💡To determine the probability that two randomly chosen cards from a standard 52-card deck are numbers (2 through 10) totaling to 15, we can break down the problem into two cases:**Case 1:** The first card is one of 6, 7, 8, or 9. There are 16 such cards (4 of each number). For each of these cards, the second card must be a specific number to sum to 15. The probability of drawing the first card is 16/52, and the probability of drawing the required second card is 4/51. Thus, the probability for this case is:[frac{16}{52} times frac{4}{51} = frac{64}{2652}]**Case 2:** The first card is a 10, and the second card is a 5. There are 4 tens and 4 fives. The probability of drawing a 10 first is 4/52, and then a 5 is 4/51. Thus, the probability for this case is:[frac{4}{52} times frac{4}{51} = frac{16}{2652}]Adding the probabilities of both cases gives the total probability:[frac{64}{2652} + frac{16}{2652} = frac{80}{2652}]Simplifying the fraction:[frac{80}{2652} = frac{10}{331}]Therefore, the probability that both cards are numbers totaling to 15 is (boxed{frac{10}{331}}).

💡Okay, so I have this problem where I'm given two quadratic equations: (a x^{2}+b x+c=0) and (-a x^{2}+b x+c=0). I need to prove that if (x_{1}) is a root of the first equation and (x_{2}) is a root of the second equation, then there exists a root (x_{3}) of the equation (frac{a}{2} x^{2}+b x+c=0) such that either (x_{1} leq x_{3} leq x_{2}) or (x_{1} geq x_{3} geq x_{2}). Hmm, okay. So, let me break this down. First, both equations are quadratic, so each should have two roots. But the problem is talking about one root from each equation, (x_{1}) and (x_{2}), and then showing that there's a root (x_{3}) of another quadratic equation that lies between them or outside of them. I think the key here is to consider the behavior of the quadratic functions and maybe use the Intermediate Value Theorem. Let me recall that the Intermediate Value Theorem says that if a function is continuous on an interval ([a, b]) and takes values (f(a)) and (f(b)) at each end of the interval, then it also takes any value between (f(a)) and (f(b)) at some point within the interval. So, maybe I can consider the function (P(x) = frac{a}{2} x^{2} + b x + c) and evaluate it at (x_{1}) and (x_{2}). If I can show that (P(x_{1})) and (P(x_{2})) have opposite signs, then by the Intermediate Value Theorem, there must be a root (x_{3}) between (x_{1}) and (x_{2}). Let me try that. First, since (x_{1}) is a root of (a x^{2} + b x + c = 0), plugging (x_{1}) into that equation gives (a x_{1}^{2} + b x_{1} + c = 0). Similarly, for (x_{2}), plugging into (-a x^{2} + b x + c = 0) gives (-a x_{2}^{2} + b x_{2} + c = 0).Now, let's compute (P(x_{1})):(P(x_{1}) = frac{a}{2} x_{1}^{2} + b x_{1} + c)But from the first equation, (a x_{1}^{2} = -b x_{1} - c). So, substituting that into (P(x_{1})):(P(x_{1}) = frac{a}{2} left( frac{-b x_{1} - c}{a} right) + b x_{1} + c)Simplifying:(P(x_{1}) = frac{-b x_{1} - c}{2} + b x_{1} + c)Combine like terms:(P(x_{1}) = left( frac{-b x_{1}}{2} + b x_{1} right) + left( frac{-c}{2} + c right))(P(x_{1}) = frac{b x_{1}}{2} + frac{c}{2})So, (P(x_{1}) = frac{1}{2}(b x_{1} + c))Similarly, let's compute (P(x_{2})):(P(x_{2}) = frac{a}{2} x_{2}^{2} + b x_{2} + c)From the second equation, (-a x_{2}^{2} + b x_{2} + c = 0), so (a x_{2}^{2} = b x_{2} + c). Substituting into (P(x_{2})):(P(x_{2}) = frac{a}{2} left( frac{b x_{2} + c}{a} right) + b x_{2} + c)Simplify:(P(x_{2}) = frac{b x_{2} + c}{2} + b x_{2} + c)Combine like terms:(P(x_{2}) = left( frac{b x_{2}}{2} + b x_{2} right) + left( frac{c}{2} + c right))(P(x_{2}) = frac{3 b x_{2}}{2} + frac{3 c}{2})So, (P(x_{2}) = frac{3}{2}(b x_{2} + c))Now, let's look at (P(x_{1})) and (P(x_{2})):(P(x_{1}) = frac{1}{2}(b x_{1} + c))(P(x_{2}) = frac{3}{2}(b x_{2} + c))Hmm, so if I can show that (P(x_{1})) and (P(x_{2})) have opposite signs, then by the Intermediate Value Theorem, there must be a root (x_{3}) between (x_{1}) and (x_{2}). But wait, is that necessarily the case? Let me think. Since (x_{1}) is a root of (a x^{2} + b x + c = 0) and (x_{2}) is a root of (-a x^{2} + b x + c = 0), the quadratics have opposite leading coefficients. That might mean that the functions cross the x-axis in different directions, so their roots could be arranged in a certain way.But I need to relate this to (P(x)). Let me consider the expressions for (P(x_{1})) and (P(x_{2})). If (b x_{1} + c) and (b x_{2} + c) have opposite signs, then (P(x_{1})) and (P(x_{2})) would also have opposite signs because they are just scaled versions. Wait, is that true? Let me see. If (b x_{1} + c) is positive, then (P(x_{1})) is positive, and if (b x_{2} + c) is negative, then (P(x_{2})) is negative. Or vice versa. So, if (b x_{1} + c) and (b x_{2} + c) have opposite signs, then (P(x_{1})) and (P(x_{2})) do too.But why would (b x_{1} + c) and (b x_{2} + c) have opposite signs? Let me think about the original equations.From the first equation, (a x_{1}^{2} + b x_{1} + c = 0), so (b x_{1} + c = -a x_{1}^{2}). Similarly, from the second equation, (-a x_{2}^{2} + b x_{2} + c = 0), so (b x_{2} + c = a x_{2}^{2}).So, (b x_{1} + c = -a x_{1}^{2}) and (b x_{2} + c = a x_{2}^{2}).Therefore, (b x_{1} + c) is equal to (-a x_{1}^{2}), which is negative if (a) is positive because (x_{1}^{2}) is always positive. Similarly, (b x_{2} + c = a x_{2}^{2}), which is positive if (a) is positive.So, if (a) is positive, then (b x_{1} + c) is negative and (b x_{2} + c) is positive. Therefore, (P(x_{1}) = frac{1}{2}(b x_{1} + c)) is negative, and (P(x_{2}) = frac{3}{2}(b x_{2} + c)) is positive.Similarly, if (a) is negative, then (b x_{1} + c = -a x_{1}^{2}) would be positive (since (-a) is positive), and (b x_{2} + c = a x_{2}^{2}) would be negative. So again, (P(x_{1})) would be positive and (P(x_{2})) would be negative.In either case, (P(x_{1})) and (P(x_{2})) have opposite signs. Therefore, by the Intermediate Value Theorem, since (P(x)) is continuous (it's a polynomial), there must be some (x_{3}) between (x_{1}) and (x_{2}) such that (P(x_{3}) = 0). Wait, but the problem states that (x_{3}) could be either between (x_{1}) and (x_{2}) or outside of them. So, does that mean I need to consider both possibilities? Or is it guaranteed to be between them?Well, the Intermediate Value Theorem only guarantees a root between (x_{1}) and (x_{2}) if (P(x_{1})) and (P(x_{2})) have opposite signs. But in this case, they do, so (x_{3}) must lie between (x_{1}) and (x_{2}). But the problem says "either (x_{1} leq x_{3} leq x_{2}) or (x_{1} geq x_{3} geq x_{2})". So, it's considering both possibilities, but in reality, depending on the order of (x_{1}) and (x_{2}), (x_{3}) will lie between them. Wait, but what if (x_{1}) and (x_{2}) are in a different order? Like, what if (x_{1}) is greater than (x_{2})? Then, the Intermediate Value Theorem would still apply, but (x_{3}) would be between (x_{2}) and (x_{1}). So, in that case, (x_{3}) would satisfy (x_{2} leq x_{3} leq x_{1}), which is the same as (x_{1} geq x_{3} geq x_{2}). Therefore, regardless of the order of (x_{1}) and (x_{2}), there exists an (x_{3}) such that (x_{3}) lies between them, which satisfies either (x_{1} leq x_{3} leq x_{2}) or (x_{1} geq x_{3} geq x_{2}). So, putting it all together, since (P(x_{1})) and (P(x_{2})) have opposite signs, there must be a root (x_{3}) of (P(x)) between (x_{1}) and (x_{2}), which is exactly what we needed to prove.I think that covers it. I should probably double-check my steps to make sure I didn't make any mistakes. Let me go through it again quickly.1. Evaluated (P(x_{1})) and (P(x_{2})) using the given roots.2. Expressed (b x_{1} + c) and (b x_{2} + c) in terms of (a) and the roots.3. Showed that depending on the sign of (a), (b x_{1} + c) and (b x_{2} + c) have opposite signs.4. Concluded that (P(x_{1})) and (P(x_{2})) have opposite signs.5. Applied the Intermediate Value Theorem to assert the existence of (x_{3}) between (x_{1}) and (x_{2}).Everything seems to check out. I think I've got it!

💡Okay, so I have this problem about a cube with edge length 2 cm. There's a dot marked in the center of one of its faces. The cube is placed on a flat surface and rolled in a straight line such that it completes one full rotation and returns to its original position with the dot again on the top face. I need to calculate the length of the path traced by the dot, expressed in the form (dpi), where (d) is a constant.Hmm, let me visualize this. The cube is rolling on a flat surface, and as it rolls, the dot moves. Since the cube is rolling in a straight line and completes one full rotation, the dot must trace some kind of circular path, right? But it's not just a simple circle because the cube is a three-dimensional object.First, I should recall that when a cube rolls on a surface, it doesn't roll like a sphere; instead, it rolls over its edges. Each time it rolls over an edge, it rotates 90 degrees. So, to complete a full rotation, it needs to roll over four edges, right? That would bring it back to its original position.Now, the dot is at the center of one of the faces. When the cube rolls, the dot moves in a circular path around the edge over which the cube is rolling. The radius of this circular path would be the distance from the dot to the edge it's rolling over. Since the cube has an edge length of 2 cm, the center of the face is 1 cm away from each edge. Wait, is that correct?Let me think. If the cube has an edge length of 2 cm, then each face is a 2 cm by 2 cm square. The center of the face is at (1,1) if we consider the face as a coordinate system from (0,0) to (2,2). So, the distance from the center to any edge is indeed 1 cm. Therefore, the radius of the circular path traced by the dot is 1 cm.But hold on, when the cube rolls over an edge, the dot is moving along a circular path with radius equal to the distance from the center of the face to the edge, which is 1 cm. So, each time the cube rolls over an edge, the dot traces a quarter-circle with radius 1 cm.Since the cube needs to complete a full rotation, it has to roll over four edges, right? Each roll over an edge corresponds to a quarter-circle. So, four quarter-circles make up a full circle. Therefore, the total path traced by the dot should be the circumference of a circle with radius 1 cm.The circumference of a circle is (2pi r), so with (r = 1) cm, the circumference is (2pi) cm. Therefore, the total path length traced by the dot is (2pi) cm.Wait, but I'm not sure if it's that simple. Let me double-check. When the cube rolls over each edge, the dot is moving along a circular path, but is the radius really 1 cm? Or is it something else?Let me consider the cube rolling over one edge. The dot is at the center of the face, so when the cube rolls over the edge, the dot moves in a circular path around that edge. The distance from the dot to the edge is 1 cm, so the radius is indeed 1 cm. Therefore, each quarter-circle has a length of (frac{1}{4} times 2pi times 1 = frac{pi}{2}) cm.Since there are four such quarter-circles in a full rotation, the total path length is (4 times frac{pi}{2} = 2pi) cm. So, that seems consistent.But wait, another thought: when the cube rolls, it's not just rotating around one edge; it's moving forward as it rolls. So, is the path traced by the dot a cycloid or something similar?Hmm, a cycloid is the path traced by a point on the rim of a circular wheel as the wheel rolls along a straight line. In this case, the cube is not a circular wheel, but a polyhedron. So, the path traced by the dot might be a type of "cubic cycloid" or something like that.But in this problem, we're told that the cube completes one full rotation and returns to its original position. So, perhaps the path is simply a circle, as I initially thought.Wait, let's think about the cube's movement. When the cube rolls over an edge, it rotates 90 degrees, and the dot moves along a quarter-circle. After four such rolls, the cube completes a full 360-degree rotation and returns to its original position. So, the dot traces four quarter-circles, each with radius 1 cm, making a total of one full circle with circumference (2pi) cm.Therefore, the length of the path traced by the dot is (2pi) cm, so (d = 2).But I'm still a bit uncertain because I'm not entirely sure if the path is a perfect circle or if it's a more complex curve. Maybe I should try to visualize or sketch it.Imagine the cube rolling over one edge. The dot starts at the center of the top face. As the cube rolls over the edge, the dot moves downward, tracing a quarter-circle. After rolling over the first edge, the dot is now at the center of the new top face. Then, as the cube rolls over the next edge, the dot moves to the next face, tracing another quarter-circle. After four such rolls, the dot returns to its original position.So, each quarter-circle is part of a circle with radius 1 cm. Therefore, the total path is indeed four quarter-circles, which make up a full circle with circumference (2pi) cm.Therefore, the length of the path traced by the dot is (2pi) cm, so (d = 2).Wait, but I just thought of something else. The cube is rolling in a straight line, so the center of the cube is moving forward as it rolls. Does that affect the path traced by the dot?Hmm, the center of the cube is moving forward, but the dot is moving in a circular path relative to the cube's rotation. So, the dot's path is a combination of the cube's forward motion and its rotational motion. But since the cube is completing a full rotation, the dot's path should still be a circle relative to the cube's center.But actually, relative to the flat surface, the dot's path is a cycloid-like curve. Wait, but since the cube is a polyhedron, it's not a smooth curve like a cycloid. Instead, it's a series of circular arcs.But in this problem, we're only concerned with the total length of the path traced by the dot, not the exact shape. So, if each roll over an edge traces a quarter-circle with radius 1 cm, and there are four such rolls, the total length is four times the length of a quarter-circle.Each quarter-circle has a length of (frac{1}{4} times 2pi times 1 = frac{pi}{2}) cm. So, four of them would be (4 times frac{pi}{2} = 2pi) cm.Therefore, the total path length is (2pi) cm, so (d = 2).I think that makes sense. The key is recognizing that each roll over an edge corresponds to a quarter-circle path for the dot, and four such rolls make up a full circle. Therefore, the total distance is the circumference of that circle.So, my final answer is (2pi), so (d = 2).

💡Okay, so I have this triangle ABC where BC is 3 cm, and angle B is π/3 radians, which is 60 degrees. There are two parts to this problem. Let me tackle them one by one.Starting with part (1): If angle A is π/4 radians, which is 45 degrees, I need to find the length of AC. Hmm, so I know two angles and one side. Since I have two angles, I can find the third angle because the sum of angles in a triangle is π radians or 180 degrees. So, angle C would be π - π/3 - π/4. Let me calculate that.First, π is approximately 3.1416, but I don't need the numerical value right now. Let me express it in fractions. π/3 is 60 degrees, π/4 is 45 degrees, so angle C is 180 - 60 - 45, which is 75 degrees or 5π/12 radians. Okay, so now I have all three angles: A = π/4, B = π/3, and C = 5π/12.Now, I need to find side AC. I remember the Law of Sines, which relates the sides and angles of a triangle. The formula is:[frac{a}{sin A} = frac{b}{sin B} = frac{c}{sin C}]In this case, side BC is 3 cm, which is opposite angle A. So, side BC is opposite angle A, which is π/4. Side AC is opposite angle B, which is π/3. So, I can set up the ratio:[frac{BC}{sin A} = frac{AC}{sin B}]Plugging in the known values:[frac{3}{sin frac{pi}{4}} = frac{AC}{sin frac{pi}{3}}]I know that sin(π/4) is √2/2 and sin(π/3) is √3/2. Let me substitute those in:[frac{3}{frac{sqrt{2}}{2}} = frac{AC}{frac{sqrt{3}}{2}}]Simplifying the denominators:[3 times frac{2}{sqrt{2}} = AC times frac{2}{sqrt{3}}]Which simplifies to:[frac{6}{sqrt{2}} = frac{2AC}{sqrt{3}}]Now, I can solve for AC. Let me cross-multiply:[6 times sqrt{3} = 2AC times sqrt{2}]Divide both sides by 2:[3 sqrt{3} = AC times sqrt{2}]Then, divide both sides by √2:[AC = frac{3 sqrt{3}}{sqrt{2}}]To rationalize the denominator, multiply numerator and denominator by √2:[AC = frac{3 sqrt{6}}{2}]So, AC is (3√6)/2 cm.Wait, let me double-check my steps. I used the Law of Sines correctly. I had BC opposite angle A, which is π/4, and AC opposite angle B, which is π/3. Plugged in the sine values correctly, did the algebra step by step. It seems right. So, I think that's the correct answer for part (1).Moving on to part (2): The area of triangle ABC is 3√3 cm². I need to find AC again. Hmm, so this time, I don't know angle A or angle C, but I know side BC is 3 cm, angle B is π/3, and the area is given.I remember that the area of a triangle can be calculated using the formula:[text{Area} = frac{1}{2}ab sin C]Where a and b are two sides, and C is the included angle between them. In this case, I can consider sides AB and BC with the included angle B.So, if I let AB = c, BC = a = 3 cm, and angle B = π/3, then the area is:[frac{1}{2} times AB times BC times sin B = 3sqrt{3}]Plugging in the known values:[frac{1}{2} times AB times 3 times sin frac{pi}{3} = 3sqrt{3}]Simplify sin(π/3) to √3/2:[frac{1}{2} times AB times 3 times frac{sqrt{3}}{2} = 3sqrt{3}]Multiply the constants:[frac{3 sqrt{3}}{4} times AB = 3sqrt{3}]Now, solve for AB:[AB = frac{3sqrt{3} times 4}{3 sqrt{3}} = 4]So, AB is 4 cm.Now, I need to find AC. Since I know sides AB and BC, and the included angle B, I can use the Law of Cosines. The Law of Cosines formula is:[c^2 = a^2 + b^2 - 2ab cos C]In this case, AC is the side opposite angle B, so:[AC^2 = AB^2 + BC^2 - 2 times AB times BC times cos B]Plugging in the known values:[AC^2 = 4^2 + 3^2 - 2 times 4 times 3 times cos frac{pi}{3}]Calculate each term:4² = 16, 3² = 9, 2×4×3 = 24, and cos(π/3) = 0.5.So,[AC^2 = 16 + 9 - 24 times 0.5]Simplify:16 + 9 = 25, and 24 × 0.5 = 12.So,[AC^2 = 25 - 12 = 13]Therefore, AC = √13 cm.Wait, let me verify. I used the area formula correctly, found AB to be 4 cm, then applied the Law of Cosines. The calculations seem straightforward. 4 squared is 16, 3 squared is 9, 16+9 is 25, 2×4×3 is 24, times 0.5 is 12, 25-12 is 13. So, AC is √13. That seems correct.So, summarizing:For part (1), AC is (3√6)/2 cm.For part (2), AC is √13 cm.**Final Answer**(1) boxed{dfrac{3sqrt{6}}{2}} cm(2) boxed{sqrt{13}} cm

💡Alright, so I have this geometry problem to solve: I need to construct triangle (ABC) given the length of side (BC = a), the length of the median (BB_1 = s_b) (where (B_1) is the midpoint of side (AC)), and the angle (angle ABB_1 = beta_2). Hmm, okay, let's break this down step by step.First, I should recall what a median is. A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. So, in this case, (B_1) is the midpoint of (AC), meaning (AB_1 = B_1C). That might be useful later on.Given that (BC = a), (BB_1 = s_b), and (angle ABB_1 = beta_2), I need to figure out how to construct triangle (ABC). I think starting by drawing a rough sketch might help visualize the problem. Let me imagine triangle (ABC) with points (A), (B), and (C). Point (B_1) is the midpoint of (AC), so it's halfway between (A) and (C).Now, I know the length of (BC) is (a), so I can draw that side first. Let me place point (B) somewhere on the paper and then mark point (C) such that the distance between (B) and (C) is (a). Okay, that's straightforward.Next, I need to consider the median (BB_1 = s_b). Since (B_1) is the midpoint of (AC), I can think of (BB_1) as a line from (B) to the midpoint of (AC). But I don't know where (A) is yet, so I can't directly draw (BB_1). Maybe I can use the angle (angle ABB_1 = beta_2) to help me.So, (angle ABB_1) is the angle at point (B) between sides (BA) and (BB_1). If I can construct this angle, I might be able to find the position of (A). Let me think about how to do that.Perhaps I can use the Law of Cosines or the Law of Sines in triangle (ABB_1). Wait, triangle (ABB_1) has sides (AB), (BB_1 = s_b), and (AB_1). But I don't know (AB) or (AB_1), so maybe that's not directly helpful. Hmm.Alternatively, maybe I can use coordinate geometry. Let me assign coordinates to the points to make it easier. Let's place point (B) at the origin ((0, 0)). Then, since (BC = a), I can place point (C) at ((a, 0)). Now, I need to find the coordinates of point (A) such that (BB_1 = s_b) and (angle ABB_1 = beta_2).Since (B_1) is the midpoint of (AC), if I can find the coordinates of (A), I can find (B_1) as the average of the coordinates of (A) and (C). Let's denote the coordinates of (A) as ((x, y)). Then, the coordinates of (B_1) would be (left(frac{x + a}{2}, frac{y + 0}{2}right) = left(frac{x + a}{2}, frac{y}{2}right)).Now, the length of (BB_1) is given as (s_b). The distance from (B) at ((0, 0)) to (B_1) at (left(frac{x + a}{2}, frac{y}{2}right)) is:[sqrt{left(frac{x + a}{2} - 0right)^2 + left(frac{y}{2} - 0right)^2} = s_b]Simplifying this equation:[left(frac{x + a}{2}right)^2 + left(frac{y}{2}right)^2 = s_b^2]Multiplying both sides by 4:[(x + a)^2 + y^2 = 4s_b^2]So, that's one equation involving (x) and (y).Next, the angle (angle ABB_1 = beta_2). To express this angle, I can use vector analysis or coordinate geometry. The angle between vectors (BA) and (BB_1) is (beta_2). The vectors (BA) and (BB_1) can be represented as:- Vector (BA = A - B = (x, y))- Vector (BB_1 = B_1 - B = left(frac{x + a}{2}, frac{y}{2}right))The angle between these two vectors is (beta_2), so using the dot product formula:[cos(beta_2) = frac{BA cdot BB_1}{|BA||BB_1|}]Calculating the dot product:[BA cdot BB_1 = x cdot frac{x + a}{2} + y cdot frac{y}{2} = frac{x(x + a) + y^2}{2}]The magnitudes are:[|BA| = sqrt{x^2 + y^2}][|BB_1| = s_b]So, putting it all together:[cos(beta_2) = frac{frac{x(x + a) + y^2}{2}}{sqrt{x^2 + y^2} cdot s_b}]Simplifying:[cos(beta_2) = frac{x(x + a) + y^2}{2s_b sqrt{x^2 + y^2}}]Now, I have two equations:1. ((x + a)^2 + y^2 = 4s_b^2)2. (cos(beta_2) = frac{x(x + a) + y^2}{2s_b sqrt{x^2 + y^2}})This seems a bit complicated, but maybe I can find a relationship between (x) and (y) from these equations.Let me denote (r = sqrt{x^2 + y^2}), which is the distance from (B) to (A), i.e., (BA). Then, the second equation becomes:[cos(beta_2) = frac{x(x + a) + y^2}{2s_b r}]But from the first equation:[(x + a)^2 + y^2 = 4s_b^2]Expanding ((x + a)^2):[x^2 + 2ax + a^2 + y^2 = 4s_b^2]But (x^2 + y^2 = r^2), so substituting:[r^2 + 2ax + a^2 = 4s_b^2]Therefore:[2ax = 4s_b^2 - r^2 - a^2]So,[x = frac{4s_b^2 - r^2 - a^2}{2a}]Now, going back to the second equation:[cos(beta_2) = frac{x(x + a) + y^2}{2s_b r}]Let's substitute (x) from the above expression:First, compute (x + a):[x + a = frac{4s_b^2 - r^2 - a^2}{2a} + a = frac{4s_b^2 - r^2 - a^2 + 2a^2}{2a} = frac{4s_b^2 - r^2 + a^2}{2a}]Now, compute (x(x + a)):[x(x + a) = left(frac{4s_b^2 - r^2 - a^2}{2a}right) left(frac{4s_b^2 - r^2 + a^2}{2a}right)]This looks like the product of ((M - N)(M + N)), which is (M^2 - N^2), where (M = 4s_b^2 - r^2) and (N = a^2). So,[x(x + a) = frac{(4s_b^2 - r^2)^2 - (a^2)^2}{(2a)^2} = frac{(4s_b^2 - r^2)^2 - a^4}{4a^2}]Now, the numerator of the second equation becomes:[x(x + a) + y^2 = frac{(4s_b^2 - r^2)^2 - a^4}{4a^2} + y^2]But (y^2 = r^2 - x^2), so substituting:[x(x + a) + y^2 = frac{(4s_b^2 - r^2)^2 - a^4}{4a^2} + r^2 - x^2]This is getting quite involved. Maybe there's a simpler way to approach this problem without diving too deep into algebra.Let me think about the geometric construction. I have side (BC = a), and I need to find point (A) such that the median (BB_1 = s_b) and (angle ABB_1 = beta_2). Maybe I can use the concept of triangle construction with given sides and angles.First, draw side (BC) with length (a). Then, construct the median (BB_1) of length (s_b). Since (B_1) is the midpoint of (AC), I can think of (B_1) as a point that is halfway between (A) and (C). So, if I can find (B_1), I can find (A) by reflecting (C) over (B_1).But how do I find (B_1)? I know that (BB_1 = s_b) and (angle ABB_1 = beta_2). Maybe I can construct triangle (ABB_1) first, given side (BB_1 = s_b), angle (angle ABB_1 = beta_2), and then find point (A).Wait, in triangle (ABB_1), I know side (BB_1 = s_b), angle (angle ABB_1 = beta_2), and I need to find side (AB). But I don't know side (AB), so maybe I can use the Law of Sines or Cosines here.Using the Law of Cosines in triangle (ABB_1):[AB_1^2 = AB^2 + BB_1^2 - 2 cdot AB cdot BB_1 cdot cos(beta_2)]But (AB_1) is half of (AC) because (B_1) is the midpoint. So, (AB_1 = frac{AC}{2}). However, I don't know (AC) yet.This seems like a circular problem because I don't know (AB) or (AC). Maybe I need another approach.Let me consider the coordinates again. I have point (B) at ((0, 0)), point (C) at ((a, 0)), and point (A) at ((x, y)). The midpoint (B_1) is at (left(frac{x + a}{2}, frac{y}{2}right)). The distance from (B) to (B_1) is (s_b), so:[sqrt{left(frac{x + a}{2}right)^2 + left(frac{y}{2}right)^2} = s_b]Squaring both sides:[left(frac{x + a}{2}right)^2 + left(frac{y}{2}right)^2 = s_b^2]Multiplying by 4:[(x + a)^2 + y^2 = 4s_b^2]This is the same equation I had before. Now, the angle (angle ABB_1 = beta_2) can be expressed using the tangent function, perhaps. The angle between (BA) and (BB_1) is (beta_2), so maybe I can find the slope of (BB_1) and relate it to the slope of (BA).The slope of (BA) is (frac{y - 0}{x - 0} = frac{y}{x}).The slope of (BB_1) is (frac{frac{y}{2} - 0}{frac{x + a}{2} - 0} = frac{y}{x + a}).The angle between two lines with slopes (m_1) and (m_2) is given by:[tan(theta) = left|frac{m_2 - m_1}{1 + m_1m_2}right|]So, in this case:[tan(beta_2) = left|frac{frac{y}{x + a} - frac{y}{x}}{1 + frac{y}{x} cdot frac{y}{x + a}}right|]Simplifying the numerator:[frac{y}{x + a} - frac{y}{x} = yleft(frac{1}{x + a} - frac{1}{x}right) = yleft(frac{x - (x + a)}{x(x + a)}right) = yleft(frac{-a}{x(x + a)}right) = -frac{ay}{x(x + a)}]The denominator:[1 + frac{y^2}{x(x + a)}]So,[tan(beta_2) = left|frac{-frac{ay}{x(x + a)}}{1 + frac{y^2}{x(x + a)}}right| = frac{frac{ay}{x(x + a)}}{1 + frac{y^2}{x(x + a)}}]Since (tan(beta_2)) is positive (angles in triangles are between 0 and 180 degrees), we can drop the absolute value.So,[tan(beta_2) = frac{ay}{x(x + a) + y^2}]But from the earlier equation, we have:[(x + a)^2 + y^2 = 4s_b^2]Expanding this:[x^2 + 2ax + a^2 + y^2 = 4s_b^2]Which can be rewritten as:[x^2 + y^2 + 2ax + a^2 = 4s_b^2]Notice that (x^2 + y^2 = r^2), the distance from (B) to (A). So,[r^2 + 2ax + a^2 = 4s_b^2]Therefore,[2ax = 4s_b^2 - r^2 - a^2]So,[x = frac{4s_b^2 - r^2 - a^2}{2a}]Now, going back to the expression for (tan(beta_2)):[tan(beta_2) = frac{ay}{x(x + a) + y^2}]But (x(x + a) + y^2) can be expressed using the earlier equation:From (x^2 + y^2 + 2ax + a^2 = 4s_b^2), we have:[x(x + a) + y^2 = x^2 + ax + y^2 = (x^2 + y^2) + ax = r^2 + ax]But from (2ax = 4s_b^2 - r^2 - a^2), we have:[ax = frac{4s_b^2 - r^2 - a^2}{2}]So,[x(x + a) + y^2 = r^2 + frac{4s_b^2 - r^2 - a^2}{2} = frac{2r^2 + 4s_b^2 - r^2 - a^2}{2} = frac{r^2 + 4s_b^2 - a^2}{2}]Therefore,[tan(beta_2) = frac{ay}{frac{r^2 + 4s_b^2 - a^2}{2}} = frac{2ay}{r^2 + 4s_b^2 - a^2}]Now, we have:[tan(beta_2) = frac{2ay}{r^2 + 4s_b^2 - a^2}]But (r^2 = x^2 + y^2), and from the earlier equation:[x = frac{4s_b^2 - r^2 - a^2}{2a}]So, substituting (x) into (r^2):[r^2 = x^2 + y^2 = left(frac{4s_b^2 - r^2 - a^2}{2a}right)^2 + y^2]This seems complicated, but maybe I can express (y) in terms of (r).From the equation:[tan(beta_2) = frac{2ay}{r^2 + 4s_b^2 - a^2}]Solving for (y):[y = frac{tan(beta_2) (r^2 + 4s_b^2 - a^2)}{2a}]Now, substitute this into the expression for (r^2):[r^2 = left(frac{4s_b^2 - r^2 - a^2}{2a}right)^2 + left(frac{tan(beta_2) (r^2 + 4s_b^2 - a^2)}{2a}right)^2]This is a quadratic equation in terms of (r^2). Let me denote (u = r^2) to simplify:[u = left(frac{4s_b^2 - u - a^2}{2a}right)^2 + left(frac{tan(beta_2) (u + 4s_b^2 - a^2)}{2a}right)^2]Expanding both terms:First term:[left(frac{4s_b^2 - u - a^2}{2a}right)^2 = frac{(4s_b^2 - u - a^2)^2}{4a^2}]Second term:[left(frac{tan(beta_2) (u + 4s_b^2 - a^2)}{2a}right)^2 = frac{tan^2(beta_2) (u + 4s_b^2 - a^2)^2}{4a^2}]So, the equation becomes:[u = frac{(4s_b^2 - u - a^2)^2 + tan^2(beta_2) (u + 4s_b^2 - a^2)^2}{4a^2}]Multiply both sides by (4a^2):[4a^2 u = (4s_b^2 - u - a^2)^2 + tan^2(beta_2) (u + 4s_b^2 - a^2)^2]This is a quadratic equation in (u), which can be solved using standard methods. However, this seems quite involved, and I'm not sure if this is the most efficient way to approach the problem.Maybe instead of using coordinates, I can use geometric construction techniques. Let me think about how to construct triangle (ABC) given (BC = a), median (BB_1 = s_b), and angle (angle ABB_1 = beta_2).First, I can draw side (BC) with length (a). Then, I need to locate point (B_1), the midpoint of (AC). Since (BB_1 = s_b), I can draw a circle centered at (B) with radius (s_b). The intersection of this circle with the line (AC) will give me point (B_1). However, I don't know where (A) is yet, so this might not be straightforward.Alternatively, I can consider triangle (BB_1C). Since (B_1) is the midpoint of (AC), triangle (BB_1C) has sides (BC = a), (BB_1 = s_b), and (B_1C = frac{AC}{2}). But I don't know (AC), so this might not help directly.Wait, maybe I can use the fact that (B_1) is the midpoint of (AC) to express (AC) in terms of (AB) and (BC). Using the Apollonius's theorem, which states that in any triangle, the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of the half third side. So,[AB^2 + BC^2 = 2BB_1^2 + 2left(frac{AC}{2}right)^2]Simplifying:[AB^2 + a^2 = 2s_b^2 + frac{AC^2}{2}]But I still have two unknowns here: (AB) and (AC). Maybe I can find another equation involving these.From the Law of Cosines in triangle (ABB_1):[AB_1^2 = AB^2 + BB_1^2 - 2 cdot AB cdot BB_1 cdot cos(beta_2)]But (AB_1 = frac{AC}{2}), so:[left(frac{AC}{2}right)^2 = AB^2 + s_b^2 - 2 cdot AB cdot s_b cdot cos(beta_2)]Simplifying:[frac{AC^2}{4} = AB^2 + s_b^2 - 2AB s_b cos(beta_2)]Now, I have two equations:1. (AB^2 + a^2 = 2s_b^2 + frac{AC^2}{2})2. (frac{AC^2}{4} = AB^2 + s_b^2 - 2AB s_b cos(beta_2))Let me denote (AB = c) and (AC = b). Then, the equations become:1. (c^2 + a^2 = 2s_b^2 + frac{b^2}{2})2. (frac{b^2}{4} = c^2 + s_b^2 - 2c s_b cos(beta_2))Now, I can solve these two equations for (c) and (b).From equation 2:[frac{b^2}{4} = c^2 + s_b^2 - 2c s_b cos(beta_2)]Multiply both sides by 4:[b^2 = 4c^2 + 4s_b^2 - 8c s_b cos(beta_2)]Now, substitute (b^2) into equation 1:[c^2 + a^2 = 2s_b^2 + frac{4c^2 + 4s_b^2 - 8c s_b cos(beta_2)}{2}]Simplify the right-hand side:[c^2 + a^2 = 2s_b^2 + 2c^2 + 2s_b^2 - 4c s_b cos(beta_2)]Combine like terms:[c^2 + a^2 = 4s_b^2 + 2c^2 - 4c s_b cos(beta_2)]Bring all terms to one side:[c^2 + a^2 - 4s_b^2 - 2c^2 + 4c s_b cos(beta_2) = 0]Simplify:[-c^2 + a^2 - 4s_b^2 + 4c s_b cos(beta_2) = 0]Multiply both sides by -1:[c^2 - a^2 + 4s_b^2 - 4c s_b cos(beta_2) = 0]Rearrange:[c^2 - 4c s_b cos(beta_2) + (4s_b^2 - a^2) = 0]This is a quadratic equation in terms of (c):[c^2 - 4s_b cos(beta_2) cdot c + (4s_b^2 - a^2) = 0]Let me denote this as:[c^2 + Bc + C = 0]where (B = -4s_b cos(beta_2)) and (C = 4s_b^2 - a^2).Using the quadratic formula:[c = frac{-B pm sqrt{B^2 - 4AC}}{2A}]Here, (A = 1), so:[c = frac{4s_b cos(beta_2) pm sqrt{(4s_b cos(beta_2))^2 - 4(1)(4s_b^2 - a^2)}}{2}]Simplify the discriminant:[(4s_b cos(beta_2))^2 - 4(4s_b^2 - a^2) = 16s_b^2 cos^2(beta_2) - 16s_b^2 + 4a^2]Factor out 4:[4[4s_b^2 cos^2(beta_2) - 4s_b^2 + a^2] = 4[4s_b^2 (cos^2(beta_2) - 1) + a^2]]Since (cos^2(beta_2) - 1 = -sin^2(beta_2)), this becomes:[4[-4s_b^2 sin^2(beta_2) + a^2] = 4(a^2 - 4s_b^2 sin^2(beta_2))]So, the discriminant is:[sqrt{4(a^2 - 4s_b^2 sin^2(beta_2))} = 2sqrt{a^2 - 4s_b^2 sin^2(beta_2)}]Therefore, the solutions for (c) are:[c = frac{4s_b cos(beta_2) pm 2sqrt{a^2 - 4s_b^2 sin^2(beta_2)}}{2} = 2s_b cos(beta_2) pm sqrt{a^2 - 4s_b^2 sin^2(beta_2)}]So,[c = 2s_b cos(beta_2) pm sqrt{a^2 - 4s_b^2 sin^2(beta_2)}]Now, since (c = AB) is a length, it must be positive. Therefore, we need to ensure that the expression under the square root is non-negative:[a^2 - 4s_b^2 sin^2(beta_2) geq 0]Which implies:[a geq 2s_b sin(beta_2)]This is a necessary condition for the existence of such a triangle.Assuming this condition is satisfied, we can proceed to find (c). Once (c) is known, we can find (b = AC) using equation 2:[frac{b^2}{4} = c^2 + s_b^2 - 2c s_b cos(beta_2)]Multiplying both sides by 4:[b^2 = 4c^2 + 4s_b^2 - 8c s_b cos(beta_2)]So,[b = sqrt{4c^2 + 4s_b^2 - 8c s_b cos(beta_2)}]Now that we have (AB = c) and (AC = b), we can construct triangle (ABC).Here's a step-by-step construction plan:1. **Draw side (BC)**: Draw segment (BC) with length (a).2. **Construct angle (angle ABB_1 = beta_2)**: At point (B), construct an angle of measure (beta_2).3. **Locate point (B_1)**: From point (B), draw a circle with radius (s_b). The intersection of this circle with the angle bisector (from step 2) will give point (B_1).4. **Find point (A)**: Since (B_1) is the midpoint of (AC), draw a line through (B_1) parallel to (BC). The intersection of this line with the extension of (AB) will give point (A).Wait, actually, that might not be accurate. Let me think again.Alternatively, once (B_1) is located, since (B_1) is the midpoint of (AC), we can find (A) by reflecting (C) over (B_1). That is, if (B_1) is the midpoint, then (A) is such that (B_1) is halfway between (A) and (C). So, if we know (B_1) and (C), we can find (A) by extending (B_1C) beyond (B_1) by the same length.So, the construction steps would be:1. Draw side (BC) with length (a).2. At point (B), construct an angle of measure (beta_2).3. From point (B), draw a circle with radius (s_b). The intersection of this circle with the angle bisector (from step 2) gives point (B_1).4. Since (B_1) is the midpoint of (AC), draw a line through (B_1) parallel to (BC). The intersection of this line with the extension of (AB) gives point (A).Wait, no. If (B_1) is the midpoint, then (A) can be found by extending (B_1C) beyond (B_1) to a point (A) such that (B_1C = B_1A).So, more accurately:1. Draw side (BC) with length (a).2. At point (B), construct an angle of measure (beta_2).3. From point (B), draw a circle with radius (s_b). The intersection of this circle with the angle bisector (from step 2) gives point (B_1).4. Draw segment (B_1C).5. Extend (B_1C) beyond (B_1) to a point (A) such that (B_1C = B_1A).This ensures that (B_1) is the midpoint of (AC).Alternatively, using coordinates, once we have point (B_1), we can find (A) as:[A = (2x_{B_1} - x_C, 2y_{B_1} - y_C)]Since (B_1) is the midpoint, so:[x_{B_1} = frac{x_A + x_C}{2} implies x_A = 2x_{B_1} - x_C][y_{B_1} = frac{y_A + y_C}{2} implies y_A = 2y_{B_1} - y_C]So, if we have coordinates for (B_1) and (C), we can find (A).Putting it all together, the construction involves:1. Drawing side (BC).2. Constructing angle (beta_2) at (B).3. Using the median length (s_b) to find (B_1).4. Using the midpoint property to find (A).This should give the desired triangle (ABC).**Final Answer**The triangle (ABC) can be constructed using the given side (BC = a), median (BB_1 = s_b), and angle (angle ABB_1 = beta_2). The construction involves drawing side (BC), constructing the angle (beta_2) at (B), locating the midpoint (B_1) using the median length, and then determining point (A) based on the midpoint property. The final result is:boxed{ABC}

💡To determine the probability that the product of three dice rolls equals 216, we start by recognizing that each die can show a number from 1 to 6. First, we factorize 216:[216 = 6^3]This means that the only combination of three numbers between 1 and 6 that multiplies to 216 is (6, 6, 6).Next, we calculate the probability of rolling a 6 on a single die, which is:[frac{1}{6}]Since the dice rolls are independent, the probability of rolling three 6s in a row is:[left(frac{1}{6}right)^3 = frac{1}{216}]Therefore, the probability that the product of the three dice rolls equals 216 is:[boxed{frac{1}{216}}]