💡Okay, so I have this problem about a 150-meter race involving three runners: A, B, and C. The problem says that runner A can beat runner B by 25 meters, and runner B can beat runner C by 15 meters. I need to figure out how many meters ahead runner A is when runner C finishes the race. The options are A) 30 meters, B) 35 meters, C) 37.5 meters, and D) 40 meters.First, I need to understand what it means when it says runner A can beat runner B by 25 meters in a 150-meter race. I think this means that when runner A finishes the 150 meters, runner B has only run 125 meters. So, in the time it takes A to run 150 meters, B runs 125 meters. Similarly, when it says runner B can beat runner C by 15 meters, that means when B finishes 150 meters, C has only run 135 meters.So, I can probably set up some ratios or equations based on their speeds. Let me denote the speeds of A, B, and C as ( v_A ), ( v_B ), and ( v_C ) respectively.From the first piece of information, when A finishes 150 meters, B has run 125 meters. The time it takes A to run 150 meters is the same as the time it takes B to run 125 meters. So, time is equal to distance divided by speed. Therefore, I can write:[frac{150}{v_A} = frac{125}{v_B}]Similarly, from the second piece of information, when B finishes 150 meters, C has run 135 meters. So, the time it takes B to run 150 meters is the same as the time it takes C to run 135 meters. Thus:[frac{150}{v_B} = frac{135}{v_C}]Now, I have two equations:1. ( frac{150}{v_A} = frac{125}{v_B} )2. ( frac{150}{v_B} = frac{135}{v_C} )I can solve these equations to find the relationship between the speeds of A, B, and C.Starting with the first equation:[frac{150}{v_A} = frac{125}{v_B}]Cross-multiplying gives:[150 v_B = 125 v_A]Dividing both sides by 150:[v_B = frac{125}{150} v_A = frac{5}{6} v_A]So, ( v_B = frac{5}{6} v_A ). That means runner B's speed is five-sixths of runner A's speed.Now, moving on to the second equation:[frac{150}{v_B} = frac{135}{v_C}]Cross-multiplying:[150 v_C = 135 v_B]Dividing both sides by 150:[v_C = frac{135}{150} v_B = frac{9}{10} v_B]So, ( v_C = frac{9}{10} v_B ). That means runner C's speed is nine-tenths of runner B's speed.But since we already know ( v_B = frac{5}{6} v_A ), we can substitute that into the equation for ( v_C ):[v_C = frac{9}{10} times frac{5}{6} v_A = frac{45}{60} v_A = frac{3}{4} v_A]So, ( v_C = frac{3}{4} v_A ). That means runner C's speed is three-fourths of runner A's speed.Now, the question is asking how many meters ahead runner A is when runner C finishes the race. So, I need to find out how much distance runner A has covered by the time runner C finishes 150 meters.First, let's find the time it takes runner C to finish the 150-meter race. Time is equal to distance divided by speed, so:[t_C = frac{150}{v_C}]But we know ( v_C = frac{3}{4} v_A ), so substituting that in:[t_C = frac{150}{frac{3}{4} v_A} = frac{150 times 4}{3 v_A} = frac{600}{3 v_A} = frac{200}{v_A}]So, the time it takes runner C to finish the race is ( frac{200}{v_A} ).Now, let's find out how far runner A has run in this time. Distance is equal to speed multiplied by time, so:[d_A = v_A times t_C = v_A times frac{200}{v_A} = 200 text{ meters}]Wait, that can't be right. The race is only 150 meters, so runner A can't have run 200 meters by the time runner C finishes. I must have made a mistake somewhere.Let me check my calculations again.Starting from ( v_C = frac{3}{4} v_A ), so ( v_A = frac{4}{3} v_C ).Then, the time for runner C to finish 150 meters is:[t_C = frac{150}{v_C}]In this time, runner A would have run:[d_A = v_A times t_C = frac{4}{3} v_C times frac{150}{v_C} = frac{4}{3} times 150 = 200 text{ meters}]Hmm, same result. But the race is only 150 meters, so runner A would have finished the race and then some. That doesn't make sense because the question is asking how many meters ahead A is when C finishes the race. So, if A has already finished, then A is ahead by the distance A ran beyond 150 meters.Wait, but in reality, runner A would have finished the race before runner C. So, when runner C finishes, runner A has already completed the race and is ahead by the distance runner A ran beyond 150 meters in the time it took runner C to finish.But according to the calculation, runner A ran 200 meters in the time runner C ran 150 meters. So, runner A is ahead by 50 meters. But that's not one of the answer choices. The options are 30, 35, 37.5, and 40 meters.Hmm, I must have misunderstood the problem. Let me try a different approach.Maybe instead of looking at the time it takes for C to finish, I should consider the time it takes for A to finish and see how much C has run in that time.Wait, but the question is asking how many meters ahead A is when C finishes the race. So, it's the distance A has run beyond 150 meters when C completes 150 meters.But according to my previous calculation, A has run 200 meters when C finishes 150 meters. So, A is ahead by 50 meters. But since 50 isn't an option, I must have messed up.Let me try to approach it differently. Maybe using the ratios of their speeds.From the first part, we have:( v_B = frac{5}{6} v_A )And ( v_C = frac{9}{10} v_B = frac{9}{10} times frac{5}{6} v_A = frac{45}{60} v_A = frac{3}{4} v_A )So, ( v_C = frac{3}{4} v_A )Now, the time it takes for C to finish 150 meters is:( t = frac{150}{v_C} = frac{150}{frac{3}{4} v_A} = frac{150 times 4}{3 v_A} = frac{600}{3 v_A} = frac{200}{v_A} )In this time, A has run:( d_A = v_A times t = v_A times frac{200}{v_A} = 200 ) meters.So, A has run 200 meters, which is 50 meters beyond the finish line. Therefore, A is ahead by 50 meters when C finishes. But again, 50 isn't an option.Wait, maybe I'm interpreting the problem incorrectly. It says "how many meters ahead A is when C finishes the race." Maybe it's asking how much distance A has covered beyond C's position when C finishes, not beyond the finish line.But that doesn't make sense because when C finishes, A has already finished. So, the lead would be how much A has run beyond the finish line, which is 50 meters.But since 50 isn't an option, perhaps I need to re-examine the initial assumptions.Let me think again. When A beats B by 25 meters in a 150-meter race, it means that when A finishes 150 meters, B has run 125 meters. Similarly, when B beats C by 15 meters, when B finishes 150 meters, C has run 135 meters.So, the speeds are in the ratios:From A to B: ( frac{v_A}{v_B} = frac{150}{125} = frac{6}{5} )From B to C: ( frac{v_B}{v_C} = frac{150}{135} = frac{10}{9} )Therefore, combining these ratios:( frac{v_A}{v_B} = frac{6}{5} ) and ( frac{v_B}{v_C} = frac{10}{9} )So, ( frac{v_A}{v_C} = frac{6}{5} times frac{10}{9} = frac{60}{45} = frac{4}{3} )Thus, ( v_A = frac{4}{3} v_C )Now, the time it takes for C to finish 150 meters is ( t = frac{150}{v_C} )In this time, A runs ( d_A = v_A times t = frac{4}{3} v_C times frac{150}{v_C} = frac{4}{3} times 150 = 200 ) meters.Again, A has run 200 meters, which is 50 meters beyond the finish line. So, A is ahead by 50 meters when C finishes.But since 50 isn't an option, I must have made a mistake in interpreting the problem.Wait, maybe the question is asking how much A is ahead of C when C finishes, not how much A has run beyond the finish line. So, perhaps it's the difference between A's position and C's position when C finishes.But when C finishes, A has already finished and is 50 meters ahead of the finish line. So, the lead is 50 meters.Alternatively, maybe the question is asking how much A is ahead of C when C finishes, meaning the distance between A and C when C finishes. But in that case, since A has finished and is 50 meters ahead, the lead is 50 meters.But again, 50 isn't an option. So, perhaps I'm misunderstanding the problem.Let me try another approach. Maybe instead of considering the time it takes for C to finish, I should consider the time it takes for A to finish and see how much C has run in that time.When A finishes 150 meters, B has run 125 meters. So, the time taken for A to finish is ( t = frac{150}{v_A} ).In this time, C has run ( d_C = v_C times t = v_C times frac{150}{v_A} ).From earlier, we have ( v_C = frac{3}{4} v_A ), so:( d_C = frac{3}{4} v_A times frac{150}{v_A} = frac{3}{4} times 150 = 112.5 ) meters.So, when A finishes, C has run 112.5 meters. Therefore, A is ahead of C by ( 150 - 112.5 = 37.5 ) meters.Ah, that makes sense. So, when A finishes, C has run 112.5 meters, meaning A is 37.5 meters ahead of C. Therefore, the answer is 37.5 meters, which is option C.I think I confused myself earlier by considering the time it takes for C to finish instead of the time it takes for A to finish. The question is asking how many meters ahead A is when C finishes the race, but actually, it's more about how much A has led C when A finishes, which is 37.5 meters.

💡Okay, so I have this problem here: I need to figure out how many ways there are to choose 3 cards from a customized deck of 48 cards. Each suit in this deck has only 12 cards. The conditions are that all three cards must be of different suits, and one of the cards must be a face card. Alright, let's break this down step by step.First, I know that in a standard deck, there are 4 suits, and each suit has 13 cards. But in this case, each suit has only 12 cards. That probably means that one card is missing from each suit, but I don't think that affects the number of face cards. Typically, face cards are Jack, Queen, King, and sometimes Ace, but I think in most cases, Ace isn't considered a face card. So, in each suit, there are 3 face cards: Jack, Queen, and King.So, the deck has 4 suits, each with 12 cards, making a total of 48 cards. I need to choose 3 cards, each from a different suit, and one of those cards has to be a face card.Let me think about how to approach this. It seems like a combinatorics problem where I need to count the number of ways to select these cards under the given constraints.First, I need to choose 3 different suits out of the 4 available. The number of ways to choose 3 suits from 4 is given by the combination formula:[binom{4}{3} = 4]So, there are 4 ways to choose which 3 suits we'll be selecting our cards from.Now, for each of these 3 suits, I need to pick one card. But one of these cards has to be a face card. So, I need to decide which of the 3 suits will provide the face card. There are 3 choices for this.Once I've chosen which suit will provide the face card, I need to select one face card from that suit. Since each suit has 3 face cards, there are 3 choices for the face card.Now, for the other two suits, I need to pick one card each, but these can be any card from those suits, not necessarily face cards. Each of these suits has 12 cards, so there are 12 choices for each.Putting it all together, the total number of ways is:[binom{4}{3} times 3 times 3 times 12 times 12]Calculating this:[4 times 3 times 3 times 12 times 12 = 4 times 3 times 3 times 144 = 4 times 9 times 144 = 36 times 144 = 5184]Wait, that seems like a lot. Let me double-check my reasoning.1. Choosing 3 suits out of 4: 4 ways.2. Choosing which of the 3 suits will have the face card: 3 ways.3. Choosing the face card from that suit: 3 ways.4. Choosing one card from each of the remaining two suits: 12 ways each, so 12 × 12 = 144.Multiplying all these together: 4 × 3 × 3 × 144 = 5184.Hmm, that seems correct. But let me think if there's another way to approach this problem to confirm.Alternatively, I could think of it as:- First, choose the face card. There are 4 suits, each with 3 face cards, so 4 × 3 = 12 face cards in total.- Then, choose 2 more cards from the remaining 3 suits, making sure they are from different suits and not face cards.Wait, no, that might complicate things because the other two cards can be any cards, not necessarily non-face cards. The only condition is that one card is a face card, and all three are from different suits.So, maybe my initial approach is better.Another way: For each of the 4 suits, calculate the number of ways where the face card comes from that suit, and the other two cards come from the remaining three suits, each from a different suit.So, for each suit:- Choose a face card: 3 choices.- Choose one card from each of the other two suits: 12 choices each.So, for one suit, it's 3 × 12 × 12 = 432.Since there are 4 suits, total ways would be 4 × 432 = 1728.Wait, that's different from 5184. Hmm, which one is correct?Let me see. In my first approach, I considered choosing 3 suits first, then choosing which of those 3 suits provides the face card. So, 4 (ways to choose 3 suits) × 3 (which suit provides the face card) × 3 (face card) × 12 × 12 (other two cards).In the second approach, I fixed a suit to provide the face card and then multiplied by 4.But in the second approach, I might be overcounting because choosing a face card from suit A and then choosing cards from suits B and C is the same as choosing a face card from suit B and then choosing cards from suits A and C, but in reality, these are different scenarios because the face card is from a different suit.Wait, no, actually, in the second approach, for each suit, I'm calculating the number of ways where that specific suit provides the face card, and the other two cards come from the remaining three suits, but ensuring that all three cards are from different suits.But in reality, when I fix a suit to provide the face card, the other two cards must come from the remaining three suits, but since we need all three cards to be from different suits, the other two cards must come from two different suits among the remaining three.Wait, so in the second approach, for each suit, the number of ways is:- Choose a face card from that suit: 3.- Choose two different suits from the remaining three: (binom{3}{2} = 3).- Choose one card from each of those two suits: 12 × 12 = 144.So, for each suit, it's 3 × 3 × 144 = 1296.Then, total ways would be 4 × 1296 = 5184.Ah, okay, so that matches the first approach. So, my initial second approach was missing the step of choosing which two suits to take the other two cards from, which is why it gave a different answer.So, the correct total number of ways is 5184.But let me think again to make sure.Total number of ways to choose 3 cards from different suits: First, choose 3 suits out of 4: 4 ways. Then, for each of those 3 suits, choose one card: 12 × 12 × 12 = 1728. So, total ways without any face card condition: 4 × 1728 = 6912.But we need only those combinations where exactly one card is a face card.So, total number of ways with one face card: ?Alternatively, total number of ways with one face card = total number of ways with 3 different suits minus the number of ways with no face cards minus the number of ways with two or three face cards.But that might be more complicated.Alternatively, think of it as:- Choose which suit provides the face card: 4 choices.- Choose a face card from that suit: 3 choices.- Choose two other suits from the remaining three: (binom{3}{2} = 3) choices.- Choose one non-face card from each of those two suits: Each suit has 12 - 3 = 9 non-face cards, so 9 × 9 = 81.So, total ways: 4 × 3 × 3 × 81 = 4 × 3 × 3 × 81 = 4 × 9 × 81 = 36 × 81 = 2916.Wait, that's different from 5184. Hmm, now I'm confused.Wait, in this approach, I'm assuming that the other two cards must be non-face cards, but the problem only requires that one card is a face card, not that the other two are non-face cards. So, actually, the other two cards can be any cards, including face cards, as long as they are from different suits.So, in my first approach, I didn't restrict the other two cards to be non-face cards, which is correct because the problem only requires one face card, not exactly one.Therefore, the correct total should be 5184, not 2916.So, to clarify:- The problem requires that all three cards are from different suits, and one of them is a face card. It doesn't specify that the other two must not be face cards. So, the other two can be any cards, including face cards, as long as they are from different suits.Therefore, the correct calculation is:- Choose 3 suits: 4 ways.- Choose which of the 3 suits provides the face card: 3 ways.- Choose the face card: 3 ways.- Choose one card from each of the other two suits: 12 × 12 = 144 ways.Total: 4 × 3 × 3 × 144 = 5184.Yes, that makes sense. So, the answer should be 5184.

💡Okay, so I have this problem here about a function ( y = f(x) ) defined on the domain ( (0, +infty) ). It says that ( f(x) < 0 ) when ( x > 1 ), and for any ( x, y in mathbb{R} ), it holds that ( f(xy) = f(x) + f(y) ). I need to solve the inequality ( f(x) + f(x - 2) geq f(8) ) and choose the correct solution set from the given options.First, let me try to understand the function ( f(x) ). The condition ( f(xy) = f(x) + f(y) ) reminds me of logarithmic functions because logarithms have the property that ( log(xy) = log x + log y ). So maybe ( f(x) ) is some kind of logarithmic function. But it's given that ( f(x) < 0 ) when ( x > 1 ). For logarithms, ( log x ) is positive when ( x > 1 ) and negative when ( 0 < x < 1 ). So this function ( f(x) ) seems to be the negative of a logarithmic function or maybe a logarithm with a different base.Wait, actually, if ( f(x) < 0 ) when ( x > 1 ), that suggests that ( f(x) ) is decreasing because as ( x ) increases beyond 1, ( f(x) ) becomes negative. So ( f(x) ) is a decreasing function. Also, since ( f(xy) = f(x) + f(y) ), it's a multiplicative function turned into an additive one, which is a key property of logarithms. So, ( f(x) ) must be a logarithmic function, but since it's negative for ( x > 1 ), it's likely ( f(x) = -k log x ) where ( k > 0 ). Or maybe ( f(x) = log_b x ) with base ( b < 1 ), because logarithms with bases less than 1 are decreasing functions.But maybe I don't need to know the exact form of ( f(x) ). The problem gives me enough properties to work with. Since ( f(xy) = f(x) + f(y) ), it's a multiplicative function, and it's negative for ( x > 1 ). So, if ( x > 1 ), ( f(x) < 0 ); if ( x = 1 ), ( f(1) = f(1 cdot 1) = f(1) + f(1) ), which implies ( f(1) = 0 ). And for ( 0 < x < 1 ), since ( x = 1/y ) where ( y > 1 ), ( f(x) = f(1/y) = f(1) - f(y) = 0 - f(y) = -f(y) ). Since ( f(y) < 0 ) when ( y > 1 ), ( f(x) = -f(y) > 0 ) when ( 0 < x < 1 ). So, ( f(x) ) is positive for ( 0 < x < 1 ), zero at ( x = 1 ), and negative for ( x > 1 ).Also, since ( f ) is a multiplicative function, it's either logarithmic or something similar. But maybe I can think of it as a logarithm with a base less than 1, which is decreasing. So, ( f(x) = log_b x ) where ( 0 < b < 1 ). That would make sense because ( log_b x ) is positive for ( 0 < x < 1 ), zero at ( x = 1 ), and negative for ( x > 1 ).But maybe I don't need to get bogged down with the exact form. Let's see what the inequality is: ( f(x) + f(x - 2) geq f(8) ). Using the property ( f(xy) = f(x) + f(y) ), this can be rewritten as ( f(x(x - 2)) geq f(8) ). So, ( f(x(x - 2)) geq f(8) ).Now, since ( f ) is a decreasing function, the inequality ( f(a) geq f(b) ) implies ( a leq b ) because as the input increases, the output decreases. So, ( x(x - 2) leq 8 ).But before that, I need to make sure that the arguments of ( f ) are within the domain ( (0, +infty) ). So, ( x > 0 ) and ( x - 2 > 0 ), which means ( x > 2 ). So, the domain for ( x ) is ( x > 2 ).So, now, I have ( x > 2 ) and ( x(x - 2) leq 8 ). Let's solve ( x(x - 2) leq 8 ):( x^2 - 2x leq 8 )( x^2 - 2x - 8 leq 0 )Let's factor this quadratic:Looking for two numbers that multiply to -8 and add to -2. Hmm, -4 and +2.So, ( (x - 4)(x + 2) leq 0 )So, the critical points are ( x = 4 ) and ( x = -2 ). Since we're dealing with ( x > 2 ), we can ignore ( x = -2 ) because it's outside our domain.Now, the inequality ( (x - 4)(x + 2) leq 0 ) holds when ( x ) is between -2 and 4. But since ( x > 2 ), the solution is ( 2 < x leq 4 ).So, the solution set is ( (2, 4] ), which is option A.Wait, but let me double-check. If ( x = 4 ), then ( x(x - 2) = 4*2 = 8 ), so ( f(8) geq f(8) ), which is equality, so that's fine. If ( x ) approaches 2 from the right, say ( x = 2.1 ), then ( x(x - 2) = 2.1*0.1 = 0.21 ), which is less than 8, so ( f(0.21) geq f(8) ). But since ( f ) is decreasing, ( 0.21 < 8 ) implies ( f(0.21) > f(8) ), so the inequality holds. So, yes, the interval ( (2, 4] ) satisfies the inequality.I think that's correct. So, the answer should be option A: ( (2, 4] ).

💡Okay, so I have this problem where I need to prove that in a finite set of points in the plane, with at least 5 points, some colored red and others blue, there exists a triangle that meets two conditions: all three vertices are the same color, and at least one of its sides doesn't have a point of the other color. Also, no three points of the same color are collinear. Hmm, let me try to wrap my head around this.First, I know that since there are at least 5 points and two colors, by the pigeonhole principle, there must be at least 3 points of one color. So, either there are at least 3 red points or at least 3 blue points. Let's assume, without loss of generality, that there are at least 3 red points. Now, since no three red points are collinear, these three red points must form a triangle. Okay, so we have a red triangle. But the problem is not just about finding a monochromatic triangle; it's also about ensuring that at least one side of this triangle doesn't have a point of the other color. So, I need to make sure that on at least one side of this red triangle, there isn't a blue point lying on it.Wait, how do I ensure that? Maybe I should think about the positions of the blue points relative to the red triangle. If all sides of the red triangle have blue points on them, then maybe I can find a contradiction or something. Let me try that approach.Suppose, for the sake of contradiction, that every side of the red triangle has at least one blue point on it. So, on side AB, there's a blue point, say P; on side BC, there's a blue point Q; and on side CA, there's a blue point R. Now, these points P, Q, R are all blue and lie on the sides of the red triangle.What can I do with these points? Maybe I can form smaller triangles or something. If I connect these blue points, maybe I can create another triangle. But wait, if I connect P, Q, and R, would that form a triangle? Hmm, not necessarily, because P, Q, R are on the sides of the original red triangle, so they might not form a triangle themselves unless they are not colinear.But the problem states that no three points of the same color are colinear. So, if P, Q, R are blue points, they can't be colinear. Therefore, connecting P, Q, and R must form a blue triangle. Okay, so now I have a blue triangle inside the red triangle.But wait, the blue triangle is entirely contained within the red triangle. Does that help me? Maybe I can use some kind of area argument or something. If the blue triangle is entirely inside the red triangle, then perhaps one of its sides must lie entirely within a side of the red triangle, but that might not necessarily be the case.Alternatively, maybe I can consider the convex hull of all the points. The convex hull is the smallest convex polygon that contains all the points. If I take the convex hull of the red points, it might form a polygon, and the blue points could be inside or on the edges.But I'm not sure if that's the right direction. Let me think differently. Maybe I can use the concept of empty triangles. An empty triangle is one where no other points lie inside it. If I can find an empty triangle of one color, then that triangle would satisfy the condition because there are no points of the other color inside it, but the problem is about points on the sides, not inside.Wait, actually, the problem says "at least one of its sides does not contain a point of the other color." So, it's about points lying on the sides, not necessarily inside the triangle. So, if I can find a triangle where at least one side doesn't have a point of the other color, that would work.Going back to my earlier assumption, if every side of the red triangle has a blue point, then we have a blue triangle inside. But maybe this blue triangle is smaller, and perhaps one of its sides doesn't have a red point on it. Hmm, not sure.Alternatively, maybe I can use Ramsey theory concepts here. Ramsey theory deals with conditions under which order must appear. In this case, we're dealing with points and colors, so maybe there's a Ramsey-type argument that ensures such a triangle exists.But I'm not too familiar with Ramsey theory, so maybe I should try a different approach. Let's think about the number of points. We have at least 5 points, so maybe we can consider cases based on the number of red and blue points.Case 1: There are at least 3 red points and at least 2 blue points.Case 2: There are at least 4 red points and 1 blue point.Case 3: There are at least 4 blue points and 1 red point.But the problem states that no three points of the same color are collinear, so in any case, we can form triangles.Wait, maybe I can use the Erdős–Szekeres theorem, which is about finding convex polygons in point sets. But I'm not sure if that's directly applicable here.Alternatively, maybe I can use the concept of duality, transforming points into lines and vice versa, but that might complicate things.Let me try to think geometrically. Suppose I have a red triangle ABC. If all sides of ABC have blue points on them, then those blue points can form another triangle inside ABC. But since ABC is the smallest red triangle, maybe the blue triangle is smaller, leading to a contradiction.Wait, that might be a way to go. If I assume that every side of the red triangle has a blue point, then those blue points form a smaller triangle inside. But if the red triangle was the smallest possible, this would be a contradiction because the blue triangle is even smaller.But how do I know that the red triangle is the smallest? Maybe I need to choose the red triangle with the smallest area. So, among all red triangles, pick the one with the smallest area. Then, if all its sides have blue points, those blue points would form a smaller triangle, which contradicts the minimality of the red triangle.Ah, that seems promising. So, formalizing this:1. By the pigeonhole principle, there are at least 3 points of one color, say red.2. Since no three red points are collinear, they form a triangle.3. Choose the red triangle with the smallest area.4. Assume, for contradiction, that every side of this red triangle has a blue point on it.5. These blue points form a smaller triangle inside the red triangle.6. This smaller triangle is also red or blue? Wait, the blue points form a blue triangle.7. But the blue triangle is entirely inside the red triangle, which was supposed to be the smallest red triangle.8. Contradiction because the blue triangle is smaller and also monochromatic.Wait, but the blue triangle is blue, not red. So, maybe the contradiction is not about the size but about the existence of another monochromatic triangle.Hmm, perhaps I need to think differently. Maybe the blue triangle would have sides that don't contain red points, which would satisfy the condition.Wait, no, the blue triangle is inside the red triangle, so its sides are chords of the red triangle's sides. Therefore, the blue triangle's sides might intersect the red triangle's sides, but they don't necessarily contain points of the other color.I'm getting confused here. Let me try to visualize it.Imagine a red triangle ABC. On each side, there's a blue point: P on AB, Q on BC, and R on CA. Connecting P, Q, R forms a blue triangle inside ABC. Now, does this blue triangle have any sides that don't contain red points? Well, the sides of the blue triangle are PQ, QR, and RP. These are chords inside ABC, so they don't extend beyond ABC. Therefore, they don't contain any red points because all red points are on the perimeter of ABC.Wait, but the blue triangle's sides are inside ABC, so they don't have any red points on them because the only red points are the vertices A, B, C. So, the sides of the blue triangle don't contain any red points. Therefore, the blue triangle satisfies the condition: all vertices are blue, and none of its sides contain red points.But wait, the problem says "at least one of its sides does not contain a point of the other color." So, if all sides of the blue triangle don't contain red points, then it's even better. So, in this case, the blue triangle would satisfy the condition.But then, why did I assume that every side of the red triangle has a blue point? Maybe that's not necessary. Maybe even if only one side of the red triangle has a blue point, I can still find a monochromatic triangle with a side without the other color.Wait, no, the problem is to find a triangle that is monochromatic and has at least one side without the other color. So, if I have a red triangle ABC, and suppose that side AB has a blue point P. Then, consider triangle APC. If APC is red, then side AC might have a blue point or not. If it doesn't, then APC is the desired triangle. If it does, then maybe we can find another triangle.But this seems too vague. Maybe I need a more systematic approach.Let me think about the dual problem. Suppose that every monochromatic triangle has all its sides containing points of the other color. Then, for every red triangle, every side has a blue point, and for every blue triangle, every side has a red point. But then, as I thought earlier, this would lead to smaller and smaller triangles, which is impossible because the number of points is finite.Therefore, such a configuration cannot exist, meaning there must be a monochromatic triangle with at least one side not containing the other color.Wait, that sounds like a proof by contradiction. Assume that every monochromatic triangle has all sides containing points of the other color. Then, by repeatedly finding smaller triangles, we get an infinite descent, which is impossible because the number of points is finite. Therefore, our assumption is wrong, and there must exist a monochromatic triangle with at least one side not containing the other color.Yes, that makes sense. So, formalizing this:1. Assume, for contradiction, that every monochromatic triangle has all its sides containing points of the other color.2. Start with a monochromatic triangle, say red, with vertices A, B, C.3. Since every side must have a blue point, let P be on AB, Q on BC, and R on CA.4. These points P, Q, R form a blue triangle.5. Now, consider the blue triangle PQR. By our assumption, every side of this blue triangle must have a red point.6. Let S be on PQ, T on QR, and U on RP.7. These points S, T, U form a red triangle.8. This red triangle is smaller than the original red triangle ABC.9. Repeating this process, we can create an infinite sequence of smaller and smaller monochromatic triangles, which is impossible because the number of points is finite.10. Therefore, our initial assumption is false, and there must exist a monochromatic triangle with at least one side not containing a point of the other color.That seems like a solid argument. So, the key idea is that if every monochromatic triangle had all sides containing points of the other color, we could keep finding smaller and smaller triangles indefinitely, which contradicts the finiteness of the point set. Hence, such a triangle must exist.I think I've got it. So, the proof relies on the pigeonhole principle to ensure there are enough points of one color, and then a contradiction argument based on infinite descent if every monochromatic triangle had sides with points of the other color.

💡First, I recognize that a standard deck of cards contains 52 cards, with half being red and half being black. This means there are 26 red cards in a complete deck.Next, I note that after some red cards were taken out, there are 16 red cards remaining. To find out how many red cards were removed, I subtract the number of remaining red cards from the original number of red cards.So, the calculation is 26 (original red cards) minus 16 (remaining red cards), which equals 10 red cards taken out.

💡First, determine the distance Hillary climbs before stopping. She stops 1000 feet short of the summit, so she climbs 5000 - 1000 = 4000 feet.Next, calculate the time it takes Hillary to climb 4000 feet at a rate of 800 feet per hour. Time = Distance / Rate = 4000 / 800 = 5 hours. Therefore, Hillary reaches her stopping point at 11:00.Eddy climbs at a rate of 500 feet per hour. By 12:00, which is 6 hours after departure, Eddy has climbed 500 * 6 = 3000 feet.At 12:00, Hillary and Eddy pass each other. This means they meet at the same point, which is 3000 feet from the base camp. Hillary has descended from her stopping point at 4000 feet to this meeting point, a distance of 1000 feet.Hillary descends this 1000 feet in the time between 11:00 and 12:00, which is 1 hour. Therefore, her descending rate is 1000 feet per hour.