🧠:Okay, let's try to figure out this grass problem. Hmm, so the grass grows at a constant rate every day, right? And the pasture can feed 20 cows for 40 days, or 35 cows for 10 days. The question is, how many days can it feed 25 cows?Alright, first, I need to understand the problem. It's about cows eating grass that's also growing. So, the total amount of grass isn't fixed because it's growing every day. The rate at which the grass is eaten by the cows must be balanced against the rate at which it's growing. If the cows eat faster than the grass grows, eventually the grass will run out, and that's when the pasture can't feed them anymore.Let me try to model this. Let's denote some variables:- Let’s say the initial amount of grass in the pasture is G (in some units, maybe units of grass).- The grass grows at a rate of r per day (units of grass per day).- Each cow eats a certain amount of grass per day. Let’s call this rate e (units of grass per cow per day).So, the total amount of grass consumed by the cows over a period of D days would be the number of cows times the eating rate times the number of days, which is N * e * D. But at the same time, the grass is growing, so the total grass available after D days is the initial grass plus the growth over D days, which is G + r * D.When the pasture can no longer feed the cows, the total grass consumed equals the total grass available. So, the equation would be:N * e * D = G + r * DWe have two different scenarios here:1. 20 cows for 40 days: 20 * e * 40 = G + r * 402. 35 cows for 10 days: 35 * e * 10 = G + r * 10We need to find D when there are 25 cows: 25 * e * D = G + r * DSo, we have two equations and two unknowns (G and r, since e is also an unknown). Wait, actually, three variables: G, r, and e. But maybe we can combine them or find a way to eliminate variables.Wait, let's see. Let me write the two equations again.First equation: 20 * 40 * e = G + 40rSecond equation: 35 * 10 * e = G + 10rSo, equation 1: 800e = G + 40rEquation 2: 350e = G + 10rNow, if we subtract equation 2 from equation 1, we can eliminate G:800e - 350e = (G + 40r) - (G + 10r)450e = 30rDivide both sides by 30: 15e = rSo, the growth rate r is 15e. That means the grass grows at a rate equivalent to 15 cows eating per day. Interesting.Now, substitute r = 15e back into one of the equations to find G. Let's take equation 2:350e = G + 10rBut r is 15e, so:350e = G + 10*15e = G + 150eSubtract 150e from both sides:350e - 150e = G => 200e = GSo, the initial grass G is 200e. That is, the initial grass is equivalent to 200 days worth of one cow's consumption.Now, we need to find D when there are 25 cows. The equation is:25e * D = G + rDSubstitute G = 200e and r = 15e:25eD = 200e + 15eDLet's divide both sides by e (assuming e ≠ 0, which makes sense):25D = 200 + 15DSubtract 15D from both sides:10D = 200Divide both sides by 10:D = 20So, the answer is 20 days. Wait, but let me check if that makes sense.Wait, when there are more cows, the pasture gets depleted faster. So 35 cows can only be fed for 10 days, 20 cows for 40 days. 25 cows is in between, so the days should be between 10 and 40? But 20 is between 10 and 40, but let me see.Alternatively, maybe I made a miscalculation. Let's verify.Given:For 20 cows: 20*40*e = 800e = G +40rBut G = 200e, r=15e, so G +40r = 200e +40*15e = 200e +600e = 800e. So that checks out.For 35 cows: 35*10*e = 350e = G +10r = 200e +150e = 350e. That works too.So, when we have 25 cows, the equation is 25e*D = 200e +15e*DWhich simplifies to 25D = 200 +15DSubtract 15D: 10D=200 => D=20. So yes, 20 days. Hmm. That seems correct.But let me think again. If 20 cows last 40 days, and 35 cows last 10 days, the difference in cows is 15 more cows leading to 30 fewer days. So per additional cow, the days decrease by 2? Not exactly linear, but maybe.But 25 cows is 5 more than 20, so 5 cows more. If the rate is linear, perhaps 40 days minus 5*2=10 days, so 30 days? But that contradicts the answer of 20. So maybe it's not linear.Alternatively, using the equations, since we have a system that models the problem, the answer is 20 days. So perhaps the answer is 20.Alternatively, let me think in terms of grass units.Suppose each cow eats 1 unit of grass per day. Then, e = 1. Then, the initial grass G = 200 units, and the growth rate r =15 units per day.So for 20 cows over 40 days:Total consumed: 20*40 = 800 units.Total grass available: 200 + 40*15 = 200 +600 =800. So that's correct.For 35 cows over 10 days:Total consumed:35*10=350.Total grass:200 +10*15=350. Correct.Now, 25 cows:Each day, 25 units eaten, but 15 units grow. So net loss per day is 25-15=10 units.The initial grass is 200 units. So, how many days until the grass is depleted?Wait, wait. Wait a second. If each day the cows eat 25 units, but the grass grows 15 units, then the net consumption per day is 10 units. So, starting from 200 units, the number of days would be 200 /10 =20 days. So that matches the previous answer.Yes, that's a simpler way to think about it. So, the pasture starts with 200 units, each day net loss is 10 units, so 200/10=20 days. Perfect, that's the same answer. So that confirms it.So the answer is 20 days. That makes sense.

✅:The pasture can feed 25 cows for 20 days. Here's the breakdown:1. Define Variables: - Let ( G ) = initial grass amount. - Let ( r ) = daily grass growth rate. - Let ( e ) = daily grass consumption per cow.2. Set Up Equations: - For 20 cows over 40 days: [ 20 cdot 40 cdot e = G + 40r quad Rightarrow quad 800e = G + 40r tag{1} ] - For 35 cows over 10 days: [ 35 cdot 10 cdot e = G + 10r quad Rightarrow quad 350e = G + 10r tag{2} ]3. Solve for ( r ) and ( G ): - Subtract Equation (2) from Equation (1): [ 450e = 30r quad Rightarrow quad r = 15e. ] - Substitute ( r = 15e ) into Equation (2): [ 350e = G + 150e quad Rightarrow quad G = 200e. ]4. Determine Days for 25 Cows: - Let ( D ) = days to deplete grass for 25 cows: [ 25e cdot D = G + rD quad Rightarrow quad 25eD = 200e + 15eD. ] - Simplify (divide by ( e )): [ 25D = 200 + 15D quad Rightarrow quad 10D = 200 quad Rightarrow quad D = 20. ]Conclusion: The pasture can sustain 25 cows for 20 days.

🔑:1. Determine the daily grass consumption rate for 20 cows over 40 days: [ text{Let } G text{ be the initial amount of grass and } R text{ be the rate at which the grass regrows daily.} text{Also, let } C = text{daily consumption rate per cow.} ] [ Rightarrow 20C(40) = G + 40R ] We denote the total initial amount of grass and daily growth as a combined factor.2. Determine the daily grass consumption rate for 35 cows over 10 days: [ Rightarrow 35C(10) = G + 10R ]3. Set up the equations based on the problem's data: [ begin{cases} 800C = G + 40R quad text{(from 20 cows for 40 days)} 350C = G + 10R quad text{(from 35 cows for 10 days)} end{cases} ]4. Eliminate (G) by subtracting the second equation from the first equation: [ 800C - 350C = (G + 40R) - (G + 10R) ] [ 450C = 30R ] [ 15C = R ]5. Substitute ( R = 15C ) into one of the original equations to find ( G ): [ 350C = G + 10(15C) ] [ 350C = G + 150C ] [ 200C = G ]6. Now, determine the number of days for which 25 cows can be fed: [ Rightarrow 25C(D) = G + DR ] Substitute ( G = 200C ) and ( R = 15C ): [ 25CD = 200C + 15CD ] Subtracted ( 15CD ) from both sides: [ 10CD = 200C ] Divided ( C ) from both sides: [ 10D = 200 ] [ D = 20 ] Conclusion:[boxed{20}]

🧠:Okay, so I need to find the product of the inradius of triangles ADB and ADC in an equilateral triangle ABC with side length 4. Point D is on BC such that BD = 1. Hmm, let's start by visualizing the problem. ABC is an equilateral triangle, so all sides are 4, and all angles are 60 degrees. Point D is on BC, and BD is 1, which means DC must be 3 since BC is 4. First, maybe I should draw a diagram. Let me imagine triangle ABC with points A at the top, B at the bottom left, and C at the bottom right. Then D is closer to B on BC because BD is only 1. So BD = 1, DC = 3. Then AD is a line from A to D, creating two smaller triangles: ADB and ADC. The question is asking for the product of the inradii (rs) of these two triangles.I remember that the formula for the inradius of a triangle is r = Area / semiperimeter. So for each triangle ADB and ADC, I need to find their areas and their semiperimeters, then compute r and s, and multiply them together.So let's break it down into steps:1. Find the coordinates of points A, B, C, D.2. Determine the lengths of AD, AB, BD, DC, AC.3. Calculate the areas of triangles ADB and ADC.4. Compute the semiperimeters of both triangles.5. Find the inradii r and s using the formula.6. Multiply r and s to get rs.Alternatively, maybe coordinate geometry is a good approach here. Let me assign coordinates to the triangle. Let's place point B at the origin (0,0), point C at (4,0), and since ABC is equilateral, point A will be at (2, 2√3). Wait, the height of an equilateral triangle with side length 4 is (√3/2)*4 = 2√3. So yes, A is at (2, 2√3). Then point D is on BC. Since BD = 1, and BC is from (0,0) to (4,0), moving along the x-axis. Therefore, D must be at (1,0).Wait, hold on. If B is at (0,0) and C is at (4,0), then BD is 1 unit from B, so D is at (1,0). Then DC is from (1,0) to (4,0), which is 3 units. That makes sense.So coordinates:- A: (2, 2√3)- B: (0,0)- C: (4,0)- D: (1,0)Now, triangles ADB and ADC. Let's find the sides of these triangles.Starting with triangle ADB:Points A(2, 2√3), D(1,0), B(0,0).Sides:- AB: already 4, but since we're in coordinate system, let's confirm. Distance between A(2, 2√3) and B(0,0): sqrt[(2-0)^2 + (2√3 - 0)^2] = sqrt[4 + 12] = sqrt[16] = 4. Correct.- BD: 1, as given.- AD: Distance between A(2, 2√3) and D(1,0). sqrt[(2-1)^2 + (2√3 - 0)^2] = sqrt[1 + 12] = sqrt[13]. So AD = √13.Similarly, for triangle ADC:Points A(2, 2√3), D(1,0), C(4,0).Sides:- AC: 4, same as AB and BC.- DC: 3, as given.- AD: same as before, √13.Wait, so triangles ADB and ADC share the side AD, which is √13. Their other sides are BD=1, AB=4, and DC=3, AC=4 respectively.Now, let's compute the areas of these triangles.For triangle ADB:We can use coordinates to calculate the area. The coordinates are A(2, 2√3), D(1,0), B(0,0). Using the shoelace formula:Area = (1/2)| (2*(0 - 0) + 1*(0 - 2√3) + 0*(2√3 - 0)) |= (1/2)| 0 + 1*(-2√3) + 0 |= (1/2)| -2√3 | = (1/2)(2√3) = √3.Alternatively, since BD is 1 and the height from A to BC is 2√3, but wait, the height from A to BC is the height of the entire triangle, which is 2√3. However, triangle ADB is not necessarily a right triangle. Wait, but maybe we can use coordinates or vectors.Alternatively, since points B and D are on the x-axis, the base BD is 1, and the height would be the y-coordinate of point A, which is 2√3. Wait, but that would be the height for triangle ABD? Wait, no. Because in triangle ABD, the base is BD = 1, and the height is the vertical distance from A to BD. Since BD is along the x-axis, the height is indeed the y-coordinate of A, which is 2√3. So area is (1/2)*base*height = (1/2)*1*2√3 = √3. That's correct. So area of ADB is √3.Similarly, for triangle ADC. The base DC is 3, and the height is still the y-coordinate of A, 2√3. So area is (1/2)*3*2√3 = 3√3. So area of ADC is 3√3.Alternatively, using shoelace formula for triangle ADC: points A(2, 2√3), D(1,0), C(4,0).Area = (1/2)| 2*(0 - 0) + 1*(0 - 2√3) + 4*(2√3 - 0) |= (1/2)| 0 + 1*(-2√3) + 4*2√3 |= (1/2)| -2√3 + 8√3 | = (1/2)|6√3| = 3√3. Correct.So areas are √3 and 3√3.Now, compute the semiperimeters.For triangle ADB:Sides AB = 4, BD = 1, AD = √13.Perimeter = 4 + 1 + √13 = 5 + √13. Semiperimeter (s1) = (5 + √13)/2.For triangle ADC:Sides AC = 4, DC = 3, AD = √13.Perimeter = 4 + 3 + √13 = 7 + √13. Semiperimeter (s2) = (7 + √13)/2.Wait, but inradius formula is Area / semiperimeter. So:r (for ADB) = Area / s1 = √3 / [(5 + √13)/2] = (2√3)/(5 + √13)Similarly, s (for ADC) = 3√3 / [(7 + √13)/2] = (6√3)/(7 + √13)Then rs = [2√3 / (5 + √13)] * [6√3 / (7 + √13)] = (2√3 * 6√3) / [(5 + √13)(7 + √13)] = (12 * 3) / [ (5 + √13)(7 + √13) ] = 36 / [ (5)(7) + 5√13 + 7√13 + (√13)^2 ] = 36 / [ 35 + 12√13 + 13 ] = 36 / [ 48 + 12√13 ]Simplify denominator: 48 + 12√13 = 12*(4 + √13). So:36 / [12*(4 + √13)] = 3 / (4 + √13)To rationalize the denominator, multiply numerator and denominator by (4 - √13):3*(4 - √13) / [ (4 + √13)(4 - √13) ] = [12 - 3√13] / [16 - 13] = [12 - 3√13]/3 = 4 - √13Wait, that can't be right. Wait:Wait, the denominator (4 + √13)(4 - √13) is 16 - (√13)^2 = 16 -13 = 3. So:[12 - 3√13] / 3 = 4 - √13. So rs = 4 - √13.But that seems a bit odd. Let me check the steps again.First, areas:ADB: √3, ADC: 3√3. Correct.Semiperimeters:ADB: (4 + 1 + √13)/2 = (5 + √13)/2. Correct.ADC: (4 + 3 + √13)/2 = (7 + √13)/2. Correct.Inradius calculations:r = √3 / [(5 + √13)/2] = 2√3 / (5 + √13)s = 3√3 / [(7 + √13)/2] = 6√3 / (7 + √13)Multiply them:rs = (2√3 * 6√3) / [ (5 + √13)(7 + √13) ] = (12 * 3) / [ ... ] = 36 / [ (5*7) + 5√13 + 7√13 + (√13)^2 ]Compute denominator:5*7 = 355√13 +7√13 = 12√13(√13)^2 =13So total denominator: 35 +12√13 +13 = 48 +12√13. Correct.So 36 / (48 +12√13) = 36 / [12*(4 + √13)] = 3 / (4 + √13). Correct.Then rationalizing:Multiply numerator and denominator by (4 - √13):3*(4 - √13) / [ (4 + √13)(4 - √13) ] = [12 -3√13]/ (16 -13) = [12 -3√13]/3 = 4 - √13.Yes, that's correct. So rs = 4 - √13.But wait, let me check if this answer makes sense. Since the original triangle has inradius, let's see, for an equilateral triangle, inradius is (side length * √3)/6. So for ABC, inradius is (4*√3)/6 = (2√3)/3 ≈ 1.1547.But rs here is 4 - √13. Let's compute √13 ≈ 3.6055, so 4 - 3.6055 ≈ 0.3945. So the product of the inradii is approximately 0.3945, which is less than the inradius of the original triangle. That seems plausible? Since the smaller triangles have smaller inradii, their product could be less. Let's see, but let's check if there's another way to approach this problem to confirm.Alternatively, maybe using trigonometry. Since ABC is equilateral, all angles are 60 degrees. Let's consider triangle ADB. We can compute angles using the Law of Cosines.For triangle ADB:Sides AB=4, BD=1, AD=√13.Law of Cosines on angle at B:AD² = AB² + BD² - 2*AB*BD*cos(angle at B)Wait, angle at B in triangle ABC is 60 degrees. Wait, but in triangle ADB, angle at B is still part of the original angle. Wait, but in the original triangle ABC, angle at B is 60 degrees. However, when we split BC into BD=1 and DC=3, the angle at B for triangle ADB is still 60 degrees? Wait, no. Because point D is along BC, so angle at B in triangle ADB is the same as angle at B in triangle ABC, which is 60 degrees. Wait, is that correct?Wait, actually, in triangle ABC, angle at B is 60 degrees. When we look at triangle ADB, the angle at B is still 60 degrees because D is on BC. So triangle ADB has angle at B = 60 degrees, sides AB=4, BD=1, and AD=√13. Similarly, triangle ADC has angle at C=60 degrees, sides AC=4, DC=3, and AD=√13.If that's the case, then maybe we can use the formula for inradius in terms of sides and angles. The inradius r = (2*Area)/(a + b + c). Wait, that's the same as Area/s, where s is semiperimeter. So same as before. But maybe there's a formula using angles?Alternatively, using formula r = (a + b - c)/2 * tan(theta/2), but I might be misremembering. Wait, for a triangle with two sides adjacent to the angle, and the included angle. Maybe not.Alternatively, using the formula r = (a sin(theta/2) ) / (1 + sin(theta/2)) ), but I need to check.Alternatively, perhaps we can use the formula for inradius in terms of sides and angles:r = (2 Area)/(a + b + c). Wait, but that's the same as before. So perhaps not helpful.Alternatively, let's compute the inradius using another method to cross-verify.For triangle ADB:We have sides AB=4, BD=1, AD=√13. Area is √3.Semiperimeter s1 = (4 +1 +√13)/2 = (5 +√13)/2.r = Area / s1 = √3 / [(5 +√13)/2] = 2√3/(5 +√13). As before.Similarly, for triangle ADC:Sides AC=4, DC=3, AD=√13. Area is 3√3.Semiperimeter s2 = (4 +3 +√13)/2 = (7 +√13)/2.s = 3√3 / [(7 +√13)/2] = 6√3/(7 +√13). As before.Multiplying r and s:rs = [2√3/(5 +√13)] * [6√3/(7 +√13)] = (12*3)/[(5 +√13)(7 +√13)] = 36/[35 +5√13 +7√13 +13] = 36/[48 +12√13] = 3/(4 +√13). Then rationalizing gives 4 -√13. So same result.Alternatively, maybe there's a smarter way using properties of the inradius. Let's think.In an equilateral triangle, if we have points dividing the base, perhaps there's a relation between the inradii of the two smaller triangles. But since the original triangle is equilateral, maybe symmetry can help, but in this case, D is not the midpoint, so asymmetry.Alternatively, coordinate geometry: perhaps we can compute the inradius by finding the equations of the angle bisectors and then the inradius. But that seems more complicated.Alternatively, using vectors. But since we already have the answer via coordinate geometry, and it seems consistent, maybe we can check with another approach.Wait, let's verify the value of rs = 4 -√13 numerically.Compute 4 -√13 ≈ 4 -3.6055 ≈0.3945.Compute r and s numerically:r = 2√3 / (5 +√13) ≈ 2*1.732 / (5 +3.6055) ≈3.464 /8.6055≈0.4025s =6√3 / (7 +√13)≈6*1.732 / (7 +3.6055)≈10.392 /10.6055≈0.980Then rs≈0.4025*0.980≈0.3945, which matches 4 -√13≈0.3945. So the calculation seems consistent.Therefore, the answer is 4 -√13. But let's check if 4 -√13 is positive. √13≈3.6055, so 4 -3.6055≈0.3945, which is positive. So the product rs is 4 -√13.But let me check once again if the step where I rationalized 3/(4 +√13) is correct.3/(4 +√13) * (4 -√13)/(4 -√13) = 3*(4 -√13)/(16 -13)=3*(4 -√13)/3=4 -√13. Yes, correct. So that step is accurate.Therefore, the product rs is indeed 4 -√13.But wait, just to ensure that there isn't a miscalculation in the initial steps. Let me recheck areas and semiperimeters.Area of ADB: √3. Since BD=1, height from A is 2√3, so area (1/2)*1*2√3=√3. Correct.Area of ADC: (1/2)*3*2√3=3√3. Correct.Semiperimeters:For ADB: (4 +1 +√13)/2=(5 +√13)/2≈(5 +3.6055)/2≈8.6055/2≈4.3027. Then r=Area/s≈1.732/4.3027≈0.4025, which matches previous.For ADC: (4 +3 +√13)/2=(7 +√13)/2≈(7 +3.6055)/2≈10.6055/2≈5.3027. Then s=Area/s≈5.196/5.3027≈0.980, as before. So rs≈0.4025*0.980≈0.394≈4 -√13. Correct.Hence, after thorough calculation, cross-verifying with numerical approximations, and checking all steps, I can confirm that rs=4 -√13.Final Answerboxed{4 - sqrt{13}}

✅:Given an equilateral triangle ( triangle ABC ) with side length 4, point ( D ) is on ( BC ) such that ( BD = 1 ). We need to find the product ( rs ) where ( r ) and ( s ) are the inradii of ( triangle ADB ) and ( triangle ADC ) respectively.1. Coordinates and Side Lengths: - Coordinates: ( A(2, 2sqrt{3}) ), ( B(0, 0) ), ( C(4, 0) ), ( D(1, 0) ). - Side lengths: ( AB = 4 ), ( BD = 1 ), ( DC = 3 ), ( AD = sqrt{(2-1)^2 + (2sqrt{3}-0)^2} = sqrt{13} ).2. Areas of Triangles: - Area of ( triangle ADB ): Using the base ( BD = 1 ) and height from ( A ), the area is ( frac{1}{2} times 1 times 2sqrt{3} = sqrt{3} ). - Area of ( triangle ADC ): Using the base ( DC = 3 ) and height from ( A ), the area is ( frac{1}{2} times 3 times 2sqrt{3} = 3sqrt{3} ).3. Semiperimeters: - Semiperimeter of ( triangle ADB ): ( frac{4 + 1 + sqrt{13}}{2} = frac{5 + sqrt{13}}{2} ). - Semiperimeter of ( triangle ADC ): ( frac{4 + 3 + sqrt{13}}{2} = frac{7 + sqrt{13}}{2} ).4. Inradii Calculation: - Inradius ( r ) of ( triangle ADB ): ( frac{sqrt{3}}{frac{5 + sqrt{13}}{2}} = frac{2sqrt{3}}{5 + sqrt{13}} ). - Inradius ( s ) of ( triangle ADC ): ( frac{3sqrt{3}}{frac{7 + sqrt{13}}{2}} = frac{6sqrt{3}}{7 + sqrt{13}} ).5. Product of Inradii: - ( rs = left( frac{2sqrt{3}}{5 + sqrt{13}} right) left( frac{6sqrt{3}}{7 + sqrt{13}} right) = frac{36}{(5 + sqrt{13})(7 + sqrt{13})} ). - Simplifying the denominator: ( (5 + sqrt{13})(7 + sqrt{13}) = 35 + 5sqrt{13} + 7sqrt{13} + 13 = 48 + 12sqrt{13} ). - ( rs = frac{36}{48 + 12sqrt{13}} = frac{3}{4 + sqrt{13}} ). - Rationalizing: ( frac{3}{4 + sqrt{13}} times frac{4 - sqrt{13}}{4 - sqrt{13}} = frac{3(4 - sqrt{13})}{16 - 13} = 4 - sqrt{13} ).Thus, the product ( rs ) is (boxed{4 - sqrt{13}}).

🔑:1. Determine the length ( AD ): - Given (triangle ABC) is equilateral with side length 4 and (D) is on (BC) such that (BD = 1). - Using the Law of Cosines in (triangle ADB), [ AD^2 = AB^2 + BD^2 - 2 cdot AB cdot BD cdot cos 60^circ ] [ AD^2 = 4^2 + 1^2 - 2 cdot 4 cdot 1 cdot cos 60^circ ] [ AD^2 = 16 + 1 - 8 cdot 0.5 = 17 - 4 = 13 ] [ AD = sqrt{13} ]2. Calculate the area of (triangle ADB): - Using the formula for the area of a triangle: [ text{Area of } triangle ADB = frac{1}{2} cdot AB cdot BD cdot sin 60^circ ] [ = frac{1}{2} cdot 4 cdot 1 cdot frac{sqrt{3}}{2} ] [ = frac{4 sqrt{3}}{4} = sqrt{3} ]3. Calculate the area of (triangle ADC): - Since (AD) divides (triangle ABC) into two parts, (triangle ADC) and (triangle ADB), and applying similar methods, [ text{Area of } triangle ABC = frac{sqrt{3}}{4} cdot text{side}^2 = frac{sqrt{3}}{4} cdot 16 = 4sqrt{3} ] [ text{Area of } triangle ADC = (text{Area of } triangle ABC) - (text{Area of } triangle ADB) ] [ = 4sqrt{3} - sqrt{3} = 3sqrt{3} ]4. Find the radii (r) and (s): - The radius of the inscribed circle of a triangle is given by: [ r = frac{text{Area}}{s} ] where (s) is the semi-perimeter of the triangle. For (triangle ADB): - Semi-perimeter: [ s_{ADB} = frac{AB + BD + AD}{2} = frac{4 + 1 + sqrt{13}}{2} = frac{5 + sqrt{13}}{2} ] - Radius (r): [ r = frac{text{Area of } triangle ADB}{s_{ADB}} times 2 = frac{2 sqrt{3}}{5 + sqrt{13}} ] For (triangle ADC): - Semi-perimeter: [ s_{ADC} = frac{AC + CD + AD}{2} = frac{4 + 3 + sqrt{13}}{2} = frac{7 + sqrt{13}}{2} ] - Radius (s): [ s = frac{text{Area of } triangle ADC}{s_{ADC}} times 2 = frac{6 sqrt{3}}{7 + sqrt{13}} ]5. Find the product (rs): - Now, calculate the product of the radii: [ rs = left(frac{2 sqrt{3}}{5 + sqrt{13}} right) left(frac{6 sqrt{3}}{7 + sqrt{13}} right) ] [ = frac{12 cdot 3}{(5 + sqrt{13})(7 + sqrt{13})} ] [ = frac{36}{35 + 5sqrt{13} + 7sqrt{13} + 13} ] [ = frac{36}{48 + 12sqrt{13}} ] [ = frac{36}{12 (4 + sqrt{13})} ] [ = frac{3}{4 + sqrt{13}} ] [ = 4 - sqrt{13} ] Conclusion:[boxed{4 - sqrt{13}}]

🧠:Okay, so I need to find the ratio MO : OA in a parallelogram ABCD where M and N are the midpoints of sides BC and CD respectively. The segments AM and BN intersect at point O. Hmm, let me start by visualizing the problem.First, I should recall that in a parallelogram, opposite sides are equal and parallel. Also, the midpoints would divide each side into two equal parts. Let me try drawing a rough sketch of parallelogram ABCD. Let's say AB is the base, BC is the right side, CD is the top, and DA is the left side. Points M is the midpoint of BC, so BM = MC. Similarly, N is the midpoint of CD, so CN = ND. Then, connecting A to M and B to N, their intersection is point O. The goal is to find the ratio MO : OA.Maybe coordinate geometry would help here. Assign coordinates to the vertices, then find equations of lines AM and BN, solve for their intersection O, and then compute the distances MO and OA. That sounds systematic. Let me try that approach.Let's place the parallelogram in a coordinate system. To make it easier, let’s assign coordinates such that point A is at the origin (0,0). Since ABCD is a parallelogram, if we let vector AB be along the x-axis, then point B can be at (b, 0) for some b > 0. Then, since opposite sides are equal and parallel, point D must be at (d, h), where h is the height of the parallelogram. Therefore, point C, which is B + D - A, would be at (b + d, h). Wait, no. Actually, in a parallelogram, the coordinates can be defined as follows: If A is (0,0), B is (a,0), D is (0,c), then C would be (a, c). But maybe the user didn't specify it's a rectangle, so perhaps sides are not necessarily axis-aligned. Hmm, maybe choosing coordinates more generally?Alternatively, let me set coordinates such that point A is (0,0), point B is (2,0), point D is (0,2), making ABCD a rhombus? Wait, but it's a general parallelogram. Let me pick coordinates more strategically to simplify calculations. Let's let A be (0,0). Let’s assume AB is along the x-axis, so B is (2,0). Then, since ABCD is a parallelogram, point D can be (0,2), so point C would be B + D - A = (2,0) + (0,2) - (0,0) = (2,2). Wait, but that makes ABCD a rectangle. Hmm, but the problem states it's a parallelogram, not necessarily a rectangle. Maybe I should use vectors instead.Alternatively, assign coordinates more generally. Let’s say point A is (0,0). Let’s assign point B as (2,0) so that AB has length 2. Then, since it's a parallelogram, point D can be (p, q), so point C is B + D = (2 + p, q). Then, coordinates:A: (0,0)B: (2,0)D: (p, q)C: (2 + p, q)Now, midpoints M and N:Point M is the midpoint of BC. Coordinates of B: (2,0), coordinates of C: (2 + p, q). So midpoint M is [(2 + (2 + p))/2, (0 + q)/2] = [(4 + p)/2, q/2].Point N is the midpoint of CD. Coordinates of C: (2 + p, q), coordinates of D: (p, q). So midpoint N is [(2 + p + p)/2, (q + q)/2] = [(2 + 2p)/2, q] = (1 + p, q).Now, equations of lines AM and BN.First, line AM connects A(0,0) to M[(4 + p)/2, q/2]. The parametric equations for AM can be written as:x = t * ( (4 + p)/2 - 0 ) = t*(4 + p)/2,y = t * ( q/2 - 0 ) = t*q/2,where t ranges from 0 to 1.Similarly, line BN connects B(2,0) to N(1 + p, q). The parametric equations for BN can be written as:x = 2 + s*( (1 + p) - 2 ) = 2 + s*(p - 1),y = 0 + s*( q - 0 ) = s*q,where s ranges from 0 to 1.We need to find the intersection point O of lines AM and BN. So, set the coordinates equal:From AM: x = t*(4 + p)/2, y = t*q/2.From BN: x = 2 + s*(p - 1), y = s*q.Therefore, equate y-coordinates:t*q/2 = s*q.Assuming q ≠ 0 (otherwise, the parallelogram would be degenerate), we can divide both sides by q:t/2 = s.So, s = t/2.Now, substitute s = t/2 into the x-coordinate equations:t*(4 + p)/2 = 2 + (t/2)*(p - 1).Multiply both sides by 2 to eliminate denominators:t*(4 + p) = 4 + t*(p - 1).Expand right-hand side:t*(4 + p) = 4 + t*p - t.Bring all terms involving t to the left:t*(4 + p) - t*p + t = 4.Factor t:t*(4 + p - p + 1) = 4Simplify inside the brackets:t*(5) = 4Therefore, t = 4/5.Then, s = t/2 = (4/5)/2 = 2/5.Therefore, the coordinates of O are:From AM: x = (4/5)*(4 + p)/2 = (4/5)*( (4 + p)/2 ) = (4*(4 + p))/10 = (16 + 4p)/10 = (8 + 2p)/5,y = (4/5)*(q/2) = (4q)/10 = (2q)/5.Alternatively, from BN: x = 2 + (2/5)*(p - 1) = 2 + (2p - 2)/5 = (10 + 2p - 2)/5 = (8 + 2p)/5,y = (2/5)*q = 2q/5. Same result.So point O has coordinates ( (8 + 2p)/5, 2q/5 ).Now, we need to find MO : OA.First, let's find coordinates of M: midpoint of BC is [(2 + (2 + p))/2, (0 + q)/2] = [(4 + p)/2, q/2].Coordinates of O: ( (8 + 2p)/5, 2q/5 ).Coordinates of A: (0,0).Compute vector MO and OA, or compute the distances.But since we need the ratio, maybe we can compute the ratio along the line AM.Since O is on AM, and we have parameter t = 4/5. Wait, in parametric terms, point O is at t = 4/5 along AM. Therefore, OA would be the distance from O to A, which is the length from t=4/5 to t=0, so OA corresponds to t=4/5. Similarly, MO is the distance from M to O, which is from t=1 (M is at t=1 on AM) to t=4/5, so that's t=1 - 4/5 = 1/5.Therefore, the ratio MO : OA would be (1/5) : (4/5) = 1 : 4.Wait, that seems straightforward. Wait, is that correct?Because in parametric terms, if O is at t = 4/5 on AM, then OA is the segment from O to A, which would be the length from t=4/5 to t=0, which is 4/5 of the total length of AM. Similarly, MO is from t=4/5 to t=1, which is 1 - 4/5 = 1/5 of the total length. Therefore, the ratio MO : OA is 1/5 : 4/5 = 1 : 4.Therefore, the ratio is 1:4.But wait, let me verify this with coordinates to ensure there is no mistake.Coordinates of A: (0,0)Coordinates of O: ( (8 + 2p)/5, 2q/5 )Coordinates of M: ( (4 + p)/2, q/2 )Compute vector OA: from O to A: (0 - (8 + 2p)/5, 0 - 2q/5 ) = ( - (8 + 2p)/5, -2q/5 )But distance OA would be sqrt[ ( (8 + 2p)/5 )^2 + ( 2q/5 )^2 ]Similarly, distance MO is sqrt[ ( (8 + 2p)/5 - (4 + p)/2 )^2 + ( 2q/5 - q/2 )^2 ]Let me compute the differences in coordinates.For x-coordinate of MO: (8 + 2p)/5 - (4 + p)/2 = ( (8 + 2p)*2 - (4 + p)*5 ) / 10= (16 + 4p - 20 - 5p ) / 10= ( -4 - p ) / 10For y-coordinate of MO: 2q/5 - q/2 = (4q - 5q)/10 = (-q)/10So distance MO is sqrt[ ( (-4 - p)/10 )^2 + ( (-q)/10 )^2 ] = sqrt[ ( (4 + p)^2 + q^2 ) / 100 ] = ( sqrt( (4 + p)^2 + q^2 ) ) / 10Similarly, distance OA is sqrt[ ( (8 + 2p)/5 )^2 + ( 2q/5 )^2 ] = ( sqrt( (8 + 2p)^2 + (2q)^2 ) ) / 5= ( sqrt( 4*(4 + p)^2 + 4q^2 ) ) / 5= ( 2*sqrt( (4 + p)^2 + q^2 ) ) / 5Therefore, the ratio MO : OA is [ ( sqrt( (4 + p)^2 + q^2 ) / 10 ) ] : [ 2*sqrt( (4 + p)^2 + q^2 ) / 5 ]Simplify:= (1/10) : (2/5) = (1/10) / (2/5) = (1/10)*(5/2) = 1/4Therefore, the ratio MO : OA is 1:4.So regardless of the values of p and q (as long as the parallelogram is non-degenerate), the ratio is always 1:4. That makes sense because the parameter t gave us that ratio directly, and the coordinate calculation confirmed it. So the answer should be boxed{dfrac{1}{4}} but wait, the problem says "Find the ratio MO : OA". So if MO : OA is 1:4, then the ratio is 1/4, which is 1:4. So the box answer should be boxed{dfrac{1}{4}}.Alternatively, maybe the question expects the ratio as MO : OA = 1 : 4, so written with a colon. But the problem says "put your final answer within boxed{}", and typically ratios can be presented either as 1:4 or 1/4. Since the question uses the colon notation, maybe boxed{1:4} is appropriate. But in math problems, sometimes fractions are preferred. Wait, let me check the problem statement again. It says "Find the ratio MO : OA". So using a colon is appropriate here.But in the instructions, the user said "put your final answer within boxed{}". Depending on convention, sometimes boxed answers for ratios can be in colon form, but LaTeX requires some workaround. For example, boxed{1:4}. Alternatively, if they expect a fraction, perhaps representing the ratio as a fraction MO/OA which is 1/4. But since the problem specifies the ratio MO : OA, writing 1:4 is correct.But to confirm, let's see. In the parametric approach, since O divides AM such that AO is 4/5 of AM and MO is 1/5 of AM, so AO = 4/5, MO = 1/5, so MO : OA = 1/5 : 4/5 = 1:4. Correct. Therefore, the answer is 1:4, boxed as boxed{1:4}.But let me check another approach using mass point geometry to verify.In mass point geometry, we can assign weights to the vertices to find the ratio.Consider parallelogram ABCD. M is the midpoint of BC, N is the midpoint of CD.Lines AM and BN intersect at O. We need the ratio MO : OA.In mass point, the idea is to assign masses to the points such that the masses are inversely proportional to the lengths.But since ABCD is a parallelogram, AB is congruent to CD, and AD congruent to BC.Let me consider triangle ABC. Wait, maybe not. Alternatively, use coordinate geometry again but with vectors.Alternatively, area ratios.But perhaps mass point.Let’s assign masses to points A and B such that the masses balance at point O.But this might get complicated. Alternatively, since M and N are midpoints, the masses can be determined accordingly.Wait, mass point might not be straightforward here because it's a parallelogram, not a triangle.Alternatively, use vector approach.Let’s express vectors in terms of the parallelogram.Let’s let vector AB = vector DC = b, and vector AD = vector BC = d.Then, coordinates:A: origin (0,0)B: bC: b + dD: dM is the midpoint of BC: ( b + b + d ) / 2 = (2b + d)/2Wait, no. Wait, point B is b, point C is b + d, so midpoint M is ( b + b + d ) / 2 = (2b + d)/2. Similarly, point N is the midpoint of CD: point C is b + d, point D is d, so midpoint N is ( b + d + d ) / 2 = ( b + 2d ) / 2.Now, line AM is the line from A (0) to M ( (2b + d)/2 ).Parametric equation for AM: r = t*(2b + d)/2, where t ∈ [0,1].Line BN is the line from B (b) to N ( ( b + 2d ) / 2 ).Parametric equation for BN: r = b + s*( ( b + 2d ) / 2 - b ) = b + s*( -b/2 + d )= b - (sb)/2 + sdSet the two parametric equations equal:t*(2b + d)/2 = b - (sb)/2 + sdThis gives two equations by equating coefficients of b and d:For b component:t*(2/2) = 1 - s/2 => t = 1 - s/2For d component:t*(1/2) = sSo from the d component: t = 2sSubstitute into the b component equation:2s = 1 - s/2Multiply both sides by 2:4s = 2 - sBring s to the left:4s + s = 2 => 5s = 2 => s = 2/5Then, t = 2s = 4/5Therefore, point O divides AM such that t = 4/5, so OA is 4/5 of AM and MO is 1/5 of AM, so the ratio MO : OA is 1:4, same as before.Therefore, regardless of the approach—coordinate geometry, parametric, or vector—the ratio is consistently 1:4. Hence, the answer is 1:4.Final Answerboxed{1:4}

✅:To find the ratio ( MO : OA ) in parallelogram ( ABCD ) where ( M ) and ( N ) are the midpoints of sides ( BC ) and ( CD ) respectively, and segments ( AM ) and ( BN ) intersect at point ( O ), we can use coordinate geometry and parametric equations.1. Assign Coordinates: - Place ( A ) at the origin ((0,0)). - Let ( B ) be ((2,0)), ( D ) be ((p, q)), and ( C ) be ((2 + p, q)).2. Midpoints: - ( M ) is the midpoint of ( BC ): coordinates (left( frac{4 + p}{2}, frac{q}{2} right)). - ( N ) is the midpoint of ( CD ): coordinates ((1 + p, q)).3. Parametric Equations: - Line ( AM ): ( x = t cdot frac{4 + p}{2} ), ( y = t cdot frac{q}{2} ) where ( t in [0, 1] ). - Line ( BN ): ( x = 2 + s cdot (p - 1) ), ( y = s cdot q ) where ( s in [0, 1] ).4. Intersection Point ( O ): - Equate ( y )-coordinates: ( frac{tq}{2} = sq ) gives ( t = 2s ). - Substitute ( t = 2s ) into ( x )-coordinates: solve to get ( t = frac{4}{5} ) and ( s = frac{2}{5} ).5. Coordinates of ( O ): - From ( AM ): ( x = frac{8 + 2p}{5} ), ( y = frac{2q}{5} ).6. Ratio Calculation: - Using parameter ( t ), ( O ) is ( frac{4}{5} ) along ( AM ). - Distance ( OA ) corresponds to ( t = frac{4}{5} ). - Distance ( MO ) corresponds to ( t = 1 - frac{4}{5} = frac{1}{5} ).Thus, the ratio ( MO : OA ) is ( frac{1}{5} : frac{4}{5} = 1 : 4 ).[boxed{1:4}]

🔑:1. Identifying midpoint segments and their properties: Consider parallelogram (ABCD) with points (M) and (N) being the midpoints of sides (BC) and (CD), respectively. We need to find the ratio (MO: OA), where (O) is the intersection of segments (AM) and (BN).2. Extending segments to find intersection: Let's extend (BN) and (AD) so that they intersect at point (E).3. Establishing triangle similarity: Note that triangle (DNE) is congruent to triangle (CNB) because they have: - Equal sides (DE = BC) and - Equal angles, all resulting from corresponding angles of parallelogram (ABCD).4. Parallelogram properties: Since (M) and (N) are midpoints: [ BM = frac{BC}{2} quad text{and} quad DN = frac{CD}{2} ] Thus, the lengths of segments (BC) and (DE) are equal, and so: [ DE = BC = 2BM ]5. Using angle bisectors and midpoint theorem: The line segment (AE) is essentially four times the length of any one of these small bisecting segments since: [ AE = 4a ] where (a) is some unit segment length.6. Analyzing triangles (BOM) and (EOA): Triangles (BOM) and (EOA) are similar by AA similarity (they have two angles equal which are the vertically opposite angles at (O) and corresponding angles being equal).7. Using similarity to find the required ratio: Therefore, by the properties of similar triangles, the ratios of their corresponding sides are equal. We can write: [ frac{MO}{OA} = frac{BM}{EA} = frac{1}{4} ]8. Conclusion: Hence, the ratio (MO:OA) is: [ boxed{frac{MO}{OA} = 1:4} ]

🧠:Okay, so I need to prove that the lines AM and AN divide the diagonal BD into three equal parts in a parallelogram ABCD where M and N are midpoints of BC and CD respectively. Let me try to visualize this first. Parallelogram ABCD: let me recall that in a parallelogram, opposite sides are equal and parallel. So AB is parallel and equal to CD, and AD is parallel and equal to BC. Points M and N are midpoints of BC and CD, so BM = MC and CN = ND. The diagonal BD is connecting vertex B to vertex D. The problem states that lines AM and AN intersect BD, dividing it into three equal segments. So, the goal is to show that BD is trisected by these two lines.Hmm, maybe using coordinate geometry could help here. Let me assign coordinates to the vertices to make it easier. Let's place point A at the origin (0,0). Since it's a parallelogram, I can denote point B as (a,0), point D as (0,b), so then point C would be (a,b). Wait, because in a parallelogram, vector AB is (a,0) and vector AD is (0,b), so adding those vectors gives the coordinates of C as (a,b). Now, point M is the midpoint of BC. The coordinates of B are (a,0) and C are (a,b), so midpoint M would be ((a + a)/2, (0 + b)/2) = (a, b/2). Similarly, point N is the midpoint of CD. Coordinates of C are (a,b) and D are (0,b), so midpoint N is ((a + 0)/2, (b + b)/2) = (a/2, b).Now, lines AM and AN: let's find their equations. First, line AM connects point A (0,0) to M (a, b/2). The parametric equations for AM can be written as x = a*t, y = (b/2)*t, where t ranges from 0 to 1.Similarly, line AN connects point A (0,0) to N (a/2, b). The parametric equations for AN can be written as x = (a/2)*t, y = b*t, where t ranges from 0 to 1.Now, diagonal BD connects point B (a,0) to D (0,b). Let me parametrize BD. Starting at B (a,0) and moving towards D (0,b), so the parametric equations can be x = a - a*s, y = 0 + b*s, where s ranges from 0 to 1. Alternatively, using a parameter s from 0 to 1, the coordinates would be (a(1 - s), b*s).We need to find the points where lines AM and AN intersect BD. Then, we can check if these intersection points divide BD into three equal parts.First, find the intersection of AM and BD. For line AM: x = a*t, y = (b/2)*t.For line BD: x = a*(1 - s), y = b*s.Set them equal to find t and s where they intersect:a*t = a*(1 - s) --> t = 1 - s.And (b/2)*t = b*s --> (1/2)*t = s.Substituting t from the first equation into the second:(1/2)*(1 - s) = s --> (1 - s)/2 = s --> 1 - s = 2s --> 1 = 3s --> s = 1/3.Therefore, s = 1/3, so the intersection point is (a*(1 - 1/3), b*(1/3)) = ( (2a/3), (b/3) ).Similarly, now find the intersection of AN and BD.For line AN: x = (a/2)*t, y = b*t.For line BD: x = a*(1 - s), y = b*s.Set them equal:(a/2)*t = a*(1 - s) --> (t/2) = 1 - s --> t = 2*(1 - s).And b*t = b*s --> t = s.Therefore, substituting t = s into t = 2*(1 - s):s = 2*(1 - s) --> s = 2 - 2s --> 3s = 2 --> s = 2/3.Thus, the intersection point is (a*(1 - 2/3), b*(2/3)) = (a*(1/3), (2b/3)).So, the two intersection points on BD are at s = 1/3 and s = 2/3. Since BD is parameterized from s=0 to s=1, these points divide BD into three equal segments: from B to first intersection (1/3), then to second intersection (another 1/3), then to D (final 1/3). Therefore, the lines AM and AN trisect BD.Wait, but let me check this again. If BD is from (a,0) to (0,b), then the length between each segment should be equal. The first intersection is at (2a/3, b/3) and the second at (a/3, 2b/3). So the distance from B to (2a/3, b/3) can be calculated. The vector from B (a,0) to (2a/3, b/3) is (-a/3, b/3). Similarly, from (2a/3, b/3) to (a/3, 2b/3) is (-a/3, b/3), and from (a/3, 2b/3) to D (0,b) is (-a/3, b/3). Each segment has the same vector difference, so they are equal in length. Therefore, BD is indeed divided into three equal parts.Alternatively, since the parameter s is divided into 1/3 intervals, and the parametrization is linear, the distances should be equal. Therefore, this proves the statement.But let me try another method using vectors to confirm.Let’s denote vectors with position vectors. Let’s take point A as the origin. Then, vector AB is denoted as vector b, and vector AD as vector d. Therefore, coordinates:- A: 0- B: b- C: b + d- D: dMidpoints M and N:- M is midpoint of BC: ( b + (b + d) ) / 2 = (2b + d)/2 = b + d/2- N is midpoint of CD: ( (b + d) + d ) / 2 = (b + 2d)/2 = b/2 + dLines AM and AN:- Line AM: from 0 to b + d/2, so parametric equation: t(b + d/2), t ∈ [0,1]- Line AN: from 0 to b/2 + d, so parametric equation: s(b/2 + d), s ∈ [0,1]Diagonal BD: from b to d, parametric equation: b + u(d - b), u ∈ [0,1]Find intersection points between AM and BD, and AN and BD.First, intersection of AM and BD.Set t(b + d/2) = b + u(d - b)Which gives:tb + (t/2)d = b + ud - ubRearranged:(t + u - 1)b + (t/2 - u)d = 0Since b and d are not parallel (they form a parallelogram), their coefficients must be zero:t + u - 1 = 0t/2 - u = 0From the second equation: u = t/2Substitute into first equation:t + t/2 - 1 = 0 --> (3t)/2 = 1 --> t = 2/3Then u = (2/3)/2 = 1/3So the intersection point on BD is at u = 1/3.Similarly, intersection of AN and BD.Set s(b/2 + d) = b + u(d - b)Which gives:(s/2)b + sd = b + ud - ubRearranged:(s/2 + u - 1)b + (s - u)d = 0Again, coefficients must be zero:s/2 + u - 1 = 0s - u = 0From the second equation: u = sSubstitute into first equation:s/2 + s - 1 = 0 --> (3s)/2 = 1 --> s = 2/3Thus, u = 2/3Therefore, the intersection point on BD is at u = 2/3.Hence, BD is divided at u = 1/3 and u = 2/3, which are equally spaced. So BD is trisected by AM and AN.Alternatively, using mass point geometry or similar triangles. Let me see if that approach works.Looking at the parallelogram, since M and N are midpoints, perhaps drawing the lines AM and AN creates triangles with certain properties. Let's consider triangles ABM and ADN.But maybe using coordinate geometry was straightforward. Alternatively, using vectors as above also confirms the result.Another approach: area ratios. If we can show that the areas created by these lines relate to the segments being divided equally. But maybe that's more complex.Alternatively, using affine transformations. Since parallelograms can be transformed into rectangles without changing the ratios, perhaps simplifying the problem. Let's imagine transforming ABCD into a rectangle with the same side lengths. The midpoints would remain midpoints, and lines AM and AN would still trisect BD. But this might not add much beyond the coordinate approach.Wait, but let's think in terms of similar triangles. For example, in triangle ABD or something. Wait, BD is a diagonal. When AM intersects BD at point P, and AN intersects BD at point Q, then BP = PQ = QD.To use similar triangles, perhaps look at the triangles formed by the intersections.Looking at line AM: starting at A, going to M (midpoint of BC). When AM intersects BD at P, maybe triangle ABP is similar to some other triangle?Alternatively, coordinates were quite effective here. The key was parametrizing the lines and solving for the parameters where they intersect BD. Both methods, coordinate and vector, gave the same result. So, this seems solid.Another way: Menelaus' theorem. Let me try applying Menelaus' theorem to triangle BDC with transversal AM.Wait, Menelaus' theorem states that for a triangle, if a line crosses the three sides (or their extensions), the product of the segment ratios is -1. Hmm.But maybe overcomplicating. Alternatively, coordinate geometry is straightforward here.Alternatively, since M and N are midpoints, the lines AM and AN might be medians in some triangles. For example, in triangle ABC, M is the midpoint of BC, so AM is a median. Similarly, in triangle ACD, N is the midpoint of CD, so AN is a median. But BD is a diagonal of the parallelogram. How do these medians interact with BD?But perhaps this is not directly helpful. Alternatively, consider that in a parallelogram, the diagonals bisect each other. So the midpoint of BD is the intersection point of the diagonals AC and BD. But here, we're dealing with lines AM and AN, not the diagonals.Wait, but in the problem, the diagonal BD is being trisected by AM and AN. So from point B to D, there are two points that divide BD into three equal parts. The previous calculations showed that these points are at 1/3 and 2/3 along BD from B.Alternatively, using vectors, the position vectors of the intersection points can be expressed as fractions of BD.Given that BD goes from B to D, the vector BD is D - B = d - b (from earlier notation). The points dividing BD into three equal parts would be at B + (1/3)(D - B) and B + (2/3)(D - B). Calculating these:First trisection point: b + (1/3)(d - b) = (2/3)b + (1/3)dSecond trisection point: b + (2/3)(d - b) = (1/3)b + (2/3)dFrom the earlier calculations using coordinates, the intersection points were (2a/3, b/3) and (a/3, 2b/3), which correspond to these vector expressions. For example, in coordinate terms, if b is (a,0) and d is (0,b), then (2/3)b + (1/3)d = (2a/3, 0) + (0, b/3) = (2a/3, b/3), which matches. Similarly, (1/3)b + (2/3)d = (a/3, 0) + (0, 2b/3) = (a/3, 2b/3), which also matches. Therefore, the intersection points are indeed the trisection points of BD.Therefore, this confirms that lines AM and AN trisect diagonal BD.I think this covers multiple approaches confirming the same result. The key idea is parameterizing the lines, finding their intersection points with the diagonal, and showing that these points divide the diagonal into three equal segments. Both coordinate geometry and vector methods lead to the same conclusion, reinforcing the validity of the proof.Final Answerboxed{BD text{ is divided into three equal parts by } AM text{ and } AN.}

✅:To prove that lines (AM) and (AN) divide the diagonal (BD) of parallelogram (ABCD) into three equal parts, we can use coordinate geometry and vector analysis.1. Assign Coordinates: - Place (A) at the origin ((0,0)). - Let (B) be ((a,0)), (D) be ((0,b)), and (C) be ((a,b)). - Midpoints (M) and (N) have coordinates (M(a, frac{b}{2})) and (N(frac{a}{2}, b)).2. Equations of Lines: - Line (AM): Parametric equations (x = at), (y = frac{b}{2}t) for (t in [0,1]). - Line (AN): Parametric equations (x = frac{a}{2}t), (y = bt) for (t in [0,1]). - Diagonal (BD): Parametric equations (x = a(1 - s)), (y = bs) for (s in [0,1]).3. Intersection of (AM) and (BD): - Set (at = a(1 - s)) and (frac{b}{2}t = bs). - Solving gives (t = 1 - s) and (frac{1}{2}t = s). - Substituting (t = 1 - s) into (frac{1}{2}t = s) yields (s = frac{1}{3}). - Intersection point: ((a cdot frac{2}{3}, frac{b}{3})).4. Intersection of (AN) and (BD): - Set (frac{a}{2}t = a(1 - s)) and (bt = bs). - Solving gives (t = 2(1 - s)) and (t = s). - Substituting (t = s) into (t = 2(1 - s)) yields (s = frac{2}{3}). - Intersection point: ((a cdot frac{1}{3}, frac{2b}{3})).5. Verification: - The intersection points divide (BD) at (s = frac{1}{3}) and (s = frac{2}{3}), which are equally spaced. - Using vector analysis, the points correspond to (frac{1}{3}) and (frac{2}{3}) along (BD), confirming trisection.Thus, the lines (AM) and (AN) trisect the diagonal (BD).[boxed{BD text{ is divided into three equal parts by } AM text{ and } AN.}]

🔑:Given points ( M ) and ( N ) are the midpoints of sides ( BC ) and ( CD ) respectively in the parallelogram ( ABCD ). We need to show that the lines ( AM ) and ( AN ) divide the diagonal ( BD ) into three equal parts.1. Identify Key Points and Properties: - ( M ) and ( N ) are midpoints. - Let ( P ) and ( Q ) be the points of intersection of diagonal ( BD ) with segments ( AM ) and ( AN ) respectively. - Let ( O ) be the intersection point of the diagonals ( AC ) and ( BD ). This is the center of the parallelogram ( ABCD ).2. Use Medians of Triangles: - ( AM ) and ( BO ) are medians of triangle ( ABC ). - Similarly, ( AN ) is a median of triangle ( ACD ).3. Applying the Centroid Theorem: - The centroid theorem states that the centroid, which is the intersection point of medians in a triangle, divides each median into segments with a ratio of ( 2:1 ). - Thus, for triangle ( ABC ), point ( P ) (where ( AM ) intersects ( BD )) divides ( BD ) such that ( BP = frac{2}{3} BO ). - Similarly, point ( Q ), being on the median ( AN ) of triangle ( ACD ), will divide ( BD ) such that ( DQ = frac{2}{3} DO ).4. Calculation of Segment Lengths: - Since ( O ) is the midpoint of ( BD ), we have: [ BO = OD = frac{1}{2} BD ] - Therefore, point ( P ) being ( frac{2}{3} BO ) from ( B ): [ BP = frac{2}{3} times frac{1}{2} BD = frac{1}{3} BD ] - Similarly, point ( Q ) being ( frac{2}{3} DO ) from ( D ): [ DQ = frac{2}{3} times frac{1}{2} BD = frac{1}{3} BD ]5. Conclusion: - It is clear that points ( P ) and ( Q ) divide ( BD ) into three equal parts, each being ( frac{1}{3} BD ). - Therefore, the lines ( AM ) and ( AN ) indeed divide the diagonal ( BD ) into three equal parts.[boxed{AM text{ and } AN text{ divide } BD text{ into three equal parts}}]

🧠:Okay, let's see. Kristina thought of a natural number, and her four friends asked questions about divisibility by 17, 19, whether it's less than 20, and divisibility by 323. She answered "yes" to exactly two of these. I need to figure out which of the options (17, 19, 34, 38, 323, 340, 400) could be the number. First, let me list the questions again:1. Julia: Divisible by 17?2. Nastya: Divisible by 19?3. Vika: Less than 20?4. Dasha: Divisible by 323?Kristina said "yes" to exactly two of these. So each number needs to satisfy exactly two "yes" answers. Let's go through each option one by one.Starting with the first option: 17.Check each question for 17:1. Divisible by 17? Yes, because 17 ÷ 17 = 1.2. Divisible by 19? No, since 17 is less than 19 and not a multiple.3. Less than 20? Yes, 17 < 20.4. Divisible by 323? 323 is 17*19, so 323 ÷17=19. So 17 divides 323, but 17 itself is not divisible by 323. Wait, the question is "Is it divisible by 323?" So for the number 17, 17 ÷ 323 is 0.052..., which is not an integer. So answer is No.So for 17: two "yes" answers (questions 1 and 3). That fits the condition. So 17 is a possible number.Next option: 19.Check the same questions:1. Divisible by 17? 19 ÷17 is 1.117... No.2. Divisible by 19? Yes, 19 ÷19=1.3. Less than 20? Yes, 19 <20.4. Divisible by 323? 19 ÷323 is ~0.0588, so No.So for 19: two "yes" answers (questions 2 and 3). That's also two. So 19 is possible.Third option: 34.Check each:1. Divisible by 17? 34 ÷17=2, yes.2. Divisible by 19? 34 ÷19≈1.789, no.3. Less than 20? 34 is greater than 20, so no.4. Divisible by 323? 34 ÷323≈0.105, no.So "yes" answers: questions 1. Only one "yes", which is less than two. So 34 doesn't fit. So not possible.Fourth option: 38.Check:1. Divisible by 17? 38 ÷17≈2.235, no.2. Divisible by 19? 38 ÷19=2, yes.3. Less than 20? 38 >20, so no.4. Divisible by 323? 38 ÷323≈0.117, no.So "yes" answers: question 2. Only one. So 38 is out.Fifth option: 323.Check each:1. Divisible by 17? 323 ÷17=19, so yes.2. Divisible by 19? 323 ÷19=17, yes.3. Less than 20? 323 is way bigger, so no.4. Divisible by 323? Yes, any number is divisible by itself.So "yes" answers: questions 1, 2, 4. That's three "yes" answers. But Kristina only answered two. So 323 is not possible.Sixth option: 340.Check:1. Divisible by 17? 340 ÷17=20, yes.2. Divisible by 19? 340 ÷19≈17.89, no.3. Less than 20? 340 >20, so no.4. Divisible by 323? 340 ÷323≈1.052, which is not an integer. So no.So "yes" answers: question 1. Only one. So 340 is not possible.Last option: 400.Check:1. Divisible by 17? 400 ÷17≈23.529, no.2. Divisible by 19? 400 ÷19≈21.052, no.3. Less than 20? 400 >20, so no.4. Divisible by 323? 400 ÷323≈1.238, no.All answers are "no". That's zero "yes", which doesn't meet the two required. So 400 is out.Wait, but wait, let me double-check some of these, especially the divisibility by 323. For example, 323 is 17*19. So any number divisible by both 17 and 19 would be divisible by 323. So if a number is divisible by 323, it must answer "yes" to both Julia and Nastya's questions. So Dasha's question (divisible by 323) would be "yes" only if both Julia and Nastya's are "yes". Conversely, if a number is divisible by both 17 and 19, then Dasha's answer is yes, but Julia and Nastya's would also be yes, so that would be three "yes" answers. Which is why 323 had three "yes" answers.But let's check another angle. For example, if a number is 17*19=323, then yes to Julia, Nastya, and Dasha. So three yeses. So if a number is 17*2=34, then yes to Julia, no to others except maybe Vika. Wait, 34: yes to Julia (divisible by 17), no to Nastya (19), no to Vika (34 <20? No), and no to Dasha (323). So only one yes. So correct.Similarly, 19: yes to Nastya and Vika. 17: yes to Julia and Vika. So those two are good.Another possible number could be a number that is less than 20 and not divisible by 17 or 19, but then Vika's answer is yes, and the others are no. That's one yes. So not sufficient.Alternatively, a number divisible by 17 and 323. Wait, but if it's divisible by 323, it must also be divisible by 19, since 323=17*19. So Dasha's yes implies Julia and Nastya's yes. Hence, Dasha's answer adds a third yes if the number is divisible by 323. So the only way for Dasha's answer to be yes is if both Julia and Nastya are yes, leading to three yeses. Hence, numbers divisible by 323 can't be answers because they would result in three yeses, which is more than allowed.Therefore, the only numbers that can have exactly two yeses are those that either:- Are divisible by 17 and less than 20 (like 17), which gives two yeses.- Are divisible by 19 and less than 20 (like 19), which also two yeses.- Or, perhaps, another combination where two answers are yes without overlapping.Wait, but what if a number is divisible by 17 and 19, but not by 323? That's impossible because 17 and 19 are primes, so their product is 323, so any number divisible by both must be divisible by 323. Hence, if a number is divisible by both 17 and 19, it must be divisible by 323, making three yeses.Another possibility: a number that's divisible by 17 and is less than 20. That's 17. Similarly, divisible by 19 and less than 20: 19. Are there other numbers less than 20 that are divisible by 17 or 19?17 and 19 are the only ones. Because 17*2=34 which is over 20, same with 19*2=38. So the only numbers less than 20 divisible by 17 or 19 are 17 and 19 themselves. Hence, those are the only two numbers that can answer yes to two questions.But wait, let's check numbers not divisible by 17 or 19 but still answer yes to two questions. For example, if a number is less than 20 and divisible by something else. But the questions are about divisibility by 17, 19, 323, and the less than 20. So if a number is say 18, which is less than 20, and not divisible by 17,19, or 323. Then it would answer yes only to Vika's question. So only one yes. Not enough.Alternatively, a number that is not less than 20, but is divisible by two of the other divisors. But if it's not less than 20, then Vika's answer is no. So possible yes answers would have to come from the other three questions: Julia, Nastya, Dasha. But if a number is divisible by both 17 and 19, then Dasha's answer is yes, making three yeses. If it's divisible by 17 and not 19, then only Julia's yes. If divisible by 19 and not 17, only Nastya's yes. If divisible by 323, which is 17*19, then three yeses. So the only way to get two yeses is if the number is divisible by either 17 or 19 and also less than 20. Hence, only 17 and 19 fit.Thus, the possible numbers are 17 and 19.

✅:Kristina's number must satisfy exactly two "yes" answers out of the four questions. Analyzing each option:1. 17 - Divisible by 17: Yes - Divisible by 19: No - Less than 20: Yes - Divisible by 323: No Result: 2 "yes" answers. Valid.2. 19 - Divisible by 17: No - Divisible by 19: Yes - Less than 20: Yes - Divisible by 323: No Result: 2 "yes" answers. Valid.3. 34 - Divisible by 17: Yes - Divisible by 19: No - Less than 20: No - Divisible by 323: No Result: 1 "yes" answer. Invalid.4. 38 - Divisible by 17: No - Divisible by 19: Yes - Less than 20: No - Divisible by 323: No Result: 1 "yes" answer. Invalid.5. 323 - Divisible by 17: Yes - Divisible by 19: Yes - Less than 20: No - Divisible by 323: Yes Result: 3 "yes" answers. Invalid.6. 340 - Divisible by 17: Yes - Divisible by 19: No - Less than 20: No - Divisible by 323: No Result: 1 "yes" answer. Invalid.7. 400 - All answers: No Result: 0 "yes" answers. Invalid.Final Answer: ○ 17 ○ 19

🔑:To solve this problem, we need to first identify all the conditions and then find the natural numbers that meet exactly two of these conditions. Let's analyze each condition step by step:1. Condition 1 (Yulia's question): The number is divisible by 172. Condition 2 (Nastya's question): The number is divisible by 193. Condition 3 (Vika's question): The number is less than 204. Condition 4 (Dasha's question): The number is divisible by 323# Analysis:- A number is divisible by 17 if it can be written in the form ( 17k ), where ( k ) is an integer.- A number is divisible by 19 if it can be written in the form ( 19m ), where ( m ) is an integer.- A number less than 20 is in the set ( {1, 2, 3, ..., 19} ).- A number is divisible by 323 if it can be written in the form ( 323n ), where ( n ) is an integer. Since ( 323 = 17 times 19 ), any number divisible by 323 is also divisible by both 17 and 19.# Step-by-Step Solution:1. Numbers satisfying two conditions: - As Dasha's question (divisible by 323) implies divisibility by both 17 and 19, any number satisfying her condition will satisfy both Yulia's and Nastya's conditions. Therefore, Dasha's condition alone cannot help us find a solution that satisfies exactly two conditions since it inherently satisfies three. - Vika's question (less than 20) includes very few numbers. We need to check which of these can be divisible by 17 or 19.2. List numbers less than 20 and check divisibility: - Numbers possible under Vika's condition are ( {1, 2, 3, ..., 19} ). - Check for divisibility by 17 and 19: [ begin{aligned} &17: quad 17 div 17 = 1 quad text{(Yes)} &19: quad 19 div 19 = 1 quad text{(Yes)} end{aligned} ]So, the numbers less than 20 that are also divisible by 17 or 19 are 17 and 19.3. Conclusion based on stated responses: - Kristina would have agreed only to one of the questions regarding being less than 20 (since either 17 or 19 is less than 20). - She would also agree to the question regarding divisibility by either 17 or 19.Thus, we identify the natural numbers she might have thought of:[boxed{17} quad text{or} quad boxed{19}]Conclusively, the numbers Kristina could have envisioned are 17 and 19 as they satisfy exactly two of the given conditions.

🧠:Okay, so I have this problem here: There are 99 positive integers, each less than 100, and they can be equal. The condition is that the sum of any 2, 3, or more of these numbers cannot be divided evenly by 100. I need to prove that all the numbers are equal. Hmm, interesting. Let me try to think through this step by step.First, let me rephrase the problem to make sure I understand it. We have 99 numbers, each between 1 and 99 (since they are positive integers less than 100). The key condition is that if you take any subset of these numbers with size 2 or more, their sum is not divisible by 100. The conclusion we need to reach is that all 99 numbers are the same. Alright, so the problem is about modular arithmetic, specifically modulo 100. The sums of subsets of size ≥2 not being divisible by 100 means that for any such subset, the sum mod 100 is not 0. So, every subset sum mod 100 is in {1, 2, ..., 99}.Let me think about what this implies. If all numbers are equal, say each number is 'a', then any subset sum of size k would be k*a. The condition would require that k*a mod 100 ≠ 0 for all k from 2 to 99. If all numbers are equal, then to satisfy this condition, we must have that a is such that k*a ≡ 0 mod 100 is impossible for k = 2, 3, ..., 99. Wait, so if all numbers are equal, then the problem reduces to choosing a single number 'a' such that multiplying it by any k from 2 to 99 doesn't give a multiple of 100. Then, if such an 'a' exists, and all numbers are equal to 'a', the condition is satisfied. So, the problem is stating that if you have 99 numbers (each less than 100) where no subset sum of size ≥2 is divisible by 100, then those numbers must all be equal. Therefore, the converse would be: if they are not all equal, then there exists some subset of size ≥2 whose sum is divisible by 100. So, the strategy here might be to assume that not all numbers are equal and then show that this leads to the existence of such a subset. Alternatively, maybe we can use the pigeonhole principle or some combinatorial argument given that there are 99 numbers, each less than 100. Let me try to explore possible directions.First, note that all numbers are less than 100, so each number is between 1 and 99. Suppose all numbers are equal. Let's say all are 'a'. Then, for k*a not to be 0 mod 100 for k = 2 to 99. So, we need that a and 100 are coprime? Wait, but if a and 100 are coprime, then multiplying by k would give 0 mod 100 only if k is a multiple of 100. But since k is from 2 to 99, that would never happen. Wait, but 100 is 2^2 * 5^2. So, if a is coprime to 100, which means a is not divisible by 2 or 5, then k*a mod 100 can only be 0 if k is divisible by 100. Since k is less than 100, that can't happen. Therefore, if all numbers are equal to a number coprime to 100, then indeed all subset sums of size ≥2 would not be divisible by 100. However, if a is not coprime to 100, say a is even or a multiple of 5, then there might be some k where k*a is divisible by 100. For example, if a is 50, then 2*a = 100, which is 0 mod 100. So, that would violate the condition. Therefore, if all numbers are equal, they must be equal to a number coprime to 100. Therefore, such numbers would be in {1, 3, 7, 9, 11, 13, 17, 19, 21, 23, 27, 29, ..., 99} (the numbers coprime to 100). But the problem doesn't state that such numbers exist; rather, it says that given the numbers satisfy the subset sum condition, prove they are all equal. So, perhaps we need to show that the only way to avoid having any subset sum divisible by 100 is to have all numbers equal and each number coprime to 100. Wait, but the problem says "each less than 100, and possibly equal". So, maybe if they are not all equal, even if they are all coprime to 100, there could be a subset sum divisible by 100. Therefore, the equality is necessary.Alternatively, maybe if the numbers are all equal and coprime to 100, then no subset sum is divisible by 100, and conversely, if they are not all equal, then regardless of their values (even if they are coprime to 100 individually), there will be a subset sum divisible by 100.So, the problem reduces to showing that in any collection of 99 numbers (each <100), if they are not all equal, then there exists a subset of size ≥2 whose sum is divisible by 100. This seems similar to the pigeonhole principle. Let me recall that in the pigeonhole principle, if you have more pigeons than pigeonholes, at least one pigeonhole must contain more than one pigeon. Maybe here, the large number of elements (99) compared to modulus 100 allows us to apply some form of pigeonhole. Alternatively, considering the Erdős–Ginzburg–Ziv theorem, which states that any 2n-1 integers have a subset of n integers whose sum is divisible by n. Wait, in our case, if we take n=100, but the numbers are less than 100. Wait, the Erdős–Ginzburg–Ziv theorem says that any 2n-1 integers have a subset of size n with sum divisible by n. But here, n would be 100, but we have 99 numbers, which is 2*50 -1. Wait, maybe not directly applicable. Wait, actually, the theorem is for any 2n-1 integers, there exists a subset of size n with sum divisible by n. So if n=50, then 99 numbers would have a subset of size 50 divisible by 50. But our modulus is 100. Hmm. Maybe not directly applicable here.Alternatively, consider modular residues. If all numbers are congruent modulo 100 to some fixed residue 'a', then any subset sum would be k*a mod 100. So, as we discussed earlier, if a is coprime to 100, then k*a mod 100 can't be 0 for k=1,...,99. But if the numbers are not all congruent modulo 100, then perhaps there exist two numbers whose sum is 0 mod 100. Wait, but even if they are not all the same, maybe you can get a subset sum. Alternatively, let's think about the residues of the numbers modulo 100. Since each number is less than 100, their residues modulo 100 are themselves. The problem states that any subset of size ≥2 has a sum not divisible by 100, which is equivalent to saying that the sum modulo 100 is not zero.Suppose that all numbers are equal to 'a' modulo 100. Then, as before, any subset sum of size k would be k*a mod 100. So, for this to never be zero for k ≥2, we must have that a is coprime to 100. Because if a shares a common factor with 100, say d>1, then choosing k=100/d would make k*a divisible by 100. But since k must be at least 2 and at most 99, we need that 100/d >99, so d=1. Therefore, a must be coprime to 100, which would ensure that k*a is not divisible by 100 for any k=1,...,99. Wait, but 100 is 2^2*5^2, so a must not be divisible by 2 or 5. Therefore, a must be coprime to 100.So, if all numbers are equal to such an 'a', then indeed, the condition holds. Now, the problem is to show that this is the only possibility, i.e., if the numbers are not all equal, then there exists a subset with sum divisible by 100.So, how to approach this? Let's assume for contradiction that not all numbers are equal. Then, there exist at least two distinct numbers. Let's say there are two numbers, a and b, with a ≠ b. Then, perhaps consider sums involving a and b. But the problem allows for up to 99 numbers, so maybe the presence of multiple numbers allows for combining them in such a way that their sum is 0 mod 100.Alternatively, consider the set of numbers as residues modulo 100. Then, we need to ensure that no two residues sum to 100, no three residues sum to 100, etc. But with 99 numbers, which is one less than 100, perhaps the pigeonhole principle applies here. Let me think.If we have 99 residues modulo 100, then by the pigeonhole principle, there must be at least one residue that is repeated. Wait, but residues can be from 1 to 99. Wait, if all residues are distinct, then there are 99 distinct residues. But since there are 99 numbers, maybe they are all distinct? But the problem says "possibly equal", so they can be equal. If they are all distinct, that would mean residues 1 to 99. Then, for example, 1 and 99 would sum to 100, which is 0 mod 100. Therefore, if residues are all distinct, then there would be pairs that sum to 100, violating the condition. Therefore, if the numbers are all distinct, the condition is violated. Therefore, the numbers cannot all be distinct.But the problem states that the numbers are "possibly equal". So, in the case where all numbers are equal, you can avoid this. But if there is any repetition, or if some numbers are different, then maybe you can find such subsets.Wait, but even if numbers are not all distinct, if there are two numbers that are different, maybe their sum is 100. For example, if there is a 1 and a 99, their sum is 100. Similarly, 2 and 98, etc. Therefore, if the set contains complementary pairs, the sum is 100, which is 0 mod 100. So, in such a case, the condition is violated. Therefore, to satisfy the condition, the set must not contain any such complementary pairs. So, if you have a number 'a', you cannot have '100 - a' in the set.But wait, even beyond pairs, maybe the sum of three numbers could be 100. For example, if you have 30, 30, 40. Then 30 + 30 + 40 = 100. So, even if there are no complementary pairs, a trio might sum to 100.Therefore, the condition is much stricter: no subset of size ≥2 sums to 100. So, the set must be structured in such a way that no combination of elements can add up to 100. How can such a set look?One way is if all elements are the same and equal to a number 'a' that is coprime to 100. Then, as discussed, any subset sum is k*a, and since a is coprime to 100, k*a mod 100 can't be zero for k=1,...,99. But if elements are different, even if each element is coprime to 100, maybe their sum could be 100. For example, 1 and 99: both coprime to 100, but their sum is 100. So, even if elements are coprime to 100, if they are different, their sum could still be 100.Therefore, perhaps the only way to avoid having any subset sum divisible by 100 is to have all elements equal and coprime to 100. Therefore, proving that if all elements are equal and coprime to 100, the condition holds, and conversely, if the elements are not all equal, then regardless of their residues, there must exist a subset sum divisible by 100. Hence, the problem statement.So, how can we formally prove that if there are 99 numbers less than 100, not all equal, then there exists a subset of size ≥2 with sum divisible by 100?Maybe we can use the pigeonhole principle on the residues. Let me consider the partial sums modulo 100. If we have a sequence of numbers, then the partial sums modulo 100 can be considered. If any partial sum is 0 modulo 100, then that subset would be the answer. However, the problem allows any subset, not just consecutive elements. So, the standard pigeonhole principle for subset sums might be applicable.The classic subset sum pigeonhole principle says that given n numbers, there exists a non-empty subset whose sum is divisible by n. But here, our modulus is 100, and we have 99 numbers. So, perhaps using the pigeonhole principle on the subset sums modulo 100. Since there are 2^99 - 1 non-empty subsets, but modulus 100 has only 100 residues. However, 2^99 is way larger than 100, so by pigeonhole, there are many subsets with the same residue. However, this line of thought might not directly lead to the conclusion, because having many subsets with the same residue doesn't immediately give a subset summing to 0. However, if we can find two subsets with the same residue, their difference would give a subset sum of 0. But since we are dealing with numbers less than 100, the difference would correspond to the symmetric difference of the subsets, and the sum would be the difference of the two subset sums. However, this requires that the two subsets are distinct and their difference is non-empty. This is the idea behind the Erdős–Ginzburg–Ziv theorem, actually.Wait, the Erdős–Ginzburg–Ziv theorem states that for any 2n-1 integers, there exists a subset of n integers whose sum is divisible by n. In our case, if we take n=100, then 2*100-1=199 integers would guarantee a subset of 100 with sum divisible by 100. But here, we have 99 integers, which is less than 199. So, maybe that theorem isn't directly applicable here.Alternatively, perhaps use induction. Suppose that for a smaller number of elements, certain properties hold, but with 99 elements, maybe we can find a subset. However, I need to think of a different approach.Let me think again. Suppose that all numbers are equal. Then, as discussed, they must be equal to a number coprime to 100, so that k*a mod 100 ≠ 0 for k=2,...,99. Now, if the numbers are not all equal, then there must be at least two distinct numbers. Let's say there are two numbers, a and b, with a ≠ b. Then, consider sums involving a and b. If a + b ≡ 0 mod 100, then we're done, as that pair would sum to 100. If not, maybe consider adding more numbers to reach a sum of 100. However, with 99 numbers, it's possible that you can combine them in various ways.Alternatively, consider the residues of the numbers. If there exists a number a, then 100 - a must not be present in the set. Because otherwise, a + (100 - a) = 100. So, for the set to not have any pairs summing to 100, it must not contain both a and 100 - a for any a. Similarly, for triples, etc. So, maybe the set is missing all the complementary residues. But since we have 99 numbers, and there are 50 complementary pairs (1 and 99, 2 and 98, ..., 49 and 51) and 50 alone (since 100 - 50 = 50). Wait, 100 is even, so 50 is its own complement. So, residues can be paired as (1,99), (2,98), ..., (49,51), and 50. So, total 49 pairs and one singleton.If the set contains 99 numbers, each less than 100, and we cannot have both a number and its complement, then the maximum size of such a set would be 50: choosing one from each complementary pair (1 to 49 and 99 to 51) plus 50. But 50 elements. However, our set has 99 numbers, which is almost double that. So, if the set cannot contain both a number and its complement, but since we have 99 numbers, which is more than 50, then by the pigeonhole principle, we must have at least one complementary pair. Therefore, this would imply that if we have more than 50 numbers, we must have a pair summing to 100. Therefore, since our set has 99 numbers, which is way more than 50, we must have multiple complementary pairs. Wait, but that seems contradictory because 99 is more than 50, but if each pair can contribute at most one element to the set, then the maximum size without any complementary pairs is 50. Therefore, if we have a set of 99 numbers, each less than 100, then it must contain at least 99 - 50 = 49 complementary pairs. Wait, maybe this is the way to go.Wait, more precisely, the maximum size of a set without any two elements summing to 100 is 50. Because there are 50 pairs: (1,99), (2,98), ..., (49,51), and 50. So, you can choose at most one from each pair, giving 49 elements plus 50, totaling 50 elements. Therefore, any set with 51 or more elements must contain at least one pair summing to 100. But in our problem, we have 99 elements, which is much larger than 51, so there must be many pairs summing to 100. Therefore, the conclusion would be that in such a set, there must exist at least one pair summing to 100, which contradicts the problem's condition. Therefore, the only way to avoid this is to have all elements equal. Wait, but if all elements are equal, say to 50, then 50 + 50 = 100, which would be a problem. Wait, but 50 is not coprime to 100. So, if all elements are equal to 50, then any two elements sum to 100, which is bad. Therefore, such a set would not satisfy the condition. Therefore, the only way to have a set of 99 numbers with no subset sum divisible by 100 is to have all elements equal to a number coprime to 100.Wait, but if all elements are equal and coprime to 100, then as we saw earlier, any subset sum k*a will not be 0 mod 100 for k=1,...,99. So, for example, if a=1, then k*1=1,2,...,99 mod 100. If a=3, then k*3 cycles through different residues. Since a is coprime to 100, 3 is coprime to 100, so the multiplicative inverse exists. Therefore, k*a mod 100 will cycle through different values and never hit 0 for k=1,...,99. Because 100 divides k*a only if 100 divides k, but k is less than 100. Therefore, if all elements are equal to a number coprime to 100, then the condition is satisfied. But how does this reconcile with the previous thought that a set of 99 elements must contain a pair summing to 100? Because if all elements are equal to a number a coprime to 100, then the sum of any two elements is 2a mod 100. Since a is coprime to 100, 2a is not 0 mod 100. Because if 2a ≡0 mod 100, then a ≡0 mod 50, which would mean a is 50, but 50 is not coprime to 100. Therefore, 2a mod 100 is not zero. Similarly, for three elements, 3a mod 100, and so on. Therefore, if all elements are equal to a coprime to 100, then all subset sums are safe.But if the elements are not all equal, even if they are all coprime to 100, then we might have two elements a and b such that a + b ≡0 mod 100. For example, 1 and 99. Therefore, the only way to avoid having such pairs is to have all elements equal. Because if you have two different elements, even if both are coprime to 100, they might add up to 100. Therefore, the conclusion is that all elements must be equal, and equal to a number coprime to 100. But the problem states "prove that all the numbers are equal". It doesn't specify that they must be equal to a number coprime to 100. However, from our reasoning, if they are equal but not coprime to 100, then certain subset sums would be divisible by 100. For example, if they are all 50, then 2*50=100. So, in that case, the condition is violated. Therefore, the problem must implicitly require that the numbers are equal and coprime to 100. But the problem statement just says "each less than 100, and possibly equal". So, the conclusion is that they must all be equal, and the fact that their subset sums are not divisible by 100 requires them to be coprime to 100, but the problem only asks to prove they are equal. Therefore, the key is that if they are not all equal, then there exists a subset sum divisible by 100. Therefore, the contrapositive is: if all subset sums of size ≥2 are not divisible by 100, then the numbers must be equal.Therefore, the main argument is that if there are two distinct numbers, then either their sum is 100, or we can form other subsets whose sums are 100. Therefore, the only way to prevent this is to have all numbers equal. But how to formalize this? Let's suppose that there are at least two distinct numbers, say a and b. If a + b ≡0 mod 100, then we are done. If not, perhaps consider the other numbers. But with 99 numbers, there might be many duplicates. Wait, but if all numbers except one are the same, for example, 98 copies of a and one copy of b. Then, we can consider sums involving different numbers of a and b. For example, 2a, 3a, ..., 98a, and then combinations with b. If a is coprime to 100, then k*a mod 100 ≠0 for any k=1,...,98. However, adding b into the mix, if b ≡ -ma mod 100 for some m, then (m+1)a + b ≡0 mod 100. Therefore, if b ≡ -ma mod 100 for some m between 1 and 99, then we can form a subset sum of m+1 a's and 1 b, but since we have 98 a's, m can be up to 98, so m+1 would be 99. Therefore, if b ≡ -98a mod 100, then 99a + b ≡99a -98a =a mod 100, which is not 0. Wait, maybe this approach isn't straightforward.Alternatively, if we have multiple distinct numbers, we can use the pigeonhole principle on their residues. For example, if there are two numbers a and b, then we can look at the set of all numbers. Since there are 99 numbers, maybe we can find a combination that cancels out. Alternatively, consider the partial sums. If we have a sequence of numbers, the partial sums modulo 100 must repeat somewhere, leading to a subset sum divisible by 100. But this is for consecutive subsets. However, the problem allows any subset.Another angle: consider the multiplicative inverses. If the numbers are all equal to a, and a is coprime to 100, then a has an inverse modulo 100. Therefore, multiplying the subset sum equation k*a ≡0 mod 100 by the inverse of a gives k ≡0 mod 100, which is impossible since k <100. Hence, this case is safe. But if numbers are different, even if they are all coprime to 100, we can't have such a guarantee.But let's step back. The problem has 99 numbers, each less than 100. If they are not all equal, there exist at least two different numbers. Let's say we have two different numbers, x and y. If x + y ≡0 mod 100, then we're done. If not, maybe consider another number z. If x + y + z ≡0 mod 100, then we're done. Otherwise, keep adding numbers until we hit a multiple of 100. But with 99 numbers, this seems likely. However, I need a more rigorous approach.Wait, another thought. Since we have 99 numbers, consider their residues modulo 100. Let's look at the cumulative sums. If we take the numbers in any order and compute the cumulative sum modulo 100, there are 99 cumulative sums. If any of these sums is 0 modulo 100, then the subset up to that point is a solution. If not, then we have 99 sums modulo 100, each in 1-99. By the pigeonhole principle, two of these sums must be equal. Say the i-th and j-th cumulative sums are equal modulo 100, with i < j. Then, the sum from i+1 to j is 0 modulo 100. Therefore, there exists a subset (the elements from i+1 to j) whose sum is divisible by 100. However, this subset has size j - i, which is at least 1. But the problem states that subsets of size ≥2 cannot sum to 0 modulo 100. Therefore, this would imply that in any 99 numbers, there is a non-empty subset (possibly of size 1) whose sum is divisible by 100. But the problem allows subsets of size ≥2. So, this approach gives a subset of size ≥1, but we need to exclude size 1. Wait, but if we adjust the argument. Suppose we have 99 numbers. Consider all subsets of size ≥1. There are 2^99 -1 subsets. By the pigeonhole principle, there must be two different subsets with the same sum modulo 100. Then, the symmetric difference of these subsets would be a non-empty subset with sum 0 modulo 100. However, the symmetric difference could be of size 1, which is a single element. Therefore, this approach might not ensure a subset of size ≥2. Alternatively, if we consider the 99 numbers, and consider the 99 partial sums s_1, s_2, ..., s_99, where s_i is the sum of the first i numbers. If any s_i ≡0 mod 100, then we have a subset of size i. If not, then by pigeonhole principle, two of them must be equal, say s_i ≡ s_j mod 100 with i < j. Then, the sum from i+1 to j is 0 mod 100. But this subset has size j - i. To ensure that j - i ≥2, we need to adjust the initial partial sums.Wait, here's a trick: consider the partial sums starting from the empty set. So, include 0 as one of the partial sums. Then, with 99 numbers, we have 100 partial sums (including 0). By pigeonhole, two of them must be congruent modulo 100. If one of them is the empty set (sum 0), then the other is a non-empty subset with sum 0 modulo 100. But this could be a subset of size 1. However, if we exclude the empty set, we have 99 partial sums, and the same problem as before. So, perhaps this approach can't guarantee a subset of size ≥2.But the problem states that subsets of size ≥2 cannot sum to 0 modulo 100. Therefore, if such subsets don't exist, then the numbers must be structured in a very specific way. Let me think again about the case when all numbers are equal. Suppose all numbers are equal to 'a', which is coprime to 100. Then, any subset sum of size k is k*a. Since a is coprime to 100, k*a ≡0 mod 100 implies that 100 divides k*a, which implies that 100 divides k (since a is coprime to 100). But k is between 1 and 99, so this is impossible. Therefore, no subset sum of any size (including size 1) is divisible by 100. Wait, but the problem allows subsets of size 1. However, the problem states that "the sum of any 2, 3, or more of these numbers cannot be divided evenly by 100". So, size 1 subsets are allowed to be divisible by 100. Wait, but each number is less than 100, so size 1 subsets cannot be divisible by 100, because each number is less than 100. So, all numbers are between 1 and 99, so their single-element subsets have sums between 1 and 99, hence not divisible by 100. Therefore, the problem's condition is automatically satisfied for subsets of size 1, and the constraint is on subsets of size ≥2. But in our previous analysis, when all numbers are equal and coprime to 100, then even subsets of size 1 are not divisible by 100 (which they aren't, as they are between 1 and 99), and subsets of size ≥2 also aren't. So, that works. However, if the numbers are not all equal, then even if each individual number is coprime to 100, there could be a subset of size ≥2 whose sum is divisible by 100. Therefore, the problem is essentially forcing the numbers to be structured in such a way that no combination of them can add up to 100. The only way this is possible is if all numbers are equal and each is coprime to 100, preventing any subset sum from being a multiple of 100. But how to prove that equality is necessary?Perhaps we can use contradiction. Assume that not all numbers are equal. Then, there are at least two distinct numbers, say a and b. If a + b ≡0 mod 100, done. If not, consider other numbers. Since there are 99 numbers, which is a lot, maybe there's a way to form a sum with some combination.Wait, another angle: if all numbers are congruent modulo 100, then they are all equal, since they are less than 100. Therefore, if they are not all equal, their residues modulo 100 are not all the same. So, there exist at least two different residues. If we can show that two different residues can be combined (possibly with others) to sum to 0 mod 100, then we have a contradiction.For example, suppose there is a residue 'a' and another residue 'b'. If we can find integers k and m such that k*a + m*b ≡0 mod 100, with k + m ≥2, then we have a subset sum of k a's and m b's. If such k and m exist, then we can form such a subset. But how to guarantee that such k and m exist? For example, if a and b are coprime to 100, then by some linear combination theorem, we can find integers k and m such that k*a + m*b ≡0 mod 100. However, the coefficients k and m would need to be non-negative and the total size k + m ≥2. Alternatively, consider the residues as elements of the additive group modulo 100. If there are two distinct elements in this group, then the subgroup generated by them would have some order dividing 100. If the subgroup is the entire group, then we can form any residue, including 0, with some combination. However, since we need non-negative coefficients (as we can't subtract numbers), this complicates things.Alternatively, think in terms of the coins problem: given denominations a and b, can we make 100 using these coins? If a and b are coprime, then yes, for sufficiently large amounts, but 100 might be achievable. However, this is not straightforward.Alternatively, consider that with 99 numbers, the number of possible subset sums is enormous, but modulo 100, there are only 100 possibilities. Therefore, by the pigeonhole principle, many subsets will have the same sum modulo 100. Then, the difference between two such subsets would be a subset sum of 0 modulo 100. However, the difference could correspond to a subset of any size, including 1. But since we need subsets of size ≥2, this might not help.Wait, let's formalize this. Suppose there are two subsets A and B such that sum(A) ≡ sum(B) mod 100. Then, the symmetric difference A Δ B is a subset whose sum is sum(A) - sum(B) ≡0 mod 100. The size of A Δ B is |A| + |B| - 2|A ∩ B|. To ensure that this size is at least 2, we need that |A| + |B| - 2|A ∩ B| ≥2. This is possible if A and B are distinct subsets of size at least 1. For example, if A and B are both size 1, their symmetric difference is size 0 or 2. If they are distinct singletons, the symmetric difference is size 2, which would be a subset of size 2 summing to 0 mod 100. But in our problem, single elements cannot sum to 0 mod 100, but two distinct elements could. Therefore, if there are two distinct elements a and b such that a + b ≡0 mod 100, then that pair is a subset of size 2 summing to 0. If not, then maybe consider larger subsets.But if no two elements sum to 0 mod 100, then perhaps look for three elements. However, with 99 elements, the likelihood is high. But how to guarantee it?Wait, let's think again about the maximum size of a set without any subset of size ≥2 summing to 0 mod 100. If we can determine that such a set can have at most 50 elements, then since our problem has 99 elements, which is much larger, it must contain such subsets. As mentioned earlier, if we consider pairs that sum to 100, like (1,99), (2,98), ..., (49,51), and the singleton 50, there are 50 such pairs. To avoid having any pair summing to 100, the set can contain at most one element from each pair. Therefore, the maximum size of such a set is 50 (49 pairs + 50). Therefore, any set with 51 or more elements must contain at least one such pair. Since our problem has 99 elements, which is way larger than 50, it must contain many pairs that sum to 100. Therefore, this would violate the problem's condition. Therefore, the only way to have a set of 99 elements with no subset of size ≥2 summing to 0 mod 100 is if it's impossible to have such pairs, which can only happen if all elements are equal and their sum cannot reach 0 mod 100 for any k ≥2.But as we saw earlier, if all elements are equal to a number 'a' coprime to 100, then k*a mod 100 ≠0 for any k=1,...,99. Therefore, such a set would satisfy the condition. However, if the set contains any two distinct elements, then the maximum size without a pair summing to 100 is 50, which is much smaller than 99. Therefore, such a set of 99 elements must contain at least one pair summing to 100. Hence, the only way to avoid this is to have all elements equal. Therefore, this leads us to the conclusion that all numbers must be equal. Because if they are not, the set would have to contain at least one pair summing to 100, which violates the condition. Therefore, by contradiction, all numbers must be equal. Additionally, as we noted earlier, if all numbers are equal, they must be coprime to 100; otherwise, a subset sum of size k=100/d (where d is gcd(a,100)) would be divisible by 100. Since the numbers are less than 100, k=100/d must be greater than or equal to 2. But since the numbers are 99 in count, the maximum possible k is 99, which would require d=1. Therefore, a must be coprime to 100. Hence, all numbers must be equal and coprime to 100. But the problem only asks to prove that all numbers are equal. The coprimality follows from the condition, but perhaps the problem assumes that if they are equal, they automatically satisfy the sum conditions. However, as we saw, if they are equal but not coprime to 100, like 50, then subsets of size 2 would sum to 100. Therefore, the problem's condition implicitly requires the numbers to be both equal and coprime to 100. But since the problem only asks to prove they are equal, the coprimality is a necessary consequence of the condition. Therefore, the final conclusion is that all numbers must be equal, and this is the only way to satisfy the given conditions.

✅:To prove that all 99 numbers must be equal under the given conditions, we analyze the constraints using modular arithmetic and combinatorial arguments:1. Modular Constraints: Each number is between 1 and 99, so residues modulo 100 are the numbers themselves. The condition that no subset of size ≥2 sums to 0 mod 100 implies strict restrictions on these residues.2. Complementary Pairs: For any number ( a ), its complement ( 100 - a ) must not exist in the set to avoid pairs summing to 100. However, with 99 numbers (exceeding the maximum 50 allowed by such pair restrictions), the set must contain complementary pairs, violating the condition unless all numbers are equal.3. Equal Numbers Case: If all numbers are equal to ( a ), subset sums are ( k cdot a mod 100 ). For these sums to never be 0, ( a ) must be coprime to 100. Otherwise, ( k cdot a equiv 0 mod 100 ) would occur for some ( k geq 2 ).4. Pigeonhole Principle: A set of 99 numbers must exceed the maximum size (50) that avoids complementary pairs, forcing at least one pair to sum to 100 unless all numbers are equal.5. Conclusion: The only way to satisfy the condition is if all numbers are equal and coprime to 100. Thus, equality is necessary to prevent any subset sum from being divisible by 100.Final Answer: All 99 numbers must be equal. This is proven by showing that any distinct pair or larger subset in a non-uniform set would violate the given modular sum condition. Hence, (boxed{text{All numbers are equal}}).

🔑:1. Define the set of given integers: Let the 99 distinct integers be ( n_1, n_2, ldots, n_{99} ).2. Assume the contrary: Suppose not all the integers are the same, i.e., there exist at least two different integers, for example, ( n_1 ne n_2 ).3. Construct the sums: Consider the following sums: [ begin{align*} S_1 &= n_1, S_2 &= n_2, S_3 &= n_1 + n_2, &;;vdots S_{100} &= n_1 + n_2 + cdots + n_{99}. end{align*} ]4. Apply the pigeonhole principle: According to the assumption, none of these sums ( S_i ) (for ( i = 1, 2, ldots, 100 )) is divisible by 100. By the Pigeonhole Principle, since there are 100 sums and only 99 possible remainders when divided by 100 (i.e., from 0 to 99, but 0 is excluded by assumption), at least two sums must leave the same remainder when divided by 100.5. Select two sums with the same remainder: Suppose ( S_k ) and ( S_l ) are two such sums where ( k > l ) and both have the same remainder when divided by 100: [ S_k equiv S_l pmod{100}. ] Therefore, [ S_k - S_l equiv 0 pmod{100}. ] This implies [ 100 mid (S_k - S_l). ]6. Simplify the difference: The difference ( S_k - S_l ) calculates to: [ S_k - S_l = (n_1 + n_2 + cdots + n_k) - (n_1 + n_2 + cdots + n_l) = n_{l+1} + n_{l+2} + cdots + n_k. ]7. Inferring a contradiction: Since both sums have the same remainder when divided by 100, [ 100 mid (n_{l+1} + n_{l+2} + cdots + n_k), ] which contradicts the given condition that sums of any ( 2 ), ( 3 ), or more are not divisible by 100.8. Conclude by contradiction: This contradiction implies that our initial assumption (that not all integers are the same) is incorrect. Therefore, all 99 integers must be the same. Conclusion[boxed{text{All the integers } n_1, n_2, ldots, n_{99} text{ must be equal.}}]