🧠:Okay, so there's this problem with 100 boxes arranged in a line, all closed. One of them has a diamond. Each box has a message saying, "the diamond is in the box to the left or to the right." We know that exactly one of the messages is true, and all the others are false. The question is, can we figure out where the diamond is by opening just one box?Hmm, let's start by breaking down the problem. First, there are 100 boxes. The diamond is in one of them. Each box's message claims that the diamond is either in the left or right neighbor. But only one message is true; the rest are false. So, if we can determine which box has the true message, we can find the diamond. But how?Let me think about the implications. If exactly one message is true, then all other 99 messages must be false. So, for the box with the true message, either the left or the right neighbor has the diamond. But for all other boxes, their statements are false. That means for those boxes, the diamond is neither in the left nor the right neighbor. So, if a box is not the one true message, then the diamond is not adjacent to it.Wait, but the diamond is in one box. Let's try to model this. Let's number the boxes from 1 to 100. Suppose the diamond is in box k. Then, the message on box k-1 would say "the diamond is in the box to the left or to the right." If the diamond is in k, then box k-1's message would be true if k-1 is saying that the diamond is either in k-2 or k. But since the diamond is in k, the message on k-1 would be "left or right" which would be true (since right neighbor is k). Similarly, box k+1 would have a message saying "left or right", which would also be true, because left neighbor is k. But the problem states that exactly one message is true. Therefore, if the diamond is in box k, both boxes k-1 and k+1 would have true messages. But that contradicts the condition that only one message is true. Therefore, the diamond cannot be in a box where it has neighbors on both sides. Wait, unless the diamond is at one end?Wait, let's consider edge cases. If the diamond is in box 1, then the only neighbor is box 2. The message on box 2 says "the diamond is in the left or right." If the diamond is in box 1, then box 2's message would be true because the left neighbor (box 1) has the diamond. However, what about box 1's message? Box 1's message says "the diamond is in the left or right." But box 1 has no left neighbor, so the message is effectively "the diamond is in box 2." Since the diamond is in box 1, that message is false. So, in this case, only box 2's message is true. Similarly, if the diamond is in box 100, then box 99's message would say "the diamond is in the left or right," which would be true because the right neighbor is box 100. But box 100's message would say "left or right," but it has no right neighbor, so it's effectively claiming the diamond is in box 99, which is false. Therefore, if the diamond is in box 1 or 100, there's exactly one true message (box 2 or box 99, respectively). But if the diamond is in any other box (say, k where 2 ≤ k ≤ 99), then both k-1 and k+1 would have true messages. But the problem states that exactly one message is true. Therefore, the diamond must be in box 1 or 100. Because in those positions, only one adjacent box exists, so only one message can be true. That seems like a possible conclusion.Wait, but the problem doesn't specify that the diamond is in an end box. So maybe the answer is that the diamond must be in box 1 or 100, and we can determine which by opening one of the end boxes. But how? Because if we open box 1 and find the diamond, great. If not, then the diamond must be in box 100. But the problem says we can open only one box. So if we open box 1 and don't find the diamond, we have to deduce it's in 100 without opening it. But the problem states "by opening the lid of just one of them," so maybe we can determine the location without necessarily finding the diamond in the opened box. Wait, but if we open a box and don't find the diamond, then we have to use the information from the message to figure out where it is.Alternatively, perhaps there's a specific box we can open that will give us enough information regardless of the diamond's position. Let's think.If the diamond is in box 1, then box 2's message is true. All others are false. If the diamond is in box 100, then box 99's message is true. All others are false. If the diamond is in any middle box, then two messages (k-1 and k+1) would be true, which contradicts the problem's condition. Therefore, the diamond must be in box 1 or 100. Therefore, the only possible positions are the first or last box. Therefore, by opening either box 1 or box 100, we can find the diamond. But wait, if we open box 1 and find the diamond, then we're done. If we open box 1 and don't find it, then it must be in box 100. Similarly for box 100. However, the problem says "by opening the lid of just one of them," so perhaps we can open a box that's not one of the ends and deduce based on the message? Wait, but the messages are on the lids, so when you open a lid, you can read the message, but the message's truth depends on the diamond's location. Wait, but the messages are fixed; each box's message is "the diamond is in the box to the left or to the right." So the messages are the same on all boxes. Wait no, each box has that message. So every box's lid says "the diamond is in the box to the left or to the right." So each box's statement is about its own neighbors. So, for example, box 5's message is "the diamond is in box 4 or 6."So, given that, the only true message is either from box 2 (if diamond is in 1) or box 99 (if diamond is in 100). All other messages are false. So, if we can determine which of these two is the true one, we can find the diamond. But how can we determine that by opening just one box? If we open box 2 and check if the diamond is there, but box 2's message is "left or right," i.e., box 1 or 3. If the diamond is in box 1, then box 2's message is true, so opening box 2 won't reveal the diamond. Wait, but if we open box 2 and find the diamond, then that would mean the diamond is in box 2, but then box 1's message would be "left or right" (box 0, which doesn't exist, or box 2). Since box 1's message is "the diamond is in box 0 or 2," which is false because the diamond is in box 2. Wait, no. If the diamond is in box 2, then box 1's message is "left or right" (box 0 or box 2). Since box 2 is the diamond's location, box 1's message would be true. Similarly, box 3's message would be "left or right" (box 2 or 4), so box 3's message is also true. But this would mean two true messages, which contradicts the problem's condition. Therefore, the diamond cannot be in box 2. Therefore, opening box 2 would not reveal the diamond. Similarly, if we open box 99, if the diamond is in box 100, then box 99's message is true, but opening box 99 won't have the diamond. Wait, this is getting confusing.Let me try a different approach. Since exactly one message is true, and all others are false, and each message is about neighbors. So, the true message must be in a position where only one of its neighbors is the diamond. That can only happen if the diamond is at one end. Because if the diamond is in box 1, then only box 2's message is true. Similarly, box 100's case. Any other position would require two messages to be true (from both neighbors). Therefore, the diamond must be in box 1 or 100. Therefore, the solution is that the diamond is in either box 1 or 100. But how do we determine which one by opening just one box?Wait, if we open box 2. If the message on box 2 is true, then the diamond is in box 1 or 3. But if the diamond is in box 3, then box 2's message is true, but box 4's message would say the diamond is in 3 or 5, which would also be true if the diamond is in 3. But that would make two true messages, which is invalid. Therefore, if box 2's message is true, the diamond must be in box 1. Similarly, if we open box 99 and check if its message is true, then the diamond is in 100. But how do we check if the message is true? Because the message is "the diamond is in left or right." To check if it's true, we need to know where the diamond is. But we don't know yet. Wait, but if we open a box, say box 50, and check its message's truthfulness. Wait, but opening box 50's lid would allow us to read the message, which is the same as all others: "the diamond is in the box to the left or to the right." But how does that help us? The message's content is the same for all boxes. The difference is which message is true. But we can't see the diamond when we open the lid; we can only open the box to check if the diamond is inside. Wait, wait, the problem says "opening the lid of just one of them." Does that mean opening the lid to read the message, or opening the box to check if the diamond is inside? The problem states, "discover where the diamond is by opening the lid of just one of them." So perhaps opening the lid refers to checking the message, not the contents. But then, if we can read the message, but all messages are the same text. Wait, but the messages are on the lids. Wait, no, each box's message is about its own neighbors. So each lid has the message "the diamond is in the box to the left or to the right." So, the message is the same on each lid, but its truth depends on the diamond's location. So, if we open a lid, we can read the message, but we already know what it says. Therefore, perhaps opening the lid here means opening the box to check if the diamond is inside. Because otherwise, reading the message on the lid doesn't give new information since all lids have the same message. The key is that only one of those messages is true. So, perhaps the problem is that you can open a box (check its content) and based on that, determine the location of the diamond using the fact that only one message is true.Let me clarify: the problem says "by opening the lid of just one of them, is it possible to discover where the diamond is." So "opening the lid" likely means opening the box to see if the diamond is inside. So you can choose to open one box, and based on whether the diamond is there or not, deduce its location. So if you open a box and find the diamond, you're done. If you don't, you have to use that information to figure out where it is.But the question is whether there exists a box such that opening it will allow you to determine the diamond's location, regardless of whether the diamond is in that box or not.Given that the diamond must be in box 1 or 100 (as deduced earlier), if we open box 2. If the diamond is in box 1, then box 2's message is true. If we open box 2 and don't find the diamond, then we know that the message on box 2 is true, which would mean the diamond is in box 1 or 3. But since we know only one message is true, if box 2's message is true, then the diamond must be in box 1 (because if it were in box 3, then box 4's message would also be true, which can't happen). Therefore, by opening box 2 and not finding the diamond, we can conclude it's in box 1. Similarly, opening box 99 and not finding the diamond would imply it's in box 100.Wait, but the problem allows us to open only one box. So if we open box 2 and don't find the diamond, we can deduce it's in box 1. Similarly, opening box 99 and not finding the diamond would lead us to box 100. But how do we choose which box to open? Because we don't know in advance whether it's in 1 or 100. So maybe there's a box we can open such that whether we find the diamond or not, we can deduce the correct location.Alternatively, maybe the solution is to open box 2. If we open box 2 and the diamond is there, then we've found it. But wait, but earlier we determined that the diamond can't be in box 2 because that would make both box 1 and box 3's messages true. Since the problem states exactly one message is true, the diamond can't be in box 2. Therefore, if we open box 2, we know the diamond isn't there, so then if box 2's message is true (which would mean diamond is in 1 or 3), but since diamond can't be in 3 (as that would require two true messages), it must be in 1. Similarly, opening box 99 would lead to 100.But how do we know whether to open box 2 or box 99? The problem requires that by opening just one box, we can discover the diamond's location. So the strategy must be to open a box such that, regardless of the outcome (finding the diamond or not), we can determine the location. However, since the diamond can't be in any box except 1 or 100, and if we open a middle box, we know it's not there, so we can ignore those. So maybe the trick is to open box 2. If we open box 2 and don't find the diamond, then since box 2's message must be the true one (because all others are false), so the diamond is in 1 or 3. But since 3 would require another true message, which is impossible, it must be in 1. Similarly, if we open box 99 and don't find the diamond, it must be in 100. But since we can only open one box, we need to choose either 2 or 99. But how do we decide which one? Unless there's a box that covers both possibilities.Wait, maybe the answer is that the diamond is in box 1 or 100, and by opening either box 1 or 100, we can find it. If we open box 1 and find the diamond, done. If not, then it must be in 100. Similarly for opening box 100. Therefore, the solution is to open one of the end boxes. If you open box 1 and it's there, great. If not, it's in 100. So by opening just one box (either 1 or 100), you can determine the location. But does this satisfy the problem's condition? The problem says "by opening the lid of just one of them, is it possible to discover where the diamond is." So yes, if you open box 1 and find the diamond, you know it's there. If you don't, you know it's in 100. Therefore, the answer is yes, and the diamond is either in box 1 or 100, so opening one of the end boxes allows you to determine which.But wait, the problem doesn't specify that you can choose which box to open based on some prior reasoning. The question is whether there exists a single box that you can open such that, after opening it, you can determine the diamond's location. So if you choose to open box 1, and if the diamond is there, you're done. If not, it's in 100. Similarly for box 100. Therefore, the answer is yes: open box 1. If the diamond is there, done. If not, it's in 100. Since only one message is true, and the diamond must be in an end box, this works.But let's verify this. Suppose the diamond is in box 1. Then, box 2's message is true (saying left or right, and left is box 1 which has the diamond). All other messages are false. If we open box 1, we find the diamond. If the diamond is in box 100, then box 99's message is true, and all others are false. If we open box 1 and don't find the diamond, then the diamond must be in box 100. Therefore, by opening box 1, we can determine the location. Similarly, opening box 100 would work. Therefore, the answer is yes, and the solution is to open an end box (either 1 or 100), and based on whether the diamond is found there, conclude its location.But wait, the problem says "discover where the diamond is" by opening just one box. So even if you don't find it in the opened box, you can deduce the location. Therefore, the answer is yes. So the strategy is to open box 1. If it's there, done. If not, it's in 100. Therefore, only one box needs to be opened.

✅:The diamond must be in either box 1 or 100. Here's why:1. Constraints Analysis: Each box claims the diamond is adjacent to it. Only one statement is true. If the diamond were in a middle box (e.g., box (k), (2 leq k leq 99)), both neighbors ((k-1) and (k+1)) would have true statements, violating the "exactly one true message" condition. Hence, the diamond must be at an end.2. Edge Cases: - Box 1: If the diamond is here, box 2's message ("left or right") is true (since the diamond is to the left of box 2). All other messages are false. - Box 100: If the diamond is here, box 99's message ("left or right") is true (since the diamond is to the right of box 99). All other messages are false.3. Verification Strategy: - Open box 1: - If the diamond is found, the problem is solved. - If not, the diamond must be in box 100. This follows because box 2's message would then be false (the diamond isn’t in box 1 or 3), leaving box 99 as the only valid truth-teller, which points to box 100. Similarly, opening box 100 and not finding the diamond would confirm it’s in box 1.Conclusion: By opening either box 1 or 100, you can definitively determine the diamond's location based on the result. If the opened box contains the diamond, that’s the location. If not, the diamond is in the opposite end box. This satisfies the problem's requirement of opening just one box. Answer: Yes, open either the first or last box. If the diamond is found, you’re done. If not, it’s in the opposite end box. Thus, the diamond is in box boxed{1} or boxed{100}, determinable by opening one end box.

🔑:Given that there are 100 identical boxes arranged in a line, it is known that one of these boxes contains a diamond. Each box has a message stating that "the diamond is in the box to the left or to the right." Exactly one message is true, and all others are false. We need to determine if it is possible to discover where the diamond is by lifting the lid of just one box.1. Analyze the problem constraints: - Only one box contains the true statement. - All other boxes contain false statements.2. Consider the boxes at the extremes: - If the diamond is not in one of the extreme boxes, then either both neighboring boxes would incorrectly point to the diamond's location (because if one message was true, then at least one other message would have to imply the diamond's wrong position, violating the given condition).3. Verify by opening an extreme box: - Select one of the extreme boxes (e.g., Box 1 or Box 100). - Let's assume we open Box 1. 4. Evaluate the message in Box 1: - If Box 1 contains the diamond, then the statement on Box 1 would have to be "the diamond is in Box 2" if it were true. This is not possible since Box 1 would then have the diamond, contradicting the statement. Thus, the statement in Box 1 should correctly be false pointing towards Box 2. - If the statement is true, the diamond's position and the message's truthfulness would not match. Therefore, Box 1 cannot have the diamond and, by elimination, the diamond must be at the other extremity, which is Box 100.5. Generalize the conclusion: - If you open Box 1 and it does not have the diamond, the diamond must be in Box 100. Alternatively, if Box 100 does not contain the diamond, opening Box 1 would reveal it. Conclusion:By opening one of the extreme boxes, we can determine with certainty where the diamond is located. If the chosen extreme box does not contain the diamond, it must be in the other extreme box.[boxed{text{Yes}}]

🧠:Okay, so I have this geometry problem here. Let me try to parse it step by step. The problem is about triangle ABC with sides AB = 2, CA = 3, and BC = 4. Then there's a point D, which is diametrically opposite to A on the circumcircle of triangle ABC. Then point E is on line AD such that D is the midpoint of AE. Then line l passes through E and is perpendicular to AE. The extensions of AB and AC intersect line l at points F and G. We need to compute the length FG.First, I need to visualize this. Let me sketch the triangle ABC with the given sides. Since AB = 2, BC = 4, and CA = 3, this isn't a right-angled triangle. Let me check if such a triangle is possible using the triangle inequality. AB + BC > AC: 2 + 4 = 6 > 3, yes. AB + AC > BC: 2 + 3 = 5 > 4, yes. BC + AC > AB: 4 + 3 = 7 > 2, yes. So, triangle ABC is valid.Next, D is diametrically opposite A on the circumcircle. So, if I imagine the circumcircle of triangle ABC, point D would be such that AD is a diameter. That means angle ABD and angle ACD are right angles? Wait, no, the angle subtended by a diameter is a right angle, so any point on the circle will form a right angle with the endpoints of the diameter. So, angle ABD and ACD would be right angles? Wait, maybe not exactly. Wait, if AD is a diameter, then any point B on the circle will have angle ABD as a right angle. But D is diametrically opposite A, so BD is perpendicular to AB? Hmm, maybe not. Let me recall the property: if AD is a diameter, then angle ABD is 90 degrees if B is on the circle. Wait, actually, if you have a circle with diameter AD, then any point B on the circle will satisfy that angle ABD is 90 degrees. So in our case, triangle ABC's circumcircle has diameter AD, so point B is on the circle, so angle ABD is 90 degrees. Similarly, angle ACD would also be 90 degrees because C is on the circle as well.Wait, but in our triangle ABC, D is diametrically opposite A. So, BD and CD would be related to AB and AC. Hmm. Maybe I can use coordinates to model this. Let me try to assign coordinates to the triangle to make calculations easier.Let me place point A at the origin (0,0). Let me denote coordinates as follows: A(0,0). Then, let me place point B somewhere on the plane. Since AB = 2, let me place B at (2,0). Then point C is such that AC = 3 and BC = 4. Let me find the coordinates of point C.So, coordinates of A(0,0), B(2,0). Let C be (x,y). Then, AC = 3 implies that x² + y² = 9. BC = 4 implies that (x - 2)² + y² = 16. Subtract the first equation from the second: (x - 2)² + y² - x² - y² = 16 - 9 => (x² -4x +4) - x² = 7 => -4x +4 =7 => -4x =3 => x = -3/4. Then, substitute x into x² + y² =9: (9/16) + y² =9 => y² = 9 - 9/16 = 135/16 => y = ±(3√15)/4. Let's take y positive, so C is (-3/4, 3√15/4).So coordinates are:A(0,0), B(2,0), C(-3/4, 3√15/4).Now, need to find the circumcircle of triangle ABC. The circumcircle can be found by determining the circumradius and the center. Alternatively, since D is diametrically opposite A, the center of the circle is the midpoint of AD. Let me denote the center as O. Since AD is a diameter, O is the midpoint of AD. Therefore, if I can find the coordinates of D, that would help. But to find D, I need to know the circumradius and the direction from A to D.Alternatively, since O is the circumcenter of triangle ABC, which can be found by perpendicular bisectors of AB and AC.Let me compute the circumcenter O.First, find the perpendicular bisector of AB. AB is from (0,0) to (2,0). The midpoint of AB is (1,0). The slope of AB is 0, so the perpendicular bisector is vertical line x=1.Now, find the perpendicular bisector of AC. AC is from (0,0) to (-3/4, 3√15/4). The midpoint of AC is (-3/8, 3√15/8). The slope of AC is (3√15/4 - 0)/(-3/4 -0) = (3√15/4)/(-3/4) = -√15. Therefore, the perpendicular bisector has slope 1/√15. So the equation is y - 3√15/8 = (1/√15)(x + 3/8).Now, find the intersection of x=1 and this line. Substitute x=1 into the equation:y - 3√15/8 = (1/√15)(1 + 3/8) = (1/√15)(11/8) = 11/(8√15)Therefore, y = 3√15/8 + 11/(8√15). Let me rationalize the denominators:3√15/8 = (3√15)/811/(8√15) = (11√15)/(8*15) = (11√15)/120So total y = (3√15)/8 + (11√15)/120 = convert to common denominator, which is 120.3√15/8 = (45√15)/120So y = (45√15 + 11√15)/120 = (56√15)/120 = (14√15)/30 = (7√15)/15Therefore, the circumcenter O is at (1, 7√15/15). Then, since AD is the diameter, D is the reflection of A over O. Since A is at (0,0), then D has coordinates (2*1 -0, 2*(7√15/15) -0) = (2, 14√15/15).So point D is (2, 14√15/15). Then, point E is such that D is the midpoint of AE. Therefore, E is such that:Coordinates of D are the midpoint of A(0,0) and E(x,y). So midpoint formula:( (0 + x)/2, (0 + y)/2 ) = (2, 14√15/15 )Therefore:x/2 = 2 => x =4y/2 =14√15/15 => y =28√15/15So point E is (4, 28√15/15).Now, line l passes through E and is perpendicular to AE. Let's find the direction of AE. AE is from A(0,0) to E(4,28√15/15). The slope of AE is (28√15/15 -0)/(4 -0) = (28√15)/15 /4 = 7√15/15. Therefore, the line l is perpendicular to AE, so its slope is the negative reciprocal: -15/(7√15) = -√15/7 (rationalizing the denominator: -15/(7√15) = - (15√15)/(7*15) = -√15/7 )So line l has slope -√15/7 and passes through E(4,28√15/15). Let's write the equation of line l.Using point-slope form:y - 28√15/15 = (-√15/7)(x -4)Let me keep this equation as it is for now.Next, we need to find the intersections F and G of the extensions of AB and AC with line l.First, find equation of AB. AB is from A(0,0) to B(2,0), so it's the x-axis. The equation is y=0. But the problem states that F and G are intersections of the extensions of AB and AC with l. However, line AB is already extended infinitely in both directions. But since E is at (4,28√15/15), which is above the x-axis, and line l is perpendicular to AE. Let me check where line l intersects AB. Wait, AB is y=0. So intersection point F is the point where l meets y=0.Wait, but let me compute it.Equation of line l: y - 28√15/15 = (-√15/7)(x -4)Set y=0:0 -28√15/15 = (-√15/7)(x -4)Multiply both sides by 7:-28√15/15 *7 = -√15(x -4)Simplify left side: -196√15/15Right side: -√15(x -4)Divide both sides by -√15:(196/15) = x -4Therefore, x =4 +196/15 = (60 +196)/15 =256/15 ≈17.066...So coordinates of F are (256/15, 0).Wait, but AB is from A(0,0) to B(2,0). The extension of AB beyond B would be the x-axis beyond x=2. But here, F is at x=256/15 ≈17.066, which is way beyond B. So F is on the extension of AB beyond B.Similarly, we need to find G, which is the intersection of the extension of AC with line l.First, find the equation of AC. AC is from A(0,0) to C(-3/4, 3√15/4). Let me compute the slope of AC: (3√15/4 -0)/(-3/4 -0) = (3√15/4)/(-3/4) = -√15. So equation of AC is y = -√15 x.But since G is on the extension of AC, which is the same line beyond C. So to find G, we need to find the intersection of line l with line AC (which is y = -√15 x).Set y = -√15 x into the equation of line l:-√15 x -28√15/15 = (-√15/7)(x -4)Multiply both sides by 15 to eliminate denominators:-15√15 x -28√15 = (-√15/7)(x -4)*15Simplify right side: (-√15/7)*15*(x -4) = (-15√15/7)(x -4)So equation becomes:-15√15 x -28√15 = (-15√15/7)(x -4)Multiply both sides by 7 to eliminate denominator:-105√15 x -196√15 = -15√15(x -4)Divide both sides by √15:-105x -196 = -15(x -4)Simplify:-105x -196 = -15x +60Bring all terms to left side:-105x -196 +15x -60 =0 => -90x -256=0 => -90x =256 => x= -256/90 = -128/45 ≈-2.844...Then y = -√15 x = -√15*(-128/45) =128√15/45.Therefore, coordinates of G are (-128/45, 128√15/45).So now we have coordinates of F(256/15, 0) and G(-128/45,128√15/45). Need to compute FG.First, compute the distance between F and G.Coordinates:F: (256/15, 0)G: (-128/45, 128√15/45)Compute Δx and Δy:Δx = (-128/45) - (256/15) = (-128/45) - (768/45) = (-896)/45Δy = 128√15/45 -0 =128√15/45Then FG = √[(Δx)^2 + (Δy)^2] = √[ ( (-896/45)^2 ) + ( (128√15/45)^2 ) ]Compute each term:First term: (-896/45)^2 = (896)^2 / (45)^2Second term: (128√15/45)^2 = (128)^2 *15 / (45)^2Let me compute them:First term:896^2: Let's compute 896^2. 800^2=640000, 96^2=9216, 2*800*96=153600. So (800+96)^2=640000 +153600 +9216= 803, 640000 +153600=793600, +9216=802,816. So 896^2=802,816. Therefore, first term:802,816 /2025.Second term:128^2=16,384. 16,384 *15=245,760. So second term=245,760 /2025.Therefore, total under the square root:(802,816 +245,760)/2025 =1,048,576 /2025Therefore, FG=√(1,048,576 /2025 )= √(1,048,576)/√(2025)= 1024/45Wait, because 1024^2=1,048,576 and 45^2=2025. So FG=1024/45.Simplify 1024/45: 1024 ÷ 45 =22.7555... But as a fraction, it's 1024/45, which can be reduced? 1024 and 45: 45=9*5, 1024 is 2^10. No common factors. So FG=1024/45.Wait, 1024 divided by 45 is 22 and 34/45. But the problem might want the answer as a fraction. So 1024/45.But let me check my calculations again because this seems straightforward, but maybe I messed up in coordinates.Wait, let's check coordinates of F and G again.Coordinates of F: Intersection of line l and AB (y=0). We solved:From line l: y -28√15/15 = (-√15/7)(x -4)Set y=0:-28√15/15 = (-√15/7)(x -4)Multiply both sides by 7:-28√15/15 *7 = -√15(x -4)Left side: -196√15/15Right side: -√15(x -4)Divide both sides by -√15:(196/15) = x -4Hence x=4 +196/15 = (60 +196)/15=256/15. So F is (256/15,0). That seems correct.Coordinates of G: Intersection of line l and AC (y= -√15 x). Solved:Substitute y= -√15 x into line l:-√15 x -28√15/15 = (-√15/7)(x -4)Multiply both sides by 15:-15√15 x -28√15 = (-15√15/7)(x -4)Multiply both sides by 7:-105√15 x -196√15 = -15√15(x -4)Divide both sides by √15:-105x -196 = -15x +60Bring variables to left:-105x +15x =60 +196-90x=256x= -256/90= -128/45. Then y= -√15*(-128/45)=128√15/45. So G is (-128/45,128√15/45). That seems correct.Then distance FG:Δx= (-128/45 -256/15)= (-128/45 -768/45)= -896/45Δy=128√15/45 -0=128√15/45Then distance squared: (896/45)^2 + (128√15/45)^2Compute 896^2=802,816; 128^2=16,384So:802,816/(45^2) + (16,384*15)/(45^2)= [802,816 +245,760]/2025=1,048,576/2025Then sqrt(1,048,576)/sqrt(2025)=1024/45. Correct.Therefore, FG=1024/45. Let me check if that's reducible. 1024 is 2^10, 45 is 3^2*5. No common factors. So the answer is 1024/45.But wait, 1024 divided by 45 is 22.7555... which is a bit strange. Let me check if my calculations for the coordinates are correct.Wait, maybe I made a mistake in coordinates of point C. Let me check again.Given triangle ABC with AB=2, BC=4, and AC=3. Coordinates: A(0,0), B(2,0). Let me recalculate point C.Coordinates of C(x,y):Distance from A: sqrt(x^2 + y^2)=3, so x² + y²=9.Distance from B: sqrt( (x-2)^2 + y² )=4, so (x-2)^2 + y²=16.Subtract first equation from second:(x-2)^2 + y² -x² -y²=16-9Expanding (x² -4x +4) -x²=7 => -4x +4=7 => -4x=3 =>x= -3/4. Correct.Then y²=9 - (9/16)= (144/16 -9/16)=135/16, so y=±(3√15)/4. So point C is (-3/4, 3√15/4). Correct.Then circumradius: found circumcenter at (1,7√15/15). Let me verify that.Midpoint of AB is (1,0), perpendicular bisector is x=1. Midpoint of AC is (-3/8, 3√15/8). The slope of AC is (3√15/4)/(-3/4)= -√15. Therefore, perpendicular bisector slope is 1/√15. Equation: y -3√15/8=(1/√15)(x +3/8). When x=1, solving for y:y=3√15/8 + (1/√15)(1 +3/8)=3√15/8 + (11/8)/√15=3√15/8 +11/(8√15). Rationalize:3√15/8= (3√15 *15)/ (8*15)=45√15/12011/(8√15)= (11√15)/(8*15)=11√15/120Total y= (45√15 +11√15)/120=56√15/120=14√15/30=7√15/15. So y=7√15/15. So circumcenter O is (1,7√15/15). Correct.Then D is diametrically opposite A, so coordinates of D: midpoint of AD is O. So since A is (0,0), then O is ( (0 + D_x)/2, (0 + D_y)/2 ). Therefore, D_x=2*1=2, D_y=2*(7√15/15)=14√15/15. So D is (2,14√15/15). Correct.Then E is such that D is midpoint of AE. So E is (4,28√15/15). Correct.Then line AE is from (0,0) to (4,28√15/15). Slope is (28√15/15)/4=7√15/15. Therefore, line l is perpendicular, slope=-√15/7. Equation of line l through E(4,28√15/15): y -28√15/15= (-√15/7)(x -4). Correct.Intersection with AB (y=0) gives x=256/15, which is F(256/15,0). Correct.Intersection with AC (y=-√15 x) gives G(-128/45,128√15/45). Correct.Distance FG computed as 1024/45. Hmm. Let me check with another approach.Alternatively, maybe using vectors or coordinate geometry properties.Alternatively, since FG is the length we need, and we have coordinates, but perhaps there's a smarter way using similar triangles or projective geometry, but given the coordinates are already found, and calculation leads to 1024/45, I think that's the answer.But to be thorough, let me verify the calculation of the distance again.Coordinates:F: (256/15, 0)G: (-128/45,128√15/45)Compute Δx = (-128/45) - (256/15) = (-128/45 - 256/15) = (-128/45 - 768/45) = (-896/45)Δy =128√15/45 -0 =128√15/45Distance FG = sqrt( (Δx)^2 + (Δy)^2 )(Δx)^2 = (896/45)^2 = (896^2)/(45^2) = 802,816 / 2025(Δy)^2 = (128√15/45)^2 = (128^2 *15)/(45^2) = (16,384 *15)/2025 =245,760 /2025Sum: 802,816 +245,760 =1,048,576. Therefore, sqrt(1,048,576/2025)=1024/45. Correct.Yes, that's correct. 1024 squared is 1,048,576, and 45 squared is 2025. So sqrt(1,048,576)/sqrt(2025)=1024/45.Thus, FG=1024/45.But let me check if this answer makes sense geometrically. The line l is perpendicular to AE, which is a relatively long line from A to E. The points F and G are intersections of AB and AC extended, which are quite far from the original triangle, so FG being a large number makes sense. 1024/45 is approximately 22.76, which seems plausible given the coordinates.Alternatively, is there another approach using properties of circles, midpoints, or similar triangles?Alternatively, since D is the midpoint of AE, then AE=2AD. Since AD is the diameter of the circumcircle, so AD=2R, where R is the radius. Therefore, AE=4R.But maybe this approach would complicate things. Alternatively, using coordinate geometry was straightforward.Alternatively, use vectors.Let me try a vector approach.Let me set A as the origin. Vector AB is (2,0), vector AC is (-3/4,3√15/4). Then, the circumcircle of ABC. The circumradius can be calculated.But perhaps this is more involved. Alternatively, since we already have coordinates and computed the answer as 1024/45, which seems consistent through multiple checks, I think that is the correct answer.Thus, the answer is boxed{dfrac{1024}{45}}.Final Answerboxed{dfrac{1024}{45}}

✅:Given triangle (ABC) with (AB = 2), (CA = 3), and (BC = 4). We need to find the length (FG) where (F) and (G) are the intersections of the extensions of (overline{AB}) and (overline{AC}) with line (l), which passes through (E) (a point such that (D) is the midpoint of (overline{AE})) and is perpendicular to (overline{AE}). 1. Coordinates of Points: - Place (A) at ((0, 0)), (B) at ((2, 0)), and find (C) using distances: - (C) coordinates: (left(-frac{3}{4}, frac{3sqrt{15}}{4}right)).2. Circumcircle and Diameter: - Find the circumcenter (O) by perpendicular bisectors: - Circumcenter (O) coordinates: (left(1, frac{7sqrt{15}}{15}right)). - Point (D) (diametrically opposite (A)) coordinates: (left(2, frac{14sqrt{15}}{15}right)).3. Point (E): - (D) is the midpoint of (AE), so (E) coordinates: (left(4, frac{28sqrt{15}}{15}right)).4. Line (l): - Line (l) through (E) with slope (-frac{sqrt{15}}{7}) (perpendicular to (AE)): - Equation: (y - frac{28sqrt{15}}{15} = -frac{sqrt{15}}{7}(x - 4)).5. Intersections (F) and (G): - Intersection (F) with (AB) (y=0): (Fleft(frac{256}{15}, 0right)). - Intersection (G) with (AC) (y = -√15 x): (Gleft(-frac{128}{45}, frac{128sqrt{15}}{45}right)).6. Distance (FG): - Calculate (Delta x) and (Delta y): - (Delta x = -frac{896}{45}) - (Delta y = frac{128sqrt{15}}{45}) - Distance (FG = sqrt{left(-frac{896}{45}right)^2 + left(frac{128sqrt{15}}{45}right)^2} = frac{1024}{45}).The final answer is (boxed{dfrac{1024}{45}}).

🔑:We are given a triangle ( triangle ABC ) with side lengths ( AB = 2 ), ( AC = 3 ), and ( BC = 4 ). Let's compute the circumradius ( R ) of this triangle and the distance ( FG ).1. Compute the Area of ( triangle ABC ): Using Heron’s formula, first calculate the semi-perimeter ( s ): [ s = frac{AB + AC + BC}{2} = frac{2 + 3 + 4}{2} = 4.5. ] Now, compute the area ( Delta ) of ( triangle ABC ): [ Delta = sqrt{s(s - AB)(s - AC)(s - BC)} = sqrt{4.5 times (4.5 - 2) times (4.5 - 3) times (4.5 - 4)}. ] Simplifying inside the square root: [ Delta = sqrt{4.5 times 2.5 times 1.5 times 0.5} = sqrt{16.875}. ] Hence, [ Delta = frac{3sqrt{15}}{4}. ]2. Calculate the Circumradius ( R ): Using the formula for the area in terms of the circumradius ( R ): [ Delta = frac{ABC}{4R}. ] Rearranging to solve for ( R ): [ R = frac{ABC}{4Delta}. ] Plugging in the values: [ R = frac{2 times 3 times 4}{4 times frac{3sqrt{15}}{4}} = frac{24}{3sqrt{15}} = frac{8}{sqrt{15}}. ] Squaring both sides: [ R^2 = left(frac{8}{sqrt{15}}right)^2 = frac{64}{15}. ]3. Considering Angle ( angle ABD = 90^circ ): Since ( D ) is diametrically opposite ( A ), ( angle ABD = 90^circ ). This makes ( BDE ) cyclic. Because ( D ) is the midpoint of ( overline{AE} ), we see that ( AD = 2R ) and ( AE = 4R ).4. Using Cyclic Quadrilaterals and Power of a Point: Since ( BDE ) and ( BDF ) are cyclic, using Power of a Point theorem: [ AB cdot AF = AD cdot AE = 2R cdot 4R = 8R^2. ] Substituting ( R^2 = frac{64}{15} ): [ AB cdot AF = 8 times frac{64}{15} = frac{512}{15}. ] Similarly, we have for ( AC cdot AG ): [ AC cdot AG = AD cdot AE = frac{512}{15}. ]5. Calculate ( FG ): Since ( triangle ABC ) and ( triangle AFG ) are similar, the ratio of the sides must be equal. Thus: [ FG = BC times frac{AF}{AC} = 4 times frac{frac{512}{30}}{3}. ] Simplifying: [ FG = 4 times frac{512}{90} = frac{2048}{90} = frac{1024}{45}. ] Conclusion:[boxed{frac{1024}{45}}]

🧠:Okay, so I need to prove that in triangle ABC with points P, Q, R on sides BC, CA, AB respectively, the ratio YX/XZ is equal to BP/PC. The points X, Y, Z are where AP intersects the circumcircles of triangles AQR, BRP, and CPQ respectively. Hmm, this seems like a problem involving cyclic quadrilaterals and maybe some properties of intersecting chords or power of a point. Let me start by drawing a diagram to visualize the problem.First, triangle ABC. Points P on BC, Q on CA, R on AB. Then, the circumcircles of AQR, BRP, and CPQ are Γ_A, Γ_B, Γ_C. The line AP intersects these circles at X, Y, Z. Wait, AP is starting at A and going through P, which is on BC. So, AP is a cevian. But the circumcircles of AQR, BRP, and CPQ – these are different from the main triangle. Let me note that Γ_A is the circumcircle of AQR, so it passes through points A, Q, R. Γ_B is the circumcircle of BRP, passing through B, R, P. Γ_C is the circumcircle of CPQ, passing through C, P, Q. So AP intersects Γ_A at X (other than A?), Γ_B at Y, and Γ_C at Z. Wait, but AP is passing through P, which is on BC. But P is also on BRP and CPQ. So does AP pass through P? Yes, because AP is from A to P. So when AP intersects Γ_B, which is the circumcircle of BRP, then since P is on both AP and Γ_B, then one intersection point is P. But the problem states that Y is the intersection of AP with Γ_B. But AP goes through P, so unless Y is another intersection point, but since BRP includes point B, R, P. If AP intersects Γ_B again at Y, then Y must be different from P. Wait, but P is on BC and on AP. So unless AP is tangent to Γ_B at P, but that might not necessarily be the case. So Y is another intersection point? Wait, but AP is a line from A through P. Since Γ_B contains point B, which is not on AP (unless AP passes through B, which it doesn't unless ABC is degenerate). Therefore, AP must intersect Γ_B at two points: one is P, and the other is Y. So Y is distinct from P? But the problem says "the line segment AP intersects the circles Γ_A, Γ_B, and Γ_C at points X, Y, and Z respectively." Wait, the line segment AP, not the entire line. So if the segment AP intersects the circles, then the intersection points must lie on the segment AP. But P is on all three circles? Let's check:Γ_A is the circumcircle of AQR. P is on BC, which is a different side, so unless AQR is constructed in a way that P is on Γ_A, which I don't think it is. Similarly, Γ_B is the circumcircle of BRP. Since P is on BRP, so P is on Γ_B. Similarly, Γ_C is the circumcircle of CPQ, so P is on Γ_C as well. Therefore, the line segment AP passes through P, which is on all three circles. But the problem states that AP intersects Γ_A, Γ_B, Γ_C at X, Y, Z respectively. So does that mean that X is another intersection point besides A? Wait, Γ_A is the circumcircle of AQR, so A is on Γ_A. Therefore, AP is a line starting at A, passing through P. So when AP intersects Γ_A, the first point is A, and maybe another point X on AP. But since AP is a line segment from A to P, maybe X is between A and P? Similarly, Γ_B is the circumcircle of BRP, so P is on Γ_B, and AP passes through P. So AP intersects Γ_B at P and maybe another point Y. But if the segment AP is considered, then Y would have to be on AP between A and P. But P is already on the segment. Wait, this is confusing. Let me parse the problem again."Suppose the line segment AP intersects the circles Γ_A, Γ_B, and Γ_C at points X, Y, and Z respectively." So line segment AP (not the whole line) intersects each circle Γ_A, Γ_B, Γ_C at points X, Y, Z. Since AP is a segment from A to P. Now, Γ_A is the circumcircle of AQR, so A is on Γ_A. Therefore, AP starts at A, which is on Γ_A, so the intersection point X is A? But the problem says "points X, Y, Z respectively," implying different points. Maybe the other intersection points? For example, AP might intersect Γ_A again at X (different from A) if AP is not tangent to Γ_A. Similarly, AP intersects Γ_B at Y (different from P) and Γ_C at Z (different from P). But if AP is a segment from A to P, then these points X, Y, Z must lie between A and P. Wait, but Γ_B and Γ_C both contain P, so AP passes through P, which is on both circles. Therefore, the intersections are at P. But the problem specifies points X, Y, Z on the segment AP. So maybe X is the second intersection of AP with Γ_A (other than A), Y is the second intersection with Γ_B (other than P), and Z is the second intersection with Γ_C (other than P). Therefore, X is between A and P, Y is between A and P, and Z is also between A and P? Wait, but the problem states that AP intersects each circle at X, Y, Z respectively. So maybe each intersection is only one point? But a line segment can intersect a circle at two points. So perhaps, for Γ_A, since A is on Γ_A, and AP is a line segment starting at A, then X is the other intersection point of AP with Γ_A (if it exists). Similarly, Γ_B and Γ_C contain P, so the other intersection point with AP (if any) would be Y and Z. But since AP is a segment from A to P, the other intersections would have to be between A and P. So X, Y, Z are all on AP between A and P. But how can AP intersect each circle at one point? Unless the problem is considering the intersections different from A and P. Let me check the exact wording: "the line segment AP intersects the circles Γ_A, Γ_B, and Γ_C at points X, Y, and Z respectively." So perhaps each circle is intersected once by AP, at X, Y, Z. But that seems unlikely unless each circle is tangent to AP at those points. But the problem doesn't mention tangency. So maybe the circles are arranged such that AP intersects each circle at two points, but on the line segment AP, only one of them is on the segment. For Γ_A, since A is on Γ_A and AP is starting at A, maybe the other intersection X is on AP beyond P? But AP is a segment from A to P, so beyond P would be outside the segment. Hmm, this is confusing. Maybe the problem is referring to the entire line AP intersecting the circles at X, Y, Z, with X on Γ_A, Y on Γ_B, Z on Γ_C, all lying on the line AP. Then, perhaps X is the other intersection of line AP with Γ_A (other than A), Y is the other intersection with Γ_B (other than P), and Z is the other intersection with Γ_C (other than P). Then, X, Y, Z would lie on the line AP, but not necessarily on the segment AP. But the problem says "the line segment AP intersects the circles...", so the intersections must be on the segment. Therefore, maybe X is between A and P on AP, and similarly Y and Z. But if Γ_B and Γ_C both contain P, then AP passes through P, which is on both circles, so the intersections would be at P. Unless there is another intersection point on the segment AP. But that would require the circles Γ_B and Γ_C to intersect AP again at another point between A and P. Let's think about Γ_B, the circumcircle of BRP. If AP intersects Γ_B at two points: P and Y. If Y is another point on AP, then Y must be either between A and P or beyond P. But since we are talking about the segment AP, Y has to be between A and P. Similarly for Z. But then, how do we know that these intersections exist? Maybe by construction, the points Q and R are chosen such that these intersections occur. The problem states that P, Q, R are points on the sides, but doesn't specify their positions. Maybe we need to assume general positions where the intersections X, Y, Z exist on segment AP. Alright, assuming that, let's proceed.So AP is a line segment from A to P. It intersects Γ_A (circumcircle of AQR) at A and X, with X between A and P. It intersects Γ_B (circumcircle of BRP) at Y and P, with Y between A and P. Similarly, intersects Γ_C (circumcircle of CPQ) at Z and P, with Z between A and P. Therefore, along AP, starting from A, we have points X, then Y, then Z, then P? Or maybe in some other order? It's not clear yet. Wait, but if X is on Γ_A, which is the circumcircle of AQR, then maybe X is related to Q and R. Similarly, Y is related to BRP, and Z to CPQ. I need to relate these points to find the ratio YX/XZ = BP/PC.This seems like it might involve Menelaus' theorem or Ceva's theorem, but since circles are involved, perhaps power of a point or cyclic quadrilaterals. Alternatively, maybe using inversion or projective geometry. Let me think step by step.First, let's recall that if four points lie on a circle, then the power of a point with respect to the circle can be expressed in terms of the distances along a line. For example, if a line intersects a circle at two points, the power of a point on the line can be calculated as the product of the distances from the point to each intersection. So, for point A with respect to Γ_A (which it's on, so power is zero), but maybe for other points.Alternatively, since X, Y, Z are on AP and on the respective circles, we can write power conditions. For example, since X is on Γ_A (circumcircle of AQR), the power of X with respect to Γ_A is zero. Similarly for Y with Γ_B and Z with Γ_C.But how to relate these to the ratio BP/PC. Let's see. The ratio BP/PC is along BC, which is the side containing P. Maybe using Ceva's theorem, but Ceva involves concurrency of lines. Alternatively, using Menelaus with a transversal cutting the sides of the triangle.Alternatively, cross ratios. Since the problem involves ratios along a line (AP) and a ratio on BC, perhaps there is a cross ratio preservation or some projective relation.Alternatively, using the theorem of intersecting chords: if two chords intersect, the products of their segments are equal. For example, in Γ_B, the circumcircle of BRP, the point Y is on AP and on Γ_B. So AY * YP = power of Y with respect to Γ_B? Wait, but Y is on Γ_B, so its power is zero. Wait, no: the power of Y with respect to Γ_B is zero because Y is on the circle. So maybe writing the power of point Y with respect to other circles.Alternatively, using angles. Since points are on circumcircles, certain angles will be equal. For example, in Γ_A (AQR), the angles at X should relate to angles in triangle AQR. Maybe some angle chasing can help.Let me attempt some angle chasing. Let's consider Γ_A, the circumcircle of AQR. Since X is on Γ_A, angle AXQ should be equal to angle ARQ because they subtend the same arc AQ. Wait, but X is on AP, so maybe looking at angles involving X, Q, R.Alternatively, since X is on Γ_A, we have that ∠AXQ = ∠ARQ (if X and R are on the same side of AQ), but I need to verify the exact configuration.Alternatively, consider the power of point X with respect to other circles. Since X is on Γ_A, its power with respect to Γ_B and Γ_C can be expressed. But this might get complicated.Wait, perhaps using radical axes. The radical axes of the circles Γ_A, Γ_B, Γ_C might intersect at a common point, but I don't see immediately how that would help.Alternatively, consider inversion. If I invert the figure with respect to a circle centered at A, maybe some of the circles will invert into lines, making the problem simpler. However, inversion can be tricky without knowing exactly which circles to invert.Alternatively, coordinate geometry. Assign coordinates to the triangle and compute the coordinates of X, Y, Z, then compute the distances. This might be a bit involved but possible. Let's consider setting coordinate system with A at (0,0), B at (1,0), C at (0,1). Then P is on BC, which goes from (1,0) to (0,1). Let’s parameterize P as (1 - t, t) for some t between 0 and 1. Similarly, Q is on AC, which is from (0,0) to (0,1); maybe parameterize Q as (0, s), and R is on AB, from (0,0) to (1,0), parameterize R as (r, 0). Then, we can write equations for the circumcircles Γ_A, Γ_B, Γ_C, find their intersections with AP (which is the line from (0,0) to (1 - t, t)), compute X, Y, Z, then compute YX/XZ and see if it equals BP/PC = t/(1 - t). This might work, but it's going to be algebra-heavy. Let me see if I can set this up.First, coordinates:Let’s set A at (0,0), B at (1,0), C at (0,1).Let’s let P divide BC in the ratio BP:PC = k:1, so coordinates of P are ( (1*1 + k*0)/(k + 1), (1*0 + k*1)/(k + 1) ) = (1/(k + 1), k/(k + 1)).Wait, but in the problem statement, BP/PC is the ratio we need to prove equal to YX/XZ. So if I let BP/PC = k, then PC = BP/k. Let’s instead let BP = k and PC = m, so that BP/PC = k/m. Then coordinates of P would be (m/(k + m), k/(k + m)). But maybe using barycentric coordinates? Alternatively, let's use a parameter t such that BP = t and PC = 1 - t, so P is (1 - t, t) in some coordinate system. Wait, no. If BC is from B(1,0) to C(0,1), then the coordinates of P can be parametrized as (1 - t, t) where t is the fraction along BC. Wait, actually, in the standard coordinate system, the line BC is from (1,0) to (0,1), so parametric equations are x = 1 - t, y = t for t in [0,1]. So point P is (1 - t, t), so BP is the length from B(1,0) to P(1 - t, t), which is sqrt( (t)^2 + (t)^2 ) = t*sqrt(2). Similarly, PC is sqrt( (1 - t)^2 + (1 - t)^2 ) = (1 - t)*sqrt(2). Therefore, BP/PC = t/(1 - t). So if I can show that YX/XZ = t/(1 - t), then that would prove the result.But maybe setting up coordinates complicates things. Let me try another approach.Since the problem involves ratios along AP and BC, maybe using Menelaus’ theorem on triangle ABC with transversal line AP? Wait, Menelaus applies to a transversal cutting the sides of the triangle, but AP is a cevian. Alternatively, use Menelaus on a different triangle.Alternatively, consider the cross ratio. The cross ratio is preserved under projection, so if we can find a projection that maps AP to BC while preserving the cross ratio, then maybe we can relate YX/XZ to BP/PC.Alternatively, use power of a point. For example, consider the power of point Y with respect to Γ_A and Γ_C. But Y is on Γ_B, so its power with respect to Γ_B is zero. Similarly, Z is on Γ_C, so power with respect to Γ_C is zero. Maybe using power of points Y and Z with respect to other circles.Alternatively, look for similar triangles. Maybe triangles formed by these intersection points and the sides of ABC.Wait, here's an idea. Since X is on Γ_A (circumcircle of AQR), Y is on Γ_B (circumcircle of BRP), and Z is on Γ_C (circumcircle of CPQ), perhaps there are cyclic quadrilaterals that we can exploit. For example, since Y is on Γ_B, points B, R, P, Y are concyclic. Therefore, ∠BYR = ∠BPR. Similarly, since Z is on Γ_C, points C, P, Q, Z are concyclic, so ∠CZQ = ∠CPQ.Alternatively, look at the power of point Y with respect to Γ_A. The power of Y with respect to Γ_A is YA * YX = YQ * YR (since Γ_A is the circumcircle of AQR). Similarly, power of Z with respect to Γ_A would be ZA * ZX = ZQ * ZR. But not sure if that helps.Wait, perhaps using harmonic division or projective geometry. The problem relates a ratio along AP to a ratio along BC, so maybe there's a harmonic bundle or something.Alternatively, use Ceva’s theorem in the form of ratios. Ceva’s theorem states that for concurrent lines, the product of certain ratios equals 1. But here, the concurrency is not obvious.Alternatively, use the theorem of intersecting lines and circles. For example, since Y is on Γ_B (circumcircle of BRP), then by power of a point, AY * YP = YB * YR. Wait, but Y is on AP and on Γ_B. Wait, power of Y with respect to Γ_B is zero, but Y is on Γ_B, so maybe that's not helpful.Wait, let's consider inversion with center A. If I invert the figure with respect to a circle centered at A, then circles passing through A (like Γ_A) will invert to lines. Let’s try that.Inversion in center A, radius r (arbitrary). Let's denote inversion as *.Under inversion:- Γ_A (circumcircle of AQR) passes through A, so inverts to a line through Q* and R*.- Similarly, Γ_B (circumcircle of BRP) inverts to a circle passing through B*, R*, P*. But since inversion maps circles not passing through the center to circles. However, B is not necessarily on Γ_B, unless B is on Γ_B, which it is. Wait, Γ_B is the circumcircle of BRP, so B is on Γ_B, so under inversion, Γ_B inverts to a line because it passes through A only if B is A, which it's not. Wait, no. Inversion centered at A: if a circle passes through A, it inverts to a line not passing through A. If a circle doesn't pass through A, it inverts to another circle. Since Γ_B is the circumcircle of BRP. If B is not A (which it isn't), then Γ_B does not pass through A, so inverts to a circle. Similarly for Γ_C.This might complicate things. Maybe instead, consider inverting with respect to a different point.Alternatively, use reciprocal directions or parametrize the points.Wait, let's get back to the problem. Let's consider the ratios on AP. Suppose we can express YX and XZ in terms of segments on AP related to BP and PC.Let me denote AP as a line segment with points A, X, Y, Z, P in that order (assuming they are arranged from A to P). Then YX = AY - AX and XZ = AZ - AX? Wait, no, if the order is A---X---Y---Z---P, then YX = Y - X, XZ = Z - X. But the exact order is not given. It depends on where the circles intersect AP.Alternatively, maybe all three points X, Y, Z lie between A and P, but in some order. For example, A---X---Y---Z---P. Then YX/XZ would be the ratio of lengths between Y and X over Z and X. But how does that relate to BP/PC?Alternatively, perhaps using mass point geometry, assigning masses at B and C such that masses are proportional to BP and PC. Then, perhaps the masses translate to the points along AP.Alternatively, use Ceva’s theorem in a clever way. Let me recall Ceva's condition: if lines from the vertices intersect the opposite sides at P, Q, R, then (BP/PC)(CQ/QA)(AR/RB) = 1. But in this problem, the points Q and R are given, but it's not clear if they satisfy Ceva's condition. The problem doesn't state that AP, BQ, CR are concurrent, so we can't assume that.Wait, but the problem states points P, Q, R are on BC, CA, AB respectively, but doesn't specify any concurrency. So Ceva might not apply here.Alternatively, use Desargues’ theorem if certain triangles are perspective.Alternatively, use spiral similarity or some other transformation that maps one circle to another.Alternatively, consider the cyclic quadrilaterals and use intersecting chords theorem.Let me think again about the cyclic quadrilaterals. For point Y on Γ_B (circumcircle of BRP), so BYRP is cyclic. Therefore, angles ∠BYR = ∠BPR. Similarly, for point Z on Γ_C (circumcircle of CPQ), so CZPQ is cyclic, hence ∠CZQ = ∠CPQ.But how to relate these angles to the ratio on AP.Alternatively, consider the cross ratios. The cross ratio (A, P; Y, Z) might be related to the cross ratio (B, C; P, some point). But I need to think about how cross ratios are preserved under projections.Alternatively, parametrize AP with a parameter λ, such that A is 0 and P is 1, then express the positions of X, Y, Z in terms of λ and compute the ratio.Wait, let's try this. Let’s parametrize AP such that A corresponds to 0 and P corresponds to 1. Let’s let the coordinate along AP be t, where t=0 is A and t=1 is P. Then, any point on AP can be represented by t ∈ [0,1]. Let’s denote X, Y, Z as points on AP with parameters x, y, z respectively, so that AX = x * AP, AY = y * AP, AZ = z * AP, and so on. Then the ratio YX/XZ would be (y - x)/(z - y). We need to show that this equals BP/PC.If we can express x, y, z in terms of BP and PC, then we can compute the ratio. To find x, y, z, we can use the fact that X, Y, Z lie on Γ_A, Γ_B, Γ_C respectively. So, for example, point X is on Γ_A (circumcircle of AQR). Therefore, the power of X with respect to Γ_A is zero. Similarly for Y and Z.But since we’re in a coordinate system along AP, maybe we can use coordinates to express the positions. Let’s assume AP has length 1 for simplicity. Then, point X is at distance x from A, Y at y, Z at z, and P at 1.To find x, since X is on Γ_A (circumcircle of AQR), we can write the condition that X lies on Γ_A. To do this, we need coordinates of Q and R. Let me assign coordinates to the triangle.Let’s set coordinate system with A at (0,0), B at (1,0), C at (0,1). Then, BC is the line from (1,0) to (0,1). Let’s let P be a point on BC, so its coordinates can be written as (1 - t, t) where t is between 0 and 1. Then BP = sqrt( (t)^2 + (t)^2 ) = t√2, and PC = sqrt( (1 - t)^2 + (1 - t)^2 ) = (1 - t)√2. So BP/PC = t/(1 - t).Points Q and R are on AC and AB respectively. Let’s parameterize Q as (0, s) on AC (from (0,0) to (0,1)), and R as (r, 0) on AB (from (0,0) to (1,0)). Now, we need to find the circumcircles Γ_A (AQR), Γ_B (BRP), Γ_C (CPQ).First, let’s find the equation of Γ_A: passing through A(0,0), Q(0, s), R(r, 0). The general equation of a circle passing through A(0,0) is x² + y² + Dx + Ey + F = 0. Since A is on the circle, F = 0. So equation becomes x² + y² + Dx + Ey = 0. Plugging in Q(0, s): 0 + s² + 0 + E*s = 0 ⇒ E = -s. Plugging in R(r, 0): r² + 0 + D*r + 0 = 0 ⇒ D = -r. Thus, the equation of Γ_A is x² + y² - r x - s y = 0.Now, AP is the line from A(0,0) to P(1 - t, t). Parametric equations for AP: x = (1 - t)λ, y = tλ, where λ ∈ [0,1]. The point X is the intersection of AP with Γ_A, different from A. So substituting x = (1 - t)λ, y = tλ into Γ_A's equation:[(1 - t)λ]^2 + [tλ]^2 - r(1 - t)λ - s tλ = 0.Expand:λ²[(1 - t)^2 + t²] - λ[r(1 - t) + s t] = 0.Factor out λ:λ [ λ{(1 - t)^2 + t²} - {r(1 - t) + s t} ] = 0.Solutions are λ = 0 (which is point A) and λ = [r(1 - t) + s t] / [ (1 - t)^2 + t² ].Therefore, the parameter x (position of X along AP) is x = [r(1 - t) + s t] / [ (1 - t)^2 + t² ].Similarly, we can find Y and Z by finding intersections of AP with Γ_B and Γ_C.First, let’s find Γ_B: circumcircle of BRP. Points B(1,0), R(r, 0), P(1 - t, t). Let’s find its equation.Using three points B(1,0), R(r,0), P(1 - t, t). The general equation of a circle is x² + y² + Dx + Ey + F = 0.For point B(1,0): 1 + 0 + D*1 + E*0 + F = 0 ⇒ 1 + D + F = 0.For point R(r,0): r² + 0 + D*r + E*0 + F = 0 ⇒ r² + D r + F = 0.For point P(1 - t, t): (1 - t)^2 + t² + D(1 - t) + E t + F = 0.Let’s set up equations:1. 1 + D + F = 0 ⇒ F = -1 - D.2. r² + D r + F = 0 ⇒ r² + D r -1 - D = 0 ⇒ D(r - 1) + (r² - 1) = 0 ⇒ D(r - 1) = -(r² - 1) ⇒ D = -(r + 1) (if r ≠ 1).So D = -(r + 1). Then F = -1 - D = -1 + r + 1 = r.Now substitute into the third equation:(1 - t)^2 + t² + D(1 - t) + E t + F = 0.Compute each term:(1 - t)^2 + t² = 1 - 2t + t² + t² = 1 - 2t + 2t².D(1 - t) = -(r + 1)(1 - t).E t = E t.F = r.So total equation:1 - 2t + 2t² - (r + 1)(1 - t) + E t + r = 0.Simplify:1 - 2t + 2t² - (r + 1) + (r + 1)t + E t + r = 0.Combine constants: 1 - (r + 1) + r = 1 - r - 1 + r = 0.Combine t terms: -2t + (r + 1)t + E t = [ -2 + r + 1 + E ] t.Quadratic term: 2t².So equation reduces to 2t² + [ -2 + r + 1 + E ] t = 0 ⇒ 2t² + (r -1 + E ) t = 0.Since this must hold for the specific t (coordinate of P), but P is on the circle by construction, so this equation must hold for any t? Wait, no, we already used the three points to determine D, E, F. However, there might be a mistake here. Let me check:Wait, when we substituted point P into the equation, we should get an equation involving E. Since we already found D and F in terms of r, the third equation allows us to solve for E.So after substituting D = -(r + 1) and F = r into the third equation:(1 - t)^2 + t² + D(1 - t) + E t + F = 0Expand:1 - 2t + 2t² - (r + 1)(1 - t) + E t + r = 0.Now, expand -(r + 1)(1 - t):= - (r + 1) + (r + 1)t.So substituting back:1 - 2t + 2t² - (r + 1) + (r + 1)t + E t + r = 0.Simplify constants: 1 - (r + 1) + r = 1 - r -1 + r = 0.Simplify terms with t: -2t + (r + 1)t + E t = [ -2 + r + 1 + E ] t = [ r - 1 + E ] t.Quadratic term: 2t².Therefore, equation becomes 2t² + (r - 1 + E ) t = 0.Since this equation must hold for the specific t where P is located, but t is a variable here. Wait, no. The circle passes through P(1 - t, t) for a fixed t. Therefore, this equation must hold for the specific value of t corresponding to point P. Therefore, we can solve for E:2t² + (r - 1 + E ) t = 0 ⇒ E = (-2t²)/t - (r - 1) = -2t - r + 1.Therefore, E = -2t - r + 1.So the equation of Γ_B is x² + y² - (r + 1)x + (-2t - r + 1)y + r = 0.Now, we need to find the intersection Y of AP with Γ_B, other than P. AP has parametric equations x = (1 - t)λ, y = tλ. Substitute into Γ_B's equation:[(1 - t)λ]^2 + [tλ]^2 - (r + 1)(1 - t)λ + (-2t - r + 1)(tλ) + r = 0.Expand:λ²[(1 - t)^2 + t²] - (r + 1)(1 - t)λ + (-2t - r + 1)tλ + r = 0.Compute each term:First term: λ²[(1 - 2t + t²) + t²] = λ²[1 - 2t + 2t²].Second term: - (r + 1)(1 - t)λ.Third term: [ -2t^2 - rt + t ]λ.Fourth term: + r.Combine the λ terms:- (r + 1)(1 - t)λ + (-2t^2 - rt + t)λ = λ [ - (r + 1)(1 - t) - 2t^2 - rt + t ].Let’s expand inside the brackets:- (r + 1)(1 - t) = - (r + 1) + (r + 1)t.Then:- (r + 1) + (r + 1)t - 2t² - rt + t = - r - 1 + r t + t - 2t² - rt + t.Combine like terms:- r - 1 + (r t - rt) + (t + t) - 2t² = - r -1 + 0 + 2t - 2t².Therefore, the entire equation becomes:λ²(1 - 2t + 2t²) + λ(- r -1 + 2t - 2t²) + r = 0.We know that P is on Γ_B and on AP, so when λ = 1, the equation should hold. Let's verify:λ = 1:1 - 2t + 2t² + (- r -1 + 2t - 2t²) + r = (1 - 2t + 2t²) + (- r -1 + 2t - 2t²) + r = (1 -1) + (-2t + 2t) + (2t² - 2t²) + (-r + r) = 0. Correct.Therefore, the other solution for λ corresponds to Y. Let’s solve the quadratic equation for λ:(1 - 2t + 2t²)λ² + (- r -1 + 2t - 2t²)λ + r = 0.We already know λ = 1 is a root, so factoring:(λ - 1)( (1 - 2t + 2t²)λ - r ) = 0.Wait, let me perform polynomial division or use the fact that if λ = 1 is a root, then:The quadratic equation can be written as (λ - 1)(Aλ + B) = 0. Expanding:Aλ² + (B - A)λ - B = 0.Comparing coefficients:A = 1 - 2t + 2t²,B - A = - r -1 + 2t - 2t²,- B = r.From -B = r ⇒ B = -r.Then, B - A = -r - (1 - 2t + 2t²) = -r -1 + 2t - 2t².Which matches the coefficient of λ. Therefore, the quadratic factors as:(λ - 1)( (1 - 2t + 2t²)λ - r ) = 0.Therefore, the other root is λ = r / (1 - 2t + 2t²).Therefore, the parameter y (position of Y along AP) is y = r / (1 - 2t + 2t²).Similarly, we need to find Z by intersecting AP with Γ_C (circumcircle of CPQ). Let’s proceed similarly.Γ_C is the circumcircle of CPQ, where C(0,1), P(1 - t, t), Q(0, s). Let’s find its equation.General equation of a circle: x² + y² + Dx + Ey + F = 0.Passing through C(0,1): 0 + 1 + 0 + E*1 + F = 0 ⇒ 1 + E + F = 0 ⇒ F = -1 - E.Passing through P(1 - t, t): (1 - t)^2 + t² + D(1 - t) + E t + F = 0.Passing through Q(0, s): 0 + s² + 0 + E s + F = 0 ⇒ s² + E s + F = 0.Substitute F = -1 - E into the equations.For Q(0, s): s² + E s -1 - E = 0 ⇒ E(s - 1) + (s² -1) = 0 ⇒ E(s - 1) = -(s² -1) ⇒ E = -(s + 1), provided s ≠ 1.Therefore, E = -(s + 1), so F = -1 - (-s -1) = s.Now substitute into the equation for P:(1 - t)^2 + t² + D(1 - t) + E t + F = 0.Compute:(1 - 2t + 2t²) + D(1 - t) - (s + 1)t + s = 0.Combine constants: 1 - 2t + 2t² + D(1 - t) - (s + 1)t + s.= 1 + s - 2t - (s +1)t + 2t² + D(1 - t).= (1 + s) + [ -2 - s -1 ]t + 2t² + D(1 - t).= (1 + s) + (-s -3)t + 2t² + D(1 - t).This must equal zero. So solve for D:D(1 - t) = - (1 + s) + (s + 3)t - 2t².Thus,D = [ - (1 + s) + (s + 3)t - 2t² ] / (1 - t).Therefore, the equation of Γ_C is x² + y² + Dx - (s + 1)y + s = 0, where D is as above.Now, find intersection Z of AP with Γ_C. AP parametric equations: x = (1 - t)λ, y = tλ. Substitute into Γ_C's equation:[(1 - t)λ]^2 + [tλ]^2 + D(1 - t)λ - (s + 1)(tλ) + s = 0.Expand:λ²[(1 - t)^2 + t²] + λ[ D(1 - t) - (s + 1)t ] + s = 0.First term: λ²(1 - 2t + 2t²).Second term: λ[ D(1 - t) - (s + 1)t ].Third term: s.Substitute D:D = [ - (1 + s) + (s + 3)t - 2t² ] / (1 - t).Thus, D(1 - t) = - (1 + s) + (s + 3)t - 2t².Therefore, the second term becomes:λ[ - (1 + s) + (s + 3)t - 2t² - (s + 1)t ].Simplify inside the brackets:- (1 + s) + (s + 3)t - (s +1)t - 2t².= - (1 + s) + [ (s + 3) - (s +1) ] t - 2t².= - (1 + s) + 2t - 2t².Therefore, the equation becomes:λ²(1 - 2t + 2t²) + λ( - (1 + s) + 2t - 2t² ) + s = 0.We know that P is on Γ_C and AP, so when λ = 1, the equation should hold. Substitute λ = 1:(1 - 2t + 2t²) + ( - (1 + s) + 2t - 2t² ) + s = (1 - 2t + 2t²) -1 - s + 2t - 2t² + s = 0. Correct.Therefore, the other solution corresponds to Z. Let’s solve the quadratic equation:(1 - 2t + 2t²)λ² + ( - (1 + s) + 2t - 2t² )λ + s = 0.Again, since λ = 1 is a root, we can factor it out:(λ - 1)( (1 - 2t + 2t²)λ - s ) = 0.Therefore, the other root is λ = s / (1 - 2t + 2t²).Thus, the parameter z (position of Z along AP) is z = s / (1 - 2t + 2t²).Now, we have parameters x, y, z along AP:x = [r(1 - t) + s t] / [ (1 - t)^2 + t² ],y = r / (1 - 2t + 2t²),z = s / (1 - 2t + 2t²).Wait, but the denominators in y and z are the same: 1 - 2t + 2t². Let's note that 1 - 2t + 2t² = (1 - t)^2 + t² - 2t(1 - t) ? Wait, no. Actually, 1 - 2t + 2t² = 2t² - 2t + 1, which is the same as 2(t² - t) + 1. Not sure if that helps.But in any case, we need to compute YX/XZ. Let's express YX and XZ in terms of these parameters.Assuming the points are ordered A---X---Y---Z---P along AP (but we need to verify the order). However, the exact order depends on the values of x, y, z. Let's see:From the expressions:x = [ r(1 - t) + s t ] / D, where D = (1 - t)^2 + t²,y = r / D',z = s / D',where D' = 1 - 2t + 2t².But D and D' are different denominators. Let me compute D and D':D = (1 - t)^2 + t² = 1 - 2t + t² + t² = 1 - 2t + 2t²,D' = 1 - 2t + 2t².Wait, D and D' are the same! So x = [ r(1 - t) + s t ] / D,y = r / D,z = s / D.Therefore, x = [ r(1 - t) + s t ] / D,y = r / D,z = s / D.So the parameters x, y, z along AP are:y = r / D,z = s / D,x = [ r(1 - t) + s t ] / D.But since all are divided by D, the relative positions depend on the numerators:Numerator of x: r(1 - t) + s t,Numerator of y: r,Numerator of z: s.Assuming all numerators are positive (since r, s, t are between 0 and 1 as they are points on the sides), the order of the points depends on the values.Suppose that r(1 - t) + s t > r and > s. Then x > y and x > z. But if that's the case, then the order would be A---Y---Z---X---P or similar. However, this is unclear without specific values.Alternatively, perhaps there is a relation between r, s, and t that we haven't considered. The problem doesn't specify any relation between P, Q, R, so they are arbitrary points on the sides. However, the problem statement must hold for any positions of P, Q, R, which suggests that the ratio YX/XZ depends only on BP/PC, regardless of Q and R. Therefore, there must be a cancellation of variables r and s in the ratio YX/XZ.Let’s compute YX and XZ. Assuming the points are ordered A---Y---Z---X---P or another order, but let's compute the absolute difference.YX = |y - x|,XZ = |x - z|.But since all are scalars along AP, we can compute:YX = | (r/D) - [ r(1 - t) + s t ] / D | = | r - r(1 - t) - s t | / D = | r t - s t | / D = t | r - s | / D.XZ = | [ r(1 - t) + s t ] / D - s / D | = | r(1 - t) + s t - s | / D = | r(1 - t) - s (1 - t) | / D = | (1 - t)(r - s) | / D.Therefore, YX/XZ = [ t | r - s | / D ] / [ (1 - t) | r - s | / D ] = t / (1 - t).Which is exactly BP/PC, since BP/PC = t/(1 - t).Therefore, the ratio YX/XZ equals BP/PC, as required.So the key was to parametrize the positions of X, Y, Z along AP, compute their parameters using the circle equations, and then find the ratio. The parameters r and s related to points Q and R canceled out in the ratio, leaving only t/(1 - t), which is BP/PC.

✅:To prove that ( frac{YX}{XZ} = frac{BP}{PC} ), we analyze the intersections of line segment ( AP ) with the circumcircles ( Gamma_A, Gamma_B, ) and ( Gamma_C ), parameterizing the positions of ( X, Y, Z ) along ( AP ).1. Parametrize Points: - Assign coordinates: ( A(0,0) ), ( B(1,0) ), ( C(0,1) ). - Let ( P ) divide ( BC ) such that ( BP/PC = t/(1-t) ). Coordinates of ( P ) are ( (1-t, t) ). - Points ( Q ) on ( AC ) and ( R ) on ( AB ) are parameterized as ( Q(0,s) ) and ( R(r,0) ).2. Equations of Circumcircles: - Γ_A (AQR): Passes through ( A(0,0) ), ( Q(0,s) ), ( R(r,0) ). Equation: ( x^2 + y^2 - rx - sy = 0 ). - Γ_B (BRP): Passes through ( B(1,0) ), ( R(r,0) ), ( P(1-t, t) ). Derived coefficients yield parametric intersection. - Γ_C (CPQ): Passes through ( C(0,1) ), ( P(1-t, t) ), ( Q(0,s) ). Derived coefficients similarly.3. Intersection Parameters: - X on Γ_A: Solve for intersection with line ( AP ). Parameter ( x = frac{r(1-t) + st}{(1-t)^2 + t^2} ). - Y on Γ_B: Solve quadratic equation, yielding parameter ( y = frac{r}{(1-t)^2 + t^2} ). - Z on Γ_C: Solve quadratic equation, yielding parameter ( z = frac{s}{(1-t)^2 + t^2} ).4. Compute Ratios: - ( YX = |y - x| = frac{|rt - st|}{(1-t)^2 + t^2} ). - ( XZ = |x - z| = frac{|(1-t)(r - s)|}{(1-t)^2 + t^2} ). - Ratio ( frac{YX}{XZ} = frac{t}{1-t} ), which equals ( frac{BP}{PC} ).Conclusion: The ratio ( frac{YX}{XZ} ) simplifies to ( frac{t}{1-t} ), directly corresponding to ( frac{BP}{PC} ). Hence, ( frac{YX}{XZ} = frac{BP}{PC} ) is proven.(boxed{dfrac{YX}{XZ} = dfrac{BP}{PC}})

🔑:1. Label the second intersection point of circles: Let's denote the second intersection point of circles Gamma_B and Gamma_C as I.2. Identify angles related to point I: From the geometry of the problem, consider the angles at the intersection points: [ angle ARI = angle IPB = angle IQC. ] These equalities hold because I is a point common to Gamma_B and Gamma_C, and lies on the circumcircles Gamma_B and Gamma_C.3. Confirm I lies on Gamma_A: Since point I intersects both circles Gamma_B and Gamma_C, as well as forming angles with points on Gamma_A, it must lie on Gamma_A. Thus, point I lies on the circumcircle of triangle triangle AQR.4. Calculate the angle involving I on Gamma_A: Since I is on Gamma_A, we have the relationship: [ angle IXZ = angle IQA = angle IPC. ]5. Prove similarity of triangles: For circles Gamma_B and Gamma_C, we employ the theorem of similarity by rotation: [ triangle IYZ sim triangle IBC. ] This implies angles in triangle IYZ correspond to angles in triangle IBC. Hence, the triangles are similar due to their respective rotations.6. Establish the ratio: From the similarity of the triangles, the following ratio holds: [ frac{YX}{XZ} = frac{BP}{PC}. ] Conclusion:The given ratio frac{YX}{XZ} is equivalent to frac{BP}{PC}. Thus, [boxed{frac{YX}{XZ} = frac{BP}{PC}}.

🧠:Okay, let's see. I need to calculate (28 × 9 + 18 × 19 + 8 × 29) divided by 14. Hmm. First, maybe I should tackle each multiplication part step by step and then add them up before dividing by 14. Let me start with the first term: 28 multiplied by 9.28 times 9... Well, 20 times 9 is 180, and 8 times 9 is 72. So adding those together: 180 + 72 = 252. So that part is 252.Next term is 18 multiplied by 19. Hmm, 18 times 20 would be 360, but since it's 19, that's one less. So 360 minus 18 is 342. Wait, let me check that again. 18 × 19. Maybe another way: 10 × 19 is 190, and 8 × 19 is 152. Then add them: 190 + 152. Let's see, 190 + 150 is 340, plus 2 more is 342. Yeah, that's right. So 18 × 19 = 342.Third term is 8 multiplied by 29. 8 times 30 is 240, minus 8 is 232. Let me confirm: 29 is 20 + 9. 8 × 20 = 160, 8 × 9 = 72. 160 + 72 = 232. Yep, that's correct.Now, add all three results together: 252 + 342 + 232. Let me do this step by step. First, 252 + 342. 200 + 300 = 500, 50 + 40 = 90, 2 + 2 = 4. So 500 + 90 is 590, plus 4 is 594. Then add 232 to 594. 500 + 200 = 700, 90 + 30 = 120, 4 + 2 = 6. So 700 + 120 is 820, plus 6 is 826. Wait, that doesn't seem right. Wait, 594 + 232. Let me add them properly:594+232= ?Starting from the right: 4 + 2 = 6, 9 + 3 = 12 (carry over 1), 5 + 2 + 1 = 8. So it's 826. Hmm, okay. So total sum is 826.Now, we need to divide 826 by 14. Let me see how to do that. Maybe simplify it first. Let's check if 14 divides into 826 evenly. Let's see. 14 × 50 is 700. Subtract 700 from 826, we get 126. Then 14 × 9 = 126. So 50 + 9 = 59. So 14 × 59 = 826. Therefore, 826 ÷ 14 = 59. So the answer should be 59.Wait, let me verify all steps again to make sure I didn't make a mistake. Because sometimes when doing arithmetic step by step, errors can creep in.First, 28 × 9: 28 × 10 is 280, minus 28 is 252. Correct. 18 × 19: 18 × 20 = 360, minus 18 = 342. Correct. 8 × 29: 8 × 30 = 240, minus 8 = 232. Correct. Summing them: 252 + 342. Let's check again: 252 + 300 = 552, then +42 is 594. Then 594 + 232: 500 + 200 = 700, 94 + 32 = 126. So 700 + 126 = 826. Correct. Then 826 ÷ 14. As above, 14 × 50 = 700, 826 - 700 = 126, 14 × 9 = 126. So total 59. Yes, that seems right.Alternatively, maybe there's a smarter way to do this by factoring out common terms before multiplying, to make the division easier. Let me see. For example, maybe some of the terms can be divided by 14 first, which could simplify the calculation.Looking at the original expression: (28×9 + 18×19 + 8×29) ÷14. Let's see if each term can be divided by 14 or parts of them. The first term is 28×9. 28 is divisible by 14 (28 ÷14=2), so 28×9 = 14×2×9 = 14×18. So that term divided by 14 is 18. Then the second term: 18×19. Hmm, 18 is not divisible by 14, but maybe factor differently. 18 = 14 + 4. So 18×19 = (14 +4)×19 = 14×19 + 4×19. Then dividing by 14 would give 19 + (4×19)/14. Similarly, third term: 8×29. 8 is not divisible by 14. So maybe this approach complicates things more. Alternatively, let's see if each term can be expressed with a 14 factor.Alternatively, maybe split each multiplication into parts that are multiples of 14. Let's try:First term: 28×9 = 14×2×9 = 14×18. So when divided by 14, that's 18.Second term: 18×19. Let's see, 18 = 14 + 4. So 18×19 = 14×19 + 4×19. Divided by 14: 19 + (4×19)/14 = 19 + (76)/14. 76 ÷14 is 5 and 6/14, which simplifies to 5 and 3/7. But this leads to fractions, which might complicate things. Maybe not the best approach.Third term: 8×29. Similarly, 8 is 14 -6. So 8×29 = 14×29 -6×29. Divided by 14: 29 - (6×29)/14. That's 29 - (174)/14 = 29 - 12.42857... Hmm, decimals. Not helpful.Alternatively, maybe it's better to compute the numerator first and then divide, as I did originally. But let's confirm once again the total sum. 252 + 342 + 232. Let me add 252 and 232 first. 252 + 232: 200 + 200 = 400, 50 + 30 = 80, 2 + 2 = 4. Total 484. Then add 342: 484 + 342. 400 + 300 = 700, 80 + 40 = 120, 4 + 2 = 6. Total 700 + 120 = 820 +6=826. Yes, that's correct. So 826 ÷14.Another way to divide 826 by 14: Let's divide step by step. 14 × 50 = 700. 826 -700=126. 14 ×9=126. So total is 50+9=59. Exactly. So the answer is 59. Seems correct.Alternatively, prime factors: 14 is 2×7. So 826 ÷2=413, then 413 ÷7. Let's compute that. 7 × 50 = 350. 413 -350=63. 7×9=63. So 50+9=59. So same result. Therefore, 413 ÷7=59. Hence, 826 ÷14=59. Yep, that checks out.So after going through the calculations multiple times and checking each step, I'm confident the result is 59.

✅:The given expression is ((28 times 9 + 18 times 19 + 8 times 29) div 14). Here's the step-by-step solution:1. Calculate each multiplication term: - (28 times 9 = 252) - (18 times 19 = 342) - (8 times 29 = 232)2. Sum the results: [ 252 + 342 + 232 = 826 ]3. Divide the total by 14: - (826 div 14 = 59) (since (14 times 59 = 826))Final Answer: (boxed{59})

🔑:1. Calculate the product of each multiplication in the expression: [ 28 times 9 = 252 ] [ 18 times 19 = 342 ] [ 8 times 29 = 232 ]2. Sum the results of the multiplications: [ 252 + 342 + 232 = 826 ]3. Divide the sum by 14: [ 826 div 14 = 59 ] Conclusion:[boxed{59}]

🧠:Okay, so I need to solve the equation xy + yz + zx = 80 where x, y, z are natural numbers. The conditions also state that none of them is smaller than x and none is larger than z. Wait, let me parse that again. It says "none of them is smaller than x" which probably means that x is the smallest among the three, and similarly, "none of them is larger than z" implies z is the largest. So x ≤ y ≤ z. That makes sense. So we need natural numbers x, y, z with x ≤ y ≤ z such that their pairwise products sum up to 80.Alright, let's start by understanding the problem. We have three variables, all natural numbers (so positive integers), ordered such that x is the smallest, z is the largest, and y is in between. The equation is symmetric in x, y, z but with the ordering constraint, maybe we can find all possible triples (x, y, z) that satisfy the equation.First, since x, y, z are natural numbers, the smallest they can be is 1. But with x ≤ y ≤ z, and considering the equation, we need to find combinations where the sum of their pairwise products is 80.Let me think about possible approaches. One way is to fix x, then find possible y and z given the constraints. Alternatively, since the variables are ordered, perhaps we can express y and z in terms of x. Let me try fixing x first.Let’s denote x as the smallest variable. Since x is a natural number, x ≥ 1. Let's start with x = 1 and see if possible.If x = 1, then the equation becomes 1*y + y*z + z*1 = 80. Simplify: y + y z + z = 80. Factor terms with y: y(1 + z) + z = 80. Let's write this as y(z + 1) + z = 80. Then, y(z + 1) = 80 - z. Therefore, y = (80 - z)/(z + 1). Since y must be a natural number and x ≤ y ≤ z, so 1 ≤ y ≤ z.So for x = 1, z must be such that (80 - z) is divisible by (z + 1). Let's find z such that (z + 1) divides (80 - z). Let's denote d = z + 1. Then, d divides 80 - (d - 1) = 81 - d. Therefore, d divides 81 - d. Which implies d divides 81. Because d divides (81 - d + d) = 81. So z + 1 divides 81. Therefore, z + 1 is a divisor of 81.Divisors of 81 are 1, 3, 9, 27, 81. Therefore, z + 1 can be 1, 3, 9, 27, 81. But z must be at least y ≥ x =1, so z must be at least 1. But z +1 =1 implies z=0, which is not natural. So the possible z +1 values are 3, 9, 27, 81. Therefore, z can be 2, 8, 26, 80.Let’s check each of these:1. z = 2: Then y = (80 - 2)/(2 + 1) = 78/3 = 26. But y =26 and z=2, but since y must be ≤ z, this is invalid because 26 > 2. So discard.2. z =8: y = (80 -8)/(8 +1) = 72/9 =8. So y=8. Here, x=1, y=8, z=8. Since y ≤ z, which is 8 ≤8, okay. So this is valid. So triple (1,8,8). Check if the sum is 1*8 +8*8 +8*1=8 +64 +8=80. Correct.3. z=26: y=(80 -26)/(26 +1)=54/27=2. So y=2. Now, check if x ≤ y ≤ z. x=1, y=2, z=26. 1 ≤2 ≤26, okay. So triple (1,2,26). Check the sum:1*2 +2*26 +26*1=2 +52 +26=80. Correct.4. z=80: y=(80 -80)/(80 +1)=0/81=0. But y=0 is not a natural number. So invalid.Thus, for x=1, we have two solutions: (1,8,8) and (1,2,26). Wait, but in the second case, when z=26, y=2, but since y must be between x=1 and z=26, 2 is in between. So that's valid.Wait, but in the first case, (1,8,8) is ordered as x=1, y=8, z=8. But since y can equal z, that's okay.But let me verify if these are the only solutions for x=1.So yes, since we derived that z +1 must divide 81, and only z=8 and z=26 give valid natural numbers for y. So moving on.Next, let's check x=2.So x=2. Then the equation becomes 2y + y z + 2 z =80. Let's factor:2y + z(y + 2) =80.Alternatively, factor terms: Maybe rearrange as y(2 + z) + 2 z =80. So y = (80 -2 z)/(2 + z). Since y must be a natural number, (80 -2 z) must be divisible by (2 + z), and y must satisfy 2 ≤ y ≤ z.So let's let d = z +2. Then 80 -2(z) =80 -2(d -2)=80 -2d +4=84 -2d. So (84 -2d)/d must be a natural number. Therefore, (84 -2d)/d = 84/d -2. So 84/d -2 must be a natural number, which implies that 84/d is an integer greater than 2. Therefore, d divides 84, and 84/d ≥3 (since y ≥2). Therefore, d must be a divisor of 84 such that d ≤84/3=28. So divisors of 84 are 1,2,3,4,6,7,12,14,21,28,42,84. But since d = z +2 ≥2 +2=4 (since z ≥ y ≥x=2), so possible d are 4,6,7,12,14,21,28.Wait, let's check:Since z ≥ y ≥2, z ≥2. So d =z +2 ≥4. Also, from (84 -2d)/d must be a natural number. Let's compute for each d:1. d=4: (84 -8)/4=76/4=19. So y=19. Then z = d -2=4-2=2. But z=2, which is less than y=19. But z must be ≥ y. So invalid.2. d=6: (84 -12)/6=72/6=12. So y=12. Then z=6-2=4. Again, z=4 < y=12. Invalid.3. d=7: (84 -14)/7=70/7=10. y=10, z=7-2=5. z=5 < y=10. Invalid.4. d=12: (84 -24)/12=60/12=5. y=5, z=12 -2=10. So z=10 ≥ y=5. Okay. So x=2, y=5, z=10. Check the equation: 2*5 +5*10 +10*2=10 +50 +20=80. Correct.5. d=14: (84 -28)/14=56/14=4. y=4, z=14 -2=12. So z=12 ≥ y=4. Check: 2*4 +4*12 +12*2=8 +48 +24=80. Correct.6. d=21: (84 -42)/21=42/21=2. y=2, z=21 -2=19. Then z=19 ≥ y=2. Check:2*2 +2*19 +19*2=4 +38 +38=80. Correct.7. d=28: (84 -56)/28=28/28=1. y=1, but x=2 and y must be ≥2. So invalid.Thus, for x=2, the valid solutions are when d=12,14,21, which correspond to triples (2,5,10), (2,4,12), (2,2,19). Wait, wait: Let me check each case again:For d=12: y=5, z=10. So x=2, y=5, z=10. Valid.For d=14: y=4, z=12. So x=2, y=4, z=12. Valid.For d=21: y=2, z=19. So x=2, y=2, z=19. Valid.Wait, so all three are valid. Let's check the equation for each:1. (2,5,10): 2*5 +5*10 +10*2=10+50+20=80. Correct.2. (2,4,12): 2*4 +4*12 +12*2=8+48+24=80. Correct.3. (2,2,19):2*2 +2*19 +19*2=4+38+38=80. Correct.Yes, all three are valid. So for x=2, we have three solutions.Next, check x=3.So x=3. The equation becomes 3y + y z +3 z=80. Let's rearrange:y(3 + z) +3 z =80. So y=(80 -3 z)/(3 + z). Since y must be a natural number and 3 ≤ y ≤ z.So (80 -3 z) must be divisible by (z +3). Let me set d=z +3. Then z =d -3. Plug into numerator:80 -3(d -3)=80 -3d +9=89 -3d. Therefore, y=(89 -3d)/d=89/d -3. So 89/d -3 must be a natural number. Hence, 89/d must be integer +3. So 89/d must be an integer, which implies that d divides 89. Since 89 is a prime number, its divisors are 1 and 89. But d = z +3 ≥3 +3=6 (since z ≥ y ≥3). So possible d is 89. Then z=89 -3=86. Then y=(89 -3*89)/89= (89 -267)/89= (-178)/89= -2. Not a natural number. So no solution for x=3.Wait, wait, let's check that again.If d divides 89, and 89 is prime, then d=1 or 89. But d must be at least 6. So only d=89 is possible. But z=89 -3=86, then y=(89 -3*86)/86 +3? Wait, no:Wait, from above: y=(89 -3d)/d. So when d=89, y=(89 -267)/89= (-178)/89=-2. Negative, invalid. So no solutions for x=3.Therefore, x=3 gives no solution.Moving on to x=4.x=4. Equation:4y + y z +4 z =80. Factor:y(4 + z) +4 z=80. So y=(80 -4 z)/(4 + z). Must be natural, and 4 ≤ y ≤ z.Set d=z +4. Then z=d -4. Plug into numerator:80 -4(d -4)=80 -4d +16=96 -4d. Then y=(96 -4d)/d=96/d -4. So 96/d -4 must be a natural number. Hence, 96/d must be integer and 96/d ≥5 (since y ≥4). Therefore, d divides 96 and 96/d ≥5 → d ≤96/5=19.2. So d ≤19. Also, since z ≥ y ≥4, z ≥4. Therefore, d=z +4 ≥4 +4=8.So divisors of 96 up to 19 and ≥8.Divisors of 96:1,2,3,4,6,8,12,16,24,32,48,96. So in the range 8 to19:8,12,16.Check each:1. d=8: y=96/8 -4=12 -4=8. Then z=d -4=4. But z=4, which is less than y=8. Invalid.2. d=12: y=96/12 -4=8 -4=4. z=12 -4=8. So z=8 ≥ y=4. Check:4*4 +4*8 +8*4=16 +32 +32=80. Correct. So triple (4,4,8).3. d=16: y=96/16 -4=6 -4=2. But y=2 <x=4. Invalid.Thus, only d=12 gives valid solution: (4,4,8).Wait, check again: x=4, y=4, z=8. Indeed, 4*4 +4*8 +8*4=16+32+32=80. Correct.Is there another possible d?Wait, what about d=24? But 24 >19.2, which we excluded. Let me confirm. The logic was that 96/d -4 ≥4 (since y ≥4). So 96/d -4 ≥4 →96/d ≥8 →d ≤12. So actually, d can be up to 12. So the possible d=8,12. But d=8 gives y=8, which is invalid. So only d=12.Therefore, only one solution for x=4: (4,4,8).Next, x=5.Equation:5y + y z +5 z =80.Factor: y(5 + z) +5 z=80. So y=(80 -5 z)/(5 + z). Must be natural, 5 ≤ y ≤ z.Set d=z +5. Then z=d -5. Then numerator becomes 80 -5(d -5)=80 -5d +25=105 -5d. Thus, y=(105 -5d)/d=105/d -5. Therefore, 105/d -5 must be natural. Hence, 105/d must be integer and ≥6 (since y ≥5). So d divides 105, and 105/d ≥6 →d ≤105/6≈17.5. So d ≤17.Also, since z ≥ y ≥5, z ≥5. So d=z +5 ≥5 +5=10.Divisors of 105 up to 17 and ≥10: 15, 21. Wait, 105's divisors:1,3,5,7,15,21,35,105. In the range 10-17:15.Check d=15:y=105/15 -5=7 -5=2. But y=2 <x=5. Invalid.Next, d=21: But d=21>17.5, so no. So no solutions for x=5.Thus, x=5 gives no solution.Next, x=6.Equation:6y + y z +6 z =80.Factor:y(6 + z) +6 z=80. So y=(80 -6 z)/(6 + z). Must be natural, 6 ≤ y ≤ z.Set d=z +6. Then z=d -6. Then numerator:80 -6(d -6)=80 -6d +36=116 -6d. Thus, y=(116 -6d)/d=116/d -6. So 116/d -6 must be a natural number ≥6. Therefore, 116/d must be integer and 116/d ≥12 (since y ≥6). So 116/d ≥12 →d ≤116/12≈9.666. So d≤9. Also, since z ≥ y ≥6, d=z +6 ≥6 +6=12. But d needs to be ≤9. Contradiction. Therefore, no solution for x=6.Similarly, x=7:Equation:7y + y z +7 z =80.Factor:y(7 + z) +7 z=80. So y=(80 -7 z)/(7 + z). Natural, 7 ≤ y ≤ z.Set d=z +7. z=d -7. Then numerator:80 -7(d -7)=80 -7d +49=129 -7d. So y=(129 -7d)/d=129/d -7. Thus, 129/d -7 must be natural and ≥7. So 129/d ≥14 →d ≤129/14≈9.214. So d ≤9. But z=d -7 ≥7. Therefore, d=z +7 ≥7 +7=14. But 14 >9.214. Contradiction. No solution.x=8:Equation:8y + y z +8 z=80.Factor:y(8 + z) +8 z=80. So y=(80 -8 z)/(8 + z). Natural, 8 ≤ y ≤ z.Set d=z +8. Then z=d -8. Numerator:80 -8(d -8)=80 -8d +64=144 -8d. y=(144 -8d)/d=144/d -8. Thus, 144/d -8 must be natural and ≥8. So 144/d ≥16 →d ≤144/16=9. Also, z=d -8 ≥8. So d ≥16. Contradiction (d≤9 and d≥16). No solution.x=9:Similarly, equation:9y + y z +9 z=80.y=(80 -9 z)/(9 + z). Since x=9, y ≥9. Let’s see. Let’s set d=z +9. Then z=d -9. Numerator:80 -9(d -9)=80 -9d +81=161 -9d. So y=(161 -9d)/d=161/d -9. Thus, 161/d -9 must be natural and ≥9. So 161/d ≥18 →d ≤161/18≈8.944. But d=z +9 ≥9 +9=18. Contradiction. No solution.Similarly, higher x values will only make the equation impossible, since x ≥10:x=10:Equation:10y + y z +10 z=80.But even if y and z are minimal (10), the sum would be 10*10 +10*z +10*z=100 +20 z. Which is already more than 80. So impossible.Thus, x can only be up to 4, since x=5 and above don't yield solutions. So compiling all solutions found so far:For x=1: (1,8,8), (1,2,26)For x=2: (2,5,10), (2,4,12), (2,2,19)For x=4: (4,4,8)Wait, let me check again for x=4, we found (4,4,8). Correct.Now, let's check if these solutions are ordered correctly, i.e., x ≤ y ≤ z.For x=1:(1,8,8): 1 ≤8 ≤8. Yes.(1,2,26): 1 ≤2 ≤26. Yes.For x=2:(2,5,10):2 ≤5 ≤10. Yes.(2,4,12):2 ≤4 ≤12. Yes.(2,2,19):2 ≤2 ≤19. Yes.For x=4:(4,4,8):4 ≤4 ≤8. Yes.So all these are ordered correctly.Now, let's check if there are any duplicates or missed solutions. Let me list all:1. (1, 2, 26)2. (1, 8, 8)3. (2, 2, 19)4. (2, 4, 12)5. (2, 5, 10)6. (4, 4, 8)Wait, that's six solutions. Let me verify again if there's any other possible x. We checked x=1,2,3,4,5,6,7,8,9,10. So all possible x are covered. So these are all solutions.But let me check another approach to ensure no solutions are missed. Maybe think of variables in order x ≤ y ≤ z and find triples where xy + yz + zx =80.Alternatively, since xy + yz + zx =80, we can write this as y(x + z) + x z =80. Maybe factor differently. But I think our previous approach is exhaustive.Alternatively, suppose we fix y. Since x ≤ y ≤ z, perhaps fixing y and looking for x and z. But given the prior thorough check, perhaps not necessary. But let me check for example y=4.If y=4, then x can be 1,2,3,4. Let's check each:x=1, y=4: Then equation:1*4 +4*z +z*1=4 +4z +z=5z +4=80 →5z=76 →z=15.2. Not integer.x=2, y=4: 2*4 +4*z +z*2=8 +4z +2z=6z +8=80 →6z=72 →z=12. So (2,4,12). Which we already have.x=3, y=4: 3*4 +4*z +z*3=12 +7z=80→7z=68→z≈9.71. Not integer.x=4, y=4:4*4 +4*z +4*z=16 +8z=80→8z=64→z=8. So (4,4,8). Which we have.Similarly, y=5:x can be 1,2,3,4,5.x=1, y=5:1*5 +5*z +z*1=5 +6z=80→6z=75→z=12.5. Not integer.x=2, y=5:2*5 +5*z +z*2=10 +7z=80→7z=70→z=10. So (2,5,10). Already in the list.x=3, y=5:3*5 +5*z +z*3=15 +8z=80→8z=65→z=8.125. Not integer.x=4, y=5:4*5 +5*z +z*4=20 +9z=80→9z=60→z≈6.666. Not integer.x=5, y=5:5*5 +5*z +z*5=25 +10z=80→10z=55→z=5.5. Not integer.So only (2,5,10) is there.Similarly, for y=8:x=1,y=8:1*8 +8*z +z*1=8 +9z=80→9z=72→z=8. So (1,8,8).x=2,y=8: equation:2*8 +8*z +z*2=16 +10z=80→10z=64→z=6.4. Not integer.Etc. So this approach also confirms existing solutions.Thus, I think we've covered all possible triples. Therefore, the solutions are:(1, 2, 26), (1, 8, 8), (2, 2, 19), (2, 4, 12), (2, 5, 10), (4, 4, 8).But wait, let's check if there's another possible ordering. For example, (1,2,26) and (1,8,8) are distinct. But if permutations were allowed, but since the problem states that none is smaller than x or larger than z, so x is the minimum and z is the maximum. So these are all ordered triples with x ≤ y ≤ z.Thus, the final solutions are the six triples listed above.

✅:The equation (xy + yz + zx = 80) with natural numbers (x leq y leq z) has the following solutions:1. Systematic Analysis: - Constraints: (x, y, z in mathbb{N}) and (x leq y leq z). - Approach: Fix (x) and solve for (y, z) under the constraints.2. Case (x = 1): - Equation: (y + yz + z = 80 implies y(z+1) + z = 80). - Solutions: - (z = 8, y = 8) ⟹ ((1, 8, 8)). - (z = 26, y = 2) ⟹ ((1, 2, 26)).3. Case (x = 2): - Equation: (2y + yz + 2z = 80 implies y(z+2) + 2z = 80). - Solutions: - (z = 10, y = 5) ⟹ ((2, 5, 10)). - (z = 12, y = 4) ⟹ ((2, 4, 12)). - (z = 19, y = 2) ⟹ ((2, 2, 19)).4. Case (x = 4): - Equation: (4y + yz + 4z = 80 implies y(z+4) + 4z = 80). - Solution: - (z = 8, y = 4) ⟹ ((4, 4, 8)).5. Higher (x geq 3, 5, 6, ldots): - No valid solutions due to constraints or non-integer results.Final Solutions:[boxed{(1, 2, 26)}, quad boxed{(1, 8, 8)}, quad boxed{(2, 2, 19)}, quad boxed{(2, 4, 12)}, quad boxed{(2, 5, 10)}, quad boxed{(4, 4, 8)}]

🔑:We are given the equation [ xy + yz + zx = 80 ]and we need to find the natural number solutions ((x, y, z)) such that ( x leq y leq z ).To determine this, follow these steps:1. Considering the constraint ( x leq y leq z ), we can formulate (x), (y), and (z) accordingly.2. Adding (x^2) to both sides of the given equation: [ xy + yz + zx + x^2 = 80 + x^2 ] This can be factored as: [ x(x + y) + z(y + x) = 80 + x^2 ] Simplifying, we get: [ (x + y)(x + z) = 80 + x^2 ]3. For fixed (x), the product (80 + x^2) must be factorizable into two natural numbers. Let’s choose (x = 1) and substitute it in the equation: [ (y + 1)(z + 1) = 81 ] The possible factorizations of 81 are: [ 1 cdot 81 = 3 cdot 27 = 9 cdot 9 ] So, this results in the solutions: [ (x, y, z) = (1, 2, 26) ] [ (x, y, z) = (1, 8, 8) ]4. The highest possible value of (x) is determined by examining (x) so that (3x^2 leq 80), i.e., (x leq sqrt{26.67} approx 5).Now, we check remaining values of (x = 2, 3, 4, 5):For (x = 2):[ (y + 2)(z + 2) = 84 ]Factorizations are:[ 4 cdot 21, 6 cdot 14, 7 cdot 12 ]This results in the solutions:[ (x, y, z) = (2, 2, 19) ][ (x, y, z) = (2, 4, 12) ][ (x, y, z) = (2, 5, 10) ]For (x = 3):[ (y + 3)(z + 3) = 89 ]Since 89 is a prime number, no factorizations exist that satisfy the requirements.For (x = 4):[ (y + 4)(z + 4) = 96 ]Factorizations are:[ 8 cdot 12 ]This results in the solution:[ (x, y, z) = (4, 4, 8) ]For (x = 5):[ (y + 5)(z + 5) = 105 ]Since 105 does not yield a solution where (x leq y leq z).5. Summarizing the solutions, we have the following 6 solutions:[ (1, 2, 26), (1, 8, 8), (2, 2, 19), (2, 4, 12), (2, 5, 10), (4, 4, 8) ]Thus, the number of solutions that satisfy the given conditions is:[ 6 ](boxed{6})

🧠:Okay, let's see. So the problem is about a three-digit number that, when you add a decimal point in the right place, becomes a decimal number. Then, this decimal number is 201.6 less than the original three-digit number. I need to find the original number. Hmm, let's break this down step by step.First, let me make sure I understand the problem correctly. We have a three-digit number, let's call it N. When we insert a decimal point somewhere in N, it becomes a decimal number. The decimal number is 201.6 less than N. So, N - (decimal number) = 201.6. Got it. The key here is figuring out where to place the decimal point in the three-digit number to make this equation true.Since N is a three-digit number, it can be written as 100a + 10b + c, where a, b, c are digits from 0 to 9, and a can't be 0 because it's a three-digit number. But maybe I don't need to break it down into digits right away. Let's think about the decimal placement.Adding a decimal point in an appropriate place—this could mean converting the three-digit number into a decimal by placing the decimal point after the first digit, after the second digit, or maybe even somewhere else? Wait, but a three-digit number has three digits, so inserting a decimal point would make it a decimal with either one or two decimal places? Or maybe more? Wait, no. Let me think.For example, if the number is 123, inserting a decimal point after the first digit gives 1.23, which is a decimal. Inserting after the second digit gives 12.3, which is also a decimal. So, there are two possible places to insert the decimal point: between the first and second digit or between the second and third digit. So, the decimal number could be either in the form of x.yz (if the decimal is placed after the first digit) or xy.z (if placed after the second digit). So, two possibilities.Therefore, the decimal number is either N divided by 100 or N divided by 10, depending on where the decimal is placed. Wait, no. Let's see. If the original number is 123, placing the decimal after the first digit gives 1.23, which is 123 divided by 100. Placing it after the second digit gives 12.3, which is 123 divided by 10. So, that's right. So, depending on where we place the decimal, the decimal number is either N divided by 10 or N divided by 100.But according to the problem, the decimal number is 201.6 less than N. So, N - (decimal number) = 201.6. Therefore, decimal number = N - 201.6. So, depending on where the decimal is placed, the decimal number is either N/10 or N/100. So, let's set up equations for both possibilities.First case: decimal number is N/10. Then, N - N/10 = 201.6. Let's compute that. N - N/10 = (9/10)N = 201.6. Then, N = 201.6 * (10/9) = 2016/9 = 224. But wait, 201.6 * 10/9 is 2016/9, which is 224. So, N would be 224. Let's check. If N is 224, then inserting a decimal after the second digit gives 22.4. Then, 224 - 22.4 = 201.6. That works. Wait, so is 224 the answer? But let's check the other case as well.Second case: decimal number is N/100. Then, N - N/100 = 201.6. So, (99/100)N = 201.6. Then, N = 201.6 * (100/99) = 20160/99. Let me compute that. 20160 divided by 99. 99*203 = 20097, 20160 - 20097 = 63. So, 203 + 63/99 = 203 + 7/11 ≈ 203.636... That's not an integer. But N must be a three-digit integer, so this case is invalid. Therefore, only the first case gives a valid integer solution. So, N is 224.Wait, but let me verify again. If N is 224, inserting a decimal after the second digit gives 22.4. Then 224 - 22.4 is indeed 201.6. So that's correct. So, the answer is 224. But let me think again if there's another possible way.Wait, the problem says "when a decimal point is added in an appropriate place." The "appropriate place" would mean that the decimal number is properly formed. For example, inserting a decimal after the third digit would just be 224., which is the same as 224, but that's not a decimal. So, the decimal must be placed between two digits, resulting in a number less than N. Therefore, the possible placements are after the first or second digit. We checked both cases, and only the second digit placement gives a valid integer. So, that's the answer.But just to make sure, let me try another example. Suppose N is 315. If we place the decimal after the second digit, we get 31.5. Then 315 - 31.5 = 283.5, which is not 201.6. So that's not the case. If we place the decimal after the first digit, 3.15, then 315 - 3.15 = 311.85, which is also not 201.6. So, 224 seems unique.Alternatively, let me check the calculation again for the first case. N - N/10 = 201.6. So, (9/10)N = 201.6. Multiply both sides by 10/9: N = 201.6 * (10/9). 201.6 divided by 9: 9*22=198, 201.6 - 198=3.6, 3.6/9=0.4. So, 22.4. Then 22.4*10=224. So yes, that's correct.Therefore, the original three-digit number is 224.

✅:The original three-digit number is 224.Step-by-Step Explanation:1. Define Variables: - Let ( N ) be the original three-digit number.2. Understand Decimal Placement: - Inserting a decimal point can convert ( N ) into a decimal either by dividing by 10 (placing after the second digit) or by 100 (placing after the first digit). For example: - If ( N = 224 ), placing the decimal after the second digit gives ( 22.4 = frac{224}{10} ). - Placing it after the first digit gives ( 2.24 = frac{224}{100} ).3. Set Up Equations for Both Cases: - Case 1 (Decimal after the second digit): [ N - frac{N}{10} = 201.6 implies frac{9N}{10} = 201.6 implies N = frac{201.6 times 10}{9} = 224 ] - Case 2 (Decimal after the first digit): [ N - frac{N}{100} = 201.6 implies frac{99N}{100} = 201.6 implies N = frac{201.6 times 100}{99} approx 203.64 quad (text{Not an integer}) ]4. Verify Validity: - Case 1 yields ( N = 224 ), a valid integer. Checking: [ 224 - frac{224}{10} = 224 - 22.4 = 201.6 quad (text{Correct}) ] - Case 2 results in a non-integer, so it’s invalid.Conclusion: The original three-digit number is 224.

🔑:Given: A three-digit number, when a decimal point is placed at the appropriate position, results in a decimal that is 201.6 less than the original three-digit number.1. Denote the three-digit number as ( x ). We know that placing the decimal point in ( x ) gives us a decimal number ( x/10 ). According to the problem condition: [ x - frac{x}{10} = 201.6 ]2. Simplify the equation: [ x - frac{x}{10} = 201.6 ]3. Combine like terms: [ frac{10x - x}{10} = 201.6 ]4. Simplify the fraction: [ frac{9x}{10} = 201.6 ]5. Solve for ( x ): - Multiply both sides by 10 to clear the fraction: [ 9x = 2016 ] - Divide both sides by 9: [ x = frac{2016}{9} ] - Perform the division: [ x = 224 ]6. Conclusion: The original three-digit number is:[boxed{224}]