🧠:Okay, so I need to figure out how many real numbers ( a ) make the quadratic equation ( x^{2} + a x + 6a = 0 ) have only integer solutions. Hmm, let me start by recalling some basics about quadratic equations. First, for a quadratic equation ( x^2 + bx + c = 0 ), the solutions can be found using the quadratic formula: ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ). But wait, in this case, the equation is given as ( x^2 + a x + 6a = 0 ), so here the coefficients are a bit different. Let me write that down properly. The equation is ( x^2 + a x + 6a = 0 ). Comparing this to the standard form ( x^2 + px + q = 0 ), I can see that ( p = a ) and ( q = 6a ). The problem states that the equation must have only integer solutions. So, for the roots to be integers, the quadratic must factor into ( (x - m)(x - n) = 0 ), where ( m ) and ( n ) are integers. Expanding this, we get ( x^2 - (m + n)x + mn = 0 ). Comparing this to the original equation ( x^2 + a x + 6a = 0 ), we can equate the coefficients:1. Coefficient of ( x ): ( -(m + n) = a )2. Constant term: ( mn = 6a )So, from the first equation, ( a = -(m + n) ), and from the second equation, ( mn = 6a ). Substituting the expression for ( a ) from the first equation into the second equation gives:( mn = 6(-m - n) )Simplifying this:( mn + 6m + 6n = 0 )Hmm, let's factor this equation. Maybe I can add 36 to both sides to make it factorable. Let's try:( mn + 6m + 6n + 36 = 36 )Now, factor the left side:( (m + 6)(n + 6) = 36 )Oh, that's clever! By adding 36 to both sides, the left side becomes a product of two terms, ( (m + 6) ) and ( (n + 6) ), which equals 36. So now, since ( m ) and ( n ) are integers, ( m + 6 ) and ( n + 6 ) must be integer divisors of 36. Therefore, all pairs of integers ( (m + 6, n + 6) ) that multiply to 36 are the factor pairs of 36. Let me list all the factor pairs of 36, both positive and negative. The positive factor pairs of 36 are:- (1, 36)- (2, 18)- (3, 12)- (4, 9)- (6, 6)And the negative factor pairs would be:- (-1, -36)- (-2, -18)- (-3, -12)- (-4, -9)- (-6, -6)So each of these pairs corresponds to ( (m + 6, n + 6) ). Therefore, for each pair, we can solve for ( m ) and ( n ), then compute ( a ) using ( a = -(m + n) ). Then, we need to check if these values of ( a ) are distinct and real. But since ( a ) is derived from integers ( m ) and ( n ), all such ( a ) will be real. The question is how many distinct ( a ) values there are.Let me go through each factor pair and compute ( m ), ( n ), and ( a ):Starting with the positive pairs:1. (1, 36): - ( m + 6 = 1 Rightarrow m = -5 ) - ( n + 6 = 36 Rightarrow n = 30 ) - ( a = -(-5 + 30) = -25 )2. (2, 18): - ( m = 2 - 6 = -4 ) - ( n = 18 - 6 = 12 ) - ( a = -(-4 + 12) = -8 )3. (3, 12): - ( m = 3 - 6 = -3 ) - ( n = 12 - 6 = 6 ) - ( a = -(-3 + 6) = -3 )4. (4, 9): - ( m = 4 - 6 = -2 ) - ( n = 9 - 6 = 3 ) - ( a = -(-2 + 3) = -1 )5. (6, 6): - ( m = 6 - 6 = 0 ) - ( n = 6 - 6 = 0 ) - ( a = -(0 + 0) = 0 )Now, the negative factor pairs:6. (-1, -36): - ( m + 6 = -1 Rightarrow m = -7 ) - ( n + 6 = -36 Rightarrow n = -42 ) - ( a = -(-7 + (-42)) = -(-49) = 49 )7. (-2, -18): - ( m = -2 - 6 = -8 ) - ( n = -18 - 6 = -24 ) - ( a = -(-8 + (-24)) = -(-32) = 32 )8. (-3, -12): - ( m = -3 - 6 = -9 ) - ( n = -12 - 6 = -18 ) - ( a = -(-9 + (-18)) = -(-27) = 27 )9. (-4, -9): - ( m = -4 - 6 = -10 ) - ( n = -9 - 6 = -15 ) - ( a = -(-10 + (-15)) = -(-25) = 25 )10. (-6, -6): - ( m = -6 - 6 = -12 ) - ( n = -6 - 6 = -12 ) - ( a = -(-12 + (-12)) = -(-24) = 24 )Wait, hold on, for the pair (-6, -6), ( m = -12 ), ( n = -12 ), so ( a = -(-12 + -12) = 24 ). Okay.So, compiling all these values of ( a ):From positive pairs:-25, -8, -3, -1, 0From negative pairs:49, 32, 27, 25, 24Now, let's list all the ( a ) values:Negative pairs give: 49, 32, 27, 25, 24Positive pairs give: -25, -8, -3, -1, 0Now, check if there are any duplicates. Let's see:Looking at the list: 49, 32, 27, 25, 24, -25, -8, -3, -1, 025 appears both in the negative pairs as 25 and in the positive pairs as -25? Wait no, 25 is from the negative pair (-4, -9), and -25 is from the positive pair (1, 36). So those are different. So no duplicates here. Each ( a ) is unique.So total number of distinct ( a ) values is 10? Let's count:From positive factor pairs: 5 valuesFrom negative factor pairs: 5 valuesTotal: 10.Wait, but wait. Let's check each pair again.Wait, in the positive factor pairs, we had (1,36), (2,18), (3,12), (4,9), (6,6). That's 5 pairs. Similarly, the negative factor pairs are (-1,-36), (-2,-18), (-3,-12), (-4,-9), (-6,-6). Another 5 pairs. Each pair gives a unique ( a ). So 10 different values of ( a ).But wait, hold on. Let me check if any of these ( a ) values are the same. For example, 25 and -25 are distinct. 24,25,27,32,49 from negative pairs. -25, -8, -3, -1, 0 from positive pairs. All different. So yes, 10 different values.But wait, hold on. When we considered the factor pairs, we considered both (1,36) and (36,1), but since multiplication is commutative, those would lead to the same product, but in our case, since we are considering ordered pairs? Wait, no, in the original factorization, when we list factor pairs, we consider all possible ordered pairs? Wait, but in our setup, we considered factor pairs where order doesn't matter because the roots m and n are interchangeable. So for example, if we have (1,36) and (36,1), they would lead to the same pair of roots m and n, just swapped. Therefore, leading to the same value of a. Wait, but in our initial listing, we only considered each factor pair once, right? For example, when we listed positive factor pairs, we didn't list both (1,36) and (36,1) separately because they are considered the same pair in terms of multiplication. So in our earlier list, each factor pair is unique in the sense that we don't repeat commutative pairs. Therefore, each of the 5 positive factor pairs and 5 negative factor pairs are distinct and each gives a unique a. Wait, but let's take an example. Suppose we take the pair (2,18) and (18,2). If we had considered (18,2), would that lead to a different a? Let's check:If (m + 6, n + 6) = (18, 2), then:m = 18 - 6 = 12n = 2 - 6 = -4Then a = -(12 + (-4)) = -8. Which is the same as when we had (2,18). So even if we swap the order, the value of a remains the same. Therefore, each unordered factor pair corresponds to a unique a. Therefore, our initial approach of considering unordered factor pairs is correct, and we don't have duplicates. Hence, 5 from positive and 5 from negative gives 10 distinct a's.But wait, hold on. Let's check the pair (-6, -6). That gives m = -12 and n = -12, so a = 24. Is there another pair that would lead to a = 24? Let me see. For example, suppose there's a factor pair (x, y) such that x * y = 36, and then m = x - 6, n = y - 6. Then, if x and y are different numbers, but swapping them doesn't change the result. But in the case of (-6, -6), it's the same as (-6, -6) swapped. So this is only one pair. So in total, all the pairs we considered lead to unique a's.Wait, but another thought: could there be different factor pairs that lead to the same a? For example, maybe one positive pair and one negative pair give the same a? Let's check:From positive pairs: a's are -25, -8, -3, -1, 0From negative pairs: a's are 49, 32, 27, 25, 24Looking at these, there's no overlap between the two sets. So all a's are unique. Therefore, the total number is 5 + 5 = 10.But wait, let me check if 25 is present in both. The negative pairs give 25, and positive pairs have -25. So 25 and -25 are different. Similarly, others don't overlap. So yes, 10 distinct a's.But hold on, let me verify with specific examples. Let's take a = 0. Then the equation becomes x² + 0x + 0 = x² = 0, which has a double root at x = 0, which is integer. So that's valid.a = -1. Then the equation is x² - x + 6*(-1) = x² - x -6 = 0. Let's factor this: (x - 3)(x + 2) = 0. Roots at 3 and -2, integers. Correct.a = -3. Equation: x² -3x + 6*(-3) = x² - 3x -18 = 0. Factoring: Let's see, factors of -18 that add to -3. That would be -6 and 3. So (x - 6)(x + 3) = 0. Roots at 6 and -3. Integers. Good.a = -8. Equation: x² -8x +6*(-8) = x² -8x -48 = 0. Factors: Let's see, two numbers that multiply to -48 and add to -8. That's -12 and 4. So (x -12)(x +4) = 0. Roots 12 and -4. Integers.a = -25. Equation: x² -25x +6*(-25) = x² -25x -150 = 0. Let's factor. Numbers that multiply to -150 and add to -25. Hmm, let's see: -30 and 5: -30*5 = -150, -30 +5 = -25. So (x -30)(x +5) = 0. Roots 30 and -5. Integers. Correct.Now for the positive a's:a = 24. Equation: x² +24x +6*24 = x² +24x +144 =0. This factors as (x +12)^2 =0. Double root at -12. Integer.a =25. Equation: x² +25x +6*25 = x² +25x +150 =0. Let's factor. Looking for two numbers that multiply to 150 and add to 25. 10 and 15: 10*15=150, 10+15=25. So (x +10)(x +15)=0. Roots -10 and -15. Integers.a =27. Equation: x² +27x +6*27 =x² +27x +162 =0. Factors: Looking for numbers that multiply to 162 and add to27. 9 and 18: 9*18=162, 9+18=27. So (x +9)(x +18)=0. Roots -9 and -18. Integers.a =32. Equation: x² +32x +6*32 =x² +32x +192=0. Factors: Numbers multiplying to 192 and adding to32. 16 and12: 16*12=192, 16+12=28. Not 32. Wait, 24 and8:24*8=192, 24+8=32. So (x +24)(x +8)=0. Roots -24 and -8. Integers.a =49. Equation: x² +49x +6*49 =x² +49x +294=0. Let's factor. Numbers that multiply to294 and add to49. Let's see, 294 divided by 6 is49. Wait, 49*6=294. So 42 and7: 42*7=294, 42+7=49. So (x +42)(x +7)=0. Roots -42 and -7. Integers.Okay, all the a's check out. Each of them gives integer roots. So all 10 values of a are valid. Therefore, the number of real numbers a is 10.But wait, just to be thorough, are there any other factor pairs of 36 that we might have missed?The factors of 36 are ±1, ±2, ±3, ±4, ±6, ±9, ±12, ±18, ±36. The factor pairs would be all combinations of two numbers whose product is 36. So we considered all positive pairs and their negatives. For example, (1,36), (2,18), (3,12), (4,9), (6,6) and their negatives (-1,-36), (-2,-18), (-3,-12), (-4,-9), (-6,-6). So that's 5 positive pairs and 5 negative pairs. Each of these gives distinct a's. Therefore, 10 values in total. I don't think we missed any factor pairs. For instance, ( -1, -36 ) gives a =49, which we accounted for, and so on. Therefore, the answer should be 10.But hold on, the problem says "real numbers a". Since a is derived from integers m and n, all these a's are real numbers. So the total number is 10. So the answer is 10. Hence, the number of real numbers a is boxed{10}.Final AnswerThe number of real numbers ( a ) is boxed{10}.

✅:To determine how many real numbers ( a ) make the quadratic equation ( x^2 + ax + 6a = 0 ) have only integer solutions, we start by noting that the quadratic must factor into ( (x - m)(x - n) = 0 ) where ( m ) and ( n ) are integers. Expanding and comparing coefficients with the given equation, we get:1. Coefficient of ( x ): ( -(m + n) = a )2. Constant term: ( mn = 6a )Substituting ( a = -(m + n) ) into ( mn = 6a ) gives:[ mn + 6m + 6n = 0 ]Adding 36 to both sides, we factorize:[ (m + 6)(n + 6) = 36 ]We list all integer factor pairs of 36, both positive and negative:- Positive pairs: ( (1, 36), (2, 18), (3, 12), (4, 9), (6, 6) )- Negative pairs: ( (-1, -36), (-2, -18), (-3, -12), (-4, -9), (-6, -6) )For each pair ((p, q)), we solve for ( m ) and ( n ):- ( m = p - 6 )- ( n = q - 6 )- Calculate ( a = -(m + n) )Positive pairs:1. ( (1, 36) rightarrow m = -5, n = 30 rightarrow a = -25 )2. ( (2, 18) rightarrow m = -4, n = 12 rightarrow a = -8 )3. ( (3, 12) rightarrow m = -3, n = 6 rightarrow a = -3 )4. ( (4, 9) rightarrow m = -2, n = 3 rightarrow a = -1 )5. ( (6, 6) rightarrow m = 0, n = 0 rightarrow a = 0 )Negative pairs:1. ( (-1, -36) rightarrow m = -7, n = -42 rightarrow a = 49 )2. ( (-2, -18) rightarrow m = -8, n = -24 rightarrow a = 32 )3. ( (-3, -12) rightarrow m = -9, n = -18 rightarrow a = 27 )4. ( (-4, -9) rightarrow m = -10, n = -15 rightarrow a = 25 )5. ( (-6, -6) rightarrow m = -12, n = -12 rightarrow a = 24 )Each factor pair results in a distinct ( a ). There are no duplicates, resulting in 10 distinct values of ( a ).Thus, the number of real numbers ( a ) is (boxed{10}).

🔑:1. Consider the quadratic equation ( x^2 + ax + 6a = 0 ). 2. Let ( m ) and ( n ) be the integer roots of the equation, where ( m leq n ). By Vieta's formulas, the sum and product of the roots are given by: [ m + n = -a ] [ mn = 6a ]3. Using the relationships, we can substitute ( -a ) and ( 6a ) into the product and sum equations: [ -6(m + n) = mn ] 4. Simplifying, we obtain another form of the equation: [ (m + 6)(n + 6) = 36 ]5. This is achieved by recognizing that: [ mn - 6m - 6n = 36 - 36 implies (m + 6)(n + 6) = 36 ]6. Factorizing 36, we get the pairs ((p, q)) such that ( pq = 36 ): [ (1 cdot 36), (2 cdot 18), (3 cdot 12), (4 cdot 9), (6 cdot 6) ] Considering ( m + 6 leq n + 6 ), list the potential ((m, n)) pairs: [ (1 - 6, 36 - 6) = (-5, 30) ] [ (2 - 6, 18 - 6) = (-4, 12) ] [ (3 - 6, 12 - 6) = (-3, 6) ] [ (4 - 6, 9 - 6) = (-2, 3) ] [ (6 - 6, 6 - 6) = (0, 0) ] Then consider the pairs ((p, q)) such that ( pq = 36 ) and listing: [ (-42, -7), (-24, -8), (-18, -9), (-15, -10), (-12, -12) ]7. Each pair ((m, n)) corresponds to different values of ( a = -(m + n) ): [ begin{align*} (-42, -7) & Rightarrow a = -(-42 + (-7)) = -(-49) = 49, (-24, -8) & Rightarrow a = -(-24 + (-8)) = -(-32) = 32, (-18, -9) & Rightarrow a = -(-18 + (-9)) = -(-27) = 27, (-15, -10) & Rightarrow a = -(-15 + (-10)) = -(-25) = 25, (-12, -12) & Rightarrow a = -(-12 + (-12)) = -(-24) = 24, (-5, 30) & Rightarrow a = -(-5 + 30) = -25, (-4, 12) & Rightarrow a = -(-4 + 12) = -8, (-3, 6) & Rightarrow a = -(-3 + 6) = -3, (-2, 3) & Rightarrow a = -(-2 + 3) = -1, (0, 0) & Rightarrow a = -(0 + 0) = 0. end{align*} ]8. Therefore, the distinct values of ( a ) are: [ 49, 32, 27, 25, 24, -25, -8, -3, -1, 0 ]9. Conclusively, there are a total of ( 10 ) distinct real numbers ( a ) that make the given quadratic equation have only integer solutions.[boxed{10}]

🧠:Alright, let's try to tackle this geometry problem. So, we have an isosceles triangle ABC with base BC. H is the midpoint of BC, which makes sense because in an isosceles triangle, the midpoint of the base is also the foot of the altitude from the apex. Then, they draw a perpendicular HE from H to side AC. O is the midpoint of HE, and we need to prove that lines AO and BE are perpendicular. First, I need to visualize the triangle. Let me sketch it mentally. ABC is isosceles with AB = AC, right? Wait, no. Wait, the base is BC, so the two equal sides would be AB and AC. Wait, actually, in an isosceles triangle, the base is the side that isn't equal. So if the base is BC, then the two legs are AB and AC. So ABC has AB = AC, and BC is the base. Therefore, vertex A is the apex, and H is the midpoint of BC. Then, from H, we draw a perpendicular to AC, which is HE. So HE is perpendicular to AC, and E is the foot of this perpendicular on AC. Then O is the midpoint of HE. We need to show that AO and BE are perpendicular. Hmm. Let me try to approach this. Maybe coordinate geometry would work here. Assign coordinates to the points and then compute the slopes of AO and BE to show their product is -1. That's a standard method for proving perpendicularity.Okay, let's set up a coordinate system. Let me place point B at (-b, 0) and point C at (b, 0) so that BC is the base lying along the x-axis, and H, the midpoint, is at (0, 0). Wait, but then the midpoint of BC is H. If we do that, then ABC is isosceles with AB = AC. Let's place point A somewhere along the y-axis. Let's say A is at (0, a) for some a > 0. So coordinates:- A: (0, a)- B: (-b, 0)- C: (b, 0)- H: midpoint of BC, so (( -b + b ) / 2, (0 + 0)/2 ) = (0, 0). Wait, that's interesting. So H is at the origin here.Then, we need to draw HE perpendicular to AC. So first, let's find the equation of side AC. Points A(0, a) and C(b, 0). The slope of AC is (0 - a)/(b - 0) = -a/b. Therefore, the line AC has equation y = (-a/b)x + a.Now, HE is a perpendicular from H(0,0) to AC. So the slope of HE is the negative reciprocal of the slope of AC. Since AC has slope -a/b, the slope of HE is b/a. Therefore, the line HE has equation y = (b/a)x.But E is the intersection of HE and AC. Let's find the coordinates of E by solving the two equations:1. y = (-a/b)x + a (equation of AC)2. y = (b/a)x (equation of HE)Set them equal:(b/a)x = (-a/b)x + aMultiply both sides by ab to eliminate denominators:b^2 x = -a^2 x + a^2 bBring terms with x to the left:b^2 x + a^2 x = a^2 bx(b^2 + a^2) = a^2 bTherefore, x = (a^2 b)/(a^2 + b^2)Then, substitute back into equation 2 to find y:y = (b/a) * (a^2 b)/(a^2 + b^2) = (a b^2)/(a^2 + b^2)Thus, E has coordinates ( (a^2 b)/(a^2 + b^2), (a b^2)/(a^2 + b^2) )Then, H is (0, 0), so HE is the segment from H(0,0) to E( (a^2 b)/(a^2 + b^2), (a b^2)/(a^2 + b^2) )Therefore, O is the midpoint of HE. So the coordinates of O will be the average of the coordinates of H and E:O_x = (0 + (a^2 b)/(a^2 + b^2))/2 = (a^2 b)/(2(a^2 + b^2))O_y = (0 + (a b^2)/(a^2 + b^2))/2 = (a b^2)/(2(a^2 + b^2))Therefore, O is at ( (a^2 b)/(2(a^2 + b^2)), (a b^2)/(2(a^2 + b^2)) )Now, we need to find the equations or slopes of lines AO and BE.First, let's compute the slope of AO. Point A is (0, a), and point O is ( (a^2 b)/(2(a^2 + b^2)), (a b^2)/(2(a^2 + b^2)) )Slope of AO: (O_y - A_y)/(O_x - A_x) = [ (a b^2/(2(a^2 + b^2)) ) - a ] / [ (a^2 b/(2(a^2 + b^2)) ) - 0 ]Simplify numerator:= [ (a b^2 - 2a(a^2 + b^2)) / (2(a^2 + b^2)) ]= [ a b^2 - 2a^3 - 2a b^2 ) / (2(a^2 + b^2)) ]= [ -2a^3 - a b^2 ) / (2(a^2 + b^2)) ]= -a(2a^2 + b^2) / (2(a^2 + b^2))Denominator:= (a^2 b)/(2(a^2 + b^2))Thus, slope of AO:[ -a(2a^2 + b^2) / (2(a^2 + b^2)) ) ] / [ a^2 b / (2(a^2 + b^2)) ) ]The denominators 2(a^2 + b^2) cancel out, as well as one 'a' from numerator and denominator:= [ - (2a^2 + b^2) ] / (a b )So slope of AO is - (2a^2 + b^2)/(a b)Now, compute the slope of BE. Point B is (-b, 0), and point E is ( (a^2 b)/(a^2 + b^2), (a b^2)/(a^2 + b^2) )Slope of BE: (E_y - B_y)/(E_x - B_x) = [ (a b^2/(a^2 + b^2) - 0 ) ] / [ (a^2 b/(a^2 + b^2) - (-b) ) ]Simplify denominator:= [ a^2 b/(a^2 + b^2) + b ]= [ a^2 b + b(a^2 + b^2) ] / (a^2 + b^2 )= [ a^2 b + a^2 b + b^3 ] / (a^2 + b^2 )= [ 2a^2 b + b^3 ] / (a^2 + b^2 )= b(2a^2 + b^2) / (a^2 + b^2 )Numerator:= a b^2 / (a^2 + b^2 )Thus, slope of BE:[ a b^2 / (a^2 + b^2 ) ] / [ b(2a^2 + b^2) / (a^2 + b^2 ) ]The (a^2 + b^2 ) terms cancel, as well as a 'b' from numerator and denominator:= a b / (2a^2 + b^2 )So slope of BE is (a b)/(2a^2 + b^2 )Now, to check if AO and BE are perpendicular, the product of their slopes should be -1.Compute slope_AO * slope_BE:[ - (2a^2 + b^2)/(a b ) ] * [ (a b)/(2a^2 + b^2 ) ] = - (2a^2 + b^2) * a b / (a b (2a^2 + b^2)) ) = -1Yes, the product is indeed -1. Therefore, AO and BE are perpendicular.Wait, that seems straightforward. Let me verify if there are any calculation errors. Let's check the coordinates of E again.Equation of AC: from (0,a) to (b,0). Slope is -a/b. Correct. Then, equation is y = -a/b x + a. Correct.Equation of HE: perpendicular from H(0,0) to AC. Slope is reciprocal and opposite sign, so b/a. So equation y = (b/a)x. Correct.Intersection E: solving (b/a)x = -a/b x + a. Multiply by ab: b² x = -a² x + a² b. Then x (b² + a²) = a² b => x = a² b / (a² + b²). Then y = (b/a)(a² b / (a² + b²)) = a b² / (a² + b²). Correct.Midpoint O: average of (0,0) and E. So x: (a² b)/(2(a² + b²)), y: (a b²)/(2(a² + b²)). Correct.Slope of AO: from (0,a) to O. So (Oy - a)/(Ox - 0) = [ (a b² / 2(a² + b²) ) - a ] / (a² b / 2(a² + b²) )Compute numerator:(a b² - 2a(a² + b²)) / 2(a² + b²) = (a b² - 2a³ - 2a b²)/denominator = (-2a³ - a b²)/denominator. Factor out -a: -a(2a² + b²)/denominator. Denominator of numerator is 2(a² + b²). Then slope is [-a(2a² + b²)/2(a² + b²)] divided by [a² b / 2(a² + b²))] = [-a(2a² + b²)] / (a² b) = - (2a² + b²)/(a b). Correct.Slope of BE: from B(-b, 0) to E. E's coordinates are (a² b/(a² + b²), a b²/(a² + b²)). So the difference in x-coordinate is a² b/(a² + b²) - (-b) = a² b/(a² + b²) + b = b(a² + (a² + b²))/(a² + b²))? Wait, wait:Wait, the denominator when adding:E_x - B_x = [a² b/(a² + b²)] - (-b) = a² b/(a² + b²) + b = [a² b + b(a² + b²)] / (a² + b²) = [a² b + a² b + b³]/(a² + b²) = [2a² b + b³]/(a² + b²) = b(2a² + b²)/(a² + b²). Correct.Difference in y-coordinate: E_y - B_y = a b²/(a² + b²) - 0 = a b²/(a² + b²). So slope is [a b²/(a² + b²)] / [b(2a² + b²)/(a² + b²)] = (a b²) / (b(2a² + b²)) = a b / (2a² + b²). Correct.Then product of slopes: [- (2a² + b²)/(a b)] * [a b / (2a² + b²)] = -1. Perfect. Therefore, AO and BE are perpendicular.Hmm, seems like coordinate geometry worked smoothly here. Maybe there's a synthetic geometry proof too, but since the coordinate method gave the answer directly, perhaps that's sufficient. Let me think if there's another way, maybe using vectors or properties of midpoints and perpendicularity.Alternatively, since O is the midpoint of HE, and H is the midpoint of BC, maybe there's some midline theorem or parallelogram properties involved. Let's see.In triangle ABC, H is the midpoint of BC, and HE is perpendicular to AC. Then O is the midpoint of HE. Maybe connecting midpoints or using similar triangles.Alternatively, considering that in coordinate geometry we proved it, maybe that's the most straightforward way. Since all steps checked out and the product of slopes is -1, the conclusion holds. Therefore, the proof is complete.Final Answerboxed{AO perp BE}

✅:To prove that lines (AO) and (BE) are perpendicular in the isosceles triangle (ABC) with base (BC), we use coordinate geometry as follows:1. Assign Coordinates: - Place (B) at ((-b, 0)), (C) at ((b, 0)), and (A) at ((0, a)). - The midpoint (H) of (BC) is at ((0, 0)).2. Equation of Line (AC): - The slope of (AC) is (-frac{a}{b}). - Equation: (y = -frac{a}{b}x + a).3. Equation of Perpendicular Line (HE): - The slope of (HE) is (frac{b}{a}) (negative reciprocal of (-frac{a}{b})). - Equation: (y = frac{b}{a}x).4. Find Intersection (E) of (HE) and (AC): - Solve (frac{b}{a}x = -frac{a}{b}x + a): - (x = frac{a^2 b}{a^2 + b^2}) - (y = frac{a b^2}{a^2 + b^2}) - Coordinates of (E): (left(frac{a^2 b}{a^2 + b^2}, frac{a b^2}{a^2 + b^2}right)).5. Midpoint (O) of (HE): - Coordinates of (O): (left(frac{a^2 b}{2(a^2 + b^2)}, frac{a b^2}{2(a^2 + b^2)}right)).6. Slope of (AO): - From (A(0, a)) to (Oleft(frac{a^2 b}{2(a^2 + b^2)}, frac{a b^2}{2(a^2 + b^2)}right)): - Slope: (-frac{2a^2 + b^2}{ab}).7. Slope of (BE): - From (B(-b, 0)) to (Eleft(frac{a^2 b}{a^2 + b^2}, frac{a b^2}{a^2 + b^2}right)): - Slope: (frac{ab}{2a^2 + b^2}).8. Product of Slopes: - Product: (left(-frac{2a^2 + b^2}{ab}right) times left(frac{ab}{2a^2 + b^2}right) = -1).Since the product of the slopes is (-1), lines (AO) and (BE) are perpendicular.[boxed{AO perp BE}]

🔑:1. Identify the relevant elements and their properties in the problem. We have an isosceles triangle ABC with AB = AC. H is the midpoint of the base BC implying H is on the perpendicular bisector of BC. A perpendicular HE is drawn from H to the lateral side AC.2. Let O be the midpoint of segment HE.3. Define D as the midpoint of BH. 4. Prove similarity of triangles triangle BHA and triangle HEA. Since H is the midpoint of BC and AB = AC, both BHA and HEA are right triangles. The right angle is at H in both triangles and A is a common vertex. Hence, triangle BHA sim triangle HEA by AA (Angle-Angle) similarity.5. From the similarity of the triangles, we know the following proportional relationships: [ frac{AD}{AO} = frac{AB}{AH} ] and their corresponding angles are equal: [ angle DAH = angle OAE. ]6. Observe that since D and O are midpoints of BH and HE respectively, triangles triangle DAO and triangle BHA are similar: [ triangle DAO sim triangle BHA. ]7. Consider the angle angle DOA. The angles angle BAH and angle OAE are equal as established through similar triangles triangle BHA sim triangle HEA. Thus: [ angle DOA = angle BAH. ]8. Since angle BAH is a right angle (90 degrees) in the isosceles triangle ABC, we have: [ angle DOA = 90^circ. ]Therefore, since angle DOA = 90^circ, the lines AO and BE are perpendicular.[boxed{AO perp BE}]

🧠:Okay, let me try to tackle this problem. So, we have a convex polygon with vertices A₁, A₂, ..., Aₙ, and there's a point O inside it. The special thing about O is that for any vertex Aᵢ, if you draw the line from O to Aᵢ, that line will also pass through another vertex Aⱼ. The task is to prove that no other point besides O in the polygon has this property. Hmm, interesting. I need to show that O is unique.First, let me make sure I understand the problem correctly. A convex polygon is one where all interior angles are less than 180 degrees, and every line segment between two vertices stays inside the polygon. The point O is inside this polygon, and for every vertex Aᵢ, the line OAᵢ must pass through another vertex Aⱼ. So, every radial line from O to a vertex also hits another vertex. Now, we need to prove that there can't be another point, say O', different from O, that also has this property. So, O is the only such point.Let me think about how to approach this. Maybe I can start with some examples. Let's take a simple convex polygon where such a point O exists. For instance, consider a regular polygon. In a regular polygon, the center is a point such that any line from the center to a vertex also passes through the opposite vertex. So, in a regular hexagon, the center O is such that OA₁ goes through A₄, OA₂ through A₅, etc. Now, in that case, is the center the only such point? If we take another point inside the regular hexagon, would it also have this property?Suppose I take a point slightly shifted from the center. Let's say in the regular hexagon, I move a bit to the right. Then, the line from this new point O' to A₁ might not pass through any other vertex. Because the symmetry is broken, the lines from O' to the vertices won't align with the opposite vertices anymore. So, maybe in the regular polygon, the center is indeed the only such point. That gives some intuition, but the problem is about any convex polygon, not just regular ones. So, I need a more general approach.The problem states that the polygon is convex and O is a point inside it with this property. So, even if the polygon isn't regular, as long as such a point O exists, it must be unique. How can I prove uniqueness?One method to prove uniqueness is to assume that there's another point O' with the same property and then reach a contradiction. So, suppose there exists another point O' ≠ O inside the polygon such that every line from O' to a vertex Aᵢ also passes through another vertex Aⱼ. Then, I need to show that this leads to some contradiction, implying that O' must equal O.Alternatively, maybe there's a way to characterize O as the intersection of certain lines or using some properties that make it unique. Let's think about what properties O must have.Since every line OAᵢ contains another vertex Aⱼ, the point O lies on the intersection of the lines AᵢAⱼ for various pairs (i, j). In a convex polygon, the intersection of these lines would be inside the polygon. If all these lines intersect at a single point O, then O is uniquely determined. But if there were another point O' lying on all these lines, then O and O' would have to coincide, assuming all these lines AᵢAⱼ are concurrent at a single point. So, if the lines AᵢAⱼ for all pairs where OAᵢ passes through Aⱼ are concurrent, then their intersection is unique, hence O is unique.Wait, but does the problem state that for every vertex Aᵢ, OAᵢ passes through another vertex Aⱼ, which might depend on i? So, for each Aᵢ, there's some Aⱼ such that OAᵢAⱼ are colinear. Then, O lies on the line AᵢAⱼ for each i. If each such line is determined by two vertices, and O is the intersection point of all these lines, then if there's another point O' also lying on all these lines, then O' must coincide with O. But this would require that all these lines AᵢAⱼ are concurrent at O. If they are concurrent, then their intersection is unique. Therefore, O is unique.But how do we know that all these lines AᵢAⱼ pass through O? Let me check. The problem states that for each Aᵢ, the line OAᵢ contains another vertex Aⱼ. So, for each i, there exists j such that Aⱼ is on OAᵢ. Therefore, O lies on the line AᵢAⱼ. But each such line is AᵢAⱼ, and O is on all those lines. Therefore, O is the intersection point of all these lines AᵢAⱼ. If all these lines intersect at O, then O is unique as the intersection point. But is that necessarily the case?Wait, in a convex polygon, if you have multiple lines connecting vertices, their intersection points can vary. However, if for every vertex Aᵢ, the line OAᵢ contains another vertex Aⱼ, then O must lie on multiple such lines. If there's another point O' that also lies on all these lines, then O' must be the same as O, provided that these lines are not all concurrent elsewhere. But how can we be sure that these lines are not concurrent at another point?Alternatively, maybe we can argue that if O and O' both lie on all these lines AᵢAⱼ, then the lines AᵢAⱼ must all pass through both O and O', which would imply that O and O' are the same point, since two distinct points can't have multiple lines passing through both unless all those lines are the same line, which is impossible in a convex polygon with n ≥ 3 vertices.Wait, let's formalize this. Suppose O and O' are two distinct points inside the polygon such that for every vertex Aᵢ, the line OAᵢ contains another vertex Aⱼ, and similarly, the line O'Aᵢ contains another vertex A_k. Then, both O and O' lie on the lines AᵢAⱼ for various i, j. If O ≠ O', then there must be two different lines AᵢAⱼ and AₖAₗ passing through both O and O'. But if two distinct lines intersect at two distinct points O and O', then those two lines must coincide. However, in a convex polygon, the lines connecting vertices are all distinct edges or diagonals, and since the polygon is convex, the diagonals don't coincide. Therefore, having two different lines AᵢAⱼ and AₖAₗ both passing through O and O' would imply that AᵢAⱼ and AₖAₗ are the same line, which is not possible unless they are the same edge or diagonal, but even then, in a convex polygon, each line is uniquely determined by two vertices. So, if two different lines intersect at two different points, that would require those lines to coincide, which they can't. Therefore, the only possibility is that O and O' coincide. Hence, O is unique.Wait, that seems promising. Let me rephrase that. If we have two distinct points O and O' both lying on multiple lines AᵢAⱼ, then unless all those lines are the same line, O and O' can't both lie on them. But in our case, the lines AᵢAⱼ are different because each line is determined by a different vertex Aᵢ and its corresponding Aⱼ. So, if O and O' are both on all those lines, then all those lines must be the same line, which is impossible. Hence, O and O' must coincide.Alternatively, maybe we can use an argument based on projective geometry or duality, but that might be overcomplicating.Another approach: consider that for each vertex Aᵢ, OAᵢ passes through another vertex Aⱼ. Then, O lies on the line AᵢAⱼ. Since the polygon is convex, the line AᵢAⱼ is either an edge or a diagonal. Now, suppose there is another point O' ≠ O with the same property. Then, O' also lies on each line AᵢAⱼ. Therefore, both O and O' lie on each line AᵢAⱼ. If two points lie on the same line, then either they are the same point, or the line is the same for all those points. But if O and O' are distinct, then all lines AᵢAⱼ must be the same line passing through both O and O'. However, in a convex polygon with at least 3 vertices, there are multiple non-parallel edges and diagonals, so they can't all be the same line. Therefore, the only possibility is that O and O' coincide. Hence, uniqueness is established.Hmm, that seems to make sense. Let me check with an example. Take a convex quadrilateral. Suppose there is a point O inside such that OA₁ passes through A₃, and OA₂ passes through A₄. Then, O is the intersection point of the diagonals A₁A₃ and A₂A₄. In a convex quadrilateral, the diagonals intersect at one point, which is unique. So, in this case, O is the intersection of the diagonals, and there's no other point where the diagonals intersect. Therefore, O is unique. So, in this case, the argument holds.Another example: a convex pentagon. Suppose there is a point O such that every line from O to a vertex passes through another vertex. Then, O must lie at the intersection of multiple diagonals. If another point O' existed with the same property, then O' would have to lie on all those diagonals, which already intersect at O, so O' would have to be O. Therefore, uniqueness follows from the concurrency of the diagonals at O.Wait, but in a general convex polygon, how do we know that the lines AᵢAⱼ for different i and j all intersect at a single point? They might not, unless the polygon has some special properties. For example, in a regular polygon, all the main diagonals pass through the center, but in an irregular convex polygon, this might not be the case. However, the problem states that such a point O exists. So, given that such a point O exists (i.e., the polygon is such that there is a point where all these lines AᵢAⱼ pass through), then we need to show that this point is unique.Therefore, perhaps the key idea is that if two points O and O' both have the property that every line from them to a vertex passes through another vertex, then O and O' must lie on all these lines AᵢAⱼ. But since two distinct points can't lie on multiple distinct lines unless those lines are the same, which is impossible in a convex polygon with n ≥ 3 vertices, then O and O' must coincide.Another way to see it: suppose O and O' are two distinct points with the given property. For each vertex Aᵢ, the line OAᵢ passes through another vertex Aⱼ, so O lies on AᵢAⱼ. Similarly, O' lies on AᵢAₖ for some k. But if O and O' are different, then for each Aᵢ, there must be two different lines AᵢAⱼ and AᵢAₖ passing through O and O' respectively. However, since the polygon is convex, each vertex Aᵢ can only be connected to other vertices through edges and diagonals. If O and O' are different, then for some Aᵢ, the lines OAᵢ and O'Aᵢ would have to pass through different vertices Aⱼ and Aₖ. But then, the lines AᵢAⱼ and AᵢAₖ are different, and O and O' lie on different lines. However, for the property to hold for all vertices, each O and O' must lie on a set of lines connecting each Aᵢ to another Aⱼ. If they are different points, these sets of lines would have to overlap in such a way that both O and O' lie on multiple lines, but this would require overlapping lines, which is not possible in a convex polygon unless O and O' are the same.Wait, maybe this is getting convoluted. Let's try to formalize it.Assume for contradiction that there exist two distinct points O and O' inside the polygon such that every line from O to a vertex passes through another vertex, and the same holds for O'. Then, for each vertex Aᵢ, OAᵢ contains some Aⱼ, and O'Aᵢ contains some Aₖ. Since the polygon is convex, the line OAᵢ is the line segment from O to Aᵢ, passing through Aⱼ. Similarly, O'Aᵢ is the line segment from O' to Aᵢ, passing through Aₖ.Now, since O ≠ O', there exists at least one vertex Aᵢ such that OAᵢ ≠ O'Aᵢ. Otherwise, if OAᵢ = O'Aᵢ for all Aᵢ, then O and O' would lie on the same lines to all vertices, which in a convex polygon would imply they are the same point (since you can triangulate the polygon and determine the point uniquely by the intersection of lines). But let's not assume that.Alternatively, pick a vertex A such that OA and O'A are different lines. Then, OA passes through another vertex B, and O'A passes through another vertex C. Since OA and O'A are different lines (as O ≠ O'), then B ≠ C. So, points O and O' lie on lines AB and AC respectively. But since the polygon is convex, AB and AC are two different edges or diagonals from A. However, both O and O' must satisfy the condition for all vertices, not just A. So, considering another vertex D, the line OD must pass through another vertex E, and O'D must pass through another vertex F. But if O and O' are different, this creates a network of lines that must intersect in a specific way, which might not be possible.Alternatively, consider the set of all lines connecting each vertex to another vertex, which pass through O. Since O is inside the polygon, these lines are chords of the polygon. Similarly for O'. If O and O' are different, there must be at least two different chords passing through O and O'. However, in a convex polygon, two different chords can intersect at most once. If O and O' are both intersection points of multiple chords, then those chords would have to intersect at both O and O', which is impossible unless O = O'.Wait, here's a better approach. Let’s consider that each line OA_i contains another vertex A_j. Therefore, O is the intersection point of several such lines A_iA_j. If another point O' also lies on all these lines, then O' must coincide with O, because two distinct points cannot lie on multiple distinct lines unless those lines are the same. Since in a convex polygon, the lines A_iA_j are distinct (each connecting a different pair of vertices), they can't all pass through two different points O and O'. Hence, O is unique.To make this more precise, suppose O and O' are two points with the given property. Then for each vertex A_i, O lies on A_iA_j for some j, and O' lies on A_iA_k for some k. If O ≠ O', then for some A_i, the line A_iA_j (containing O) and the line A_iA_k (containing O') must be different. Therefore, A_i has two different lines A_iA_j and A_iA_k passing through O and O' respectively. However, since the polygon is convex, these two lines can't coincide unless j = k, which would mean O and O' are on the same line A_iA_j. But even then, if O and O' are distinct points on the same line A_iA_j, then for another vertex A_m, the line OA_m must pass through some vertex A_n, and O'A_m must pass through some vertex A_p. If A_n ≠ A_p, then lines A_mA_n and A_mA_p are different, and both O and O' must lie on these lines. But this would require that lines A_mA_n and A_mA_p intersect at both O and O', which is impossible unless O = O'.Therefore, by considering different vertices and the lines associated with them, we can see that having two distinct points O and O' with the given property would require multiple lines to intersect at two distinct points, which is impossible in a convex polygon. Thus, O must be unique.Another angle: think of O as a point through which many chords of the polygon pass. Each chord connects two vertices. If another point O' also has chords passing through it for every vertex, then O and O' must be the same point because the intersection of these chords is unique. In other words, if all these chords pass through both O and O', then O and O' must coincide.Yes, that seems to sum it up. In a convex polygon, two different points cannot both be the common intersection of all these vertex-connecting lines. Therefore, O is the only such point.Final Answerboxed{O}

✅:To prove that no other point besides ( O ) has the property that any line ( OA_i ) contains another vertex ( A_j ) in a convex polygon ( A_1 ldots A_n ), we can proceed as follows:1. Assumption for Contradiction: Suppose there exists another point ( O' ) inside the polygon such that every line ( O'A_i ) also contains another vertex ( A_k ).2. Lines Through Vertices: For each vertex ( A_i ), the lines ( OA_i ) and ( O'A_i ) must pass through another vertex ( A_j ) and ( A_k ) respectively. This means both ( O ) and ( O' ) lie on the lines ( A_iA_j ) and ( A_iA_k ).3. Concurrency of Lines: If ( O ) and ( O' ) are distinct, they must lie on multiple lines ( A_iA_j ) and ( A_iA_k ). However, in a convex polygon, two distinct lines connecting different pairs of vertices can intersect at most once. 4. Uniqueness of Intersection: If ( O ) and ( O' ) were distinct, they would have to lie on the intersection of multiple lines ( A_iA_j ). This is impossible because two distinct points cannot be the intersection of multiple distinct lines unless those lines coincide, which they do not in a convex polygon.5. Conclusion: The assumption that ( O ) and ( O' ) are distinct leads to a contradiction. Therefore, no such point ( O' ) can exist, and ( O ) must be unique.Thus, the unique point ( O ) is the only point with the given property.[boxed{O}]

🔑:Step 1: Given a convex polygon ( A_1 A_2 ldots A_n ), we have a point ( O ) inside the polygon such that any line segment ( O A_i ) contains another vertex ( A_j ). This indicates that for each vertex ( A_i ), there exists a corresponding vertex ( A_j ) such that ( O ) lies on ( A_i A_j ).Step 2: Since each ( O A_i ) line contains another vertex ( A_j ), the vertices of the polygon can be paired up such that lines ( A_i A_j ) pass through ( O ). These lines are diagonals of the polygon passing through the common point ( O ).Step 3: Because each pair of vertices ((A_i, A_j)) is connected by a diagonal passing through ( O ), the total number of vertices ( n ) must be even. This is because splitting ( n ) vertices into pairs leaves no vertex unpaired, ensuring each diagonal pairs exactly two vertices.Step 4: In an even-sided polygon, if vertices are paired as described, there must be ( m ) such pairs, where ( m ) is half of the total number of vertices ( n ) (( m = frac{n}{2} )).Step 5: For each diagonal ( A_i A_j ) passing through ( O ), there must be an equal number of vertices on either side of the diagonal. This symmetric configuration is only possible if ( O ) is positioned at the unique intersection point of all these diagonals within the polygon.Step 6: Therefore, ( O ) is the intersection point of all diagonals linking pairs of opposite vertices. This point must naturally divide the polygon into two equal parts on each diagonal, which means ( O ) is the unique central point where all diagonals intersect.Step 7: The property that any line ( O A_i ) contains another vertex ( A_j ) cannot be satisfied by any other point except ( O ). This is because ( O ) is the only point ensuring that any diagonal ( A_i A_j ) includes exactly two vertices divided by ( O ).Conclusion: Thus, ( O ) is the only point inside the convex polygon with the stated property, where each vertex is part of a diagonal passing through ( O ).[boxed{O text{ is unique with this property}}]

🧠:Okay, so I need to prove that for positive real numbers x and y where xy = 4, the inequality 1/(x+3) + 1/(y+3) ≤ 2/5 holds. And also find when equality occurs. Hmm, let's start by understanding the problem.First, given that xy = 4, and x, y are positive. So both x and y are greater than zero, and their product is 4. The inequality involves the sum of reciprocals of (x+3) and (y+3). The goal is to show this sum is at most 2/5 and determine when equality happens.Maybe I can use some inequality techniques here. Since there's a condition xy = 4, perhaps substituting y in terms of x (or vice versa) might simplify things. Let's try that. Let me express y as 4/x, since xy = 4, so y = 4/x. Then substitute into the expression.So, substituting y = 4/x into the sum:1/(x + 3) + 1/(4/x + 3) = 1/(x + 3) + 1/(3 + 4/x).Simplify the second term: 1/(3 + 4/x) = x/(3x + 4). So the entire expression becomes 1/(x + 3) + x/(3x + 4). Let's write that as:1/(x + 3) + x/(3x + 4). Now, we need to show that this sum is ≤ 2/5.So, the problem reduces to proving that for x > 0,f(x) = 1/(x + 3) + x/(3x + 4) ≤ 2/5.Alternatively, maybe there's another approach. Since the expression is symmetric in x and y (if we swap x and y, the equation remains the same because xy = 4), perhaps we can assume that x = y? Wait, but if x = y, then x^2 = 4, so x = 2, y = 2. Let's check what the sum becomes in that case.For x = y = 2:1/(2 + 3) + 1/(2 + 3) = 1/5 + 1/5 = 2/5. So that's exactly the upper bound given in the inequality. So maybe equality occurs when x = y = 2. That's a good sign. But we need to confirm that for all other x and y with xy = 4, the sum is less than or equal to 2/5.Alternatively, perhaps we can use the method of Lagrange multipliers, but that might be overcomplicating. Let's think about using the AM-HM inequality or Cauchy-Schwarz.Alternatively, since we have two variables constrained by xy = 4, maybe substituting variables in terms of t and 4/t. Wait, that's similar to substituting y = 4/x.Alternatively, perhaps consider the function f(x) as above and find its maximum for x > 0.Yes, let's try that approach. Let's take f(x) = 1/(x + 3) + x/(3x + 4). Compute its derivative and find critical points.First, compute f'(x):The derivative of 1/(x + 3) is -1/(x + 3)^2.For the second term, x/(3x + 4). Let's use the quotient rule: derivative is [ (1)(3x + 4) - x(3) ] / (3x + 4)^2 = [3x + 4 - 3x] / (3x + 4)^2 = 4/(3x + 4)^2.So f'(x) = -1/(x + 3)^2 + 4/(3x + 4)^2.Set f'(x) = 0 to find critical points:-1/(x + 3)^2 + 4/(3x + 4)^2 = 0Which implies 4/(3x + 4)^2 = 1/(x + 3)^2Take square roots (since denominators are positive):2/(3x + 4) = 1/(x + 3)Cross-multiplying:2(x + 3) = 3x + 42x + 6 = 3x + 4Subtract 2x + 4 from both sides:6 - 4 = 3x - 2x2 = xSo critical point at x = 2. Then y = 4/x = 2. So this is the point we checked earlier. Now, we need to confirm if this is a maximum.Compute the second derivative or test intervals around x = 2.Alternatively, take x slightly less than 2 and x slightly more than 2 and see if f'(x) changes from positive to negative.Take x = 1.9:f'(1.9) = -1/(1.9 + 3)^2 + 4/(3*1.9 + 4)^2Compute denominators:1.9 + 3 = 4.9, so (4.9)^2 ≈ 24.013*1.9 = 5.7, 5.7 + 4 = 9.7, (9.7)^2 ≈ 94.09So f'(1.9) ≈ -1/24.01 + 4/94.09 ≈ -0.0417 + 0.0425 ≈ 0.0008. So positive.Take x = 2.1:f'(2.1) = -1/(2.1 + 3)^2 + 4/(3*2.1 + 4)^22.1 + 3 = 5.1, (5.1)^2 ≈ 26.013*2.1 = 6.3, 6.3 + 4 = 10.3, (10.3)^2 ≈ 106.09So f'(2.1) ≈ -1/26.01 + 4/106.09 ≈ -0.0384 + 0.0377 ≈ -0.0007. Negative.So derivative changes from positive to negative at x = 2, which implies that x = 2 is a local maximum. Since it's the only critical point, it's the global maximum. Therefore, the maximum value of f(x) is f(2) = 2/5, so for all x > 0, f(x) ≤ 2/5, which proves the inequality. Equality occurs when x = y = 2.Alternatively, maybe there's a simpler algebraic approach without calculus. Let's try to manipulate the inequality.We need to show that 1/(x + 3) + 1/(y + 3) ≤ 2/5, given that xy = 4.Let me denote S = x + y and P = xy = 4. Then, since x and y are positive, S ≥ 2√(xy) = 2√4 = 4 by AM ≥ GM. So S ≥ 4.Now, let's express the left-hand side in terms of S and P.1/(x + 3) + 1/(y + 3) = [ (y + 3) + (x + 3) ] / ( (x + 3)(y + 3) )= (x + y + 6) / (xy + 3x + 3y + 9 )Since xy = 4, this becomes:(S + 6) / (4 + 3S + 9) = (S + 6)/(3S + 13)So the inequality becomes:(S + 6)/(3S + 13) ≤ 2/5Multiply both sides by (3S + 13), which is positive since S ≥ 4, so denominator is positive.Thus:S + 6 ≤ (2/5)(3S + 13)Multiply both sides by 5:5(S + 6) ≤ 2(3S + 13)5S + 30 ≤ 6S + 26Subtract 5S + 26:30 - 26 ≤ 6S - 5S4 ≤ SWhich is true since S ≥ 4. So the inequality holds, and equality occurs when S = 4. But S = x + y = 4, and since xy = 4, equality in AM ≥ GM occurs when x = y. Therefore, x = y = 2.Therefore, the inequality holds, and equality occurs at x = y = 2.Wait, that seems a much cleaner approach. Let me verify the steps again.Expressing the sum as (S + 6)/(3S + 13) and then showing (S + 6)/(3S + 13) ≤ 2/5 leads to S ≥ 4, which is true because AM ≥ GM gives S ≥ 4. Therefore, the inequality holds with equality when S = 4, i.e., when x = y = 2.Yes, that works. So this approach avoids calculus and uses algebraic manipulation with substitution based on S and P.Alternatively, let's check if this is valid. Starting from 1/(x + 3) + 1/(y + 3) ≤ 2/5.As we have:1/(x + 3) + 1/(y + 3) = [ (y + 3) + (x + 3) ] / ( (x + 3)(y + 3) )= (x + y + 6) / (xy + 3x + 3y + 9 )Since xy = 4,= (x + y + 6) / (4 + 3x + 3y + 9 )= (x + y + 6)/(3x + 3y + 13 )Let S = x + y, then:= (S + 6)/(3S + 13)So the inequality becomes:(S + 6)/(3S + 13) ≤ 2/5Cross-multiplying (since denominator is positive):5(S + 6) ≤ 2(3S + 13)5S + 30 ≤ 6S + 26Subtract 5S and 26:4 ≤ SWhich is exactly the AM ≥ GM result, since S = x + y ≥ 2√(xy) = 4. Therefore, S ≥ 4, so 4 ≤ S holds, meaning the original inequality is true. Equality occurs when S = 4, which, by AM = GM, occurs when x = y = 2.Therefore, the maximum value of the expression is indeed 2/5, achieved when x = y = 2. Thus, the inequality holds for all positive real numbers x and y with xy = 4, and equality occurs only at x = y = 2.This seems solid. Let me check with an example. Suppose x = 1, then y = 4.Compute 1/(1 + 3) + 1/(4 + 3) = 1/4 + 1/7 ≈ 0.25 + 0.1429 ≈ 0.3929, which is less than 2/5 = 0.4. So that's correct.Another example: x = 4, y = 1. Same as above.Another test case: x = 8, y = 0.5. Then:1/(8 + 3) + 1/(0.5 + 3) = 1/11 + 1/3.5 ≈ 0.0909 + 0.2857 ≈ 0.3766 < 0.4.If x approaches infinity, y approaches 0. Then 1/(x + 3) approaches 0, and 1/(y + 3) approaches 1/3 ≈ 0.3333, which is still less than 0.4.If x approaches 0, y approaches infinity, similar result.At x = 2, y = 2, as we saw, sum is exactly 2/5. So that's the maximum. Therefore, the proof is correct.So, the answer is that the inequality holds, and equality occurs when x = y = 2.Final AnswerThe inequality holds for all positive real numbers ( x ) and ( y ) with ( xy = 4 ), and equality occurs if and only if ( x = y = 2 ). Thus, the final answer is (boxed{x = 2 text{ and } y = 2}).

✅:Given positive real numbers ( x ) and ( y ) such that ( xy = 4 ), we need to prove that the inequality[frac{1}{x+3} + frac{1}{y+3} leq frac{2}{5}]holds and determine when equality occurs.First, we express ( y ) in terms of ( x ) using the given condition ( xy = 4 ), which gives ( y = frac{4}{x} ). Substituting ( y = frac{4}{x} ) into the inequality, we get:[frac{1}{x + 3} + frac{1}{frac{4}{x} + 3} = frac{1}{x + 3} + frac{x}{3x + 4}]We denote this expression as ( f(x) ). To find the maximum value of ( f(x) ), we consider the sum in terms of ( S = x + y ) and ( P = xy = 4 ). Using the AM-GM inequality, we know ( S geq 2sqrt{xy} = 4 ).Rewriting the left-hand side of the inequality in terms of ( S ):[frac{1}{x + 3} + frac{1}{y + 3} = frac{(y + 3) + (x + 3)}{(x + 3)(y + 3)} = frac{x + y + 6}{xy + 3x + 3y + 9} = frac{S + 6}{3S + 13}]We need to show:[frac{S + 6}{3S + 13} leq frac{2}{5}]Cross-multiplying (since the denominator is positive):[5(S + 6) leq 2(3S + 13)]Simplifying:[5S + 30 leq 6S + 26 implies 4 leq S]This inequality holds because ( S geq 4 ) by AM-GM. Equality occurs when ( S = 4 ), which happens if and only if ( x = y ). Given ( xy = 4 ), we find ( x = y = 2 ).Thus, the inequality holds for all positive real numbers ( x ) and ( y ) with ( xy = 4 ), and equality occurs if and only if ( x = y = 2 ).[boxed{x = 2 text{ and } y = 2}]

🔑:Given the conditions for the positive real numbers (x) and (y) such that (xy = 4), we need to prove ( frac{1}{x+3} + frac{1}{y+3} leq frac{2}{5} ) and determine when equality occurs.1. Homogenization: We substitute (x = a^2) and (y = b^2) with (a, b > 0). The condition (xy = 4) translates to (a^2 b^2 = 4) which simplifies to (ab = 2). The inequality becomes: [ frac{1}{a^2 + 3} + frac{1}{b^2 + 3} leq frac{2}{5} ]2. Simplifying the inequality: Using ( ab = 2 ), multiply both terms by (ab) to obtain: [ frac{ab}{a^2 + 3ab} + frac{ab}{b^2 + 3ab} leq frac{2}{5} ] Substitute (ab = 2): [ frac{2}{a^2 + 2 cdot 3} + frac{2}{b^2 + 2 cdot 3} leq frac{2}{5} ]3. Simplify further: [ frac{2}{a^2 + 6} + frac{2}{b^2 + 6} leq frac{2}{5} ] Dividing the inequality by 2: [ frac{1}{a^2 + 6} + frac{1}{b^2 + 6} leq frac{1}{5} ]4. Let ( t = 2a + 3b ) and ( s = 2b + 3a ): Note that: [ 5a = 3s - 2t quad text{and} quad 5b = 3t - 2s ] The inequality becomes: [ frac{t}{2a+3b} + frac{s}{2b+3a} leq frac{2}{5} ] Simplifying, if we assume (t) and (s) then: [ frac{5b}{2a+3b} + frac{5a}{2b+3a} leq 2 ]5. Transform and interpret inequality: Given: [ frac{3t-2s}{t} + frac{3s-2t}{s} leq 2 ] Combine the fractions: [ frac{3t}{t} - frac{2s}{t} + frac{3s}{s} - frac{2t}{s} Rightarrow 3 - 2 left( frac{s}{t} right) + 3 - 2 left( frac{t}{s} right) leq 2 ] Rearrange: [ 6 - 2 left(frac{s}{t} + frac{t}{s}right) leq 2 ] Simplify: [ left(frac{s}{t} + frac{t}{s}right) geq 2 ] This inequality is true by the AM-GM inequality: [ frac{s}{t} + frac{t}{s} geq 2 ]6. Conclusion: Equality holds precisely when: [ frac{s}{t} = frac{t}{s} Rightarrow s = t ] This means (2a + 3b = 2b + 3a), subsequently implying (a = b). Given (ab = 2), this means (a = b = sqrt{2}). Consequently: [ x = y = 2 ] Hence, we conclude that equality holds when (x = y = 2).[boxed{2}]

🧠:Okay, let's try to figure out this problem step by step. So, we have two conditions that the coordinates (x, y) must satisfy. The first one is y² - yx² = 0, and the second one is x⁵ + x⁴ = 0. I need to find all the points (x, y) that satisfy both equations simultaneously. Hmm, let's start with the second equation because it only involves x. Maybe solving that first will simplify things.Looking at the second equation: x⁵ + x⁴ = 0. Let me factor this equation. Both terms have x⁴ in common, right? So if I factor out x⁴, I get x⁴(x + 1) = 0. Now, for a product to be zero, either factor must be zero. So, either x⁴ = 0 or x + 1 = 0. Solving x⁴ = 0, that would mean x = 0, since any number to the fourth power is zero only if the number itself is zero. And solving x + 1 = 0 gives x = -1. So the solutions for x are x = 0 and x = -1. Alright, so now that we have possible x-values, we can substitute these into the first equation to find the corresponding y-values. Let's start with x = 0. Plugging x = 0 into the first equation: y² - y*(0)² = 0. Simplifying that, the second term becomes zero, so we get y² = 0. Taking the square root of both sides, y = 0. So when x = 0, y must also be 0. Therefore, one point is (0, 0).Now, let's check x = -1. Substituting x = -1 into the first equation: y² - y*(-1)² = 0. Since (-1)² is 1, this simplifies to y² - y*1 = 0, which is y² - y = 0. Let's factor this equation. Taking y common, we get y(y - 1) = 0. So, either y = 0 or y - 1 = 0, which means y = 1. Therefore, when x = -1, y can be either 0 or 1. That gives us two more points: (-1, 0) and (-1, 1).Wait, let me double-check these results to make sure I didn't make a mistake. Starting with the second equation: x⁵ + x⁴ = x⁴(x + 1) = 0. Yes, that's correct. So x = 0 (with multiplicity 4) and x = -1. Then substituting into the first equation.For x = 0: y² = 0 implies y = 0. That makes sense. So (0,0) is definitely a solution.For x = -1: y² - y*(1) = y² - y = 0. Factoring gives y(y - 1) = 0, so y = 0 or y = 1. Therefore, points (-1, 0) and (-1, 1). That seems right.Is there any other possible x-value that I might have missed? Let me check the second equation again. x⁵ + x⁴ = 0. Factoring x⁴(x + 1) = 0. The solutions are x = 0 (with multiplicity 4) and x = -1. Since x⁴ = 0 only gives x = 0, regardless of the multiplicity. So there are only two real solutions for x: 0 and -1. Therefore, no other x-values to consider.Now, let's verify if these points actually satisfy both equations. Let's start with (0, 0). Plugging into the first equation: 0² - 0*0² = 0 - 0 = 0, which works. For the second equation: 0⁵ + 0⁴ = 0 + 0 = 0, which is also good.Next, (-1, 0): First equation: 0² - 0*(-1)² = 0 - 0 = 0. Second equation: (-1)^5 + (-1)^4 = -1 + 1 = 0. Perfect.Then (-1, 1): First equation: 1² - 1*(-1)^2 = 1 - 1*1 = 1 - 1 = 0. Second equation: same as before, -1 + 1 = 0. So both equations are satisfied here as well.Therefore, all three points (0,0), (-1,0), and (-1,1) are solutions. Wait a second, let me think if there could be other possibilities. For example, sometimes when equations have higher multiplicities or if we divide by a variable, we might lose solutions. But in this case, for the first equation, y² - yx² = 0, we factored it as y(y - x²) = 0. So either y = 0 or y = x². Did we consider both cases when substituting the x-values?Hold on, maybe I oversimplified earlier. Let me revisit the first equation. The first equation is y² - yx² = 0. Let's factor this properly: y(y - x²) = 0. So either y = 0 or y = x². So actually, for each x, y can be 0 or x². But earlier, I substituted the x-values into the equation and solved for y. Let's see if that approach is equivalent. When x is given, substituting into the equation gives y² - yx² = 0, which is a quadratic in y. Solving for y gives y = 0 or y = x². So, for each x, there are two possible y-values: 0 or x². However, in our case, we found x-values first from the second equation, which are x = 0 and x = -1. Then substituting these x-values into the first equation gives possible y-values. So for x = 0: y = 0 or y = (0)² = 0. So both cases give y = 0, which is why we only have one point here. For x = -1: y = 0 or y = (-1)² = 1. So that's why we have two points here. Therefore, the approach is correct. So the three points are indeed (0,0), (-1,0), and (-1,1). But let's think again: when we solved the second equation first, we restricted x to 0 or -1, and then found the corresponding y-values. But is there another way to approach this? Maybe solving the equations simultaneously. Alternatively, we could solve the first equation for y in terms of x and substitute into the second equation. Let's try that. From the first equation: y² - yx² = 0. So y(y - x²) = 0, leading to y = 0 or y = x². If y = 0, then substitute y = 0 into the second equation. Wait, but the second equation is only in terms of x: x⁵ + x⁴ = 0. So even if y = 0, x still has to satisfy x⁵ + x⁴ = 0. So when y = 0, x can be 0 or -1. That gives us points (0, 0) and (-1, 0). If y = x², then substitute y = x² into the second equation. But the second equation is still x⁵ + x⁴ = 0. So x must be 0 or -1. Then, for x = 0, y = 0² = 0. So that's the point (0, 0). For x = -1, y = (-1)² = 1. So that gives the point (-1, 1). Therefore, the same three points result. So whether we solve for x first or consider the cases for y first, we end up with the same solutions. Therefore, the set of points is (0, 0), (-1, 0), and (-1, 1). I think that's all. Let me just confirm once more. Let me check if there's any other x that could satisfy the second equation. The equation x⁵ + x⁴ = 0 is equivalent to x⁴(x + 1) = 0. The real solutions are x = 0 and x = -1. There are complex solutions as well, but since we're dealing with the coordinate plane, which is real numbers, we only consider real x and y. So no other x-values. Therefore, these three points are the only solutions. Final AnswerThe set of points is (boxed{{(0, 0), (-1, 0), (-1, 1)}}).

✅:To find the set of points on the coordinate plane where the coordinates (x) and (y) satisfy the given conditions, we start by analyzing each equation step by step.1. Solving the second equation (x^5 + x^4 = 0): - Factor the equation: (x^4(x + 1) = 0). - This gives the solutions (x = 0) (with multiplicity 4) and (x = -1).2. Substituting the values of (x) into the first equation (y^2 - yx^2 = 0): - For (x = 0): - Substitute (x = 0) into the equation: (y^2 - y cdot 0^2 = 0). - This simplifies to (y^2 = 0), giving (y = 0). - Thus, the point is ((0, 0)). - For (x = -1): - Substitute (x = -1) into the equation: (y^2 - y cdot (-1)^2 = 0). - This simplifies to (y^2 - y = 0). - Factor the equation: (y(y - 1) = 0), giving (y = 0) or (y = 1). - Thus, the points are ((-1, 0)) and ((-1, 1)).3. Verification: - Check if the points ((0, 0)), ((-1, 0)), and ((-1, 1)) satisfy both equations: - For ((0, 0)): Both equations are satisfied. - For ((-1, 0)): Both equations are satisfied. - For ((-1, 1)): Both equations are satisfied.After thorough analysis and verification, the set of points that satisfy both conditions is:[boxed{{(0, 0), (-1, 0), (-1, 1)}}]

🔑:1. Start by analyzing the first equation ( y^2 - y x^2 = 0 ). [ y^2 - y x^2 = 0 ] Factor out ( y ) from the equation: [ y(y - x^2) = 0 ] This gives us two solutions: [ y = 0 quad text{or} quad y = x^2 ] These correspond to two geometric shapes on the coordinate plane: - ( y = 0 ) (the x-axis) - ( y = x^2 ) (a parabola)2. Next, analyze the second equation ( x^5 + x^4 = 0 ). [ x^5 + x^4 = 0 ] Factor out ( x^4 ) from the equation: [ x^4(x + 1) = 0 ] This gives us two solutions: [ x^4 = 0 quad text{or} quad x + 1 = 0 ] Which further translates to: [ x = 0 quad text{or} quad x = -1 ] These correspond to vertical lines: - ( x = 0 ) (the y-axis) - ( x = -1 ) (a vertical line one unit to the left of the y-axis)3. Using these results, find the intersection of the solutions of the two sets of conditions. - For ( y = 0 ) (the x-axis): - If ( x = 0 ), then the point is ( (0, 0) ). - If ( x = -1 ), then the point is ( (-1, 0) ). - For ( y = x^2 ) (the parabola): - If ( x = 0 ), substituting in ( y = x^2 ), we get the point ( (0, 0) ). - If ( x = -1 ), substituting in ( y = x^2 ), we get ( y = 1 ), thus the point ( (-1, 1) ). Conclusion:The points that satisfy both equations are:1. ( (0,0) )2. ( (-1,0) )3. ( (-1,1) )[boxed{(0,0), (-1,0), (-1,1)}]

🧠:Okay, so I need to prove that the sum S of all products of 1000 numbers out of 1959 positive numbers that add up to 1 is less than 1. Hmm, let's start by understanding the problem.First, we have 1959 positive numbers a₁, a₂, ..., a₁₉₅₉, and their sum is 1. Then, we're looking at all possible combinations of 1000 of these numbers, multiplying them together, and summing all those products. That sum is S, and we need to show that S < 1.Alright, let me recall some combinatorial identities. The sum of all possible products of k distinct terms from n variables is the elementary symmetric sum of degree k, denoted as e_k(a₁, a₂, ..., a_n). So in this case, S = e₁₀₀₀(a₁, a₂, ..., a₁₉₅₉). The problem states that this sum is less than 1.Given that all a_i are positive and sum to 1, maybe we can use some inequality related to symmetric sums. I remember that for symmetric sums, when variables are non-negative, there are inequalities like Newton's inequalities, or perhaps using AM ≥ GM.Wait, Newton's inequalities relate elementary symmetric sums of different degrees. For example, they might relate e_k / C(n, k) to e_{k-1}/C(n, k-1), etc. But I need to recall the exact form. Newton's inequalities state that for each k, (e_k / C(n, k))^2 ≥ (e_{k-1}/C(n, k-1)) * (e_{k+1}/C(n, k+1)). But how would that help here?Alternatively, maybe I can use the fact that the product of numbers is maximized when all numbers are equal, but since the sum is fixed, the maximum product under sum constraint would be when variables are equal. But here, we are dealing with products of subsets, not all variables. Hmm.Wait, if all the a_i are equal, then each a_i = 1/1959. Then, each product of 1000 terms would be (1/1959)^1000, and the number of such terms is C(1959, 1000). So S would be C(1959, 1000) * (1/1959)^1000. If I can show that this is less than 1, and that this is the maximum possible S, then maybe that would work.But is this the case? If the variables are unequal, does the symmetric sum e_k increase or decrease? I think that symmetric sums are maximized when variables are equal. Wait, actually, for fixed sum, the elementary symmetric sums are maximized when all variables are equal. Is that a theorem?Yes, I recall that the elementary symmetric sums are Schur-concave, which would imply that they are maximized when the variables are equal. So if all variables are equal, then e_k is maximized. Therefore, in this case, if we set all a_i = 1/1959, then S would be equal to C(1959, 1000)*(1/1959)^1000, and if we can show that this is less than 1, then since any other distribution of a_i would give a smaller S, then S < 1 in general.Therefore, the strategy would be to compute S when all a_i are equal and show that this is less than 1, thereby proving the inequality.So let's compute C(1959, 1000)*(1/1959)^1000. But computing such a huge binomial coefficient is not feasible directly. However, maybe we can approximate it or use an inequality.Alternatively, note that the sum S is equal to the coefficient of x¹⁰⁰⁰ in the generating function (1 + a₁x)(1 + a₂x)...(1 + a₁₉₅₉x). Since all a_i sum to 1, maybe we can bound this generating function.But if all a_i are equal to 1/1959, then the generating function becomes (1 + x/1959)^1959. Then, the coefficient of x¹⁰⁰⁰ is C(1959, 1000)*(1/1959)^1000. So again, we need to estimate this.Alternatively, using the inequality that (1 + 1/n)^n < e, but here we have (1 + x/1959)^1959, which for x = 1 would approach e. But here, x is a variable, but when expanding, the coefficients are related to moments.Wait, perhaps using the binomial theorem. The sum S is equal to the sum over all subsets of size 1000 of the product of a_i's. If all a_i are equal, then S = C(1959, 1000)*(1/1959)^1000. We can compare this to 1.But how can we estimate this value? Let's note that C(n, k) is largest when k is around n/2. Here, n=1959, so n/2 is approximately 979.5. So 1000 is a bit larger than half of 1959. So C(1959, 1000) is a very large number, but multiplied by (1/1959)^1000, which is a very small number. The question is whether the product is less than 1.Alternatively, perhaps using logarithms. Let's compute ln(C(1959, 1000)) - 1000*ln(1959). If this is less than 0, then the original term is less than 1.To approximate ln(C(n, k)), we can use Stirling's formula: ln(n!) ≈ n ln n - n + (ln n)/2 - (ln(2π))/2. Similarly for ln(k!) and ln((n - k)!).So, ln(C(1959, 1000)) = ln(1959!) - ln(1000!) - ln(959!). Applying Stirling's approximation:ln(1959!) ≈ 1959 ln 1959 - 1959 + (ln 1959)/2 - (ln(2π))/2Similarly,ln(1000!) ≈ 1000 ln 1000 - 1000 + (ln 1000)/2 - (ln(2π))/2ln(959!) ≈ 959 ln 959 - 959 + (ln 959)/2 - (ln(2π))/2Therefore,ln(C(1959, 1000)) ≈ [1959 ln 1959 - 1959 + (ln 1959)/2 - (ln(2π))/2] - [1000 ln 1000 - 1000 + (ln 1000)/2 - (ln(2π))/2] - [959 ln 959 - 959 + (ln 959)/2 - (ln(2π))/2]Simplify term by term:First, the 1959 ln 1959 - 1000 ln 1000 - 959 ln 959Then, the constants: -1959 + 1000 + 959 = 0The logarithmic terms: (ln 1959)/2 - (ln 1000)/2 - (ln 959)/2The constants involving ln(2π)/2: - (ln(2π))/2 + (ln(2π))/2 + (ln(2π))/2 = (ln(2π))/2Wait, let's check:Original expression:ln(C(1959, 1000)) ≈ [1959 ln 1959 - 1959 + (ln 1959)/2 - (ln(2π))/2] minus [1000 ln 1000 - 1000 + (ln 1000)/2 - (ln(2π))/2] minus [959 ln 959 - 959 + (ln 959)/2 - (ln(2π))/2]So expanding this:= 1959 ln 1959 - 1959 + (ln 1959)/2 - (ln(2π))/2 - 1000 ln 1000 + 1000 - (ln 1000)/2 + (ln(2π))/2 - 959 ln 959 + 959 - (ln 959)/2 + (ln(2π))/2Now, combining the terms:The coefficients:1959 ln 1959 - 1000 ln 1000 - 959 ln 959Constants:-1959 + 1000 + 959 = 0Logarithmic terms:(ln 1959)/2 - (ln 1000)/2 - (ln 959)/2And the constants involving ln(2π):- (ln(2π))/2 + (ln(2π))/2 + (ln(2π))/2 = (ln(2π))/2Therefore, overall:ln(C(1959, 1000)) ≈ 1959 ln 1959 - 1000 ln 1000 - 959 ln 959 + (ln 1959 - ln 1000 - ln 959)/2 + (ln(2π))/2Now, let's compute the main terms:Let me write:Term1 = 1959 ln 1959 - 1000 ln 1000 - 959 ln 959Term2 = (ln 1959 - ln 1000 - ln 959)/2Term3 = (ln(2π))/2 ≈ (1.837877)/2 ≈ 0.9189385So we need to compute Term1 + Term2 + Term3.First, compute Term1:Let me note that 1959 = 1000 + 959.Let’s denote x = 1000, y = 959, so that x + y = 1959.Then Term1 = (x + y) ln(x + y) - x ln x - y ln yWhich is equal to x [ln(x + y) - ln x] + y [ln(x + y) - ln y]= x ln(1 + y/x) + y ln(1 + x/y)This is similar to the entropy expression.But let's compute the numerical value.First, x = 1000, y = 959.Compute ln(1959) ≈ ln(2000 - 41) ≈ ln(2000) - 41/2000 ≈ 7.6009024595 - 0.0205 ≈ 7.5804Wait, let me use calculator-like approximations.But maybe using exact values:ln(1959) ≈ ln(2000) - ln(2000/1959) ≈ 7.6009024595 - ln(1.021) ≈ 7.6009 - 0.0208 ≈ 7.5801Similarly,ln(1000) = 6.907755278ln(959) ≈ ln(1000 - 41) ≈ ln(1000) - 41/1000 ≈ 6.907755 - 0.041 ≈ 6.866755Therefore,Term1 = 1959 * 7.5801 - 1000 * 6.907755 - 959 * 6.866755Compute each term:1959 * 7.5801 ≈ 1959 * 7.5 + 1959 * 0.0801 ≈ 14692.5 + 156.9159 ≈ 14849.41591000 * 6.907755 = 6907.755959 * 6.866755 ≈ (960 - 1) * 6.866755 ≈ 960 * 6.866755 - 6.866755 ≈ 6592.0848 - 6.866755 ≈ 6585.217Therefore, Term1 ≈ 14849.4159 - 6907.755 - 6585.217 ≈ 14849.4159 - 13492.972 ≈ 1356.4439Now, Term2 = (ln 1959 - ln 1000 - ln 959)/2 ≈ (7.5801 - 6.907755 - 6.866755)/2 ≈ (7.5801 - 13.77451)/2 ≈ (-6.19441)/2 ≈ -3.097205Term3 ≈ 0.9189385Therefore, total approximation:ln(C(1959,1000)) ≈ 1356.4439 - 3.097205 + 0.9189385 ≈ 1356.4439 - 2.1782665 ≈ 1354.2656Then, ln(S) when all a_i equal is ln(C(1959,1000)) - 1000 ln(1959) ≈ 1354.2656 - 1000 * 7.5801 ≈ 1354.2656 - 7580.1 ≈ -6225.8344Therefore, S ≈ e^{-6225.8344} which is an extremely small number, way less than 1. Hence, when all a_i are equal, S is astronomically small, much less than 1. Therefore, if S is maximized when all a_i are equal, then S < 1.But wait, this seems contradictory. If all a_i are equal, then S is very small. But maybe when some a_i are larger and others are smaller, S can be larger? Wait, but I thought the symmetric sum is maximized when variables are equal. Hmm.Wait, maybe I got that wrong. Let me check. Suppose we have two variables, a and b, with a + b = 1. Then, e₁ = a + b = 1, e₂ = ab. ab is maximized when a = b = 0.5, right? So in that case, symmetric sum e₂ is maximized when variables are equal. Similarly, for three variables, the product abc is maximized when a = b = c = 1/3, given that a + b + c = 1.But in our problem, the sum of a_i is 1, and we are looking at e_k, the sum of products of k variables. So if k is fixed, and sum of variables is fixed, then e_k is maximized when all variables are equal? That seems to be the case based on the two-variable example.But here, with 1959 variables and k=1000, which is more than half of 1959. Wait, but in the case of n variables and k > n/2, does the symmetric sum still get maximized when variables are equal?Wait, let's take a small example. Let n=3, sum a + b + c =1. Compare e₂ when variables are equal vs when one variable is larger.If all equal, e₂ = 3*(1/3)^2 = 1/3 ≈ 0.333.If one variable is 1 - 2ε and others are ε each, then e₂ = (1 - 2ε)*ε + ε*(1 - 2ε) + ε*ε = 2ε(1 - 2ε) + ε² = 2ε -4ε² + ε² = 2ε -3ε².For small ε, this is approximately 2ε, which can be made larger than 1/3? Wait, when ε approaches 0, e₂ approaches 0, so actually making variables unequal in this case reduces e₂. Hence, even for k > n/2, the maximum e_k occurs at equal variables.Wait, but n=3, k=2: maximum e₂ is at equal variables. Hmm.Wait, perhaps for all k, the maximum e_k under sum constraint is achieved when all variables are equal.Yes, I think this is a result from the theory of symmetric functions. The elementary symmetric sums are Schur-concave, which means that they are maximized when the variables are equal, given a fixed sum. So, the maximum value of e_k(a₁,...,a_n) under the constraint that a₁ + ... + a_n = 1 is achieved when all a_i = 1/n.Therefore, in our problem, S = e₁₀₀₀(a₁,...,a₁₉₅₉) ≤ e₁₀₀₀(1/1959, ..., 1/1959). Then, if we can show that e₁₀₀₀(1/1959, ..., 1/1959) < 1, then S < 1 in general.But as calculated before, when all a_i are equal, S is extremely small, like e^{-6225}, which is way less than 1. Therefore, the inequality holds. Therefore, S < 1.But the problem states to prove S < 1, which seems almost trivial if the maximum possible S is already so tiny. Maybe there's a different approach here.Wait, maybe I made a mistake in assuming that maximum e_k is achieved at equal variables? Wait, let me check with n=4, k=3.Suppose we have four variables a, b, c, d with a + b + c + d =1. What's the maximum of e₃ = abc + abd + acd + bcd.If all variables are equal, e₃ = 4*(1/4)^3 = 4*(1/64) = 1/16 ≈0.0625.If we set three variables to ε and one variable to 1 - 3ε. Then, e₃ = 3*ε²*(1 - 3ε) + ε³.For ε approaching 0, e₃ ≈ 3ε². If we set ε = 1/4, then 1 - 3ε = 1 - 3/4 =1/4. Then, e₃ = 3*(1/4)^2*(1/4) + (1/4)^3 = 3/64 +1/64=4/64=1/16, same as equal case.If we take ε = 0.1, then 1 - 3*0.1 =0.7. Then, e₃=3*(0.1)^2*0.7 + (0.1)^3=3*0.01*0.7 +0.001=0.021 +0.001=0.022, which is less than 0.0625.If we take ε=0.25, same as before. If we take another distribution, say two variables as 0.4 and two variables as 0.1. Then, e₃ = (0.4*0.4*0.1)*2 + (0.4*0.1*0.1)*2 = 0.016*2 +0.004*2=0.032 +0.008=0.04, which is still less than 0.0625.Alternatively, set two variables to 0.5 and the other two to 0. Then, e₃ = 0 (since two variables are 0). That's even worse.Alternatively, set one variable to 0.9 and the rest to 0.0333. Then, e₃ = C(3,2)*(0.0333)^2*0.9 + (0.0333)^3 ≈ 3*(0.00111)*0.9 + 0.000037 ≈ 0.003 +0.000037≈0.003, which is still less than 0.0625.So in this case, even for k=3, n=4, the maximum e₃ is achieved when all variables are equal. Therefore, it seems that the maximum of e_k is achieved when all variables are equal, regardless of k.Therefore, in our original problem, since S = e₁₀₀₀(a₁,…,a₁₉₅₉) ≤ e₁₀₀₀(1/1959,…,1/1959), and e₁₀₀₀(1/1959,…,1/1959) = C(1959,1000)*(1/1959)^1000, which is extremely small, so S <1.But maybe the problem is expecting a different approach, perhaps using generating functions or probabilistic methods.Alternatively, note that S is the probability that a randomly selected subset of 1000 variables all happen to be included, but that doesn't directly make sense.Alternatively, think of the generating function. The product (1 + a₁)(1 + a₂)...(1 + a_n) expands to 1 + e₁ + e₂ + ... + e_n. If we set each a_i to a_i x, then the generating function becomes product(1 + a_i x), and e_k is the coefficient of x^k. So S is the coefficient of x¹⁰⁰⁰ in product(1 + a_i x).But how can we bound this coefficient?Alternatively, note that for any x > 0, we have:product(1 + a_i x) ≥ sum_{k=0}^{1959} e_k x^kBut this might not help directly.Alternatively, use the inequality that product(1 + a_i) ≤ e^{sum a_i}, which is 1 + sum a_i ≤ e^{sum a_i}. But here, product(1 + a_i x) ≤ e^{x sum a_i} = e^x.So, product(1 + a_i x) ≤ e^x for all x >0.But then, the coefficient of x¹⁰⁰⁰ in product(1 + a_i x) is S, which is less than the coefficient of x¹⁰⁰⁰ in e^x, which is 1/1000!.But 1/1000! is much less than 1, so that would also give S < 1. But this seems too crude.Wait, but the inequality product(1 + a_i x) ≤ e^{x sum a_i} holds term-wise? Not exactly. The inequality 1 + a_i x ≤ e^{a_i x} holds for each term, since e^{a_i x} ≥ 1 + a_i x. Therefore, the product over i of (1 + a_i x) ≤ product over i of e^{a_i x} = e^{x sum a_i} = e^x.Therefore, the entire generating function is ≤ e^x. Therefore, each coefficient e_k ≤ 1/k! (since e^x = sum_{k=0}^infty x^k /k! )But wait, that would mean that e_k ≤ 1/k! , but in reality, for our case, e_k is much larger than 1/k! because the variables are scaled. Wait, no, the coefficients in e^x are 1/k! for the expansion, but in our generating function, the coefficients are e_k. If e_k x^k ≤ e^x for all x >0, but this isn't necessarily true. The inequality product(1 + a_i x) ≤ e^x holds for all x >0, but coefficient-wise, it's not straightforward.However, we can use the fact that for any positive x, S x^{1000} ≤ product(1 + a_i x) ≤ e^x.Therefore, S x^{1000} ≤ e^x for all x >0.Therefore, S ≤ e^x / x^{1000} for all x >0.To find the minimal upper bound, we can minimize e^x / x^{1000} over x >0.Let’s compute the minimum of f(x) = e^x / x^{1000}.Take natural logarithm: ln f(x) = x - 1000 ln x.To find the minimum, take derivative: d/dx (x - 1000 ln x) = 1 - 1000/x.Set to zero: 1 - 1000/x =0 => x = 1000.Therefore, the minimum occurs at x =1000, so the minimal value of f(x) is e^{1000}/1000^{1000}.Therefore, S ≤ e^{1000}/1000^{1000}.But e^{1000}/1000^{1000} = (e / 1000)^{1000} ≈ (2.718 / 1000)^{1000} ≈ (0.002718)^{1000}, which is again an extremely small number, much less than 1. Therefore, S <1.But this approach gives a very loose upper bound, even smaller than the previous one, but it's sufficient to prove S <1.Alternatively, maybe use induction or some combinatorial argument.But given that both approaches—using the maximum when variables are equal and using the generating function bound—lead to S being astronomically small, the inequality S <1 is certainly true.Wait, but the problem says "prove that S <1", and maybe there's a more elegant way rather than computing these bounds. Let me think.Consider that the sum of all products of 1000 terms is equal to the elementary symmetric sum of degree 1000. There is an identity related to the sum of all elementary symmetric sums: the product (1 + a₁)(1 + a₂)...(1 + a_n) = 1 + e₁ + e₂ + ... + e_n. Therefore, the sum S is e₁₀₀₀, and the total product is 1 + e₁ + e₂ + ... + e₁₉₅₉.But since all a_i are positive, each e_k is positive. Therefore, 1 + e₁ + ... + e₁₉₅₉ = product_{i=1 to 1959} (1 + a_i). But the product is greater than 1 + sum a_i =1 +1=2, by the inequality product(1 + a_i) ≥ 1 + sum a_i for non-negative a_i. So, 1 + S + ... + e₁₉₅₉ ≥ 2. Therefore, the total sum of all e_k is at least 1. But this doesn't directly help, since S is just one term.Alternatively, note that if we consider the sum of all e_k for k=0 to 1959, it's equal to product(1 + a_i) = product(1 + a_i). Since each a_i ≤1 (since sum a_i=1 and each a_i is positive), then (1 + a_i) ≤2, so product(1 + a_i) ≤2¹⁹⁵⁹. But this is a huge number, and doesn't help.Alternatively, use convexity. The function f(t) = ln(product(1 + a_i t)) is convex, and we can use some properties of convex functions to bound the coefficients.Alternatively, think probabilistically. Imagine that each a_i is the probability of some event. Wait, but sum a_i =1, so they can be considered as probabilities of a discrete distribution. Then, the sum S is the sum over all size 1000 subsets of the product of their probabilities. But I'm not sure how to interpret this.Alternatively, consider the complement: instead of choosing 1000 elements, choose 959 elements. Since the sum of all products of 1000 elements is equal to the sum of all products of 959 elements (since each product of 1000 elements corresponds to leaving out 959 elements). Therefore, S = e₁₀₀₀ = e_{1959 - 1000} = e₉₅₉. There's a symmetry in elementary symmetric sums: e_k = e_{n -k} if we replace each a_i with a_i / (1 + a_i), but not sure.Wait, actually, for a set of variables a₁, ..., a_n, the elementary symmetric sums satisfy e_k(a₁, ..., a_n) = e_{n -k}(1/a₁, ..., 1/a_n) * (a₁...a_n), but this might not be helpful here.Alternatively, use the inequality between elementary symmetric sums and power sums. Maybe using Maclaurin's inequality, which states that (e_k / C(n, k))^{1/k} ≤ (e_{k-1}/C(n, k-1))^{1/(k-1)}}But starting from e₁ =1, so (e₁ / C(n, 1)) =1/n. Then, (e₂ / C(n, 2))^{1/2} ≤ 1/n, so e₂ ≤ C(n, 2)/n² = (n(n -1)/2)/n² = (n -1)/(2n) ≈ 1/2 for large n. Continuing this, e_k / C(n, k) ≤ (1/n)^k, so e_k ≤ C(n, k)/n^k. Wait, but C(n, k)/n^k is exactly e_k when all a_i =1/n. So, this is again the same result as before, that e_k ≤ C(n, k)/n^k, which is exactly the case when all a_i are equal. Therefore, Maclaurin's inequality gives us that e_k ≤ C(n, k)/n^k, hence S = e₁₀₀₀ ≤ C(1959, 1000)/1959¹⁰⁰⁰. And since we have already established that this is much less than 1, the result follows.Therefore, applying Maclaurin's inequality directly gives us the required result. Maclaurin's inequality states that for positive real numbers a₁, a₂, ..., a_n,(e_k / C(n, k))^{1/k} ≤ (e₁ / n)In our case, e₁ =1, so (e_k / C(n, k))^{1/k} ≤ 1/n, which implies e_k ≤ C(n, k)/n^k. Therefore, S = e₁₀₀₀ ≤ C(1959, 1000)/1959¹⁰⁰⁰ <1.But to show that C(1959, 1000)/1959¹⁰⁰⁰ <1, which is equivalent to C(1959, 1000) <1959¹⁰⁰⁰. But C(1959, 1000) =1959!/(1000! 959!) and 1959¹⁰⁰⁰ is much larger. However, given that when all a_i are equal, the product term C(1959, 1000)*(1/1959)^1000 is extremely small, which is S in that case. So the inequality e_k ≤ C(n, k)/n^k holds, and since the RHS is very small, S <1.Therefore, using Maclaurin's inequality gives the result immediately without needing to compute the exact value.Hence, the proof is as follows:By Maclaurin's inequality, for positive real numbers a₁, a₂, ..., a_n summing to 1, the elementary symmetric sum of degree k satisfiese_k ≤ (binom{n}{k} left(frac{1}{n}right)^k).Applying this to our case with n=1959 and k=1000,S = e₁₀₀₀ ≤ (binom{1959}{1000} left(frac{1}{1959}right)^{1000}).Now, note that(binom{1959}{1000} left(frac{1}{1959}right)^{1000} = frac{1959!}{1000! cdot 959!} cdot frac{1}{1959^{1000}}).This expression represents the probability of selecting 1000 distinct objects out of 1959 with replacement, but actually, it's the term from the binomial distribution. However, regardless of interpretation, since each term in the product (1 + a₁)(1 + a₂)...(1 + a_{1959}) is less than or equal to e^{a_i} as before, and the entire product is less than e^1, summing all e_k gives the total product less than e, hence each e_k is less than e. But this line of thought is unnecessary because Maclaurin's inequality directly gives the required bound.Moreover, since (binom{n}{k} leq n^k /k!), we have(binom{1959}{1000} leq frac{1959^{1000}}{1000!}).Therefore,(binom{1959}{1000} left(frac{1}{1959}right)^{1000} leq frac{1}{1000!}).Since 1000! is a huge number, much larger than 1, hence S ≤ 1/1000! <1.Therefore, S <1, as required.Alternatively, since all terms in the expansion of (1 + 1/n)^n are positive, and their sum is (1 + 1/n)^n <e, but this also might not directly apply.But in any case, multiple approaches confirm that S is bounded above by a quantity much less than 1, hence S <1.Therefore, the conclusion is that S <1, proven via Maclaurin's inequality or by considering the maximum of the symmetric sum when variables are equal.

✅:To prove that ( S < 1 ), where ( S ) is the sum of all products of 1000 numbers from the set ( {a_1, a_2, ldots, a_{1959}} ) with ( a_i > 0 ) and ( sum_{i=1}^{1959} a_i = 1 ), we utilize Maclaurin's inequality. This inequality states that for positive real numbers and any ( 1 leq k leq n ):[left( frac{e_k}{binom{n}{k}} right)^{1/k} leq frac{e_1}{n},]where ( e_k ) is the elementary symmetric sum of degree ( k ). Given ( e_1 = sum_{i=1}^{1959} a_i = 1 ), applying Maclaurin's inequality yields:[e_k leq binom{n}{k} left( frac{e_1}{n} right)^k = binom{1959}{1000} left( frac{1}{1959} right)^{1000}.]The term ( binom{1959}{1000} left( frac{1}{1959} right)^{1000} ) represents the symmetric sum when all ( a_i = frac{1}{1959} ). This value is extremely small due to the vast difference between the combinatorial coefficient and ( 1959^{1000} ). Specifically, using Stirling's approximation or recognizing ( binom{1959}{1000} leq frac{1959^{1000}}{1000!} ), we find:[binom{1959}{1000} left( frac{1}{1959} right)^{1000} leq frac{1}{1000!} ll 1.]Thus, ( S leq frac{1}{1000!} < 1 ), proving ( S < 1 ).Answer: boxed{S < 1}

🔑:Consider the 1959 positive numbers denoted by ( a_1, a_2, ldots, a_{1959} ) such that the sum of these numbers is 1. We aim to prove that the sum of the products of every combination of 1000 of these numbers, denoted by ( S ), is less than 1. 1. Step 1: Expressing the Sum Raised to the 1000th Power We start by using the power sum of these numbers: [ 1 = left( sum_{i=1}^{1959} a_i right)^{1000} ] This can be expanded as: [ left( sum_{i=1}^{1959} a_i right)^{1000} = sum frac{1959!}{k_1! cdot k_2! cdots k_{1959}!} a_1^{k_1} cdot a_2^{k_2} cdots a_{1959}^{k_{1959}} ] Here, the sum is taken over all non-negative integer combinations ( k_1, k_2, ldots, k_{1959} ) such that their sum is ( k_1 + k_2 + cdots + k_{1959} = 1000 ).2. Step 2: Understanding the Expression Within the expansion: - Each term (frac{1959!}{k_1! cdot k_2! cdots k_{1959}!} a_1^{k_1} cdot a_2^{k_2} cdot ldots cdot a_{1959}^{k_{1959}}) represents a specific combination of the (a_i) terms, with the multinomial coefficient accounting for the different ways these combinations can be selected.3. Step 3: Relation to the Combinations of Products The relevant part of the expansion we need to consider involves terms where the exponents ( k_i ) are either (0) or (1), because we are interested in the sum of products of 1000 out of 1959 numbers: [ S = sum_{1 leq i_1 < i_2 < cdots < i_{1000} leq 1959} a_{i_1} a_{i_2} cdots a_{i_{1000}} ] This sum ( S ) is one part of the full expansion of (left( sum_{i=1}^{1959} a_i right)^{1000}).4. Step 4: Comparing the Sum of Products The sum ( S ) is specifically captured within the expansion and it is part of the overall sum defined by the multinomial expansion. Hence, [ S < frac{1}{1959!} ]5. Conclusion: Given that the sum of the relevant products is ( S ) and noting the factorial comparison, [ frac{1}{1959!} < 1, ] it follows directly that [ S < frac{1}{1959!} < 1. ]Therefore, we conclude the proof by stating [boxed{S<1}.]