after inscribing the square PQRS, we have two smaller triangles: one between the square and vertex B, and another between the square and vertex C.Let me denote the triangle adjacent to B as triangle BPS, and the triangle adjacent to C as triangle CQR.These two smaller triangles are similar to the original triangle ABC.Let me verify this. Triangle BPS has a right angle at P, since BP is along AB (horizontal) and PS is vertical (if the square is aligned such that PQ is horizontal and QR is vertical). Wait, no, the square is not necessarily axis-aligned. Wait, in our coordinate system, AB is horizontal and AC is vertical, but the square is rotated.Alternatively, since ABC is a right triangle and the square is inscribed with one side on the hypotenuse, the two smaller triangles formed (BSP and QRC) should be similar to ABC.Assuming similarity, let's denote:- Triangle BSP ~ triangle ABC- Triangle QRC ~ triangle ABCLet’s let’s denote the side length of the square as s.In triangle BSP, the side corresponding to AB (the horizontal leg) would be BP = AB - AP = a - pSimilarly, the vertical leg would be some length, but maybe this is not straightforward.Alternatively, let's use the similarity ratios.For triangle BSP:If triangle BSP ~ triangle ABC, then the ratio of sides is proportional.In triangle ABC, AB = a, AC = a tan(theta), BC = a sec(theta)In triangle BSP, suppose the legs are BP and PS, with BP along AB and PS perpendicular to AB (since it's a right triangle). However, PS is part of the square, so PS = s.Therefore, triangle BSP has legs BP and PS = s, hypotenuse BS.Since triangle BSP ~ triangle ABC, the ratio of corresponding sides should be equal.Therefore, BP / AB = PS / AC = BS / BCSo BP / a = s / (a tan(theta)) = BS / (a sec(theta))From BP / a = s / (a tan(theta)), we get BP = s / tan(theta)Similarly, BS / (a sec(theta)) = s / (a tan(theta)) => BS = (a sec(theta)) * (s / (a tan(theta))) = s / sin(theta)Similarly, for triangle QRC ~ triangle ABC:The legs would be QC and QR = s, with hypotenuse RC.Therefore, QC / AC = QR / AB = RC / BCQC / (a tan(theta)) = s / a = RC / (a sec(theta))From QC / (a tan(theta)) = s / a => QC = s tan(theta)Similarly, RC = (a sec(theta)) * (s / a) = s sec(theta)Now, the hypotenuse BC is equal to BS + SR + RC. Wait, BC is the original hypotenuse, length a sec(theta). But in the divided parts, from B to S is BS, from S to R is SR = s (the side of the square), and from R to C is RC.Wait, but in reality, BC is divided into BS, SR, and RC. Wait, no, the square's side SR is along BC, so BC is divided into two parts: from B to S, then from S to R, and then from R to C. Wait, but in the square, RS is a side of length s. Wait, but BC is a straight line, and SR is along BC with length s. However, BS and RC are not along BC; rather, BS and RC are the hypotenuses of the smaller triangles.Wait, perhaps my initial assumption is incorrect. If triangle BSP is similar to ABC, then BS would be the hypotenuse of triangle BSP, not a segment along BC. Therefore, my previous approach may be flawed.Alternatively, perhaps the length along BC is BS' and S'C, where S' is the projection or something else. This is getting confusing. Maybe I need to correct this.Wait, perhaps it's better to consider that when you inscribe a square in a right triangle with one side on the hypotenuse, the two smaller triangles formed are similar to the original triangle.Yes, I have read before that this is the case. So the triangles on either side of the square are similar to the original triangle.Therefore, triangle BPQ ~ triangle ABC, and triangle QRC ~ triangle ABC.Wait, no. Wait, in our case, the square is PQRS with P on AB, Q on AC, and RS on BC. Therefore, the two smaller triangles would be APQ and ... Wait, APQ is the triangle formed by A, P, Q. But APQ is a smaller right triangle with legs AP and AQ. However, since PQ is the side of the square, AP = p, AQ = q, and PQ = s. So APQ is a right triangle with legs p and q, and hypotenuse s. But this triangle is not necessarily similar to ABC unless p/q = AB/AC = a / (a tan(theta)) = 1 / tan(theta) = cot(theta). So if p/q = cot(theta), then APQ ~ ABC. But in general, unless the square is placed such that p/q = cot(theta), which may not be the case.Alternatively, perhaps the triangles adjacent to the square, which are BSP and QRC, are similar to ABC.Wait, if we consider triangle BSP: it has a right angle at P, since BP is along AB and SP is perpendicular to AB (assuming the square has SP perpendicular to AB). But in reality, the square's side SP is not necessarily perpendicular to AB. Wait, no. In the square PQRS, adjacent sides are perpendicular. So PQ is from P to Q, QR is from Q to R, RS is from R to S, and SP is from S to P. So PQ is one side, QR is the next, etc.Given that PQ is from P on AB to Q on AC, QR is from Q to R on BC, RS is along BC to S, and SP is from S back to P.In this case, the angles at P and Q are 90 degrees because it's a square. So angle at P between SP and PQ is 90 degrees, and angle at Q between PQ and QR is 90 degrees.Therefore, triangle SPB is a right triangle with right angle at P, and triangle RQC is a right triangle with right angle at Q.Wait, maybe not. Let me clarify:- At point P, we have the square's vertices S, P, and Q connected. Wait, no. The square is PQRS, so the connections are P connected to Q, Q connected to R, R connected to S, S connected to P.Therefore, at point P: the sides SP and PQ meet at 90 degrees.Similarly, at point Q: sides PQ and QR meet at 90 degrees.Therefore, triangle SPA is the triangle outside the square at vertex P, but actually, point S is on BC, so the triangle involving P would be triangle SPP? Wait, maybe not.Alternatively, perhaps the area of triangle ABC can be expressed as the sum of the areas of square PQRS and the two smaller triangles APQ and QRC.Wait, no. APQ is part of the original triangle, but if PQ is a side of the square, then APQ would actually be a quadrilateral? Wait, maybe this approach isn't correct.Alternatively, consider that the entire triangle ABC is divided into three parts: the square PQRS, the triangle BPQ, and the triangle SRC.Wait, no. Let me try to visualize it again.Point P is on AB, point Q is on AC. The square PQRS has side PQ between P and Q, then QR goes from Q to R on BC, RS goes along BC to S, and SP goes back to P.Therefore, the area of ABC is equal to the area of the square plus the areas of triangles APQ, BSP, and SRC.Wait, maybe not. Let me think.Actually, when you inscribe the square in the triangle, the square touches all three sides: AB, AC, and BC. Therefore, the remaining areas are three small triangles, but in this case, since the square is touching AB, AC, and BC, perhaps there are two small triangles left: one between the square and vertex B, and another between the square and vertex C.But in our case, since the square has one vertex on AB (P), one on AC (Q), and two on BC (R and S), the area of ABC would be the area of the square plus the areas of triangles BPQ and QRC.Wait, triangle BPQ: point B, point P, point Q. But Q is on AC, so BPQ is a triangle with vertices at B, P, and Q. However, this might not be a right triangle. Similarly, QRC has vertices at Q, R, and C.Alternatively, maybe the two triangles are BSP and QRC, where S is between B and R on BC.This is getting confusing. Perhaps it's better to go back to the coordinate system.Let me define:- Coordinates: - A: (0,0) - B: (a,0) - C: (0, a tan(theta)) - Square PQRS: - P: (p,0) on AB - Q: (0,q) on AC - R and S on BCWe need to define the coordinates of R and S such that PQRS is a square.Vector PQ is from P(p,0) to Q(0,q): (-p, q)Vector QR is from Q(0,q) to R(x_R, y_R): (x_R, y_R - q)Vector RS is from R(x_R, y_R) to S(x_S, y_S): (x_S - x_R, y_S - y_R)Vector SP is from S(x_S, y_S) to P(p,0): (p - x_S, - y_S)Since PQRS is a square, all sides must have length s, and adjacent sides must be perpendicular.Therefore:1. |PQ| = |QR| = |RS| = |SP| = s2. PQ • QR = 0, QR • RS = 0, RS • SP = 0, SP • PQ = 0Starting with PQ • QR = 0:PQ • QR = (-p, q) • (x_R, y_R - q) = -p x_R + q(y_R - q) = 0QR • RS = (x_R, y_R - q) • (x_S - x_R, y_S - y_R) = x_R(x_S - x_R) + (y_R - q)(y_S - y_R) = 0Similarly for the other dot products, but this might get too complex. Let's focus on the first two conditions.Additionally, RS is along BC, so the vector RS should be parallel to BC. The direction vector of BC is (-a, a tan(theta)), so RS should be a scalar multiple of (-a, a tan(theta)).Since RS is a side of the square, its length is s, so:sqrt( (x_S - x_R)^2 + (y_S - y_R)^2 ) = sAnd direction vector (x_S - x_R, y_S - y_R) = k*(-a, a tan(theta)), where k is a scalar such that |k*(-a, a tan(theta))| = s.The length of BC's direction vector is sqrt(a² + (a tan(theta))²) = a sec(theta). Therefore, k*a sec(theta) = s => k = s / (a sec(theta)) = (s cos(theta))/aTherefore, the vector RS is:(x_S - x_R, y_S - y_R) = ( -a*(s cos(theta)/a ), a tan(theta)*(s cos(theta)/a ) ) = ( -s cos(theta), s sin(theta) )Therefore, coordinates:x_S = x_R - s cos(theta)y_S = y_R + s sin(theta)Moreover, since S lies on BC, its coordinates must satisfy the equation of BC: y_S = -tan(theta) x_S + a tan(theta)Substitute x_S and y_S:y_R + s sin(theta) = -tan(theta)(x_R - s cos(theta)) + a tan(theta)But y_R = -tan(theta) x_R + a tan(theta) (since R is on BC), so substitute that:(-tan(theta) x_R + a tan(theta)) + s sin(theta) = -tan(theta)(x_R - s cos(theta)) + a tan(theta)Simplify:Left side: -tan(theta) x_R + a tan(theta) + s sin(theta)Right side: -tan(theta) x_R + s tan(theta) cos(theta) + a tan(theta)Note that tan(theta) cos(theta) = sin(theta), so right side becomes:-tan(theta) x_R + s sin(theta) + a tan(theta)Therefore, both sides are equal:- tan(theta) x_R + a tan(theta) + s sin(theta) = -tan(theta) x_R + s sin(theta) + a tan(theta)Which is an identity. Therefore, this condition doesn't provide new information.Now, moving on. We need to use the other conditions.We have:1. From PQ: sqrt(p² + q²) = s2. PQ • QR = 0 => -p x_R + q(y_R - q) = 03. QR vector: (x_R, y_R - q), which must satisfy |QR| = s4. R lies on BC: y_R = -tan(theta) x_R + a tan(theta)5. Coordinates of S: x_S = x_R - s cos(theta), y_S = y_R + s sin(theta)6. SP vector: (p - x_S, - y_S) must have length s and be perpendicular to RS vector (-s cos(theta), s sin(theta))First, let's handle condition 6. The vector SP is (p - x_S, - y_S). Since SP must have length s:sqrt( (p - x_S)^2 + (- y_S)^2 ) = sAnd since SP must be perpendicular to RS, their dot product is zero:(p - x_S)(-s cos(theta)) + (- y_S)(s sin(theta)) = 0Simplify:-s cos(theta)(p - x_S) - s sin(theta) y_S = 0Divide both sides by -s:cos(theta)(p - x_S) + sin(theta) y_S = 0But x_S = x_R - s cos(theta), y_S = y_R + s sin(theta). Substitute these:cos(theta)(p - (x_R - s cos(theta))) + sin(theta)(y_R + s sin(theta)) = 0Simplify:cos(theta)(p - x_R + s cos(theta)) + sin(theta)(y_R + s sin(theta)) = 0Expand:cos(theta)(p - x_R) + cos(theta) s cos(theta) + sin(theta) y_R + sin(theta) s sin(theta) = 0Combine terms:cos(theta)(p - x_R) + sin(theta) y_R + s [ cos²(theta) + sin²(theta) ] = 0But cos²(theta) + sin²(theta) = 1, so:cos(theta)(p - x_R) + sin(theta) y_R + s = 0Now, from condition 4, y_R = -tan(theta) x_R + a tan(theta). Substitute this into the equation:cos(theta)(p - x_R) + sin(theta)(-tan(theta) x_R + a tan(theta)) + s = 0Simplify:cos(theta) p - cos(theta) x_R - sin(theta) tan(theta) x_R + sin(theta) a tan(theta) + s = 0Note that sin(theta) tan(theta) = sin(theta) * (sin(theta)/cos(theta)) = sin²(theta)/cos(theta)Therefore:cos(theta) p - cos(theta) x_R - (sin²(theta)/cos(theta)) x_R + a sin(theta) tan(theta) + s = 0Combine the x_R terms:- x_R [ cos(theta) + sin²(theta)/cos(theta) ] + cos(theta) p + a sin(theta) tan(theta) + s = 0Factor out x_R:- x_R [ (cos²(theta) + sin²(theta)) / cos(theta) ] + cos(theta) p + a sin(theta) tan(theta) + s = 0Since cos²(theta) + sin²(theta) = 1:- x_R / cos(theta) + cos(theta) p + a sin(theta) tan(theta) + s = 0Multiply through by cos(theta) to eliminate denominator:- x_R + cos²(theta) p + a sin(theta) tan(theta) cos(theta) + s cos(theta) = 0Simplify terms:- x_R + cos²(theta) p + a sin(theta) * sin(theta) + s cos(theta) = 0Because tan(theta) cos(theta) = sin(theta):So:- x_R + cos²(theta) p + a sin²(theta) + s cos(theta) = 0Solve for x_R:x_R = cos²(theta) p + a sin²(theta) + s cos(theta)Now, recall from condition 2:- p x_R + q(y_R - q) = 0But y_R - q = (-tan(theta) x_R + a tan(theta) - q)Wait, we need to express this equation in terms of p, q, and s.But from condition 1: p² + q² = s²And from QR's length:sqrt(x_R² + (y_R - q)^2) = sLet me try to substitute x_R from the previous result into these equations.We have:x_R = cos²(theta) p + a sin²(theta) + s cos(theta)And y_R = -tan(theta) x_R + a tan(theta)Thus, y_R - q = -tan(theta) x_R + a tan(theta) - qWe also have from condition 2:- p x_R + q(y_R - q) = 0Substituting x_R and y_R - q:- p [ cos²(theta) p + a sin²(theta) + s cos(theta) ] + q [ -tan(theta) x_R + a tan(theta) - q ] = 0This is getting very complicated. Maybe there's a better substitution.Alternatively, since we have multiple equations involving p, q, x_R, y_R, and s, perhaps we can find a relationship between s and theta.Alternatively, let's assume that the square's side s can be expressed in terms of a and theta. Let's try to find such an expression.In similar problems, the side length of a square inscribed in a right triangle with one side on the hypotenuse is given by s = (a b c) / (a² + b² + a b), where a and b are the legs, and c is the hypotenuse. But in our case, the legs are a and a tan(theta), and hypotenuse is a sec(theta). Let me check if this formula applies.Using the formula:s = (a * a tan(theta) * a sec(theta)) / (a² + (a tan(theta))² + a * a tan(theta))Simplify numerator and denominator:Numerator: a^3 tan(theta) sec(theta)Denominator: a² + a² tan²(theta) + a² tan(theta) = a² [1 + tan²(theta) + tan(theta)] = a² [ sec²(theta) + tan(theta) ]Therefore:s = [ a^3 tan(theta) sec(theta) ] / [ a² (sec²(theta) + tan(theta)) ] = [ a tan(theta) sec(theta) ] / [ sec²(theta) + tan(theta) ]Simplify denominator:sec²(theta) + tan(theta) = 1 + tan²(theta) + tan(theta)Therefore:s = [ a tan(theta) sec(theta) ] / [ 1 + tan(theta) + tan²(theta) ]But we need to verify if this formula is applicable here. This formula is for a square inscribed with one side on the hypotenuse, but depending on the orientation, the formula might differ.Alternatively, perhaps we can derive it for our case.Let me consider the standard formula for the side length of a square inscribed in a right triangle with legs of lengths m and n, and hypotenuse p. The side length s of the square is given by s = (m n p) / (m² + n² + m n).In our case, legs m = a, n = a tan(theta), hypotenuse p = a sec(theta). Plugging into the formula:s = (a * a tan(theta) * a sec(theta)) / (a² + (a tan(theta))² + a * a tan(theta)) = (a^3 tan(theta) sec(theta)) / (a² (1 + tan²(theta) + tan(theta))) ) = (a tan(theta) sec(theta)) / (1 + tan(theta) + tan²(theta))Which matches the earlier result. Therefore, this formula might be applicable here. Assuming this is correct, then s = [ a tan(theta) sec(theta) ] / [ 1 + tan(theta) + tan²(theta) ]Simplify tan(theta) sec(theta):tan(theta) sec(theta) = (sin(theta)/cos(theta)) * (1/cos(theta)) = sin(theta)/cos²(theta)Therefore, s = [ a sin(theta)/cos²(theta) ] / [ 1 + tan(theta) + tan²(theta) ]Simplify denominator:1 + tan(theta) + tan²(theta) = 1 + tan(theta) + tan²(theta)Express in terms of sin and cos:= 1 + (sin(theta)/cos(theta)) + (sin²(theta)/cos²(theta)) = [cos²(theta) + sin(theta) cos(theta) + sin²(theta)] / cos²(theta)But cos²(theta) + sin²(theta) = 1, so:= [1 + sin(theta) cos(theta)] / cos²(theta)Therefore, s = [ a sin(theta)/cos²(theta) ] / [ (1 + sin(theta) cos(theta)) / cos²(theta) ] = [ a sin(theta) ] / [ 1 + sin(theta) cos(theta) ]So s = a sin(theta) / (1 + sin(theta) cos(theta))Therefore, the side length of the square is s = (a sin(theta)) / (1 + sin(theta) cos(theta))Now, having found s in terms of a and theta, we can compute the ratio S_ABC / S_PQRS.The area of triangle ABC is (a * a tan(theta)) / 2 = (a² tan(theta))/2.The area of the square is s² = [a² sin²(theta)] / (1 + sin(theta) cos(theta))²Therefore, the ratio is:[ (a² tan(theta))/2 ] / [ a² sin²(theta) / (1 + sin(theta) cos(theta))² ] = [ tan(theta) / 2 ] * [ (1 + sin(theta) cos(theta))² / sin²(theta) ]Simplify tan(theta) = sin(theta)/cos(theta):= [ (sin(theta)/cos(theta)) / 2 ] * [ (1 + sin(theta) cos(theta))² / sin²(theta) ]= [ 1/(2 cos(theta)) ] * [ (1 + sin(theta) cos(theta))² / sin(theta) ]= (1 + sin(theta) cos(theta))² / (2 sin(theta) cos(theta))Let me write this ratio as:Ratio = [ (1 + sin(theta) cos(theta))² ] / [ 2 sin(theta) cos(theta) ]Simplify numerator and denominator:Let’s expand the numerator:(1 + sin(theta) cos(theta))² = 1 + 2 sin(theta) cos(theta) + sin²(theta) cos²(theta)Denominator: 2 sin(theta) cos(theta)Therefore:Ratio = [1 + 2 sin(theta) cos(theta) + sin²(theta) cos²(theta)] / [2 sin(theta) cos(theta)]Split the fraction:= [1 / (2 sin(theta) cos(theta))] + [2 sin(theta) cos(theta) / (2 sin(theta) cos(theta))] + [sin²(theta) cos²(theta) / (2 sin(theta) cos(theta))]Simplify each term:First term: 1/(2 sin(theta) cos(theta)) = 1/(sin(2 theta)) using identity sin(2 theta) = 2 sin(theta) cos(theta)Second term: 1Third term: sin(theta) cos(theta)/2 = (1/2) sin(theta) cos(theta)Therefore:Ratio = (1/sin(2 theta)) + 1 + (1/2) sin(theta) cos(theta)But this seems messy. Maybe there's a better way to simplify.Alternatively, let's set u = sin(theta) cos(theta). Note that u = (sin(2 theta))/2.Then, the ratio becomes:Ratio = (1 + 2u + u²) / (2u) = ( (1 + u)^2 ) / (2u )Since u = sin(theta) cos(theta).Therefore, Ratio = (1 + u)^2 / (2u) where u = sin(theta) cos(theta)Let’s denote u = (sin(2 theta))/2, so:Ratio = (1 + (sin(2 theta)/2))² / (2*(sin(2 theta)/2)) = (1 + (sin(2 theta)/2))² / (sin(2 theta))Expand the numerator:(1 + (sin(2 theta)/2))² = 1 + sin(2 theta) + (sin²(2 theta))/4Therefore, Ratio = [1 + sin(2 theta) + (sin²(2 theta))/4] / sin(2 theta)Split into terms:= 1/sin(2 theta) + 1 + (sin(2 theta))/4Hmm, this seems to complicate things further. Maybe another substitution.Let’s let v = sin(2 theta). Then, since theta is between 0 and pi/2 (as it's a right triangle angle), 2 theta is between 0 and pi, so sin(2 theta) is between 0 and 1 (since theta is acute, 2 theta is less than pi). Wait, theta is angle B in the right triangle, so theta is between 0 and pi/2. Therefore, 2 theta is between 0 and pi. But sin(2 theta) reaches maximum 1 at theta = pi/4.But perhaps we can express the ratio in terms of v:Ratio = (1 + v/2)^2 / (2*(v/2)) = (1 + v/2)^2 / vExpand numerator:= (1 + v + v²/4) / v = 1/v + 1 + v/4So Ratio = 1/v + 1 + v/4 where v = sin(2 theta) and v ∈ (0, 1]We need to find the minimum of this expression with respect to v in (0,1].Let’s denote f(v) = 1/v + 1 + v/4. Find the minimum of f(v) for v ∈ (0,1].To find the minimum, take derivative f’(v) = -1/v² + 0 + 1/4Set derivative to zero:-1/v² + 1/4 = 0 => 1/v² = 1/4 => v² = 4 => v = 2But v ∈ (0,1], so v = 2 is outside the domain. Therefore, the minimum occurs at the boundary v =1 or as v approaches 0.But as v approaches 0, 1/v approaches infinity, so f(v) approaches infinity.At v=1, f(v)=1/1 +1 +1/4=1 +1 +0.25=2.25=9/4.Therefore, the minimum of f(v) in (0,1] is 9/4, achieved at v=1.But v=sin(2 theta)=1 implies 2 theta=pi/2, so theta=pi/4=45 degrees.Therefore, the minimum value of the ratio S_ABC / S_PQRS is 9/4=2.25 when theta=45 degrees.But let's verify this result.If theta=45 degrees, then triangle ABC is an isosceles right triangle with legs AB=AC=a. Then BC=a√2.The inscribed square would have side length s. Using the formula derived earlier:s = a sin(theta) / (1 + sin(theta) cos(theta)).When theta=45 degrees, sin(theta)=cos(theta)=√2/2.So s = a*(√2/2) / (1 + (√2/2)*(√2/2)) = (a√2/2) / (1 + 0.5) = (a√2/2) / (1.5) = (a√2/2) * (2/3) = a√2/3.Then area of square s² = (a² * 2)/9 = 2a²/9.Area of triangle ABC= (a * a)/2= a²/2.Ratio= (a²/2) / (2a²/9)= (1/2)/(2/9)= (1/2)*(9/2)=9/4=2.25, which matches our earlier result.Therefore, the minimum ratio is 9/4 when theta=45 degrees.But the problem states that theta can vary, and we need to find the minimum value of the ratio. Therefore, the answer is 9/4, achieved when theta=45 degrees.Final AnswerThe minimum value of the ratio is boxed{dfrac{9}{4}}.

✅:Given a right triangle ( ABC ) with (angle A = 90^circ), ( |AB| = a ), and (angle B = theta), we need to find the minimum value of the ratio (frac{S_{triangle ABC}}{S_{PQRS}}) where ( PQRS ) is an inscribed square with one side ( RS ) on the hypotenuse ( BC ).1. Sides of Triangle ( ABC ): - ( AB = a ) - ( AC = a tan(theta) ) - ( BC = a sec(theta) )2. Area of Triangle ( ABC ): [ S_{triangle ABC} = frac{1}{2} times AB times AC = frac{1}{2} times a times a tan(theta) = frac{a^2 tan(theta)}{2} ]3. Side Length of the Inscribed Square ( PQRS ): - Using the formula for the side length ( s ) of a square inscribed in a right triangle with legs ( a ) and ( a tan(theta) ): [ s = frac{a sin(theta)}{1 + sin(theta) cos(theta)} ]4. Area of the Square ( PQRS ): [ S_{PQRS} = s^2 = left( frac{a sin(theta)}{1 + sin(theta) cos(theta)} right)^2 ]5. Ratio Calculation: [ frac{S_{triangle ABC}}{S_{PQRS}} = frac{frac{a^2 tan(theta)}{2}}{left( frac{a sin(theta)}{1 + sin(theta) cos(theta)} right)^2} = frac{frac{a^2 tan(theta)}{2}}{frac{a^2 sin^2(theta)}{(1 + sin(theta) cos(theta))^2}} = frac{tan(theta) (1 + sin(theta) cos(theta))^2}{2 sin^2(theta)} ] Simplifying using (tan(theta) = frac{sin(theta)}{cos(theta)}): [ frac{frac{sin(theta)}{cos(theta)} (1 + sin(theta) cos(theta))^2}{2 sin^2(theta)} = frac{(1 + sin(theta) cos(theta))^2}{2 sin(theta) cos(theta)} ]6. Minimizing the Ratio: - Let ( u = sin(theta) cos(theta) ). The ratio becomes: [ frac{(1 + u)^2}{2u} ] - Let ( v = sin(2theta) ), then ( u = frac{v}{2} ). The ratio transforms to: [ frac{(1 + frac{v}{2})^2}{v} = frac{1 + v + frac{v^2}{4}}{v} = frac{1}{v} + 1 + frac{v}{4} ] - Minimizing this expression for ( v in (0, 1] ), we find the minimum occurs at ( v = 1 ) (when (theta = 45^circ)), giving: [ frac{1}{1} + 1 + frac{1}{4} = 1 + 1 + 0.25 = frac{9}{4} ]The minimum value of the ratio (frac{S_{triangle ABC}}{S_{PQRS}}) is (boxed{dfrac{9}{4}}).

🔑:We are given a right triangle (ABC) with (angle A = 90^circ) where (AB = a) and (angle B = theta). Inside this triangle, there is an inscribed square (PQRS) such that:- One side (RS) lies on the hypotenuse (BC),- Point (P) lies on (AB),- Point (Q) lies on (AC).We aim to find the minimum value of the ratio (frac{S_{triangle ABC}}{S_{PQRS}}).1. Identify the areas: - The area of (triangle ABC): [ S_{triangle ABC} = frac{1}{2} cdot AB cdot AC ] Given (AB = a) and the height from (A) to (BC) is (a tan theta) (since height to the hypotenuse is given by the tangent of (theta)): [ S_{triangle ABC} = frac{1}{2} cdot a cdot (a tan theta) = frac{1}{2} a^2 tan theta ] - The area of (PQRS): To find this, consider that the square (PQRS) is inscribed in (triangle ABC), such that (sin(2theta)) relates to the side length of the square. The side length (s) of the square depends on the triangle's sides and angle (theta). The side length of the square inscribed in a right triangle where one side lies on the hypotenuse can be represented as: [ s = frac{2a sin theta cos theta}{sin 2 theta + 2} ] Thus, the area of the square is: [ S_{PQRS} = s^2 = left( frac{2a sin theta cos theta}{sin 2 theta + 2} right)^2 = frac{4a^2 (sin theta cos theta)^2}{(sin 2 theta + 2)^2} ] Since (sin theta cos theta = frac{1}{2} sin 2theta), we can rewrite this as: [ S_{PQRS} = frac{4a^2 left( frac{1}{2} sin 2theta right)^2}{(sin 2theta + 2)^2} = frac{a^2 sin^2 2theta}{(sin 2theta + 2)^2} ]2. Formulate the ratio: [ frac{S_{triangle ABC}}{S_{PQRS}} = frac{frac{1}{2} a^2 tan theta}{frac{a^2 sin^2 2theta}{(sin 2theta + 2)^2}} ] Simplify this ratio: [ frac{S_{triangle ABC}}{S_{PQRS}} = frac{frac{1}{2} tan theta (sin 2theta + 2)^2}{sin^2 2theta} ] Rewrite (frac{tan theta}{sin 2theta}) using (tan theta = frac{ sin theta}{ cos theta}) and (sin 2 theta = 2 sin theta cos theta): [ frac{S_{triangle ABC}}{S_{PQRS}} = frac{frac{1}{2} cdot frac{sin theta}{cos theta} cdot (sin 2 theta + 2)^2}{sin^2 2 theta} ] Substitute ( sin 2 theta = 2 sin theta cos theta): [ frac{S_{triangle ABC}}{S_{PQRS}} = frac{ (sin 2 theta + 2)^2}{4 sin 2 theta} cdot frac{1}{sin 2 theta} ] Simplify further: [ frac{S_{triangle ABC}}{S_{PQRS}} = frac{ (sin 2 theta + 2)^2}{4 sin 2 theta} ]Let's denote ( x = sin 2 theta ). Then the equation becomes: [ y = 1 + frac{sin^2 2 theta + 4}{4 sin 2 theta} ]3. Analyze the function behavior: - The function (y = frac{(sin 2 theta + 2)^2}{4 sin 2 theta}) must be minimized over the interval ([0, 1]). - For ( x = 0 ) or ( x = 1 ): [ y = 1 + frac{sin^2 2 theta + 4}{4 sin 2 theta} ] Simplifies and taking critical points: - After suitable evaluation within the interval ([-pi/2, pi/2)), - The minimum value of (y) is 9/4. Conclusion:The minimum value of the ratio (frac{S_{triangle ABC}}{S_{PQRS}}) is[ boxed{frac{9}{4}} ]

🧠:Alright, let's tackle this problem. We need to find a unique function ( f ) on the positive integers with positive integer values such that ( f(m + f(n)) = n + f(m + 95) ) for all ( m, n ). Then, compute the sum ( f(1) + f(2) + ldots + f(19) ).First, functional equations can be tricky, but usually, plugging in specific values for variables can help us find patterns or properties of the function. Let's start by understanding the given equation:( f(m + f(n)) = n + f(m + 95) )This holds for all positive integers ( m ) and ( n ). Our goal is to determine ( f(n) ).Since the equation must hold for all ( m ), perhaps setting specific values for ( m ) or ( n ) can simplify things. Let's try setting ( m = 1 ):( f(1 + f(n)) = n + f(1 + 95) )( f(1 + f(n)) = n + f(96) )Hmm, this relates the value of ( f ) at ( 1 + f(n) ) to ( n ) plus a constant ( f(96) ). Maybe if we can express ( f(n) ) in terms of ( n ), that would help. Let's hypothesize that ( f ) is a linear function. Suppose ( f(n) = an + b ), where ( a ) and ( b ) are constants. Let's plug this into the original equation:Left-hand side (LHS):( f(m + f(n)) = a(m + f(n)) + b = a(m + an + b) + b = am + a^2n + ab + b )Right-hand side (RHS):( n + f(m + 95) = n + a(m + 95) + b = n + am + 95a + b )Set LHS = RHS:( am + a^2n + ab + b = n + am + 95a + b )Cancel out similar terms on both sides:( a^2n + ab = n + 95a )Now, equate coefficients for like terms. The coefficients of ( n ) must be equal:( a^2 = 1 )So ( a = 1 ) or ( a = -1 ). But since ( f ) has positive integer values, ( a ) must be positive. Hence, ( a = 1 ).Now, substitute ( a = 1 ) into the remaining terms:( ab = 95a )Since ( a = 1 ):( b = 95 times 1 = 95 )Therefore, the function ( f(n) = n + 95 ) is a candidate. Let's verify this in the original equation:LHS: ( f(m + f(n)) = f(m + n + 95) = (m + n + 95) + 95 = m + n + 190 )RHS: ( n + f(m + 95) = n + (m + 95 + 95) = n + m + 190 )They are equal. So ( f(n) = n + 95 ) satisfies the equation. Wait, but the problem states that there is a unique function. So is this the only solution? Let's check if our assumption of linearity was necessary or if there could be non-linear solutions.Suppose ( f ) is not linear. Let's see if we can derive properties of ( f ) without assuming linearity. Let's try to find if ( f ) is injective or surjective.First, check injectivity. Suppose ( f(a) = f(b) ). Then for any ( m ):( f(m + f(a)) = f(m + f(b)) )Which implies:( a + f(m + 95) = b + f(m + 95) )Thus, ( a = b ). Therefore, ( f ) is injective.Next, check surjectivity. Let's see if every integer can be expressed as ( f(n) ). Suppose not, but we need more information. Let's see if we can express ( f ) in terms of itself.Looking back at the equation:( f(m + f(n)) = n + f(m + 95) )Let’s set ( m = k - f(n) ). Wait, but ( m ) must be a positive integer, so ( k - f(n) ) must be positive. Maybe not the best approach. Alternatively, let's fix ( n ) and vary ( m ).Alternatively, set ( n = 1 ). Then:( f(m + f(1)) = 1 + f(m + 95) )This looks similar to a functional equation that could be shifted. Let’s denote ( c = f(1) ). Then:( f(m + c) = f(m + 95) + 1 )This suggests that shifting the argument by ( c ) increases the function value by 1 compared to shifting by 95. Interesting. Let's explore this equation.Let’s denote ( d = c - 95 ). Then:( f(m + 95 + d) = f(m + 95) + 1 )This resembles a recurrence relation. If we set ( m' = m + 95 ), then:( f(m' + d) = f(m') + 1 )This suggests that for any ( m' ), adding ( d ) to the argument increases the function value by 1. Which would imply that ( f ) is linear with slope ( 1/d ). But since we have integer arguments and integer values, ( d ) must divide 1. Wait, because if ( f(m' + d) = f(m') + 1 ), then repeating this ( k ) times gives ( f(m' + k d) = f(m') + k ). For this to hold for all ( m' ), ( d ) must be 1, because otherwise, if ( d > 1 ), then between ( m' ) and ( m' + d ), the function would have to increase by 1, but we don't know about intermediate values. Wait, but the equation only tells us about increments of ( d ), not necessarily between. Hmm, perhaps ( d = 1 ), leading to ( f(m + 1) = f(m) + 1 ), meaning ( f ) is linear with slope 1. Let's check.If ( d = 1 ), then ( c = 95 + d = 96 ). So ( f(1) = 96 ). Then from the equation ( f(m + 96) = f(m + 95) + 1 ). So starting from some base, each increment of the argument by 1 increases the function by 1. If this holds for all ( m ), then ( f(m + 95 + k) = f(m + 95) + k ). If we can show that ( f(m + 1) = f(m) + 1 ) for all ( m geq some base ), then maybe ( f ) is linear.But we need to confirm this. Let's suppose that ( f(m + 1) = f(m) + 1 ) for all ( m geq 95 ). Wait, but actually, the equation ( f(m + 96) = f(m + 95) + 1 ) is given for all ( m geq 1 ). Therefore, replacing ( m ) with ( m - 95 ), we get ( f(m + 1) = f(m) + 1 ) for all ( m geq 96 ). So starting from ( m = 96 ), the function is linear with slope 1. But what about values before 96?Wait, but we know that ( f ) is injective. Suppose ( f ) is linear beyond a certain point, but not before. Let's see if we can extend this.From the original equation ( f(m + f(n)) = n + f(m + 95) ), let's set ( m = 1 ):( f(1 + f(n)) = n + f(96) )But from the previous result, ( f(96) = f(95) + 1 ). If ( f(m + 1) = f(m) + 1 ) for ( m geq 95 ), then ( f(95) = f(94) + 1 ), but we don't know about ( f(94) ). However, this seems to create a chain reaction. Let's think.Alternatively, let's assume that ( f(n) = n + c ) for some constant ( c ). Then:LHS: ( f(m + f(n)) = f(m + n + c) = m + n + c + c = m + n + 2c )RHS: ( n + f(m + 95) = n + (m + 95 + c) = m + n + 95 + c )Equate LHS and RHS:( m + n + 2c = m + n + 95 + c )Thus, ( 2c = 95 + c Rightarrow c = 95 ). Therefore, ( f(n) = n + 95 ). This works, as we saw earlier.But the problem states there is a unique function. So is this the only solution? We need to confirm that no non-linear solutions exist.Suppose there exists a non-linear solution. Let's assume ( f ) is not linear. However, since ( f ) is injective (as proven earlier), and from the equation ( f(m + f(n)) = n + f(m + 95) ), maybe we can show that ( f(n) = n + 95 ) is the only possibility.Another approach: Let's consider defining ( g(n) = f(n) - 95 ). Then substituting into the original equation:( f(m + f(n)) = n + f(m + 95) )Expressed in terms of ( g ):( f(m + g(n) + 95) = n + f(m + 95) )But ( f(k) = g(k) + 95 ), so:Left-hand side:( g(m + g(n) + 95) + 95 )Right-hand side:( n + g(m + 95) + 95 )Subtract 95 from both sides:( g(m + g(n) + 95) = n + g(m + 95) )This looks similar to a Cauchy-like equation. Let's try to see if ( g ) is linear. Suppose ( g(n) = n + d ). Wait, but then:Left-hand side:( g(m + (n + d) + 95) = m + n + d + 95 + d = m + n + 95 + 2d )Right-hand side:( n + g(m + 95) = n + m + 95 + d )Equating:( m + n + 95 + 2d = n + m + 95 + d )Which implies ( 2d = d Rightarrow d = 0 ). But then ( g(n) = n ), so ( f(n) = n + 95 ). So this again gives the linear solution.Alternatively, maybe ( g ) is additive. Suppose ( g(n) = kn ). Then:Left-hand side:( g(m + kn + 95) = k(m + kn + 95) = km + k^2n + 95k )Right-hand side:( n + g(m + 95) = n + k(m + 95) = n + km + 95k )Equate:( km + k^2n + 95k = n + km + 95k )Cancel terms:( k^2n = n )Thus, ( k^2 = 1 Rightarrow k = 1 ) (since ( k ) must be positive). Therefore, ( g(n) = n ), so ( f(n) = n + 95 ). So again, the same solution.This suggests that the only possible solution is linear. Hence, ( f(n) = n + 95 ) is the unique function.Now, we need to compute ( f(1) + f(2) + ldots + f(19) ).Since ( f(k) = k + 95 ), each term ( f(k) = k + 95 ). So the sum from ( k = 1 ) to ( 19 ) is:( sum_{k=1}^{19} (k + 95) = sum_{k=1}^{19} k + sum_{k=1}^{19} 95 )Compute each sum separately:First sum: ( sum_{k=1}^{19} k = frac{19 times 20}{2} = 190 )Second sum: ( 19 times 95 = 1805 )Add them together: ( 190 + 1805 = 1995 )Therefore, the sum ( f(1) + f(2) + ldots + f(19) = 1995 ).But wait, let's double-check the arithmetic:19 times 95: 95*20 = 1900, subtract 95 gives 1900 - 95 = 1805. Correct.Sum of 1 to 19: 19*20/2 = 190. Correct.190 + 1805 = 1995. Yes.So the final answer should be 1995. However, let me just confirm that the function we found is indeed unique. We assumed linearity, but also checked injectivity and derived that the function must be linear. So unless there's a non-linear function that satisfies the equation, which we couldn't find, and the problem states there is a unique function, so this must be the case. Therefore, the answer is 1995.

✅:To solve the functional equation ( f(m + f(n)) = n + f(m + 95) ) for all positive integers ( m, n ), we start by hypothesizing a linear function ( f(n) = n + c ). Substituting into the equation:- Left-hand side (LHS): ( f(m + f(n)) = f(m + n + c) = (m + n + c) + c = m + n + 2c )- Right-hand side (RHS): ( n + f(m + 95) = n + (m + 95 + c) = m + n + 95 + c )Equating LHS and RHS gives ( 2c = 95 + c ), leading to ( c = 95 ). Thus, ( f(n) = n + 95 ) is a solution. To confirm uniqueness, we show ( f ) must be injective and derive a recurrence relation by setting ( n = 1 ):- ( f(m + f(1)) = 1 + f(m + 95) )- Letting ( f(1) = 96 ), we get ( f(m + 96) = f(m + 95) + 1 ), implying ( f ) increases by 1 for each increment in its argument beyond 95. This extends to all arguments, confirming ( f(n) = n + 95 ).Finally, compute the sum ( f(1) + f(2) + ldots + f(19) ):- Each term ( f(k) = k + 95 )- Sum: ( sum_{k=1}^{19} (k + 95) = sum_{k=1}^{19} k + sum_{k=1}^{19} 95 = 190 + 1805 = 1995 )Answer: (boxed{1995})

🔑:Let's define the function ( f ) on the positive integers with positive integer values such that:[f(m + f(n) ) = n + f(m + 95) quad forall m, n.]1. Define a new function: Let's define a new function ( F(n) ) based on ( f(n) ): [ F(n) = f(n) - 95. ] This implies: [ f(n) = F(n) + 95. ]2. Substitute into the given functional equation: Substitute ( f(n) ) into the original equation: [ f(m + f(n)) = f(m + F(n) + 95) = n + f(m + 95). ] Using the definition ( f(k) = F(k) + 95 ): [ f(m + F(n) + 95) = n + f(m + 95). ] Simplify using ( f(k) = F(k) + 95 ): [ F(m + F(n) + 95) + 95 = n + F(m + 95) + 95. ] Cancel out the 95 terms: [ F(m + F(n) + 95) = n + F(m + 95). ]3. Align indices: Notice that ( F(m + 95) = F(m) ) since there's no defined dependency on ( 95 ): [ F(m + F(n)) = n + F(m). ] This is valid for ( n geq 1 ) and ( m geq 1 ).4. Recursive relationship: Assume: [ F(m + F(n)) = n + F(m). ] Substitute ( k = h + n ) and ( l = m ): [ F(h + n + F(m)) = F(h + F(m + F(n))). ]5. Analyze the shifted function: Given ( F(h + n) = F(h) + F(n) ) for ( h geq 96, n geq 1 ), we deduce the additivity: [ F(h + n) = F(h) + F(n). ]6. Induction step: We show by induction that ( F(n) = n cdot F(1) ): - Base case: For ( n = 1 ): [ F(1) = F(1). ] - Inductive step: Suppose ( F(k) = k cdot F(1) ) for ( k ). Then: [ F(k + 1) = F(k) + F(1) = k cdot F(1) + F(1) = (k + 1) cdot F(1). ] This is valid for all ( k geq 1 ).7. Characterize the function: We have: [ F(n) = n cdot F(1). ]8. Examine function values: Consider ( f(n) ): [ n + F(1) mid F(m + 95) = n = F(m + 95) ] Test ( 1 ): [ (96 + F(m + F(n))) = F(n + (n + F(m))) = n cdot F(1). ]9. Solve for ( F(1) ): If ( F(1) = 1 ), then: [ F(n) = n cdot 1 = n. ] This leads to: [ f(n) = n + 95. ]9. Test for any contradictions: We check ( F(1) ): If ( F(1)= -1 ), then ( f ) non-positive integers. So, ( F(95) = 0). Verification show only ( F(1)=1 ),10. Sum computation: Finally calculate ( f(1) +f (n cdots 19 ): [ f(1 +2+ +19) boxed{1-1995}. ]We have boxed conclude to computation the final results : ( boxed{1995}).

🧠:Okay, so I need to solve the inequality sin x sin 2x < sin 3x sin 4x for x between 0 and π/2. Hmm, let me think about how to approach this. Maybe I can start by expanding or simplifying both sides using trigonometric identities. Let me recall some product-to-sum formulas because multiplying sines can sometimes be converted into sums or differences of cosines, which might make the inequality easier to handle.The product-to-sum formula for sine functions is: sin A sin B = [cos(A - B) - cos(A + B)] / 2. Let me apply that to both sides of the inequality.First, the left side: sin x sin 2x. Applying the formula here:sin x sin 2x = [cos(x - 2x) - cos(x + 2x)] / 2 = [cos(-x) - cos(3x)] / 2. But cos(-x) is the same as cos x, so this simplifies to [cos x - cos 3x] / 2.Now the right side: sin 3x sin 4x. Applying the same formula:sin 3x sin 4x = [cos(3x - 4x) - cos(3x + 4x)] / 2 = [cos(-x) - cos(7x)] / 2. Again, cos(-x) is cos x, so this becomes [cos x - cos 7x] / 2.So substituting back into the inequality:[cos x - cos 3x] / 2 < [cos x - cos 7x] / 2.Since both sides are divided by 2, I can multiply both sides by 2 to eliminate the denominators:cos x - cos 3x < cos x - cos 7x.Now, subtract cos x from both sides:-cos 3x < -cos 7x.Multiply both sides by -1, which reverses the inequality:cos 3x > cos 7x.So the inequality simplifies to cos 3x > cos 7x. Now, I need to find the values of x in (0, π/2) where cos 3x is greater than cos 7x.Hmm, how do I solve cos 3x > cos 7x? Let me recall that cosine is a decreasing function from 0 to π and increasing from π to 2π, but since the arguments here are 3x and 7x, and x is between 0 and π/2, the arguments 3x and 7x will be between 0 and 3π/2 and 0 and 7π/2 respectively. But 7π/2 is 3.5π, so the 7x term goes beyond 2π. But perhaps there's a way to compare these cosines.Alternatively, maybe bring all terms to one side:cos 3x - cos 7x > 0.Use the sum-to-product identity for cos A - cos B. The formula is:cos A - cos B = -2 sin[(A + B)/2] sin[(A - B)/2].Applying this to cos 3x - cos 7x:cos 3x - cos 7x = -2 sin[(3x + 7x)/2] sin[(3x - 7x)/2] = -2 sin(10x/2) sin(-4x/2) = -2 sin(5x) sin(-2x).But sin(-2x) is -sin 2x, so:-2 sin 5x (-sin 2x) = 2 sin 5x sin 2x.Therefore, cos 3x - cos 7x = 2 sin 5x sin 2x.So the inequality becomes 2 sin 5x sin 2x > 0.Since 2 is positive, we can divide both sides by 2:sin 5x sin 2x > 0.So now, the inequality reduces to sin 5x sin 2x > 0. So we need to find where the product of sin 5x and sin 2x is positive. Remember that a product of two numbers is positive if both are positive or both are negative.Given that x is in (0, π/2), let's analyze the signs of sin 5x and sin 2x in this interval.First, sin 2x: For x in (0, π/2), 2x is in (0, π). In this interval, sin 2x is always positive because sine is positive in (0, π). So sin 2x > 0 for all x in (0, π/2). Therefore, the sign of the product sin 5x sin 2x depends solely on the sign of sin 5x. Since sin 2x is positive, the product is positive if and only if sin 5x is positive.Therefore, the inequality sin 5x sin 2x > 0 simplifies to sin 5x > 0.So now, the problem reduces to solving sin 5x > 0 for x in (0, π/2).But let me verify that conclusion. Since sin 2x is positive in (0, π/2), then the product sin 5x sin 2x is positive exactly when sin 5x is positive. Therefore, the original inequality is equivalent to sin 5x > 0 in (0, π/2). Correct.So now, find all x in (0, π/2) where sin 5x > 0.Recall that sin θ > 0 when θ is in (0, π) modulo 2π. So we need to find all x in (0, π/2) such that 5x is in (0, π) + 2π k for some integer k where 5x is in that interval.But x is in (0, π/2), so 5x is in (0, 5π/2). Let's analyze the intervals where sin 5x > 0.Sin θ is positive in (0, π) and (2π, 3π), etc. But since 5x can go up to 5π/2, which is 2.5π, so in the interval (0, 5π/2), sin θ is positive in (0, π) and (2π, 5π/2). Wait, 2π is approximately 6.28, and 5π/2 is 7.85, but 5x is up to 5*(π/2) = 5π/2. Wait, 5π/2 is 2.5π, which is 7.85398... Wait, no, 5π/2 is 2.5π, which is between 2π (≈6.28) and 3π (≈9.42). Wait, no, π is approximately 3.14, so π/2 is 1.57, so 5*(π/2) is 7.85, which is 2.5π. Wait, 5π/2 is 2.5π? Wait, π/2 is 0.5π, so 5*(π/2) is 2.5π. Yes, correct.So 5x is in (0, 2.5π). So sin 5x is positive in (0, π) and negative in (π, 2π), and positive again in (2π, 3π), but 5x only goes up to 2.5π. Therefore, sin 5x is positive in (0, π) and (2π, 2.5π). So translating back to x:When 5x is in (0, π), that is x in (0, π/5). And when 5x is in (2π, 2.5π), which is x in (2π/5, 2.5π/5) = (2π/5, π/2). Therefore, sin 5x > 0 when x is in (0, π/5) union (2π/5, π/2).Therefore, the solution to sin 5x > 0 in (0, π/2) is x ∈ (0, π/5) ∪ (2π/5, π/2).But wait, we need to check if 2π/5 is less than π/2. Let's compute π/5 ≈ 0.628 radians, 2π/5 ≈ 1.257 radians, π/2 ≈ 1.571 radians. Yes, 2π/5 is about 1.257, which is less than π/2 (1.571). Therefore, the intervals are indeed (0, π/5) and (2π/5, π/2).Therefore, the original inequality sin x sin 2x < sin 3x sin 4x holds for x in (0, π/5) ∪ (2π/5, π/2).But wait, let me verify this conclusion step by step to ensure there are no mistakes.First, we transformed the original inequality using product-to-sum formulas correctly:Left side became [cos x - cos 3x]/2, right side became [cos x - cos 7x]/2. Then subtracting and simplifying led us to cos 3x > cos 7x. Then using the identity for cos A - cos B gave us 2 sin 5x sin 2x > 0. Since sin 2x is positive in the interval, we concluded sin 5x > 0, leading to the intervals (0, π/5) and (2π/5, π/2). This seems correct.But let me test some specific points in those intervals to see if the inequality holds.Take x = π/6 ≈ 0.523. Is π/6 in (0, π/5)? π/5 ≈ 0.628, so π/6 ≈0.523 is in (0, π/5). Then check the original inequality:Left side: sin(π/6) sin(2*(π/6)) = sin(π/6) sin(π/3) = (1/2)*(√3/2) = √3/4 ≈0.433.Right side: sin(3*(π/6)) sin(4*(π/6)) = sin(π/2) sin(2π/3) = 1*(√3/2) = √3/2 ≈0.866. So 0.433 < 0.866 is true. So the inequality holds here.Now take x = π/4 ≈0.785. π/4 is 0.785, which is between 2π/5 ≈1.257 and π/2? Wait, π/4 is approximately 0.785, which is less than π/5≈0.628? Wait no, π/4 is approximately 0.785, which is greater than π/5≈0.628. Wait, 2π/5≈1.257 is greater than π/2≈1.571. Wait, no, 2π/5 is approximately 1.257, and π/2 is approximately 1.571, so between them is x=1.257 to 1.571. So π/4 is 0.785, which is between 0 and π/5≈0.628? No, π/4 is larger than π/5. Wait π≈3.14, so π/5≈0.628, π/4≈0.785, 2π/5≈1.257, π/2≈1.571.So x=π/4 is in (π/5, 2π/5), which according to our solution is where the inequality does not hold. Let's check:Left side: sin(π/4) sin(2*(π/4)) = sin(π/4) sin(π/2) = (√2/2)*1 = √2/2 ≈0.707.Right side: sin(3*(π/4)) sin(4*(π/4)) = sin(3π/4) sin(π) = (√2/2)*0 = 0. So 0.707 < 0? No, this is false. So the inequality does not hold here, which matches our solution.Now, take x=3π/10≈0.942, which is in (2π/5≈1.257, π/2≈1.571)? Wait, 3π/10≈0.942 is less than 2π/5≈1.257, so it's still in (π/5, 2π/5). Let's check another x in (2π/5, π/2). Let's take x=3π/5≈1.885. Wait, but x is supposed to be less than π/2≈1.571. So 3π/5≈1.885 is outside the interval. Let me pick x=4π/10≈1.257, which is 2π/5. Wait, 2π/5 is exactly the lower bound. Let me take x=1.3 radians (approximately 74.5 degrees), which is within (2π/5≈1.257, π/2≈1.571).Compute left side: sin(1.3) sin(2*1.3) ≈ sin(1.3) sin(2.6). Let's compute these values:sin(1.3) ≈0.9636, sin(2.6)≈0.5155. So left side≈0.9636*0.5155≈0.496.Right side: sin(3*1.3) sin(4*1.3)= sin(3.9) sin(5.2). Compute these:sin(3.9)≈sin(3.9 - π)≈sin(3.9 - 3.1416)≈sin(0.7584)≈0.688. Wait, but 3.9 radians is in the third quadrant (π≈3.14, 3π/2≈4.712), so sin(3.9) is negative. Wait, sin(3.9)=sin(π + 0.7584)= -sin(0.7584)≈-0.688.Similarly, sin(5.2)=sin(5.2 - 2π)=sin(5.2 -6.283)=sin(-1.083)= -sin(1.083)≈-0.875.Therefore, right side≈(-0.688)*(-0.875)=0.602. So left side≈0.496 < 0.602≈right side. So the inequality holds here, which is in (2π/5, π/2). Correct.Now, let's check a point in (π/5, 2π/5). Let's take x=π/3≈1.047 radians. π/5≈0.628, 2π/5≈1.257, so π/3≈1.047 is in between.Left side: sin(π/3) sin(2π/3)= (√3/2)*(√3/2)= 3/4=0.75.Right side: sin(3*(π/3)) sin(4*(π/3))= sin(π) sin(4π/3)= 0*(-√3/2)=0. So 0.75 < 0? No, which is false. So the inequality doesn't hold here, as expected.So based on testing these points, our solution seems correct.Therefore, the solution to the inequality sin x sin 2x < sin 3x sin 4x in the interval (0, π/2) is x ∈ (0, π/5) ∪ (2π/5, π/2).But to present the final answer, we can write it in terms of intervals:0 < x < π/5 and 2π/5 < x < π/2.So the answer is the union of these two intervals.Final AnswerThe solution to the inequality is boxed{left(0, frac{pi}{5}right) cup left(frac{2pi}{5}, frac{pi}{2}right)}.

✅:To solve the inequality (sin x sin 2x < sin 3x sin 4x) for (0 < x < frac{pi}{2}), we start by using product-to-sum identities on both sides.First, apply the product-to-sum formula to both sides:[sin x sin 2x = frac{cos x - cos 3x}{2}][sin 3x sin 4x = frac{cos x - cos 7x}{2}]Substituting these into the inequality gives:[frac{cos x - cos 3x}{2} < frac{cos x - cos 7x}{2}]Multiply both sides by 2 to eliminate the denominators:[cos x - cos 3x < cos x - cos 7x]Subtract (cos x) from both sides:[-cos 3x < -cos 7x]Multiply both sides by -1 and reverse the inequality:[cos 3x > cos 7x]Using the sum-to-product identity for (cos 3x - cos 7x):[cos 3x - cos 7x = 2 sin 5x sin 2x]This transforms the inequality into:[2 sin 5x sin 2x > 0]Since (sin 2x > 0) for (0 < x < frac{pi}{2}), the inequality simplifies to:[sin 5x > 0]We need to find where (sin 5x > 0) for (0 < x < frac{pi}{2}). The sine function is positive in the intervals ((0, pi)), ((2pi, 3pi)), etc. Since (5x) ranges from (0) to (frac{5pi}{2}), we consider the intervals where (sin 5x) is positive:- (0 < 5x < pi) which translates to (0 < x < frac{pi}{5})- (2pi < 5x < frac{5pi}{2}) which translates to (frac{2pi}{5} < x < frac{pi}{2})Thus, the solution to the inequality is:[boxed{left(0, frac{pi}{5}right) cup left(frac{2pi}{5}, frac{pi}{2}right)}]

🔑:To solve the inequality [sin x sin 2x < sin 3x sin 4x]in the interval [0 < x < frac{pi}{2},]we start by utilizing trigonometric identities to simplify the terms. Let's begin:1. Use product-to-sum identities: [ sin A sin B = frac{1}{2} [ cos (A - B) - cos (A + B) ]. ]2. Apply the identity to both sides: [ sin x sin 2x = frac{1}{2} [ cos(x - 2x) - cos(x + 2x) ] = frac{1}{2} [ cos(-x) - cos(3x) ] = frac{1}{2} [ cos x - cos 3x ], ] [ sin 3x sin 4x = frac{1}{2} [ cos(3x - 4x) - cos(3x + 4x) ] = frac{1}{2} [ cos(-x) - cos 7x ] = frac{1}{2} [ cos x - cos 7x ]. ]3. Rewrite the inequality: Since both sides of the original inequality have (frac{1}{2}), we can ignore this constant factor: [ cos x - cos 3x < cos x - cos 7x. ]4. Simplify the inequality: [ -cos 3x < -cos 7x quad text{or} quad cos 3x > cos 7x. ]5. Analyzing cosine function: Since (cos x) is a decreasing function in the interval (0 < x < frac{pi}{2}): [ 3x < 7x quad text{implies} quad text{the inequality is always satisfied in this interval.} ]6. Combine conditions: Let's analyze if there are any specific ranges for (x) where (sin 2x) and (sin 5x) are positive. Since [ sin 2x > 0 text{when} 0 < 2x < pi, text{which simplifies to} 0 < x < frac{pi}{2}, ] and [ sin 5x > 0 text{when} 0 < 5x < pi text{or} 2pi < 5x < frac{5pi}{2}, ] this simplifies to [ 0 < x < frac{pi}{5} quad text{or} quad frac{2pi}{5} < x < frac{pi}{2}. ]7. Final ranges: So the inequality (sin x sin 2x < sin 3x sin 4x) is valid in: [ 0 < x < frac{pi}{5} quad text{or} quad frac{2pi}{5} < x < frac{pi}{2}. ] Conclusion:[boxed{0 < x < frac{pi}{5} text{or} frac{2pi}{5} < x < frac{pi}{2}}]

🧠:Okay, let's try to figure out the seller's loss here. So, the buyer takes 10 rubles worth of goods and pays with a 25 ruble note. The seller doesn't have change, so he goes to the neighbor to exchange the 25 rubles. Then the neighbor later finds out the 25 was counterfeit, so the seller has to give back 25 rubles. Hmm, let's break it down step by step.First, the buyer gives the seller a counterfeit 25 ruble note. Since it's fake, that 25 rubles is essentially worthless. But the seller doesn't know that yet. He uses that fake money to get real change from the neighbor. So he gives the neighbor the counterfeit 25, and the neighbor gives him real 25 rubles in change. Then the seller gives the buyer the goods worth 10 rubles and the change of 15 rubles (since 25 - 10 = 15). So at this point, the seller has given away 10 rubles in goods and 15 rubles in cash, totaling 25 rubles. But he got 25 rubles from the neighbor, which he thought was real but turned out to be fake.Wait, so the seller used the counterfeit money to get real money from the neighbor. Then when the neighbor finds out the money was fake, the seller has to return 25 rubles. So the seller initially got 25 real rubles from the neighbor, then had to give back 25 real rubles. So that cancels out, right? But he also gave the buyer 15 rubles in change and 10 rubles worth of goods. So the total loss would be 15 + 10 = 25 rubles.Wait, but let me check that again. Let's track each party's gains and losses. The buyer started with counterfeit money and left with 10 rubles of goods and 15 rubles in change. So the buyer gained 25 rubles worth of value. The neighbor initially gave 25 rubles to the seller and then got it back when the seller returned the 25. So the neighbor breaks even. The seller gave away 10 rubles in goods and 15 rubles in cash (total 25) and had to return 25 rubles to the neighbor. But where did the 25 rubles come from that he returned? He had taken 25 from the neighbor, which was real, so when he returns it, he's out 25. So the seller's total loss is the 25 he gave to the buyer (10 goods +15 change) plus the 25 he had to return to the neighbor? Wait, no, because the 25 he got from the neighbor was used to give the 15 change. So maybe it's only 25 total?Wait, let's think of it in terms of the seller's assets. Initially, the seller has the goods worth 10 and some cash (unknown). The buyer gives fake 25, which is worthless. Seller takes fake 25 to neighbor, gets real 25. Then gives buyer 15 change and 10 goods. So seller's assets now: the original cash minus 15 (because he gave 15 as change) plus the 25 from the neighbor. But then the neighbor asks for 25 back, so he has to give that back. So his cash flow is: +25 (from neighbor) -15 (to buyer) -25 (back to neighbor) = -15. And he lost the 10 rubles goods. So total loss is 15 +10 =25. That matches.Alternatively, another way: The counterfeit money is a red herring. The seller effectively gave away 10 goods and 15 cash, and had to cover the 25 that the neighbor lost, but since the neighbor got their 25 back, the seller's only loss is the 25 he gave to the buyer. Because the neighbor didn't lose anything; the seller compensated the neighbor. So the seller's loss is 25 rubles.But wait, some people might argue differently. Let me make sure. The key is that the seller received nothing of value from the buyer (since the 25 was fake). So the seller gave away 10 goods and 15 change (total 25) and had to repay the neighbor 25. But the 25 he repaid was from his own pocket, right? Because the 25 he got from the neighbor was used to give the 15 change. So he had 25 from the neighbor, gave 15 to buyer, kept 10. Then when he has to repay the neighbor 25, he uses his own 10 plus another 15? Wait, no. Let's track the actual cash.Seller starts with, say, X rubles in cash.1. Buyer gives fake 25. Seller has X cash + fake 25.2. Seller exchanges fake 25 with neighbor for real 25. Now seller has X +25 (real) cash, neighbor has fake 25.3. Seller gives buyer 15 change and 10 goods. So seller's cash is X +25 -15 = X +10. Goods inventory is down by 10.4. Neighbor finds out the 25 was fake, seller has to return 25 to neighbor. So seller's cash becomes X +10 -25 = X -15. Goods inventory still down by 10.So compared to the start, seller's cash is down by 15, and goods down by 10. Total loss 25. So yes, the total loss is 25 rubles.Another angle: The seller lost the value of the goods (10) and the cash given as change (15), and had to compensate the neighbor for the counterfeit 25. Wait, but he compensated the neighbor by giving back the 25 he took, which was actually the neighbor's own money. Wait, no. Let's see:The neighbor gave the seller 25 real rubles in exchange for the fake 25. Then when the neighbor realizes it's fake, the seller has to give back 25 real rubles. So the neighbor's net is zero (gave 25, got back 25). The seller's net from the neighbor is +25 then -25, so zero. But in between, the seller used that 25 to give 15 to the buyer. So the seller's own cash was used to give the 15. Wait, maybe not. Let's model it step by step:1. Seller's initial cash: S.2. Buyer gives fake 25. Seller now has S + fake 25.3. Seller exchanges fake 25 with neighbor for real 25. Seller's cash: S +25 (real). Neighbor's cash: N -25 (gave to seller) + fake 25.4. Seller gives buyer 15 change and 10 goods. Seller's cash: S +25 -15 = S +10. Goods: G -10.5. Neighbor discovers fake, seller returns 25. Seller's cash: S +10 -25 = S -15. Neighbor's cash: N -25 +25 (back) = N.So seller ends up with cash S -15 and goods G -10. So total loss is 15 cash +10 goods =25.Yes, that's consistent. So the answer should be 25 rubles.

✅:The seller's total loss is 25 rubles. Here's the breakdown:1. Goods Given: The seller provided goods worth 10 rubles to the buyer.2. Change Provided: The seller gave 15 rubles in change (25 rubles received - 10 rubles for goods).3. Compensation to Neighbor: The seller had to return 25 rubles to the neighbor after the counterfeit was discovered.However, the 25 rubles borrowed from the neighbor were already used to give the 15 rubles change, so the seller effectively lost:- 10 rubles (goods) + 15 rubles (change from their own funds) = 25 rubles total loss.The compensation to the neighbor cancels out the temporary 25 rubles received earlier, leaving the seller’s net loss at 25 rubles.

🔑:Given information:- The buyer took goods worth 10 rubles and gave 25 rubles to the seller.- The seller, lacking change, exchanged the 25 rubles with the neighbor.- The neighbor later discovered that the 25 rubles were counterfeit.- The seller returned 25 rubles to the neighbor.Now, we need to determine the loss incurred by the seller.1. The buyer receives goods worth 10 rubles and 15 rubles in change from the seller, totaling: [ 10 , text{rubles (goods)} + 15 , text{rubles (change)} = 25 , text{rubles} ]2. The neighbor initially provides the seller with 25 rubles in exchange for the counterfeit 25 ruble note.3. Upon discovering the counterfeit, the seller returns 25 rubles to the neighbor.To calculate the total loss to the seller:- Goods worth 10 rubles are given to the buyer.- An additional 15 rubles are given to the buyer as change.- The counterfeit 25 rubles were returned to the neighbor.Hence, the total loss can be expressed as:[10 , text{rubles (goods)} + 15 , text{rubles (change)} = 25 , text{rubles}]Hence, the total net loss for the seller is:[boxed{25 , text{rubles}}]

🧠:Okay, so I need to figure out how many different "friendly hands" (which are combinations of 4 cards that add up to 24) have a largest number of 7. Let me start by understanding the problem step by step.First, the example given is Wang Liang's cards: 7, 7, 7, 3. Their sum is 7 + 7 + 7 + 3 = 24. So this is one such friendly hand where the largest number is 7. Now, the question is asking how many different such combinations exist where 7 is the largest number in the set, and the total sum is 24.Let me break down the constraints:1. Four cards: Each combination has exactly four numbers.2. Sum to 24: The total of these four numbers must be 24.3. Largest number is 7: So all numbers in the combination must be ≤7, and at least one of them must be 7.4. Different combinations: The problem specifies "different friendly hands," so I need to consider the uniqueness of the combinations. However, since the example includes three 7s and a 3, which are presumably from a standard deck of cards, I need to clarify if we're considering the cards as distinct (like actual playing cards where there are four suits) or if we're just considering numerical values regardless of suits. The problem statement doesn't mention suits, so I think it's about numerical combinations. So, combinations are considered the same if they have the same multiset of numbers. For example, three 7s and a 3 is one combination, regardless of the order or suits. So, the count is based on the multiset.Therefore, the problem reduces to finding the number of multisets of four integers (each ≤7) where the sum is 24, and at least one of the integers is 7 (since it's the largest). Since the largest number is 7, all other numbers must be ≤7, and we need to have at least one 7.Let me formalize the problem:Find the number of non-negative integer solutions (since cards can have numbers, but in reality, card numbers are positive integers. Wait, actually, cards have positive integers. So each number must be at least 1. Wait, but in the example, there is a 3. So the numbers are positive integers. So each card is a positive integer, and we have four cards, each ≥1, each ≤7, with at least one card equal to 7, and the sum is 24.So the problem is: find the number of quadruples (a, b, c, d) where each a, b, c, d ∈ {1, 2, 3, 4, 5, 6, 7}, at least one of a, b, c, d is 7, and a + b + c + d = 24.Additionally, since we are considering combinations (multisets), the order doesn't matter. So different orderings of the same numbers are considered the same hand. Therefore, we need to count the number of multisets {w, x, y, z} where each element is between 1 and 7 inclusive, at least one element is 7, and the sum is 24.But wait, the problem might not specify whether the cards are unique. In many card games, a hand is a combination of cards where each card is unique (i.e., no duplicates unless there are multiple suits). However, in the example given, there are three 7s and a 3, which suggests that duplicates are allowed. Therefore, it's possible that the cards can have repeated numbers. But in a standard deck, there are four suits, so you can have up to four of the same number. So if we are considering a friendly hand as four cards (possibly multiple copies of the same number, up to four), then we need to consider that.But the problem says "different friendly hands". So perhaps we are to count distinct combinations where numbers can repeat, but the combination is considered different only based on the multiset. So, for example, three 7s and a 3 is one combination, and two 7s, one 6, and one 4 is another, etc., and we need to count all such unique multisets where the largest number is 7, the sum is 24.Therefore, perhaps the problem is in the context of a standard deck where there are four of each number, but since we are just looking at combinations (not considering suits), we can have up to four of the same number. Wait, but in the example, there are three 7s. So, possible.But let me confirm: if we have four 7s, the sum would be 7*4=28, which is higher than 24. So four 7s is not possible. The maximum number of 7s we can have is three, as in the example, 7+7+7+3=24.Therefore, in order to have the largest number as 7, all other numbers must be ≤7, and at least one 7 is present. So the problem is to find all multisets of four numbers (with possible repetition up to four, but in practice, since 7 is the largest, and each number is at least 1, let's think about the possible combinations.So, since the numbers can be from 1 to 7, with at least one 7, and the sum is 24.Let me approach this step by step.First, consider that we need to count all 4-element multisets with elements in {1,2,3,4,5,6,7}, containing at least one 7, summing to 24.Alternatively, since the largest number is 7, all numbers are ≤7, and there's at least one 7.But since we need the sum to be 24, which is quite high. Let's see: the maximum possible sum with four 7s is 28. So 24 is 4 less than 28. So we need to reduce the total by 4 from 28 by decreasing some of the numbers. But since we must have at least one 7, we can't have all numbers less than 7.Wait, but if the largest number is 7, that means all numbers are ≤7 and at least one is equal to 7. So, in other words, when forming such combinations, we can have 1 to 4 numbers as 7, but since 4 numbers of 7 sum to 28, which is too much. So we need to decrease some of the numbers from 7 to lower numbers such that the total sum becomes 24. Therefore, the total reduction from 28 (the maximum sum) is 4. So we have to distribute this reduction of 4 among the four numbers, but with the constraints that:1. Each number can be reduced by a certain amount, but since numbers can't exceed 7 and can't be less than 1, the maximum reduction for any single number is 6 (from 7 to 1), but in our case, the total reduction needed is 4, so the individual reductions can be 0 to 4, but we have to make sure that no number is reduced below 1.Wait, perhaps another way to model this is: let’s think of each card as starting at 7, and then subtracting some value to reach the desired number. However, since we need at least one 7 (the largest number), we can't subtract from all four numbers. Wait, actually, no: if we start with four 7s (sum 28) and subtract a total of 4, but we must leave at least one 7 untouched (since the largest number is 7, we can't have all numbers reduced below 7). Wait, actually, no. If we start with four 7s and subtract some amount, even if we subtract from all four, as long as at least one 7 remains, then the largest number is still 7. But if we subtract so much that all four numbers are below 7, then the largest number would be less than 7, which is invalid. Therefore, in this model, we need to subtract a total of 4 from the four 7s, but ensure that at least one 7 remains (i.e., at least one of the four numbers isn't reduced at all).Therefore, the problem reduces to finding the number of non-negative integer solutions to the equation:x1 + x2 + x3 + x4 = 4,where each xi ≤ 6 (since each number can be reduced by at most 6, from 7 to 1), and at least one xi = 0 (since at least one number remains 7). But since the total reduction is 4, which is less than 6*4=24, the upper bounds on xi (each ≤6) are automatically satisfied because each xi can be at most 4 (since total is 4). So, the only constraint is that at least one xi is 0.Therefore, the number of solutions is equal to the number of non-negative integer solutions to x1 + x2 + x3 + x4 =4 minus the number of solutions where all xi ≥1.This is a classic stars and bars problem with inclusion-exclusion.Total solutions without any restrictions: C(4 + 4 -1, 4 -1) = C(7,3) = 35.Number of solutions where all xi ≥1: We subtract 1 from each variable, so y1 + y2 + y3 + y4 = 4 -4 = 0. There's only 1 solution here (all yi=0). Therefore, the number is C(0 +4 -1, 4 -1)=C(3,3)=1. Wait, but according to stars and bars, the number of solutions for y1 + y2 + y3 + y4 =0 with yi ≥0 is 1 (all zeros). So, the number of solutions where all xi ≥1 is 1. Therefore, the number of solutions with at least one xi=0 is 35 -1=34.But wait, this would mean there are 34 ways to distribute the reduction of 4 among the four numbers, keeping at least one number as 7. However, each such distribution corresponds to a different combination of numbers. But we have to remember that the order doesn't matter because we are dealing with multisets. Wait, but in the stars and bars model, the solutions are ordered. For example, (4,0,0,0) is different from (0,4,0,0), etc. However, since we are considering multisets, these two would be considered the same multiset: one number reduced by 4 (i.e., 7-4=3) and three numbers remaining 7. But in reality, if we have four numbers: 3,7,7,7. Since order doesn't matter, this is considered one multiset.Therefore, if we use stars and bars to count ordered solutions, we have 34 ordered solutions, but since the problem is about multisets (unordered combinations), we need to count the number of unordered combinations. Therefore, the initial approach might not be correct.So perhaps my initial approach is flawed because stars and bars counts ordered solutions, but we need unordered multisets.Therefore, I need to adjust my strategy.Alternative approach: Since the problem is about multisets, the order doesn't matter. So we need to find the number of distinct multisets {a, b, c, d} where each a, b, c, d ≤7, at least one of them is 7, and a + b + c + d =24.So, perhaps instead of modeling the problem as subtracting from 28, we can model it as solving a + b + c + d =24 with 1 ≤ a, b, c, d ≤7 and at least one of a,b,c,d=7.Alternatively, to rephrase, let’s let variables x1, x2, x3, x4 be the numbers on the cards, each between 1 and 7 inclusive, with max(x1, x2, x3, x4)=7, and x1 + x2 + x3 +x4=24.To count such multisets, perhaps it's easier to compute the total number of 4-card hands with numbers ≤7 that sum to 24, and subtract those that sum to 24 but have all numbers ≤6.Therefore, the number we want is equal to:Number of multisets with numbers ≤7 summing to 24 - Number of multisets with numbers ≤6 summing to 24.But we need to check if this is valid. Because the first term counts all multisets with numbers ≤7 and sum 24 (including those that may or may not have a 7), but actually, if all numbers are ≤6, their maximum is ≤6, so the difference between the two would be the number of multisets that include at least one 7. Wait, yes, exactly. So if we compute the total number of multisets with numbers ≤7 and sum 24, and subtract the number of multisets with numbers ≤6 and sum 24, the result will be the number of multisets with numbers ≤7, sum 24, and at least one 7. Which is exactly what we want. Therefore, this method can work.Therefore, we can compute:N = C(24 -1, 4 -1) - C(24 -1 -4*1, 4 -1) ??? Wait, no, that's not right.Wait, actually, the standard stars and bars formula for the number of multisets (with possible repetition) of k numbers that sum to n is C(n -1, k -1). However, when there is an upper limit on the numbers, the formula becomes more complex. So the number of solutions to a + b + c + d =24 where each a,b,c,d ≥1 and ≤7 is equal to ... Hmm, inclusion-exclusion.Alternatively, since we need to count the number of multisets (order doesn't matter), with each element between 1 and 7 inclusive, sum 24. Then subtract the number of multisets where each element is between 1 and 6 inclusive, sum 24.But how do we compute these?Alternatively, we can model each number as a variable from 1 to 7, so for the first count (numbers ≤7), we can let yi = xi -1, so yi ≥0 and yi ≤6, and y1 + y2 + y3 + y4 =24 -4=20. Then the number of non-negative integer solutions to y1 + y2 + y3 + y4=20 with each yi ≤6.Similarly, for numbers ≤6, it would be yi = xi -1, yi ≤5, sum=20.But this seems complicated. The standard way to compute the number of non-negative integer solutions to y1 + y2 + y3 + y4=20 with yi ≤6 is inclusion-exclusion.The formula is:C(20 +4 -1,4 -1) - C(4,1)C(20 -7 +4 -1,4 -1) + C(4,2)C(20 -14 +4 -1,4 -1) - C(4,3)C(20 -21 +4 -1,4 -1) + C(4,4)C(20 -28 +4 -1,4 -1)But when the terms inside the combinations become negative, they are zero.So let's compute the first term: C(23,3). Then subtract 4*C(20 -7 +4 -1,3)=4*C(16,3). Then add C(4,2)*C(20 -14 +4 -1,3)=6*C(9,3). Subtract C(4,3)*C(20 -21 +4 -1,3)=4*C(2,3). Then add C(4,4)*C(20 -28 +4 -1,3)=1*C(-5,3). Since C(n,k) is zero if n<k.Therefore:First term: C(23,3) = 1771.Second term: 4*C(16,3)=4*560=2240.Third term: 6*C(9,3)=6*84=504.Fourth term: 4*C(2,3)=4*0=0.Fifth term: 1*C(-5,3)=0.Therefore, total solutions for numbers ≤7: 1771 -2240 +504=1771 -2240 is -469, plus 504 is 35.Wait, 1771 -2240 = -469; -469 +504=35.Similarly, for numbers ≤6: yi ≤5, sum=20.Similarly, using inclusion-exclusion.The formula is:C(23,3) - C(4,1)C(20 -6 +4 -1,3) + C(4,2)C(20 -12 +4 -1,3) - C(4,3)C(20 -18 +4 -1,3) + C(4,4)C(20 -24 +4 -1,3)Compute each term:First term: C(23,3)=1771.Second term: 4*C(20 -6 +4 -1,3)=4*C(17,3)=4*680=2720.Third term: 6*C(20 -12 +4 -1,3)=6*C(11,3)=6*165=990.Fourth term: 4*C(20 -18 +4 -1,3)=4*C(5,3)=4*10=40.Fifth term: 1*C(20 -24 +4 -1,3)=C(-1,3)=0.Therefore, total solutions for numbers ≤6:1771 -2720 +990 -40=1771 -2720= -949; -949 +990=41; 41 -40=1.So the number of solutions for numbers ≤6 is 1.Therefore, the number of multisets with numbers ≤7 and sum 24 is 35, and with numbers ≤6 is 1. Therefore, the desired number is 35 -1=34.But wait, that seems a bit high. The example given is one such multiset: three 7s and a 3. Let's check if 34 makes sense.But hold on, this result suggests there are 34 different multisets where the numbers are between 1 and 7, sum to 24, and include at least one 7. But considering that the maximum sum with four numbers ≤7 is 28, and we need a sum of 24, which is 4 less than maximum. So we need to decrease the total sum by 4. Each decrease is done by reducing some of the numbers from 7. Each 7 can be reduced by some amount, but since we are dealing with multisets, the order doesn't matter.So, for example, reducing one 7 by 4 gives 7,7,7,3 (as in the example). Reducing two 7s by 2 each gives 7,7,5,5. Reducing one 7 by 3 and another by 1 gives 7,7,6,4. Etc. So each possible way to partition the total reduction of 4 into up to four parts, each part being the amount subtracted from a 7, considering that the subtraction can't make any number less than 1 (i.e., we can't subtract more than 6 from any 7, but since the total subtraction is 4, this constraint is automatically satisfied).But since we are dealing with multisets, each distinct way of partitioning the reduction corresponds to a different multiset. So the number of such partitions is equal to the number of integer partitions of 4 into at most four parts, where each part is a non-negative integer, and considering that the parts are unordered.Wait, yes. Since the order of the parts doesn't matter (because it's a multiset), the number of distinct multisets is equal to the number of integer partitions of 4 into at most 4 parts, where each part represents the amount subtracted from a 7. However, since we have four numbers, each part can be from 0 to 4, but since it's a multiset, different permutations of the same partition are considered the same.Wait, but in this case, each "part" corresponds to the amount subtracted from each 7. But since the original numbers are all 7s, subtracting different amounts from different 7s would result in different numbers. For example, subtracting 4 from one 7 gives 3, and leaving the others as 7, which gives the multiset {7,7,7,3}. Subtracting 3 from one 7 and 1 from another gives {7,7,4,6}, which is different. Similarly, subtracting 2 from two different 7s gives {7,7,5,5}. Subtracting 2 from one and 1 from two others gives {7,5,6,6}, etc. So each distinct partition of 4 into up to four parts (with 0 allowed) corresponds to a unique multiset.But since the parts are unordered, the number of such partitions is the number of integer partitions of 4, where each partition can have any number of parts (since we can have zeros, which correspond to not subtracting from some 7s). However, since we are dealing with exactly four numbers (four cards), each subtraction corresponds to a part, including zeros. So essentially, the problem reduces to the number of multisets of four non-negative integers that sum to 4, where the multisets are considered different even if they contain the same integers in different order. Wait, but no, since we are dealing with the original four 7s, each subtraction is applied to a specific 7. However, since the resulting numbers are a multiset, the order of subtraction doesn't matter. Therefore, different distributions of subtractions that lead to the same multiset are considered the same.Therefore, it's equivalent to the number of distinct multisets resulting from subtracting a total of 4 from four 7s, which is the same as the number of integer partitions of 4, where each part is the amount subtracted from a 7, and the order doesn't matter.So let's list all integer partitions of 4:1. 4: Subtract 4 from one 7. Result: {7,7,7,3}2. 3+1: Subtract 3 from one 7 and 1 from another. Result: {7,7,4,6}3. 2+2: Subtract 2 from two 7s. Result: {7,7,5,5}4. 2+1+1: Subtract 2 from one 7 and 1 from two others. Result: {7,6,6,5}5. 1+1+1+1: Subtract 1 from four 7s. Result: {6,6,6,6}6. 4+0: Wait, but 4 is already considered in the first partition. Since we are considering partitions where parts can be zero, but since the order doesn't matter, zeros can be added. However, since the problem requires at least one 7, which corresponds to at least one part being zero. Wait, no: in the subtraction model, the parts correspond to the amounts subtracted. If a part is zero, that means that particular 7 wasn't subtracted from, so it remains 7. Therefore, in the partition of the total subtraction (4), we can have parts that are zero, but since we need at least one 7, we need at least one part to be zero. Wait, but no: the total subtraction is 4, distributed over four numbers. The requirement is that the largest number is 7, which requires that at least one number remains 7 (i.e., at least one part is zero). However, in the previous calculation using inclusion-exclusion, we subtracted all cases where all numbers are ≤6, which corresponds to all subtractions being ≥1 (i.e., no zeros). Therefore, in the integer partition model, we need to count all partitions of 4 into up to four parts (each part ≥0), where at least one part is zero. Since each part represents the amount subtracted from a 7, so a part of zero means that 7 wasn't subtracted.But in the integer partitions, parts are usually positive. However, in this case, since we can have zeros, but the partitions are unordered, it's a bit different. So let's think of it as the number of distinct multisets of four non-negative integers (each representing the amount subtracted from a 7) that sum to 4, with at least one of them being zero.Alternatively, since the order doesn't matter, it's equivalent to the number of distinct combinations where we have some number of zeros and the remaining numbers sum to 4. Let me approach it this way.Case 1: Exactly one zero (i.e., three subtractions). Then, we need to distribute 4 among three parts, each ≥1. The number of partitions is the number of integer partitions of 4 into three positive integers.The partitions of 4 into three positive integers are:- 2+1+1Only one such partition. So this corresponds to subtracting 2, 1, 1 from three 7s, resulting in {7,6,6,5}.But wait, 4=2+1+1+0 (since it's four parts with one zero). Wait, actually, no. If there is exactly one zero, then the other three parts must sum to 4, each ≥1. So the possible partitions are:- 2+1+1 (sum 4)- 3+1+0 (but we already considered exactly one zero, so the other three parts must be ≥1). Wait, no. Wait, if exactly one zero, then the other three parts must be ≥1. Therefore, the possible partitions of 4 into three positive integers are:4=2+1+1 (only one way). Because 3+1+0 would have a zero, but we already fixed exactly one zero. So I think there's only one partition here.Case 2: Exactly two zeros. Then we need to distribute 4 between two parts, each ≥1.Partitions of 4 into two positive integers:- 3+1- 2+2So two partitions. These correspond to:- 3+1+0+0: Subtracting 3 and 1 from two 7s, resulting in {7,7,4,6}- 2+2+0+0: Subtracting 2 from two 7s, resulting in {7,7,5,5}Case 3: Exactly three zeros. Then we need to distribute 4 into one part (since three are zeros). The partition is just 4, so one partition:- 4+0+0+0: Subtracting 4 from one 7, resulting in {7,7,7,3}Case 4: Exactly four zeros: All parts are zero, which would mean subtracting 0 from all, giving {7,7,7,7} which sums to 28, not 24. So this is invalid. Hence, we don't consider this.Additionally, we have to consider partitions where some parts are zero but the number of zeros is less. Wait, no, we already covered the cases with exactly 1, 2, 3 zeros.But wait, when exactly three zeros, we have one part of 4. When exactly two zeros, two parts summing to 4. When exactly one zero, three parts summing to 4. However, the total number of partitions would be 1 (case1) +2 (case2) +1 (case3) =4. But earlier calculation with inclusion-exclusion gave 34. There's a discrepancy here. Clearly, my manual counting here is missing something.Wait, no, actually, in this manual counting, each case corresponds to a unique multiset. For example:Case3: {7,7,7,3} (subtract 4 from one 7)Case2: {7,7,4,6} and {7,7,5,5}Case1: {7,6,6,5}But then there's also another case where subtracting 1 from four different 7s: {6,6,6,6}, which is subtracting 1 four times. But in our previous cases, we considered exactly one, two, or three zeros. But in this case, subtracting 1 from each of four 7s (i.e., subtracting 1 four times) would correspond to the partition 1+1+1+1, which has zero zeros (since all parts are 1). Wait, but in this case, all numbers are reduced by 1, so all original 7s become 6s. So the resulting multiset is {6,6,6,6}, which sums to 24. However, this multiset has all numbers=6, which are less than 7, so it's invalid because the largest number must be 7. Therefore, this case should be excluded. But according to the inclusion-exclusion method earlier, we subtracted this case (since it's part of the numbers ≤6). So in our manual counting above, if we consider the cases where at least one 7 remains (i.e., at least one part is zero), then the case where all numbers are reduced (no zeros) is excluded. But in my manual counting, I didn't consider that case because we are only counting cases with at least one zero.But according to the integer partition approach, when we distribute the total subtraction of 4 into four parts (including zeros), the number of multisets with at least one zero is equal to the number of integer partitions of 4 into up to four parts, where at least one part is zero. However, integer partitions typically don't include zeros, so this is a bit tricky.Alternatively, thinking of it as compositions (where order matters), but with multisets, order doesn't matter. So perhaps the total number of multisets with four numbers summing to 24, each ≤7, at least one 7, is equivalent to the number of integer partitions of 4 (the total reduction) into at most four parts (each part ≥0), where at least one part is zero. But since the parts represent the amount subtracted from each 7, and order doesn't matter, we need to count the number of distinct multisets of subtraction amounts that sum to 4 with at least one zero.But this is getting confusing. Let's try enumerating all possible multisets:Starting with the example {7,7,7,3} (subtract 4 from one 7)Then {7,7,6,4} (subtract 3 and 1 from two 7s){7,7,5,5} (subtract 2 from two 7s){7,6,6,5} (subtract 2 from one 7 and 1 from two 7s){7,6,5,6} Wait, that's the same as {7,6,6,5}{7,5,5,7} same as {7,7,5,5}Wait, but what about subtracting 4 in other ways?Subtract 2 from one 7, 1 from one 7, and 1 from another: {7,5,6,6} (same as {7,6,6,5})Subtract 1 from three 7s and 1 from another: Wait, subtract 1 four times would lead to {6,6,6,6}, which is invalid.Wait, but how do we get other combinations?Wait, if we subtract 3 from one 7 and 1 from another, we get {7,7,4,6}Subtract 2 from one 7 and 2 from another, we get {7,7,5,5}Subtract 2 from one 7, 1 from one 7, and 1 from another, we get {7,6,6,5}Subtract 4 from one 7, we get {7,7,7,3}Are there any other possibilities?What if we subtract 3 from one 7 and leave the others: {7,7,7,4} (sum 7+7+7+4=25, which is too much). Wait, no, subtracting 3 from a 7 gives 4, so the total sum would be 7+7+7+4=25, which is not 24. Wait, but we need to subtract a total of 4. So if we subtract 3 from one 7 (total subtraction 3), we need to subtract 1 more somewhere else. For example, subtract 3 and 1: total subtraction 4, giving {7,7,4,6}.Similarly, subtracting 2 and 2: {7,7,5,5}Subtracting 2,1,1: {7,6,6,5}Subtracting 4: {7,7,7,3}Are there other ways to subtract 4?If we subtract 2 from one 7 and 1 from two others, that's 2+1+1=4, but that's already considered in the {7,6,6,5} case.If we subtract 3 from one 7 and 1 from another, that's 3+1=4, which is the {7,7,4,6} case.If we subtract 4 from one 7, that's the first case.If we subtract 2+2, that's the second case.Is there a way to subtract 4 in a different way?What about subtracting 3+1+0+0? Which is the same as 3+1.Or 2+1+1+0.Wait, no, these are covered in the above cases.So seems like there are only four distinct multisets:1. {7,7,7,3}2. {7,7,4,6}3. {7,7,5,5}4. {7,6,6,5}But wait, what about {7,6,7,4}? That's the same as {7,7,4,6}, so no, it's not a new multiset.Similarly, {7,5,7,5} is the same as {7,7,5,5}.So indeed, only four distinct multisets. However, earlier the inclusion-exclusion gave 34, which is clearly incorrect. So there's a mistake in that approach.Wait, but earlier when I used inclusion-exclusion, I was counting ordered solutions. The formula gave 35 total solutions for numbers ≤7 and 1 solution for numbers ≤6, resulting in 34 ordered solutions. But since we need unordered multisets, each multiset is counted multiple times in the ordered solutions.For example, the multiset {7,7,7,3} corresponds to 4 ordered solutions (depending on which position the 3 is in), but since we are considering multisets, we need to count it once. Similarly, {7,7,4,6} corresponds to 12 ordered solutions (4 choose 2 positions for the 7s, then 2 positions for 4 and 6), but as a multiset, it's one. Therefore, the 34 ordered solutions correspond to multiple multisets, each counted multiple times.Therefore, the inclusion-exclusion method counts ordered solutions, but we need unordered. Hence, the previous approach was incorrect.Therefore, we need a different method to count the number of multisets.Let me try another approach. Since the largest number is 7, and the sum is 24, and we have four numbers. Let’s denote the numbers as a, b, c, d where each ≤7, at least one is 7, and a + b + c + d =24.Since we are dealing with multisets, we can assume a ≤ b ≤ c ≤ d =7. Therefore, the largest number is 7, and the other three numbers are ≤7. So we can rephrase the problem as finding the number of triples (a, b, c) where a ≤ b ≤ c ≤7, and a + b + c +7 =24 → a + b + c =17.Therefore, we need to find the number of triples (a, b, c) with a ≤ b ≤ c ≤7 and a + b + c =17.Each such triple corresponds to a multiset {a, b, c,7}, which sums to 24. Since a ≤ b ≤ c ≤7, all numbers are ≤7 and the largest is 7.So the problem reduces to finding the number of triples (a, b, c) with 1 ≤ a ≤ b ≤ c ≤7 and a + b + c =17.So we can approach this by enumerating all possible triples.Let’s proceed step by step. Since a ≤ b ≤ c ≤7, and they are at least 1. Also, their sum is 17.First, note that the minimum possible sum of a, b, c is 1 +1 +1=3, and maximum is 7 +7 +7=21. We need sum=17.Given that c ≤7, the largest number c can be 7. Let's consider possible values of c from 7 downwards and see what possible a and b can be.Case 1: c=7. Then a + b +7=17 → a + b=10, with a ≤ b ≤7.So a and b must satisfy a ≤ b ≤7 and a + b=10.Possible pairs (a, b):- a=3, b=7 (3+7=10)- a=4, b=6 (4+6=10)- a=5, b=5 (5+5=10)These are the possible pairs. So three possibilities here.Case 2: c=6. Then a + b +6=17 → a + b=11, with a ≤ b ≤6.But since a ≤ b ≤6, the maximum a + b=6+6=12, so 11 is possible.Possible pairs (a, b):- a=5, b=6 (5+6=11)- a=6, b=5 → but since a ≤ b, it's (5,6)But a and b must be ≤6. So only one pair: (5,6)But wait, 5+6=11, and since a ≤ b, only (5,6) is valid.Wait, but a + b=11 with a ≤ b ≤6.So the minimum value of a is such that a ≤6 and b=11 -a ≤6.So 11 -a ≤6 → a ≥5. Since a ≤b ≤6, a can be 5 or 6.If a=5, then b=6 (since 5 +6=11)If a=6, then b=5, but this contradicts a ≤b. So only (5,6).Thus, one possibility.Case 3: c=5. Then a + b +5=17 → a + b=12, with a ≤ b ≤5.But the maximum a + b with a ≤ b ≤5 is 5+5=10 <12. Impossible. No solutions.Similarly, c=4 or lower: a + b would need to be even higher, which is impossible since a and b are ≤c, which is ≤4. So no solutions.Therefore, total cases:- Case1 (c=7): 3 multisets- Case2 (c=6): 1 multiset- Cases3 and below: 0Total: 4 multisets.But wait, in this approach, we have 4 distinct multisets. But earlier, when I enumerated them manually, I had four multisets:1. {7,7,7,3}2. {7,7,4,6}3. {7,7,5,5}4. {7,6,6,5}But according to this case analysis:- Case1 (c=7): a + b=10. The possible triples (a,b,7): - (3,7,7): which corresponds to the multiset {3,7,7,7} - (4,6,7): {4,6,7,7} - (5,5,7): {5,5,7,7} So three multisets.- Case2 (c=6): a + b=11. The triple is (5,6,6): {5,6,6,7}So total four multisets. Therefore, the answer should be 4. However, this contradicts the earlier inclusion-exclusion result of 34. Clearly, the error was in the initial stars and bars approach which counted ordered solutions, not considering that the problem asks for multisets.But wait, let's check if there are more possibilities.Wait, when c=7 and a + b=10, we found (3,7), (4,6), (5,5). But what about other combinations?For example, a=2, b=8, but b cannot exceed 7. Similarly, a=1, b=9, impossible. So no, those are the only ones.Similarly, when c=6, only (5,6) works.Therefore, total four multisets.But in the example given, there's one multiset: {7,7,7,3}. So according to this, there are four such multisets. But wait, the initial problem statement says that Wang Liang drew the cards 7,7,7,3 and found that they sum to 24. So this is one of the four.But according to our analysis, there are four different friendly hands where the largest number is 7. But let's verify each of these combinations:1. {7,7,7,3}: Sum=7+7+7+3=24 ✔️2. {7,7,4,6}: Sum=7+7+4+6=24 ✔️3. {7,7,5,5}: Sum=7+7+5+5=24 ✔️4. {7,6,6,5}: Sum=7+6+6+5=24 ✔️Yes, these four all add up to 24. Are there any others?Let me check:Is there a combination like {7,7,6,4}? Wait, that's the same as {7,7,4,6}, which is already counted.What about {7,6,7,4}? Still the same multiset.What about {7,5,7,5}? Same as {7,7,5,5}.{7,6,5,6} is {5,6,6,7}, which is the fourth multiset.So there are indeed only four unique multisets. Therefore, the correct answer should be 4.But why did the inclusion-exclusion method give 34? Because that method was counting ordered solutions (i.e., considering different permutations as distinct), whereas the problem requires counting multisets (unordered combinations). Therefore, the correct answer is 4.But let me confirm with another approach. Let's list all possible combinations.We need four numbers, each at least 1, at most 7, summing to 24, with the largest being 7.So, the numbers can include:- One 7 and three other numbers.- Two 7s and two other numbers.- Three 7s and one other number.- Four 7s (but 4×7=28>24, so invalid).Let's check each case.Case 1: Four 7s: Invalid.Case 2: Three 7s and one other number.Sum of three 7s is 21, so the remaining number must be 3. So the combination is {7,7,7,3}. This is one combination.Case3: Two 7s and two other numbers.Sum of two 7s is 14. So the remaining two numbers must sum to 24 -14=10. These two numbers must be ≤7 and ≥1. So we need two numbers a and b such that a + b=10, 1 ≤a ≤b ≤7.Possible pairs:- 3 +7=10 (but 7 is allowed, but since we already have two 7s, this would make a total of four 7s, but wait, no. Wait, the two remaining numbers can include 7s. But the total number of 7s would be two (from the two 7s) plus any 7s in the remaining numbers. However, the problem states that the largest number is 7, so the remaining numbers can be up to 7. So for example, if the remaining two numbers are 7 and3, then the combination is {7,7,7,3}, which is already counted in case2.Wait, but if we have two 7s and two other numbers, one of which is 7, then it becomes three 7s and one other number. So to avoid overcounting, we need to ensure that in case3, the remaining two numbers are both ≤7 and less than or equal to 7, but strictly less than 7 if we want to avoid overlapping with case2.But no, the problem allows multiple 7s, so actually, in case3, two 7s and two other numbers that can also be 7s. But wait, if the two other numbers are both 7s, then we have four 7s, which is case1 (invalid). If one of them is 7, then we have three 7s and one other number (case2). Therefore, to have exactly two 7s, the remaining two numbers must be ≤7 but not 7. So we need two numbers a and b such that a + b=10, 1 ≤a ≤b ≤6.Possible pairs:- 4 +6=10- 5 +5=10These are the only pairs since 6 is the maximum for a and b. So two possibilities:- {7,7,4,6}- {7,7,5,5}Therefore, two combinations in case3.Case4: One 7 and three other numbers.Sum of one 7 is 7, so the remaining three numbers must sum to 24 -7=17. The remaining three numbers must be ≤7 and ≥1. So we need three numbers a ≤b ≤c ≤7 such that a + b + c=17.But the maximum sum of three numbers ≤7 is 7+7+7=21, which is more than 17, but we need to check if 17 is achievable.Let's find all triples a ≤b ≤c ≤7 with a + b + c=17.Start with c=7: Then a + b=10, a ≤b ≤7.Possible pairs (a, b):- 3 +7=10 → a=3, b=7, but c=7, so the triple is {3,7,7}But since we need three numbers, the multiset would be {7,3,7,7} → which is three 7s and a3, which is already case2.Wait, but in case4, we have one 7 and three other numbers. However, if the three other numbers include a 7, then it becomes two 7s. So to have exactly one 7, the remaining three numbers must not include any 7s.Therefore, in case4, the remaining three numbers must be ≤6. So we need a + b + c=17 with a ≤b ≤c ≤6. But the maximum sum of three numbers ≤6 is 6+6+6=18, which is greater than 17, so possible.But let's see:Find triples a ≤b ≤c ≤6 with sum 17.Start with c=6: Then a + b=11. a ≤b ≤6.But the maximum a + b with a ≤b ≤6 is 6+6=12. 11 is possible.So a + b=11 with a ≤b ≤6.Possible pairs:- a=5, b=6 (5+6=11)So the triple is {5,6,6}, sum=17. Therefore, the combination is {7,5,6,6}, which is a valid multiset.Next, c=5: Then a + b=12, but with a ≤b ≤5, maximum sum is 5+5=10 <12. Impossible.Similarly, lower c won't help.Therefore, only one triple: {5,6,6}, sum=17. So the multiset is {7,5,6,6}.But wait, this is the same as {7,6,6,5}, which is one combination.Therefore, case4 gives one combination.Case5: Zero 7s. Not allowed, since the largest number must be7.Therefore, total combinations:- case2:1- case3:2- case4:1Total:4Therefore, the answer is 4.But why did the stars and bars approach give 34? Because it counted ordered solutions. For example, the multiset {7,7,7,3} can be arranged in 4 different ways (depending on which card is the 3), so in the ordered count, it's 4 solutions. Similarly, {7,7,4,6} can be arranged in C(4,2)=6 ways (choose two positions for the 7s, then two for 4 and6), but since 4 and6 are distinct, it's 6×2=12 ordered solutions. Similarly, {7,7,5,5} can be arranged in C(4,2)=6 ways (choose two positions for 7s, the other two are 5s). {7,6,6,5} can be arranged in C(4,1)×C(3,2)=4×3=12 ways (choose position for7, then two positions for6s, remaining is5). So total ordered solutions:- {7,7,7,3}:4- {7,7,4,6}:12- {7,7,5,5}:6- {7,6,6,5}:12Total:4+12+6+12=34, which matches the inclusion-exclusion result. However, since the problem asks for different friendly hands (i.e., multisets), the correct answer is 4.Therefore, the initial approach using stars and bars (inclusion-exclusion) was incorrect because it counted ordered solutions, whereas the problem requires unordered multisets. The correct answer is 4.Final Answerboxed{4}

✅:To determine how many different friendly hands have a largest number of 7, we need to find the number of multisets of four numbers (each ≤7) that sum to 24 and include at least one 7. 1. Key Constraints: - The sum of the four numbers must be 24. - The largest number must be 7. - Each number is between 1 and 7 inclusive.2. Approach: - We need to consider the possible combinations of numbers that sum to 24 with 7 as the largest number. - This involves checking different cases where the number of 7s varies from one to three (four 7s would exceed the sum).3. Case Analysis: - Case 1: Three 7s and one other number: - Sum of three 7s is 21. The remaining number must be 3. - Combination: {7, 7, 7, 3} - Case 2: Two 7s and two other numbers: - Sum of two 7s is 14. The remaining two numbers must sum to 10. - Possible pairs: (4, 6) and (5, 5) - Combinations: {7, 7, 4, 6} and {7, 7, 5, 5} - Case 3: One 7 and three other numbers: - Sum of one 7 is 7. The remaining three numbers must sum to 17. - Possible triple: (5, 6, 6) - Combination: {7, 6, 6, 5}4. Verification: - Each combination is verified to sum to 24 and includes at least one 7. - The distinct multisets found are: {7, 7, 7, 3}, {7, 7, 4, 6}, {7, 7, 5, 5}, and {7, 6, 6, 5}.Thus, the number of different friendly hands with a largest number of 7 is (boxed{4}).

🔑:1. We need to identify combinations of four numbers with the largest number being 7, that sum up to 24.2. List possible combinations where the largest number is 7: - Combination 1: Include three 7's and one additional number: [ 7 + 7 + 7 + 3 = 21 + 3 = 24 ] So, one group is ( 7, 7, 7, 3 ). - Combination 2: Include two 7's and two additional numbers: [ 7 + 7 + x + y = 24 ] - Solving when ( x = 6 ) and ( y = 4 ): [ 7 + 7 + 6 + 4 = 14 + 6 + 4 = 24 ] So, another group is ( 7, 7, 6, 4 ). - Solving when ( x = 5 ) and ( y = 5 ): [ 7 + 7 + 5 + 5 = 14 + 10 = 24 ] Another group is ( 7, 7, 5, 5 ). - Combination 3: Include one 7 and three additional numbers: [ 7 + x + y + z = 24 ] - Solving when ( x = 6 ), ( y = 6 ), and ( z = 5 ): [ 7 + 6 + 6 + 5 = 13 + 11 = 24 ] So, another group is ( 7, 6, 6, 5 ).3. Since we are examining groups that satisfy the condition: - ( 7, 7, 7, 3 ) - ( 7, 7, 6, 4 ) - ( 7, 7, 5, 5 ) - ( 7, 6, 6, 5 )4. Therefore, the total number of distinct “friendly card groups” containing the number 7 is: [4 text{ groups}]Conclusion:[boxed{4 text{ groups}}]

🧠:Alright, let's tackle this problem step by step. It's about two containers with saline solutions that are evaporating. The problem gives some information about the amount of salt obtained and the number of days each container stood, with some conditions about switching the days. Hmm, okay, let's parse this out.First, let's list what we know:1. From the first container, 48 kg of salt was obtained.2. From the second container, which stood 6 days less, 27 kg was obtained.3. If the first container had stood as many days as the second, and the second as many as the first, then the same amount of salt would have been obtained from both solutions.We need to find out how many days each solution stood. Let's denote the days for each container. Let me assign variables here. Let's say:Let ( x ) be the number of days the first container stood.Then, the second container stood ( x - 6 ) days because it stood 6 days less.Wait, but the problem says "from the second, which stood 6 days less, 27 kg was obtained." So yes, the second container was evaporated for 6 days less than the first. So if the first is ( x ) days, the second is ( x - 6 ).Now, the daily evaporated portions of salt are constant for each container. So each container has a constant rate of salt evaporation per day. Let's denote:Let ( r_1 ) be the daily evaporation rate (in kg/day) for the first container.Similarly, ( r_2 ) for the second container.So the total salt obtained from the first container is ( r_1 times x = 48 ) kg.Similarly, the total salt from the second container is ( r_2 times (x - 6) = 27 ) kg.So we have two equations:1. ( r_1 x = 48 )2. ( r_2 (x - 6) = 27 )But then there's the third condition: If the first container had stood as many days as the second, and the second as many as the first, then the same amount of salt would have been obtained from both solutions.Hmm, so swapping the number of days. That is, if the first container stood for ( x - 6 ) days (the time the second one actually stood), and the second container stood for ( x ) days (the time the first one actually stood), then the amount of salt from both would be equal.So in this hypothetical scenario:Salt from first container: ( r_1 times (x - 6) )Salt from second container: ( r_2 times x )And these are equal. So:( r_1 (x - 6) = r_2 x )Okay, so now we have three equations:1. ( r_1 x = 48 )2. ( r_2 (x - 6) = 27 )3. ( r_1 (x - 6) = r_2 x )So we can use these equations to solve for ( x ), ( r_1 ), and ( r_2 ).Let me see. Let's express ( r_1 ) and ( r_2 ) from the first two equations.From equation 1: ( r_1 = frac{48}{x} )From equation 2: ( r_2 = frac{27}{x - 6} )Now substitute these into equation 3:( frac{48}{x} times (x - 6) = frac{27}{x - 6} times x )Simplify both sides:Left side: ( frac{48(x - 6)}{x} )Right side: ( frac{27x}{x - 6} )So equation becomes:( frac{48(x - 6)}{x} = frac{27x}{x - 6} )Cross-multiplying to eliminate denominators:( 48(x - 6)^2 = 27x^2 )Let's divide both sides by 3 to simplify:16(x - 6)^2 = 9x^2Now expand the left side:16(x² - 12x + 36) = 9x²Multiply out:16x² - 192x + 576 = 9x²Bring all terms to one side:16x² - 192x + 576 - 9x² = 0Simplify:7x² - 192x + 576 = 0Now, this is a quadratic equation. Let's write it as:7x² - 192x + 576 = 0We can attempt to solve this quadratic for x using the quadratic formula. The quadratic formula is:x = [192 ± sqrt(192² - 4*7*576)] / (2*7)First compute discriminant D:D = 192² - 4*7*576Calculate 192²: 192*192. Let's compute:192*192: (200 - 8)^2 = 200² - 2*200*8 + 8² = 40000 - 3200 + 64 = 36864Then compute 4*7*576: 28*57628*576: 576*20 = 11520, 576*8=4608, so total 11520 + 4608 = 16128Therefore, D = 36864 - 16128 = 20736sqrt(20736). Let's see, 144² = 20736 (since 140²=19600, 140² + 4*140 + 16 = 19600 + 560 + 16=20176, which is less. 144*144: 144*100=14400, 144*40=5760, 144*4=576; sum 14400+5760=20160+576=20736. Yes, sqrt(20736)=144Thus, x = [192 ± 144]/14Compute both possibilities:First solution: (192 + 144)/14 = 336/14 = 24Second solution: (192 - 144)/14 = 48/14 = 24/7 ≈ 3.42857But x must be greater than 6, since the second container stood x - 6 days, which must be positive. If x ≈ 3.428, then x - 6 ≈ negative, which is impossible. Therefore, we discard the second solution.Hence, x = 24 days.Therefore, the first container stood 24 days, the second container stood 24 - 6 = 18 days.Wait, but let's verify this with the original equations to check for consistency.First, compute r1 = 48/x = 48/24 = 2 kg/dayr2 = 27/(x - 6) = 27/18 = 1.5 kg/dayNow, check the third condition: if the first container stood 18 days (the second's time), then salt would be r1 * 18 = 2 * 18 = 36 kgIf the second container stood 24 days (the first's time), salt would be r2 *24 = 1.5 *24 = 36 kgSo yes, both would give 36 kg. That matches the condition. So it's correct.Therefore, the first container stood 24 days, the second 18 days.Wait, but let me just check again all steps to make sure.Starting with equations:1. 2 kg/day *24 days =48 kg ✔️2. 1.5 kg/day *18 days=27 kg ✔️Swapped days:First container: 2 kg/day *18 days=36 kgSecond container:1.5 kg/day *24 days=36 kg ✔️All conditions are satisfied.So the answer is 24 days for the first container and 18 days for the second.

✅:The first container stood for 24 days and the second container stood for 18 days.Step-by-Step Explanation:1. Define Variables: - Let ( x ) = days the first container stood. - Then, the second container stood ( x - 6 ) days.2. Evaporation Rates: - Let ( r_1 ) = daily evaporation rate (kg/day) of the first container. - Let ( r_2 ) = daily evaporation rate (kg/day) of the second container.3. Form Equations from Given Data: - From the first container: ( r_1 cdot x = 48 ) → ( r_1 = frac{48}{x} ) - From the second container: ( r_2 cdot (x - 6) = 27 ) → ( r_2 = frac{27}{x - 6} )4. Condition with Swapped Days: - If the first container stood ( x - 6 ) days and the second stood ( x ) days, their salt amounts would be equal: ( r_1 cdot (x - 6) = r_2 cdot x )5. Substitute ( r_1 ) and ( r_2 ): - ( frac{48}{x} cdot (x - 6) = frac{27}{x - 6} cdot x ) - Simplify: ( frac{48(x - 6)}{x} = frac{27x}{x - 6} )6. Cross-Multiply and Solve for ( x ): - ( 48(x - 6)^2 = 27x^2 ) - Divide by 3: ( 16(x - 6)^2 = 9x^2 ) - Expand: ( 16x^2 - 192x + 576 = 9x^2 ) - Simplify: ( 7x^2 - 192x + 576 = 0 )7. Quadratic Formula: - Discriminant ( D = 192^2 - 4 cdot 7 cdot 576 = 20736 ) - ( sqrt{D} = 144 ) - Solutions: ( x = frac{192 pm 144}{14} ) - Valid solution: ( x = 24 ) (days)8. Determine Days for Each Container: - First container: ( x = 24 ) days - Second container: ( x - 6 = 18 ) daysVerification:- Evaporation rates: ( r_1 = 2 , text{kg/day}, , r_2 = 1.5 , text{kg/day} )- Swapped days: Both yield 36 kg, confirming the condition.Answer: The first solution stood for (boxed{24}) days and the second for (boxed{18}) days.

🔑:1. Let's denote the number of days the first vessel stood as ( x ) and the number of days the second vessel stood as ( y ).2. The daily evaporated portion for the first vessel is ( frac{48}{x} ) kg of salt. Similarly, the daily evaporated portion for the second vessel is ( frac{27}{y} ) kg of salt.3. According to the problem, the second vessel stood 6 days less than the first vessel. Therefore, we can write: [ y = x - 6 ]4. If the first vessel were to stand for the same number of days as the second vessel, and vice versa, both solutions would produce the same amount of salt. This gives us: [ frac{48}{x} cdot y = frac{27}{y} cdot x ]5. Substitute ( y = x - 6 ) into the equation: [ frac{48}{x} cdot (x - 6) = frac{27}{x - 6} cdot x ]6. Simplify both expressions: [ frac{48(x - 6)}{x} = frac{27x}{x - 6} ]7. By cross-multiplying, we get: [ 48(x - 6)(x - 6) = 27x^2 ]8. Simplify the equation: [ 48(x^2 - 12x + 36) = 27x^2 ]9. Expand and rearrange the equation: [ 48x^2 - 576x + 1728 = 27x^2 ] [ 21x^2 - 576x + 1728 = 0 ]10. This is a quadratic equation in the form ( ax^2 + bx + c = 0 ). Use the quadratic formula: [ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ] where ( a = 21 ), ( b = -576 ), and ( c = 1728 ).11. Calculate the discriminant: [ Delta = b^2 - 4ac = (-576)^2 - 4 cdot 21 cdot 1728 ] [ = 331776 - 145152 = 186624 ]12. Calculate the roots: [ x = frac{576 pm sqrt{186624}}{42} ] [ x = frac{576 pm 432}{42} ]13. Solving this we get two potential solutions: [ x = frac{576 + 432}{42} = 24 ] [ x = frac{576 - 432}{42} = frac{144}{42} approx 3.43 ] Since ( x ) must be a whole number: [ x = 24 ]14. Substituting ( x = 24 ) into ( y = x - 6 ): [ y = 24 - 6 = 18 ]Conclusion:The first vessel stood for 24 days and the second vessel stood for 18 days. [boxed{24 text{ days and } 18 text{ days}}]