🧠:Alright, let's see. I need to find the value of ( x - sqrt{x^2 - 1} ) given that ( x + sqrt{x^2 - 1} = a^{frac{m - n}{2mn}} ). Hmm, okay. So, the equation given is ( x + sqrt{x^2 - 1} = a^{frac{m - n}{2mn}} ). And I need to find ( x - sqrt{x^2 - 1} ). First, I remember that expressions like ( x + sqrt{x^2 - 1} ) and ( x - sqrt{x^2 - 1} ) often have a reciprocal relationship. Let me think. If I let ( y = x + sqrt{x^2 - 1} ), then maybe ( x - sqrt{x^2 - 1} ) is something like ( 1/y )? Let me check that.Suppose ( y = x + sqrt{x^2 - 1} ). Then, multiplying ( y ) by ( x - sqrt{x^2 - 1} ), we get:( y cdot (x - sqrt{x^2 - 1}) = (x + sqrt{x^2 - 1})(x - sqrt{x^2 - 1}) ).Using the difference of squares formula, this is ( x^2 - (sqrt{x^2 - 1})^2 = x^2 - (x^2 - 1) = 1 ).So, ( y cdot (x - sqrt{x^2 - 1}) = 1 ), which implies that ( x - sqrt{x^2 - 1} = 1/y ).Therefore, if ( x + sqrt{x^2 - 1} = a^{frac{m - n}{2mn}} ), then ( x - sqrt{x^2 - 1} ) should be the reciprocal of that, which is ( a^{-frac{m - n}{2mn}} ). But let's simplify the exponent. The negative exponent would flip the fraction, so:( a^{-frac{m - n}{2mn}} = a^{frac{n - m}{2mn}} ). Looking at the options, option C is ( a^{frac{n - m}{2mn}} ), which matches this. But let me double-check to make sure I didn't make a mistake.Wait, let me confirm the reciprocal relationship again. If ( y = x + sqrt{x^2 -1} ), then indeed ( x - sqrt{x^2 -1} = 1/y ). So, since ( y = a^{frac{m -n}{2mn}} ), then ( 1/y = a^{-frac{m -n}{2mn}} ). Let me handle the exponent carefully. The exponent is ( -frac{m - n}{2mn} ). Let's distribute the negative sign:( -frac{m - n}{2mn} = frac{ -m + n }{2mn } = frac{ n - m }{2mn } ). Yes, that's correct. So the reciprocal is ( a^{frac{n - m}{2mn}} ), which is option C. Let me check the options again.Options:(A) ( a^{frac{2mn}{m-n}} )(B) ( a^{frac{2mn}{n-m}} )(C) ( a^{frac{n-m}{2mn}} )(D) ( a^{frac{n+m}{2mn}} )Yes, option C is exactly what I derived. Hmm. Wait, but I need to make sure that there isn't another way to interpret the problem. Let me go through the steps again.Given that ( x + sqrt{x^2 - 1} = a^{frac{m - n}{2mn}} ), and we need ( x - sqrt{x^2 -1} ).We established that ( x - sqrt{x^2 -1} = 1/(x + sqrt{x^2 -1}) ). Therefore, substituting the given value:( x - sqrt{x^2 -1} = 1/a^{frac{m - n}{2mn}} = a^{-frac{m - n}{2mn}} ).Simplifying the exponent:( -frac{m - n}{2mn} = frac{n - m}{2mn} ).Therefore, the result is ( a^{frac{n - m}{2mn}} ), which is option C. That seems right.Alternatively, maybe the problem is expecting some substitution or manipulation with x. Let me try solving for x.Let me denote ( y = x + sqrt{x^2 -1} = a^{frac{m -n}{2mn}} ). Then, as before, ( x - sqrt{x^2 -1} = 1/y ).But perhaps if I want to express x in terms of y, I can add the two equations:( y + 1/y = [x + sqrt{x^2 -1}] + [x - sqrt{x^2 -1}] = 2x ).Therefore, ( x = frac{1}{2}(y + 1/y) ). But I don't know if that helps here because we were only asked for ( x - sqrt{x^2 -1} ), which is 1/y. So maybe going through x is unnecessary. Alternatively, if I set ( t = x + sqrt{x^2 -1} ), then ( x - sqrt{x^2 -1} = 1/t ). So given t, we can directly write the other expression as reciprocal. Therefore, the answer is just the reciprocal of the given expression. Since the given is ( a^{frac{m -n}{2mn}} ), reciprocal is ( a^{-frac{m -n}{2mn}} ), which simplifies to ( a^{frac{n - m}{2mn}} ). So that's option C. I think that's solid. Let me check if any of the other options could be equivalent. For example, option B is ( a^{frac{2mn}{n - m}} ), which is different because the exponent is positive 2mn over (n - m), which is not the same as (n - m)/(2mn). Similarly, option C is (n - m)/(2mn), which is correct.Wait, just to make sure, is there any exponent rule I might have messed up? Let's confirm:Reciprocal of ( a^k ) is ( a^{-k} ). So if the given is ( a^{frac{m -n}{2mn}} ), reciprocal is ( a^{-frac{m -n}{2mn}} = a^{frac{n -m}{2mn}} ). Yes, that's correct. Therefore, the correct answer is option C. I think that's the conclusion here. The key insight was recognizing the reciprocal relationship between ( x + sqrt{x^2 -1} ) and ( x - sqrt{x^2 -1} ). Once you note that, it's straightforward to take the reciprocal of the given expression and simplify the exponent accordingly. Another way to think about it is by solving for ( sqrt{x^2 -1} ). Let's say from the original equation:Let ( y = x + sqrt{x^2 -1} = a^{frac{m -n}{2mn}} ). Then, ( sqrt{x^2 -1} = y - x ). Squaring both sides:( x^2 -1 = (y - x)^2 = y^2 - 2xy + x^2 ).Subtract ( x^2 ) from both sides:( -1 = y^2 - 2xy ).Then, rearranged:( 2xy = y^2 + 1 ).Therefore, ( x = frac{y^2 + 1}{2y} ). But maybe this approach complicates things. Alternatively, if we have ( x = frac{y + 1/y}{2} ), as before. But again, since we already established the reciprocal relationship, this seems redundant. Alternatively, let's suppose I didn't remember the reciprocal relationship. Let me try to compute ( x - sqrt{x^2 -1} ) from the given equation. Let's denote ( S = x + sqrt{x^2 -1} = a^{frac{m -n}{2mn}} ). Then, we can write:( x = frac{S + frac{1}{S}}{2} ), as per hyperbolic identities or the earlier reasoning. Then, ( sqrt{x^2 -1} = frac{S - frac{1}{S}}{2} ).Therefore, ( x - sqrt{x^2 -1} = frac{S + 1/S}{2} - frac{S - 1/S}{2} = frac{S + 1/S - S + 1/S}{2} = frac{2/S}{2} = 1/S ). So, indeed, it's 1/S. Therefore, ( x - sqrt{x^2 -1} = 1/S = a^{-frac{m -n}{2mn}} = a^{frac{n -m}{2mn}} ), which is option C. So, multiple approaches confirm that the answer is option C. I don't see any mistakes in the reasoning. Let me check the exponent simplification once again:Original exponent in S: ( frac{m - n}{2mn} ).Reciprocal exponent: ( - frac{m - n}{2mn} = frac{n - m}{2mn} ). Yes, that's correct.So, the answer should be option C.Final Answerboxed{C}

✅:Given the equation ( x + sqrt{x^2 - 1} = a^{frac{m - n}{2mn}} ), we need to find the value of ( x - sqrt{x^2 - 1} ).First, we recognize that the expressions ( x + sqrt{x^2 - 1} ) and ( x - sqrt{x^2 - 1} ) are reciprocals of each other. If we let ( y = x + sqrt{x^2 - 1} ), then multiplying ( y ) by ( x - sqrt{x^2 - 1} ) gives:[y cdot (x - sqrt{x^2 - 1}) = (x + sqrt{x^2 - 1})(x - sqrt{x^2 - 1}) = x^2 - (x^2 - 1) = 1]This implies:[x - sqrt{x^2 - 1} = frac{1}{y}]Given ( y = a^{frac{m - n}{2mn}} ), the reciprocal ( frac{1}{y} ) is:[frac{1}{y} = a^{-frac{m - n}{2mn}}]Simplifying the exponent:[-frac{m - n}{2mn} = frac{n - m}{2mn}]Thus, the value of ( x - sqrt{x^2 - 1} ) is:[a^{frac{n - m}{2mn}}]The correct answer is (boxed{C}).

🔑:1. We start with the given equation: [ x + sqrt{x^2 - 1} = a^{frac{m-n}{2mn}} ]2. Consider the mathematical identity: [ left(x + sqrt{x^2 - 1}right)left(x - sqrt{x^2 - 1}right) = x^2 - (sqrt{x^2 - 1})^2 = x^2 - (x^2 - 1) = 1 ]3. From the above identity, we get: [ x - sqrt{x^2 - 1} = frac{1}{x + sqrt{x^2 - 1}} ]4. Substitute ( x + sqrt{x^2 - 1} = a^{frac{m-n}{2mn}} ) into the equation: [ x - sqrt{x^2 - 1} = frac{1}{a^{frac{m-n}{2mn}}} ]5. Using the property of exponents, we know: [ frac{1}{a^{frac{m-n}{2mn}}} = a^{-frac{m-n}{2mn}} ]6. Since ( a^{-frac{m-n}{2mn}} = a^{frac{n-m}{2mn}} ) (because changing the sign of the exponent changes the order of the subtraction): [ x - sqrt{x^2 - 1} = a^{frac{n-m}{2mn}} ]7. Therefore, the expression for ( x - sqrt{x^2 - 1} ) is: [ x - sqrt{x^2 - 1} = a^{frac{n-m}{2mn}} ] Conclusion:The correct answer is:[boxed{text{(C) } a^{frac{n-m}{2mn}}}]

🧠:Okay, let's see. The problem is to find the range of values for 'a' such that the function f(x) = e^x - 1 - x - a x^2 is monotonically increasing when x is greater than or equal to 0. Hmm. Monotonically increasing means that the derivative of the function is non-negative for all x ≥ 0. So my first thought is to compute the derivative of f(x) and then find the conditions on 'a' that make this derivative non-negative for all x ≥ 0.Let me start by finding the derivative of f(x). The derivative of e^x is e^x, the derivative of -1 is 0, the derivative of -x is -1, and the derivative of -a x^2 is -2a x. So putting that all together, the first derivative f'(x) should be e^x - 1 - 2a x.So the condition we need is that f'(x) = e^x - 1 - 2a x ≥ 0 for all x ≥ 0. Now, we need to find all 'a' such that this inequality holds for every x ≥ 0. Hmm. To approach this, I might need to analyze the behavior of f'(x). Let's see.First, let's check the value at x = 0. Plugging x = 0 into f'(x), we get e^0 - 1 - 2a*0 = 1 - 1 - 0 = 0. So at x = 0, the derivative is zero. That's okay because the function is monotonically increasing if the derivative is non-negative, and starting at zero is allowed. Now, we need to ensure that for all x > 0, the derivative stays non-negative.So, we need e^x - 1 - 2a x ≥ 0 for all x > 0. Let's rearrange this inequality to solve for 'a'. Let's isolate 'a' terms:e^x - 1 - 2a x ≥ 0=> e^x - 1 ≥ 2a xAssuming x > 0, we can divide both sides by 2x:(e^x - 1)/(2x) ≥ aTherefore, a must be less than or equal to (e^x - 1)/(2x) for all x > 0. To find the maximum possible value of 'a' that satisfies this inequality for all x > 0, we need to find the minimum value of (e^x - 1)/(2x) over x > 0. Because if 'a' is less than or equal to the minimum of that expression, then the inequality will hold for all x > 0.So, the problem reduces to finding the minimum of the function g(x) = (e^x - 1)/(2x) for x > 0. Once we find the minimum value, say m, then a must be ≤ m.To find the minimum of g(x), we can take its derivative and set it equal to zero. Let's compute g'(x). First, let me note that g(x) = (e^x - 1)/(2x). Let's compute the derivative:g'(x) = [ ( derivative of numerator * denominator - numerator * derivative of denominator ) / (denominator)^2 ] * 1/2, since there's a 1/2 factor from the original function.Wait, actually, since the 2x is in the denominator, let's write g(x) = (e^x - 1)/(2x). Therefore, the derivative g'(x) would be:First, factor out the 1/2: g(x) = (1/2) * (e^x - 1)/x. Then the derivative of that is (1/2) * derivative of (e^x - 1)/x.So, derivative of (e^x - 1)/x is [ (e^x * x - (e^x - 1) * 1 ) ] / x^2.Therefore, g'(x) = (1/2) * [ (e^x x - e^x + 1) / x^2 ].Set this derivative equal to zero to find critical points. So:(1/2) * [ (e^x x - e^x + 1) / x^2 ] = 0Multiply both sides by 2x^2:e^x x - e^x + 1 = 0So, we need to solve e^x x - e^x + 1 = 0.Let me factor out e^x:e^x (x - 1) + 1 = 0So, e^x (x - 1) = -1Therefore, e^x (1 - x) = 1Hmm. This equation seems a bit tricky. Let's denote h(x) = e^x (1 - x). We need to solve h(x) = 1.Let me analyze h(x). When x = 0, h(0) = e^0 (1 - 0) = 1. So x=0 is a solution. Wait, but x=0 is in our domain? Wait, but we are considering x > 0 here because we divided by x when we derived the expression for 'a', so x > 0. But when x approaches 0 from the right, let's see the limit of g(x) as x approaches 0.Wait, actually, when x approaches 0, (e^x -1)/x approaches 1, since e^x ≈ 1 + x + x^2/2 + ..., so (e^x -1)/x ≈ (x + x^2/2 + ...)/x ≈ 1 + x/2 + ... which approaches 1 as x→0. Therefore, the limit of g(x) as x approaches 0 is (1)/(2*1) = 1/2. Wait, actually, let me compute the limit as x approaches 0 of (e^x -1)/(2x):Using L’Hospital’s Rule, since both numerator and denominator approach 0 as x→0. So:lim_{x→0} (e^x -1)/(2x) = lim_{x→0} (e^x)/2 = 1/2.So as x approaches 0, g(x) approaches 1/2. At x=0, the original function f'(0) = 0, but we are considering x>0.Now, we saw that h(x) = e^x (1 - x). At x=0, h(0)=1, and h(x)=1 is achieved at x=0. But we are looking for solutions in x>0. Let's check x=1: h(1) = e^1 (1 -1) = 0. So h(1)=0. Then, h(x) decreases from x=0 to x=1? Wait, let's compute the derivative of h(x):h(x) = e^x (1 - x). Then h'(x) = e^x (1 - x) + e^x (-1) = e^x (1 - x -1) = e^x (-x). So h'(x) = -x e^x. Therefore, for x >0, h'(x) <0. So h(x) is decreasing for x>0. Therefore, h(x) starts at 1 when x=0, decreases to 0 at x=1, and continues decreasing for x>1. So the equation h(x)=1 is only satisfied at x=0. But x=0 is the boundary point.Therefore, in the domain x>0, the equation h(x)=1 has no solution, since h(x) is strictly decreasing from 1 to -infinity as x increases from 0 to infinity. Wait, wait, h(x) = e^x (1 -x). As x approaches infinity, 1 -x approaches -infinity, and e^x approaches infinity, but the product? It would go to -infinity because the exponential dominates. So h(x) approaches -infinity as x→infty.But in any case, h(x) =1 only occurs at x=0 in the domain x≥0. Therefore, in x>0, h(x) <1. Therefore, the equation h(x)=1 has no solution for x>0. That suggests that the derivative of g(x) we computed earlier, which was proportional to h(x) -1, is actually always negative for x>0? Wait, let me check again.Wait, earlier, we set the derivative of g(x) to zero and arrived at the equation e^x (x -1) +1 =0. Which simplifies to e^x (1 -x) =1. But h(x) = e^x (1 -x). So h(x)=1 only at x=0, but in x>0, h(x) <1. Therefore, e^x (1 -x) <1 for x>0. Therefore, the equation e^x (1 -x) =1 has no solution in x>0. Therefore, the derivative of g(x) is never zero in x>0. Wait, but this can't be right because when we analyze the function g(x) = (e^x -1)/(2x), as x approaches 0 from the right, g(x) approaches 1/2, and as x approaches infinity, let's see what happens to g(x):As x approaches infinity, e^x grows exponentially, so (e^x -1)/(2x) ≈ e^x/(2x) which goes to infinity. So g(x) approaches infinity as x→infty. Therefore, if the derivative is always positive, then g(x) is increasing on x>0, which would mean that its minimum is at x approaching 0, which is 1/2. But that contradicts the earlier computation where the derivative was:g'(x) = (1/2) [ (e^x x - e^x +1 ) / x^2 ].Wait, since h(x) = e^x (1 -x) +1 = e^x x - e^x +1? Wait, let's check again. The derivative was:g'(x) = (1/2) * [ (e^x x - e^x +1 ) / x^2 ]Wait, so the numerator is e^x x - e^x +1. Let's factor that:e^x (x -1) +1. So if we let x=0, that becomes e^0 (0 -1) +1 = (-1) +1=0. Then, for x>0, we can analyze the sign of e^x (x -1) +1.Let me consider x=1: e^1 (1 -1) +1 =0 +1=1>0.x=0.5: e^0.5 (0.5 -1) +1= e^{0.5}*(-0.5)+1≈1.6487*(-0.5)+1≈-0.8243+1≈0.1757>0.x=2: e^2 (2 -1)+1= e^2 +1≈7.389 +1≈8.389>0.Wait, but when x approaches 0 from the right, e^x (x -1) +1 ≈ (1 +x + ...)(x -1) +1 ≈ (x -1) + x(x -1) + ... +1 ≈ (x -1) +1 + higher terms ≈ x + higher terms. So as x approaches 0+, the numerator is approximately x, which is positive. Therefore, for all x>0, the numerator e^x (x -1) +1 is positive? Then g'(x) is positive for all x>0? If that's the case, then g(x) is increasing on x>0. Therefore, its minimum value is at x approaching 0, which is 1/2, and it increases to infinity as x approaches infinity. Therefore, the minimum value of g(x) is 1/2. Therefore, in order to have a ≤ g(x) for all x>0, a must be ≤1/2.But wait, let me verify with some test values. Let's pick a=1/2. Then f'(x)=e^x -1 -2*(1/2)x= e^x -1 -x. So f'(x)=e^x -1 -x. We need to check if this is non-negative for all x≥0.We know that e^x can be expanded as 1 +x +x^2/2 +x^3/6 +..., so e^x -1 -x = x^2/2 +x^3/6 +... which is non-negative for all x≥0. So when a=1/2, f'(x) is indeed non-negative for all x≥0. Therefore, a=1/2 is acceptable.Now, suppose a>1/2. Let's take a=0.6. Then f'(x)=e^x -1 -2*0.6 x= e^x -1 -1.2x. Let's check at x=0: f'(0)=1 -1 -0=0. At x approaching 0, the derivative is zero, but what about for small x>0? Let's compute f'(x) for small x. Using the Taylor series: e^x ≈1 +x +x^2/2. Then f'(x)≈1 +x +x^2/2 -1 -1.2x = (x -1.2x) +x^2/2= -0.2x +x^2/2. For very small x, the linear term dominates, so near x=0, f'(x)≈-0.2x, which is negative. Therefore, when a=0.6>1/2, the derivative becomes negative for small x>0, which means the function is decreasing there, contradicting the requirement of being monotonically increasing. Hence, a cannot be greater than 1/2.On the other hand, if a<1/2, say a=0.4. Then f'(x)=e^x -1 -0.8x. Let's check its behavior. At x=0, f'(0)=0. For small x>0, using Taylor series: e^x -1 -0.8x≈x +x^2/2 -0.8x=0.2x +x^2/2>0. So the derivative is positive for small x. As x increases, since e^x grows exponentially and -0.8x is linear, the derivative will definitely become positive and stay positive. Hence, for a<1/2, the derivative remains non-negative for all x≥0. Therefore, the critical value is a=1/2.But wait, we need to confirm that for a=1/2, the derivative is non-negative for all x≥0. As we saw before, when a=1/2, f'(x)=e^x -1 -x. The function e^x -1 -x is known to be non-negative for all x≥0. In fact, the Taylor series of e^x is 1 +x +x^2/2 +x^3/6 +..., so subtracting 1 and x gives x^2/2 +x^3/6 +... which is always non-negative for x≥0. Therefore, a=1/2 is indeed the boundary case.Therefore, the range of values for 'a' is all real numbers less than or equal to 1/2. So a ≤1/2.But wait, let's double-check by considering another approach. Suppose we consider the second derivative of f(x). If the first derivative f'(x) is non-decreasing, then perhaps we can ensure that once it starts increasing, it stays positive. Wait, but the problem is only about f(x) being monotonically increasing, which depends solely on the first derivative being non-negative.Alternatively, another way to approach this is to note that for f'(x) = e^x -1 -2a x ≥0 for all x≥0. We can consider the minimum of f'(x). The minimum occurs where its derivative, f''(x), is zero. Let's compute f''(x): the derivative of f'(x) is e^x - 2a. Setting f''(x)=0 gives e^x -2a=0 => x=ln(2a). This critical point exists only if 2a>0 => a>0. So, if a>0, then the critical point is at x=ln(2a). We need to check whether at this critical point, f'(x) is non-negative.So, substituting x=ln(2a) into f'(x):f'(ln(2a)) = e^{ln(2a)} -1 -2a ln(2a) = 2a -1 -2a ln(2a)For f'(x) to be non-negative at this critical point, we need:2a -1 -2a ln(2a) ≥0Let's denote t=2a. Then, the inequality becomes t -1 -t ln t ≥0, where t>0.We need to find t such that t -1 -t ln t ≥0.Let me define h(t) = t -1 -t ln t. We need h(t) ≥0.Let's analyze h(t):Compute h(1): 1 -1 -1*ln1=0 -0=0.Compute h'(t): derivative of t is 1, derivative of -1 is 0, derivative of -t ln t is - [ ln t + t*(1/t) ] = - (ln t +1 )Thus, h'(t) =1 - (ln t +1 ) = -ln t.Therefore, h'(t) = -ln t.So, when t>1, ln t >0, so h'(t) <0. When t<1, ln t <0, so h'(t) >0. Therefore, h(t) has a maximum at t=1, where h(1)=0. Therefore, h(t) ≤0 for all t>0, with equality only at t=1. Wait, but if h(t) ≤0 everywhere, except at t=1 where it is zero, that suggests that the inequality t -1 -t ln t ≥0 holds only when t=1. So, h(t) ≥0 only when t=1.But t=1 implies 2a=1 => a=1/2. Therefore, when a=1/2, the critical point at x=ln(2a)=ln1=0, which is the boundary. So in this case, the minimum of f'(x) occurs at x=0, where f'(0)=0. For any other a>0, the critical point x=ln(2a) would be such that h(t)=2a -1 -2a ln(2a) <0, meaning that f'(x) would be negative at that critical point, violating the requirement. Therefore, only when a=1/2, the minimum value of f'(x) is zero at x=0, and for a<1/2, since there is no critical point (because when a<1/2, 2a<1, so ln(2a) is negative, but x must be ≥0, so the critical point x=ln(2a) is not in the domain x≥0). Therefore, for a<1/2, f''(x)=e^x -2a. Since a<1/2, 2a<1. Therefore, f''(x)=e^x -2a > e^0 -2a =1 -2a >0 (since a<1/2, 2a<1). Therefore, for a<1/2, f''(x) >0 for all x≥0, which means f'(x) is increasing on x≥0. Since f'(0)=0 and f'(x) is increasing, then f'(x) ≥0 for all x≥0. Thus, for a ≤1/2, the function f(x) is monotonically increasing on x≥0.Wait, this seems to align with our previous conclusion. So combining both approaches, we see that the critical value is a=1/2. For a ≤1/2, f'(x) ≥0 for all x≥0. Therefore, the range of values for 'a' is (-∞, 1/2].But wait, hold on. Let's check a negative value of 'a'. Suppose a is negative. For example, a=-1. Then f'(x)=e^x -1 -2*(-1)x= e^x -1 +2x. Since e^x grows exponentially and 2x is linear, clearly e^x -1 +2x is always positive for x≥0. Therefore, if a is negative, the term -2a x becomes positive, making f'(x) larger, which is good for the function to be increasing. Therefore, all a ≤1/2 are allowed, including negative a.Hence, the range is a ≤1/2.Therefore, after going through these steps, the answer is that a must be less than or equal to 1/2.

✅:To determine the range of values for ( a ) such that ( f(x) = mathrm{e}^x - 1 - x - a x^2 ) is monotonically increasing for ( x geq 0 ), we analyze the first derivative:[f'(x) = mathrm{e}^x - 1 - 2a x.]For ( f(x) ) to be monotonically increasing, ( f'(x) geq 0 ) for all ( x geq 0 ). Rearranging the inequality:[mathrm{e}^x - 1 geq 2a x quad implies quad a leq frac{mathrm{e}^x - 1}{2x} quad text{for all } x > 0.]Define ( g(x) = frac{mathrm{e}^x - 1}{2x} ). The critical value of ( a ) corresponds to the minimum of ( g(x) ) over ( x > 0 ). To find this minimum:1. Limit as ( x to 0 ): Using L’Hôpital’s Rule, [ lim_{x to 0} frac{mathrm{e}^x - 1}{2x} = lim_{x to 0} frac{mathrm{e}^x}{2} = frac{1}{2}. ]2. Behavior as ( x to infty ): ( g(x) to infty ), since ( mathrm{e}^x ) dominates.3. Derivative Analysis: Compute ( g'(x) ): [ g'(x) = frac{mathrm{e}^x (x - 1) + 1}{2x^2}. ] Setting ( g'(x) = 0 ) leads to ( mathrm{e}^x (1 - x) = 1 ), which has no solution for ( x > 0 ). Further analysis shows ( g(x) ) is increasing for ( x > 0 ), confirming its minimum at ( x to 0 ), which is ( frac{1}{2} ).Thus, ( a leq frac{1}{2} ) ensures ( f'(x) geq 0 ) for all ( x geq 0 ). Verifying:- For ( a = frac{1}{2} ), ( f'(x) = mathrm{e}^x - 1 - x geq 0 ) (non-negative by Taylor expansion).- For ( a > frac{1}{2} ), ( f'(x) ) becomes negative near ( x = 0 ).- For ( a < frac{1}{2} ), ( f'(x) ) remains non-negative.Final Answer: The range of ( a ) is ( (-infty, frac{1}{2}] ), so ( boxed{(-infty, frac{1}{2}]} ).

🔑:1. Given the function ( f(x) = mathrm{e}^x - 1 - x - a x^2 ), we need to find the range of ( a ) such that ( f(x) ) is monotonically increasing when ( x geq 0 ).2. To determine whether ( f(x) ) is increasing, we first compute the first derivative: [ f'(x) = mathrm{e}^x - 1 - 2ax ]3. For ( f(x) ) to be monotonically increasing, we require ( f'(x) geq 0 ) for all ( x geq 0 ). Evaluating this at ( x = 0 ): [ f'(0) = mathrm{e}^0 - 1 - 2a cdot 0 = 1 - 1 - 0 = 0 ] This shows that ( f'(0) = 0 ), and hence, the inequality ( f'(x) geq 0 ) must be verified for ( x > 0 ).4. For ( x > 0 ): [ mathrm{e}^x - 1 - 2ax geq 0 implies 2ax leq mathrm{e}^x - 1 implies a leq frac{mathrm{e}^x - 1}{2x} ]5. Define ( g(x) = frac{mathrm{e}^x - 1}{x} ). Investigate the behavior of ( g(x) ): [ g(x) = frac{mathrm{e}^x - 1}{x} ]6. Compute the derivative of ( g(x) ): [ g'(x) = frac{xmathrm{e}^x - (mathrm{e}^x - 1)}{x^2} = frac{xmathrm{e}^x - mathrm{e}^x + 1}{x^2} = frac{(x-1)mathrm{e}^x + 1}{x^2} ]7. Define ( p(x) = (x-1)mathrm{e}^x + 1 ). Now compute ( p'(x) ): [ p'(x) = frac{d}{dx}[(x-1)mathrm{e}^x + 1] = (x-1)mathrm{e}^x + mathrm{e}^x = xmathrm{e}^x > 0 text{ for } x > 0 ] Since ( p'(x) > 0 ) for ( x > 0 ), ( p(x) ) is increasing.8. Since ( p(0) = (0-1)mathrm{e}^0 + 1 = 0 ), it follows ( p(x) > 0 ) for ( x > 0 ). Therefore, ( g'(x) > 0 ); thus, ( g(x) ) is increasing.9. Also, [ g(0) = lim_{x to 0} frac{mathrm{e}^x - 1}{x} = lim_{x to 0} mathrm{e}^x = 1 ]10. Since ( g(x) ) is increasing and ( g(0) = 1 ), for all ( x geq 0 ), ( g(x) geq 1 ). Thus, [ a leq frac{g(x)}{2} leq frac{1}{2} ]11. Hence, the range of ( a ) for which ( f(x) geq 0 ) is: [ boxed{(-infty, frac{1}{2}]} ]

🧠:Okay, so I need to figure out the minimum number of rooks required to attack all the white cells on a 9x9 checkerboard pattern where the corners are white. Let me start by visualizing the board. Since it's a checkerboard, the colors alternate, and the corners being white means that the first cell (1,1) is white, then (1,2) is black, (1,3) white, and so on. This pattern alternates every row as well, so row 2 starts with black, then white, etc. First, let me confirm the total number of white cells. In a 9x9 checkerboard, each row alternates starting with white or black. Since 9 is odd, each row will have 5 of one color and 4 of the other. The first row (row 1) starts with white, so it has 5 white cells and 4 black cells. The next row (row 2) starts with black, so it has 4 white cells and 5 black cells. This alternates for each row. Therefore, over 9 rows, the number of white cells is (5 + 4 + 5 + 4 + 5 + 4 + 5 + 4 + 5) = 5*5 + 4*4 = 25 + 16 = 41. Wait, no, that's not right. Wait, actually, each pair of rows would be 5 + 4 = 9 cells. Since there are 9 rows, we have 5 rows starting with white and 4 rows starting with black? Wait, no. Let me recalculate.Wait, rows 1, 3, 5, 7, 9 (odd rows) start with white, so each of these has 5 white cells. Rows 2, 4, 6, 8 (even rows) start with black, so each of these has 4 white cells. Therefore, total white cells are 5 rows * 5 white cells + 4 rows * 4 white cells = 25 + 16 = 41. Yes, that's correct. So there are 41 white cells that need to be under attack.Now, the problem is to place the minimum number of rooks such that every white cell is in the same row or column as at least one rook. Rooks attack all cells in their row and column, so each rook can potentially cover multiple white cells. The challenge is to place rooks in such a way that their lines of attack (rows and columns) cover all 41 white cells with as few rooks as possible.Hmm. Let me think. Since each rook covers an entire row and an entire column, maybe we can model this as a covering problem where we need to cover all white cells by selecting a set of rows and columns such that every white cell is in at least one selected row or column. Then, the number of rooks needed would be the minimum number of rows and columns needed to cover all white cells. However, each rook can be placed at the intersection of a selected row and column. Wait, but actually, if we choose a row and a column, the rook is placed at their intersection. But the problem is that if you have a rook in a row, it covers all the cells in that row, but if you have a rook in a column, it covers all the cells in that column. However, when you place a rook on a particular cell, it's covering both its row and column. So perhaps we can model this as a set cover problem where the universe is the 41 white cells, and each possible rook placement corresponds to covering all white cells in its row and column.But set cover is NP-hard, so we need a heuristic or perhaps find a pattern here. Alternatively, maybe there's a way to use the checkerboard pattern to our advantage.First, let's note that the white cells are arranged such that in each row, they are on the odd columns for odd rows and even columns for even rows. Wait, no. Let's see. Since the first row starts with white at (1,1), then alternates. So row 1: columns 1,3,5,7,9 (white). Row 2: columns 2,4,6,8 (white). Row 3: same as row 1. Row 4: same as row 2, etc. Therefore, the white cells in odd rows are on odd columns, and white cells in even rows are on even columns.So, for odd rows (1,3,5,7,9), the white cells are columns 1,3,5,7,9. For even rows (2,4,6,8), the white cells are columns 2,4,6,8.So, in total, the white cells are:- For odd rows (5 rows), 5 columns each (odd columns): 5x5=25 white cells.- For even rows (4 rows), 4 columns each (even columns): 4x4=16 white cells.Total 41 white cells, as before.Therefore, to cover all white cells, we need to cover all these cells either by covering their rows or their columns. But each rook can cover a row and a column. The problem is similar to hitting set or set cover where we need to cover all elements (white cells) by selecting sets (rows and columns) such that each white cell is in at least one selected row or column. The objective is to minimize the number of selected rows and columns. However, each rook placed at the intersection of a row and column counts as one, but selecting a row and a column requires a rook at their intersection. Wait, no. Wait, if we select a row, we can cover all white cells in that row. If we select a column, we can cover all white cells in that column. So, in effect, to cover all white cells, we need to select a set of rows and columns such that every white cell is either in a selected row or a selected column. Then, the minimal number of rooks needed would be the minimal number of rows plus columns needed, but with the caveat that each rook is placed at the intersection of a selected row and column. Wait, but actually, no, the rooks can be placed anywhere, but each rook contributes to covering its row and column. So perhaps the problem is equivalent to covering all white cells with rows and columns, where each rook can cover one row and one column. Therefore, the minimal number of rooks required is the minimal number of rows and columns such that their union covers all white cells, and each rook is placed in the intersection of a selected row and column. However, since a rook can cover both a row and a column, perhaps the minimal number is the minimal size of a set of rows and columns covering all white cells, divided by 2? No, not exactly. Because each rook can cover one row and one column. If we select a row, we can cover all white cells in that row. Similarly for a column. But if a rook is placed in a cell, it covers both that row and column. So if we have a rook in row R and column C, then row R and column C are covered. So the problem reduces to choosing a set of rows and columns such that all white cells are in at least one of the chosen rows or columns, and the number of rooks needed is equal to the number of rows plus columns chosen, but each rook can be placed at the intersection of a chosen row and column. Wait, but that might not be possible if there's overlap. Hmm, this is getting a bit confusing.Alternatively, perhaps think of it as a bipartite graph where one partition is the set of rows and the other is the set of columns. Then, edges represent white cells: a white cell at (i,j) is represented as an edge between row i and column j. Then, the problem reduces to finding the minimum vertex cover in this bipartite graph. A vertex cover is a set of rows and columns such that every edge (white cell) is incident to at least one vertex in the set. The Konig's theorem states that in bipartite graphs, the size of the minimum vertex cover equals the size of the maximum matching. Therefore, if we can model the white cells as a bipartite graph and find its maximum matching, we can determine the minimum vertex cover, which would give the minimal number of rows and columns needed to cover all white cells. Then, the number of rooks needed would be equal to the size of the minimum vertex cover. However, since each rook is placed at an intersection of a row and column, perhaps each rook can cover one row and one column. Wait, but vertex cover can include both rows and columns. For example, if we select two rows and three columns, the vertex cover size is 5, so we need five rooks? But how?Wait, no. Wait, Konig's theorem says that in bipartite graphs, the size of the minimum vertex cover is equal to the size of the maximum matching. But a vertex cover is a set of vertices (rows and columns) such that every edge (white cell) is adjacent to at least one of them. The minimum vertex cover would thus be the minimum number of rows and columns needed to cover all white cells. Therefore, the minimal number of rooks required would be equal to the size of this vertex cover, because each vertex in the cover is either a row or a column, and each rook can be placed at the intersection of a row and column. However, to cover both a row and a column, you need a rook in that row or column. Wait, maybe not. If you select a row, you don't need a rook in that row, but you need to have all columns that have white cells in that row? Wait, no, this is getting mixed up.Let me try to clarify. Let's model the problem as a bipartite graph with rows on one side and columns on the other. An edge connects row i and column j if cell (i,j) is white. Then, a vertex cover is a set of rows and columns such that every white cell is in a selected row or column. The minimum vertex cover would give the minimal number of rows and columns needed. But each rook can cover a row and a column. However, if we have a vertex cover of size k, can we place k rooks such that each rook is in a row or column from the vertex cover? For example, if a vertex cover includes a row, we can place a rook in that row (in any column) to cover the row, and similarly for a column. But if a rook is placed in a covered row and a covered column, then it can cover both. Wait, perhaps the minimum number of rooks needed is equal to the size of the vertex cover. Because each rook can be placed in a row or column from the vertex cover. Wait, but if the vertex cover includes both a row and a column, you can place a rook at their intersection, thereby covering both with one rook. Therefore, if the vertex cover has R rows and C columns, you can place R + C - overlap rooks, where overlap is the number of intersections between the selected rows and columns. But this complicates things.Alternatively, perhaps the minimum number of rooks needed is equal to the maximum of the number of rows and columns needed. Wait, no. Let's think with an example. Suppose we have a vertex cover with 5 rows and 5 columns. Then, if we place a rook at each intersection of these rows and columns, but that would be 25 rooks, which is too many. But perhaps you can place rooks in such a way that each rook covers a row and a column from the vertex cover. For example, if you have R rows and C columns in the vertex cover, you can place a rook in each row (at any column in the vertex cover) and a rook in each column (at any row in the vertex cover). But if the rows and columns overlap, you can combine them. For example, if a rook is placed in a row from the vertex cover and a column from the vertex cover, then it covers both. Therefore, the minimal number of rooks needed is the minimal number of rooks such that every row and column in the vertex cover has at least one rook in it. That is equivalent to covering the vertex cover set (which is a mix of rows and columns) with rooks such that each selected row has at least one rook in it and each selected column has at least one rook in it. The minimal number of rooks required to do this is the maximum of the number of rows and columns in the vertex cover. Wait, is that correct?Wait, suppose the vertex cover has R rows and C columns. Then, we need to place rooks such that each of the R rows has a rook (which can be in any column, but preferably in one of the C columns), and each of the C columns has a rook (which can be in any row, preferably in one of the R rows). Therefore, the minimal number of rooks is the minimal number that can cover all R rows and C columns. This is equivalent to the problem of covering R rows and C columns with rooks, which can be done with max(R, C) rooks. For example, if R >= C, place a rook in each of the R rows, each in one of the C columns (since C columns can cover the columns). Similarly, if C >= R, place a rook in each of the C columns, each in one of the R rows. Therefore, the minimal number of rooks needed is max(R, C). But since the vertex cover has size R + C, but we can cover it with max(R, C) rooks, this would mean that the minimal number of rooks is at least the ceiling of (R + C)/2, but actually, it's max(R, C). Wait, for example, if R = 5 and C = 5, then max(R, C) = 5, which is better than R + C = 10. But how? If we have 5 rows and 5 columns in the vertex cover, then we can place 5 rooks, each in a different row and column (like a diagonal), covering all 5 rows and 5 columns. Therefore, the minimal number of rooks is indeed max(R, C) if R and C are part of a vertex cover.But this is only possible if the selected rows and columns can be covered by a matching. Wait, maybe there's a relation here with Konig's theorem. Konig's theorem states that in bipartite graphs, the size of the maximum matching equals the size of the minimum vertex cover. Wait, no, it's the other way: Konig's theorem says that in bipartite graphs, the size of the maximum matching equals the size of the minimum vertex cover. Wait, no, actually Konig's theorem states that in bipartite graphs, the size of the minimum vertex cover plus the size of the maximum matching equals the number of vertices in one partition. Wait, no, let me recall. Konig's theorem states that in bipartite graphs, the size of the maximum matching is equal to the size of the minimum vertex cover. Wait, actually, no. Wait, let me check.No, Konig's theorem says that in bipartite graphs, the size of the maximum matching equals the size of the minimum vertex cover. Wait, that can't be, because in a bipartite graph, the maximum matching and minimum vertex cover are related but not necessarily equal. Wait, actually, Konig's theorem states that in bipartite graphs, the size of the minimum vertex cover is equal to the size of the maximum matching. No, wait, maybe not. Let me look it up mentally. Konig's theorem: In any bipartite graph, the number of edges in a maximum matching equals the number of vertices in a minimum vertex cover. Wait, no. Wait, actually, Konig's theorem states that in bipartite graphs, the maximum matching size is equal to the minimum vertex cover size. Hmm, maybe.Alternatively, maybe the theorem is that in bipartite graphs, the maximum matching size plus the maximum independent set size equals the total number of vertices. Wait, no. I might be confusing different theorems. Let me think again.Actually, Konig's theorem states that in bipartite graphs, the size of the maximum matching equals the size of the minimum vertex cover. Yes, that's correct. So if we have a bipartite graph, the minimum number of vertices needed to cover all edges (vertex cover) is equal to the maximum number of edges that can be chosen without overlapping vertices (matching). Therefore, in our problem, the bipartite graph is between rows and columns with edges representing white cells. Then, the minimum vertex cover (minimum number of rows and columns to cover all white cells) is equal to the maximum matching in this graph.Therefore, if I can find the maximum matching in this bipartite graph, that will give me the minimum vertex cover size. Then, as per earlier reasoning, the number of rooks needed would be equal to the maximum of the number of rows and columns in the vertex cover. Wait, but maybe the vertex cover size is equal to the maximum matching, which might be different. Wait, no. According to Konig's theorem, the minimum vertex cover size equals the maximum matching size. So, the minimum number of rows and columns needed to cover all white cells is equal to the maximum matching size in this bipartite graph.Therefore, if I can compute the maximum matching, that will give me the minimum vertex cover size. Then, as per the previous reasoning, the number of rooks needed is max(R, C), where R is the number of rows and C the number of columns in the vertex cover. But since R + C = vertex cover size, and vertex cover size = maximum matching size, then the number of rooks needed is at least ceiling(vertex cover size / 2). Wait, no. Wait, suppose the vertex cover is composed of R rows and C columns, then R + C = vertex cover size. But to cover these R rows and C columns with rooks, we need at least max(R, C) rooks, since you need at least one rook per row and one per column. However, if the rows and columns can be arranged such that each rook covers one row and one column, the minimal number of rooks needed is the minimal number that can cover all R rows and C columns. Which is the minimal number such that all R rows are covered and all C columns are covered. This is equivalent to the problem of covering R rows and C columns with the fewest rooks, which is max(R, C). Because if you have more rows than columns, you can place a rook in each row in a different column, but since there are more rows, you need at least R rooks. Similarly, if there are more columns, you need C rooks. However, since R + C = vertex cover size, and vertex cover size is equal to the maximum matching size, then the minimal number of rooks is max(R, C). But since R and C can vary as long as R + C = vertex cover size, we need to choose R and C such that max(R, C) is minimized. That would occur when R and C are as equal as possible. For example, if vertex cover size is k, then the minimal max(R, C) is ceil(k / 2). Because if k is even, you can split into R = k/2, C = k/2, so max(R, C) = k/2. If k is odd, ceil(k / 2). Therefore, the minimal number of rooks needed would be ceil(k / 2), where k is the minimum vertex cover size, which is equal to the maximum matching size.Therefore, the problem reduces to finding the maximum matching in the bipartite graph of rows and columns connected by white cells, then taking the ceiling of half that size. Wait, no. Wait, let me get back. Let me outline the steps again:1. Model the white cells as a bipartite graph between rows and columns. Each white cell (i,j) is an edge between row i and column j.2. Find the maximum matching in this bipartite graph. Let the size be m.3. By Konig's theorem, the minimum vertex cover size is m.4. The vertex cover can be composed of some rows and columns. Let R be the number of rows and C be the number of columns in the vertex cover, so R + C = m.5. To cover these R rows and C columns with rooks, the minimal number of rooks needed is the minimal value of max(R, C) over all possible R and C such that R + C = m.6. The minimal value of max(R, C) given R + C = m is ceil(m / 2). For example, if m is even, split R = C = m/2, so max(R, C) = m/2. If m is odd, ceil(m / 2).Therefore, the minimal number of rooks required is ceil(m / 2), where m is the maximum matching size in the bipartite graph.Therefore, the problem now becomes: find the maximum matching in the bipartite graph where rows and columns are connected if the cell is white. Then take the ceiling of half that number.So, first, let's analyze the structure of the bipartite graph. Each odd row (rows 1,3,5,7,9) is connected to all odd columns (columns 1,3,5,7,9). Each even row (rows 2,4,6,8) is connected to all even columns (columns 2,4,6,8). Therefore, the bipartite graph consists of two disconnected components: one connecting odd rows to odd columns, and another connecting even rows to even columns.Therefore, the maximum matching in the entire graph is the sum of the maximum matchings in each component.First, let's consider the odd component: 5 rows connected to 5 columns, each row connected to all 5 columns. This is a complete bipartite graph K_{5,5}. The maximum matching here is 5, since we can match each row to a distinct column.Second, the even component: 4 rows connected to 4 columns, each row connected to all 4 columns. This is a complete bipartite graph K_{4,4}. The maximum matching here is 4.Therefore, the total maximum matching size is 5 + 4 = 9. Therefore, by Konig's theorem, the minimum vertex cover size is 9. Then, the minimal number of rooks needed is ceil(9 / 2) = 5 (since 9 / 2 = 4.5, ceil(4.5) = 5). Wait, but let me confirm.Wait, the maximum matching is 9. Therefore, the minimum vertex cover is 9. Then, if the vertex cover can be split into R rows and C columns such that R + C = 9, the minimal max(R, C) is ceil(9 / 2) = 5. However, perhaps this isn't directly applicable because the vertex cover is split across the two components. Let's see.Wait, the bipartite graph is split into two components: odd and even. Each component's vertex cover is separate. For the odd component (K_{5,5}), the minimum vertex cover is 5 (by Konig's theorem, since the maximum matching is 5). Similarly, for the even component (K_{4,4}), the minimum vertex cover is 4. Therefore, the total vertex cover size is 5 + 4 = 9. Therefore, the overall minimum vertex cover is 9. Now, in each component, the vertex cover can be either all rows or all columns. For example, in K_{5,5}, a vertex cover could be all 5 rows or all 5 columns. Similarly, for K_{4,4}, a vertex cover could be all 4 rows or all 4 columns. However, to minimize the number of rooks, we need to choose the vertex cover in each component such that the total number of rows and columns can be covered by as few rooks as possible.But if we choose rows in one component and columns in the other, we can potentially overlap the rook placements. Wait, but the components are separate: odd rows and columns vs even rows and columns. Therefore, a rook placed in an odd row and odd column cannot cover any even rows or columns, and vice versa. Therefore, the two components are independent. Therefore, the minimal number of rooks needed is the sum of the minimal rooks needed for each component.In the odd component, the minimum vertex cover is 5. Therefore, in the odd component, we can cover it with either 5 rows, 5 columns, or a combination. However, to minimize the number of rooks, we can choose to cover it with 5 rows (which requires 5 rooks, one in each row), or 5 columns (5 rooks, one in each column), or a mix. However, if we use a mix, since it's a K_{5,5}, any row and column cover must have at least 5 elements. Therefore, the minimal number of rooks needed for the odd component is 5. Similarly, for the even component, the minimum vertex cover is 4, so the minimal number of rooks needed is 4. Therefore, total rooks would be 5 + 4 = 9. But that contradicts the previous reasoning. Wait, so there's a mistake here.Wait, actually, the problem is that the vertex cover for each component is 5 and 4. However, to cover those vertex covers with rooks, if the vertex cover for the odd component is 5 rows, then you need 5 rooks in those rows (each in any column). But since the columns in the odd component are also 5, if we instead choose 5 columns, then you need 5 rooks in those columns. Similarly for the even component. However, if we can choose a combination of rows and columns for the vertex cover, then perhaps we can cover them with fewer rooks.Wait, in the odd component, suppose we select R rows and C columns such that R + C = 5 (the vertex cover size). Then, the minimal number of rooks needed to cover these R rows and C columns is max(R, C). To minimize max(R, C), we need R and C as balanced as possible. For R + C = 5, the most balanced is R = 2, C = 3 or R = 3, C = 2, giving max(R, C) = 3. Wait, but is that possible? Let me see.For example, if in the odd component (K_{5,5}), we select 2 rows and 3 columns. Then, to cover these 2 rows and 3 columns with rooks, we need max(2, 3) = 3 rooks. Place rooks in the 3 columns, each in one of the 2 rows. Wait, but you can place 2 rooks in the 2 rows (each in one of the 3 columns), and 1 more rook in the remaining column (in one of the rows). But actually, if you have 2 rows and 3 columns, you can place 3 rooks: one in each column, each in one of the two rows. Since two rows, three columns, you can place a rook in column 1, row 1; column 2, row 1; column 3, row 2. Then, columns 1,2,3 are covered by the rooks, and rows 1 and 2 are covered. Wait, but we have 5 rows and 5 columns in the odd component. If we select 2 rows and 3 columns, we need to ensure that all edges (white cells) in the odd component are covered. But if we select 2 rows and 3 columns, are all white cells in the odd component covered? No, because there are 5 rows and 5 columns. If we select 2 rows, then the white cells in the other 3 rows are not covered by those rows. Similarly, selecting 3 columns would cover the white cells in those columns. But the intersection of the uncovered rows and uncovered columns would still have white cells. Therefore, the vertex cover must cover all edges, so if you choose R rows and C columns, then every edge (white cell) must be in either a selected row or column. Therefore, in the odd component, if you choose R rows and C columns such that R + C = 5, then the remaining (5 - R) rows and (5 - C) columns must have no edges between them. But in K_{5,5}, every row is connected to every column. Therefore, if you have any remaining rows and columns, there are edges between them. Therefore, the only way to cover all edges is to choose all rows or all columns. Wait, that can't be. Wait, no. If we choose R rows and C columns such that R + C = 5, then any edge (i,j) in the graph is either in one of the R rows or in one of the C columns. But in K_{5,5}, any edge not in the R rows must be in the C columns. However, if we have an edge (i,j) where i is not in the R rows and j is not in the C columns, then it's not covered, which is a contradiction. Therefore, in a complete bipartite graph K_{n,n}, the only vertex covers are those where you select all n rows, all n columns, or a combination where R + C >= n. Wait, no. Wait, in K_{n,n}, to cover all edges, you need to select at least n vertices (rows or columns), since if you select fewer than n, then there will be at least one row and one column not selected, and their edge is uncovered. Therefore, the minimum vertex cover in K_{n,n} is n. Therefore, in our case, the odd component is K_{5,5}, so minimum vertex cover is 5, which can be achieved by selecting all 5 rows or all 5 columns. Similarly, the even component is K_{4,4}, so minimum vertex cover is 4.Therefore, going back, the total minimum vertex cover size is 5 + 4 = 9. Since these are separate components, we can't overlap the coverage. Therefore, the minimal number of rooks required is 5 (for the odd component) + 4 (for the even component) = 9. But that can't be, because earlier reasoning suggested that using Konig's theorem and splitting into rows and columns could give a lower number. Wait, but in this case, since the components are separate, we have to cover each component separately. For each component, since the vertex cover is all rows or all columns, we need to cover each component with either all its rows or all its columns. However, if we choose to cover a component with all its rows, we need to place rooks in those rows. Each rook placed in a row also covers a column. But since the components are separate (odd and even), the columns covered in the odd component are odd columns, and in the even component are even columns. Therefore, placing a rook in an odd row (to cover it) must be placed in an odd column (to cover a column in the odd component), and similarly for even. Therefore, for the odd component, if we cover it by selecting all 5 rows, we need 5 rooks, each in those rows (each in any column). However, if we place them in different columns, we can cover 5 columns as well. Similarly, if we choose to cover the odd component by selecting all 5 columns, we need 5 rooks in those columns. But since we can choose, perhaps there's a smarter way.Wait, actually, for each component, if we choose to cover it with rows, we need as many rooks as the number of rows, each placed in any column. However, if we can place those rooks in columns that are also covering another component, but since the components are separate, the columns in the odd component are different from those in the even component. Therefore, there is no overlap. Therefore, covering the odd component requires 5 rooks, and the even component requires 4 rooks, totaling 9. But this seems high. The answer is probably lower.Wait, but perhaps we can do better by covering some rows and columns in each component, thereby reducing the total number.Wait, let's take the odd component (K_{5,5}). If we select 3 rows and 2 columns, total vertex cover size 5. Then, to cover these 3 rows and 2 columns with rooks, we can place rooks at the intersections of these rows and columns. For example, place a rook in each of the 3 rows, each in one of the 2 columns. This would require 3 rooks. Then, place 2 rooks in the 2 columns, but since the 3 rows already cover those columns, we might not need additional rooks. Wait, no. Wait, the rows and columns in the vertex cover need to be covered by rooks. So if we have 3 rows and 2 columns in the vertex cover, we need to place rooks such that each of these 3 rows has a rook (covering the row) and each of these 2 columns has a rook (covering the column). The minimal number of rooks needed is the maximum of 3 and 2, which is 3. For example, place a rook in each of the 3 rows, each in one of the 2 columns. This covers all 3 rows and 2 columns. Therefore, with 3 rooks. Similarly, for the even component (K_{4,4}), if we choose 2 rows and 2 columns, total vertex cover size 4. Then, the number of rooks needed is max(2,2) = 2. Therefore, total rooks would be 3 + 2 = 5.But wait, this contradicts the earlier statement that the minimum vertex cover for K_{5,5} is 5. So what's the catch here? If in the odd component, selecting 3 rows and 2 columns would only cover the edges (white cells) in those rows and columns. However, since it's a complete bipartite graph, any row not in the vertex cover must be covered by a column, and any column not in the vertex cover must be covered by a row. Therefore, if we select 3 rows and 2 columns, we need to ensure that all edges are covered. But in K_{5,5}, the edges not in the 3 rows must be covered by the 2 columns. However, there are 5 - 3 = 2 rows not selected, and 5 - 2 = 3 columns not selected. The edges between these 2 rows and 3 columns would not be covered by any rows or columns in the vertex cover. Therefore, such a vertex cover (3 rows + 2 columns) would not cover all edges. Therefore, my previous reasoning was flawed. In a complete bipartite graph K_{n,n}, any vertex cover must have at least n vertices. Therefore, you can't have a vertex cover smaller than n. Therefore, in K_{5,5}, the minimum vertex cover is indeed 5, and similarly for K_{4,4} it's 4. Therefore, we can't split them into rows and columns to get a smaller vertex cover.Therefore, returning, the total minimum vertex cover size is 5 + 4 = 9. Therefore, to cover these 9 rows and columns (split between the two components), we need to place rooks such that each selected row and column has at least one rook. However, since the components are separate, the rooks for the odd component must be placed on odd rows and odd columns, and those for the even component on even rows and even columns. Therefore, the rooks are in separate parts of the board and don't interfere with each other.But how many rooks do we need? For the odd component's vertex cover, which is 5 rows or 5 columns, we need to place 5 rooks. Similarly, for the even component's vertex cover, which is 4 rows or 4 columns, we need 4 rooks. Therefore, in total, 5 + 4 = 9 rooks. But this seems like a lot. The question is asking for the minimum number of rooks. Is there a better way?Wait, perhaps I'm misunderstanding the relationship between vertex cover and rook placement. Let's think differently. Instead of trying to model it as a bipartite graph, maybe there's a pattern or arrangement of rooks that can cover all white cells efficiently.Given that the white cells are on odd rows and odd columns, and even rows and even columns, maybe we can use the fact that placing a rook on an odd row and odd column can cover a row and column of white cells, similarly for even.But since odd rows have white cells on odd columns, a rook on an odd row and odd column can cover all white cells in that row and column. Similarly, a rook on an even row and even column can cover all white cells in that row and column. Therefore, if we place rooks on the main diagonals of the odd and even blocks.For example, consider the odd component: 5 rows and 5 columns. If we place rooks along the main diagonal of the odd component, i.e., (1,1), (3,3), (5,5), (7,7), (9,9), then each rook covers a row and a column in the odd component. This would cover all white cells in the odd component, because each white cell in an odd row is in one of the odd columns, so each white cell is in the same row or column as one of these rooks.Similarly, for the even component: 4 rows and 4 columns. Place rooks along the main diagonal of the even component: (2,2), (4,4), (6,6), (8,8). Each rook covers a row and a column in the even component, thereby covering all white cells there.Therefore, total rooks needed are 5 + 4 = 9. However, this is the same as before. But maybe there's a way to overlap the coverage?Wait, but the odd and even components are separate. A rook on an odd row and column cannot cover any even row or column, and vice versa. Therefore, overlapping is not possible. Therefore, the minimal number of rooks needed is indeed 9. But that seems high, as the board is 9x9 and you can cover all cells with 9 rooks (one per row), but here we only need to cover white cells. But maybe you can do better.Wait, another approach. Let's think about covering all white rows and columns. The white cells are in odd rows (with odd columns) and even rows (with even columns). If we can cover all odd rows and even columns, or even rows and odd columns, maybe we can cover all white cells.But white cells are in:- Odd rows, odd columns.- Even rows, even columns.Therefore, if we cover all odd rows, then all white cells in odd rows are covered. Then, we need to cover the white cells in even rows (which are in even columns). Similarly, if we cover all even columns, then the white cells in even rows are covered. So, covering all odd rows and even columns would cover:- All white cells in odd rows (via the rows).- All white cells in even rows and even columns (via the columns).Therefore, total rooks needed would be the number of odd rows plus the number of even columns. There are 5 odd rows and 4 even columns (since columns 2,4,6,8). So 5 + 4 = 9 rooks. Again, 9. But wait, but each rook can cover both a row and a column. If we place rooks at the intersections of odd rows and even columns, then each rook covers an odd row and an even column. However, there are 5 odd rows and 4 even columns. If we place a rook at each intersection of an odd row and an even column, that would be 5*4=20 rooks, which is more. Alternatively, if we place rooks in such a way that each rook is in an odd row or an even column, but not necessarily both. Wait, but to cover all odd rows, you need at least 5 rooks (one per row). To cover all even columns, you need at least 4 rooks (one per column). However, if you can place some rooks in both an odd row and an even column, you can cover a row and a column with one rook. Therefore, the minimal number of rooks needed is the maximum of the number of odd rows and even columns. Wait, no. Let me think.Suppose we want to cover all 5 odd rows and 4 even columns. The minimal number of rooks needed to cover all 5 rows and 4 columns is 5, since you can place a rook in each odd row, and choose their columns to be any of the even columns. However, there are only 4 even columns. Therefore, by the pigeonhole principle, two rooks would have to be placed in the same column. But a rook in a column covers that column, so placing multiple rooks in the same column is redundant. Wait, but if we place 5 rooks in the 5 odd rows, each in one of the 4 even columns, then two rooks will share a column. However, each rook covers its row and column. Therefore, even if two rooks are in the same column, they still cover their respective rows. The columns are already being covered by the rooks. Therefore, to cover all 5 odd rows, you need at least 5 rooks. Each of these rooks will also cover their respective columns. However, if the columns are even, then they cover the even columns. Therefore, by placing 5 rooks in odd rows and even columns, you cover all 5 odd rows and 4 even columns. Therefore, the total number of rooks needed is 5. Similarly, if you place 4 rooks in even columns and odd rows, but there are 5 odd rows, so you need 5 rooks. Therefore, this gives 5 rooks. However, does this cover all white cells?The white cells are:- In odd rows and odd columns: covered by the odd rows (since we're covering all odd rows).- In even rows and even columns: covered by the even columns (since we're covering all even columns).Therefore, placing 5 rooks in odd rows and even columns covers all white cells. Similarly, placing 4 rooks in even columns and odd rows might not cover all odd rows, but if you place 5 rooks in odd rows (regardless of column), they cover all odd rows (and hence all white cells in odd rows), and the columns they are placed in will cover some white cells in even rows. However, the white cells in even rows are in even columns. Therefore, if the rooks in odd rows are placed in even columns, then those columns will cover the white cells in even rows. Therefore, yes, by placing 5 rooks in odd rows and even columns, you cover all white cells.Wait, let me confirm. If a rook is in an odd row and an even column, it covers:- All cells in the odd row (which are white at odd columns).- All cells in the even column (which are white in even rows).Therefore, the white cells in odd rows and odd columns are covered by the rook's row. The white cells in even rows and even columns are covered by the rook's column. But each rook is in an odd row and even column, so the column is even, hence covering white cells in even rows and that column. If we have rooks in all odd rows and all even columns, then yes, all white cells are covered.But to have all even columns covered, you need at least one rook in each even column. Since there are 4 even columns, you need at least 4 rooks. However, you have 5 odd rows to cover. If you place 4 rooks in the even columns, each in a different even column and different odd row, then you cover 4 odd rows and all 4 even columns. The remaining odd row needs a rook, which can be placed in any column. If you place it in an even column, you already have a rook there, so that column is already covered. If you place it in an odd column, then that rook's column is odd, which doesn't cover any white cells in even rows (since those are in even columns). However, the remaining odd row's white cells are in odd columns, which are already covered by the row coverage. Therefore, to cover the fifth odd row, you can place a rook in an odd column, which covers the row (all its white cells) and the odd column. But the odd column's white cells are in odd rows, which are already covered by the rows. Therefore, placing a rook in an odd column for the fifth odd row is redundant for column coverage but necessary for row coverage. However, this rook would add coverage to the odd column. But since we are only required to cover white cells, and the odd column's white cells are in odd rows, which are already covered by the row coverage. Therefore, the column coverage from this rook is unnecessary. Therefore, it's sufficient to place the fifth rook in any column (even or odd), but placing it in an even column allows covering an extra column, but since we already have all even columns covered by the first four rooks, it's redundant. Therefore, the fifth rook can be placed in an odd column, but it only needs to cover the row. However, since the problem requires that all white cells are under attack, the rook in the fifth odd row can be placed in an odd column, which is a white cell, but it's already covered by the row. Wait, no. If we place a rook in the fifth odd row, even if it's in an odd column, it covers the entire row (which is necessary) and the entire column. The column coverage is for all cells in that column, including white cells in even rows. But the white cells in even rows are only in even columns. Therefore, covering an odd column doesn't help with covering any white cells. Therefore, the fifth rook's column coverage is irrelevant. Therefore, the minimal way is to place four rooks in even columns (covering four odd rows and all four even columns) and one rook in the fifth odd row (in any column, even or odd). However, the fifth rook must be placed in the fifth odd row to cover that row's white cells. The column it's placed in doesn't matter for coverage, but to minimize the number of rooks, we can place it in an even column, but since we already have rooks in all even columns, does that help? Wait, placing the fifth rook in an even column would mean that column already has a rook, so we don't need an additional rook for that column. However, we already have four even columns covered by four rooks. The fifth rook in an even column would mean two rooks in one even column. This doesn't increase coverage, but it's necessary to cover the fifth odd row. Therefore, the minimal number of rooks is 5: four in even columns (each in a distinct even column and distinct odd row), and one in the fifth odd row, which can be placed in any column (preferably an even column, but it's already covered). However, the fifth rook must be in the fifth odd row, but the column can be arbitrary. However, since the even columns are already covered by the first four rooks, placing the fifth rook in an even column is redundant for column coverage but necessary for row coverage. Therefore, total rooks needed is 5. Similarly, if we start by placing rooks in odd rows and even columns, we can cover all odd rows and all even columns with 5 rooks.Therefore, this suggests that the minimal number of rooks needed is 5. Let's check if this works.Place rooks at (1,2), (3,4), (5,6), (7,8), (9, any column). Wait, but columns 2,4,6,8 are even columns. If we place the fifth rook in column 9 (an odd column), then it doesn't cover any even column. But we need to cover all even columns. Wait, we need to have all even columns (2,4,6,8) covered. Therefore, the first four rooks must be in these columns. Therefore, we need to place four rooks in even columns and different odd rows. Then, the fifth rook must be in the remaining odd row (row 9) and can be placed in any column. However, to cover the white cells in row 9, which are in columns 1,3,5,7,9, the rook in row 9 will cover all of row 9's white cells, regardless of the column it's placed in. However, if we place it in an even column, say column 2, then column 2 is already covered by the rook in row 1. Therefore, placing it in an even column is redundant for column coverage but necessary for row coverage. Alternatively, placing it in an odd column will cover that column, but since odd columns are only in odd rows, which are already covered by their own rooks. Therefore, the fifth rook can be placed anywhere in row 9. Therefore, with five rooks, we can cover all odd rows (and their white cells) and all even columns (and their white cells). Therefore, this covers all white cells. Let's verify:- White cells in odd rows (1,3,5,7,9): Covered by the rooks in those rows.- White cells in even rows (2,4,6,8): These are in even columns (2,4,6,8). Covered by the rooks in those columns.Therefore, yes, all white cells are covered. Therefore, the minimal number of rooks needed is 5. Similarly, we could place rooks in even rows and odd columns. Let's see:If we place rooks in even rows (2,4,6,8) and odd columns (1,3,5,7,9). There are 4 even rows and 5 odd columns. To cover all even rows and all odd columns, we need to place 5 rooks. Place four rooks in even rows and four odd columns, and one more rook in the remaining odd column and any even row. However, there are only four even rows. Therefore, we can place four rooks in even rows and different odd columns, and one more rook in the fifth odd column and an even row. But there are only four even rows, so the fifth rook would have to be in an even row already covered. Therefore, this might not work. For example, place rooks at (2,1), (4,3), (6,5), (8,7), and (2,9). This covers all even rows and all odd columns. However, the rook at (2,9) covers even row 2 and odd column 9. This covers all white cells in even rows (since they are in even columns, but we're covering odd columns). Wait, no. Wait, white cells in even rows are in even columns, so covering odd columns doesn't help. Therefore, this approach is incorrect.Therefore, covering even rows and odd columns would not cover the white cells in even rows (which are in even columns). Therefore, the correct approach is to cover odd rows and even columns. Therefore, the minimal number of rooks is 5.But wait, let me think again. When we place a rook in an odd row and an even column, it covers that odd row (covering all white cells in that row) and that even column (covering all white cells in that column, which are in even rows). Therefore, by placing rooks in all odd rows and even columns, we cover all white cells. But to cover all even columns, we need at least four rooks (since there are four even columns). However, there are five odd rows. Therefore, we need to place a rook in each odd row (five rooks), and each rook can be placed in an even column. However, there are only four even columns, so by the pigeonhole principle, two rooks must share a column. But each rook in a column covers that column. Therefore, even if two rooks are in the same even column, the column is still covered. Therefore, we can place the five rooks as follows:- Row 1, Column 2- Row 3, Column 4- Row 5, Column 6- Row 7, Column 8- Row 9, Column 2 (or any other even column)This way, all five odd rows are covered, and all four even columns (2,4,6,8) are covered. The rook in row 9, column 2 covers row 9 and column 2 (already covered by the rook in row 1, column 2). However, the coverage is still achieved. Therefore, all white cells are under attack:- White cells in odd rows (covered by their respective rows).- White cells in even rows (covered by the even columns, each of which has at least one rook).Yes, this works. Therefore, the minimal number of rooks needed is 5.But earlier, when we considered the bipartite graph approach, we ended up with 9 rooks, which contradicts this. Therefore, where was the mistake?The mistake was in assuming that the vertex cover had to be entirely rows or entirely columns within each component. However, by allowing a mix of rows and columns across components, we can achieve a lower total. Wait, no, because the components are separate. The odd component's rows and columns are separate from the even component's. Therefore, when we place a rook in an odd row and even column, it belongs to both components. However, in reality, the rook is placed on a white cell only if it's in an odd row and odd column or even row and even column. A rook placed in an odd row and even column is on a black cell, but the problem doesn't require the rook to be on a white cell. The question only requires that all white cells are under attack, regardless of where the rooks are placed. Therefore, rooks can be placed on any cell, black or white, as long as all white cells are attacked.Therefore, the earlier bipartite graph approach was incorrect because it assumed that rooks had to be placed on white cells, but the problem allows rooks to be placed on black cells as well. Therefore, the bipartite graph model where edges are white cells is not directly applicable, because rooks can be placed anywhere, attacking entire rows and columns regardless of the cell color.Therefore, the bipartite graph approach was a red herring because it assumed rooks had to be placed on white cells, which is not the case. The problem allows rooks to be placed on any cell. Therefore, the earlier reasoning leading to 5 rooks is correct.Therefore, the minimal number of rooks required is 5. Let's confirm this with another approach.Imagine placing a rook in each odd row (rows 1,3,5,7,9), each in an even column (columns 2,4,6,8). Since there are 5 odd rows and 4 even columns, one even column will have two rooks. For example:- Rook at (1,2)- Rook at (3,4)- Rook at (5,6)- Rook at (7,8)- Rook at (9,2)These five rooks cover:- All odd rows (1,3,5,7,9), thus covering all white cells in odd rows.- All even columns (2,4,6,8), thus covering all white cells in even rows (which are in even columns).Therefore, all white cells are under attack. Hence, 5 rooks suffice.Is it possible to do it with fewer than 5 rooks? Let's see. Suppose we try with 4 rooks. Each rook can cover one row and one column. To cover all white cells, which are in 5 odd rows and 4 even columns, we need to cover all 5 odd rows and 4 even columns. If we have only 4 rooks, they can cover at most 4 rows and 4 columns. However, there are 5 odd rows to cover, so at least one odd row would remain uncovered. Therefore, 4 rooks are insufficient. Hence, 5 is the minimum.Therefore, the final answer is 5.

✅:The minimum number of rooks required to attack all white cells on a 9×9 checkerboard is 5.Step-by-Step Explanation:1. Board Analysis: - The board alternates colors starting with white at the corners. - Odd rows (1,3,5,7,9) have white cells in odd columns (1,3,5,7,9). - Even rows (2,4,6,8) have white cells in even columns (2,4,6,8). - Total white cells: 25 (in odd rows) + 16 (in even rows) = 41.2. Coverage Strategy: - To cover all white cells, ensure every white cell is in a rook's row or column. - Odd row white cells can be covered by placing rooks in odd rows. - Even row white cells can be covered by placing rooks in even columns.3. Optimal Placement: - Place rooks at intersections of odd rows and even columns. - 5 odd rows (need coverage) and 4 even columns (need coverage). - Use 5 rooks: one in each odd row, strategically placed in even columns.4. Example Configuration: - Rook positions: (1,2), (3,4), (5,6), (7,8), (9,2). - Covers all odd rows (via rows 1,3,5,7,9) and all even columns (via columns 2,4,6,8).5. Verification: - White cells in odd rows: Covered by the rooks' rows. - White cells in even rows: Covered by the rooks' columns (even columns 2,4,6,8).Conclusion: Using 5 rooks placed as described ensures all white cells are attacked, and fewer than 5 rooks cannot cover all required rows and columns. Thus, the minimum number is 5.boxed{5}

🔑:1. Chessboard Analysis: - A 9x9 chessboard consists of 81 cells. - These cells are alternatively colored black and white. - Corner cells and other cells in the same color pattern make all four corners white.2. Evaluate Rooks' Influence on White Cells: - A rook controls a row and a column, meaning all cells in its row and column are under attack. - Since the board has white cells only on white diagonals, a rook on any white cell will influence certain white cells in the row and column of its position. - The maximum number of white cells in any line (row or column) is 5.3. Estimate Minimum Rooks Needed: - Ideal placement would ensure each white cell is attacked with the least number of rooks. - Assuming no overlap, a rook can attack no more than 5 white cells in its row and 5 in its column. - A numerical lower bound: - On the 9x9 board, there are frac{81}{2} = 40.5, but practically 40 or 41 white cells. - With 4 rooks each attacking 5 white cells, theoretical maximum is 4 times 10 = 40 cells. - Conclude that 4 rooks cannot control all white cells.4. Visualization and Positioning: - Place rooks on the chessboard in such a way that maximizes the influence of each rook. - For example, if we place rooks at: (1,1), (3,3), (5,5), (7,7), (9,9) - Here, each rook covers maximum possible white cells without unnecessary overlaps.5. Verification: - Given rook positioning at (1,1), it targets all white cells in row 1 and column 1 similarly for other rooks. - This pattern will cover the board efficiently. Conclusion:The minimum number of rooks required to cover all the white cells on a ( 9 times 9 ) chessboard following the chess rules is:(boxed{5})

🧠:Okay, let's try to solve this problem step by step. So, there are nine cards numbered from 1 to 9, and four people each picked two cards. The remaining card is the one we need to find. Each person made a statement about their cards, and all statements are true. Let's break down each statement and see what possible pairs they could have.First, Aleksa says her numbers add up to 6. So, we need to find all possible pairs of distinct numbers between 1 and 9 that add up to 6. Let's list them:Possible pairs for Aleksa (sum to 6):- 1 + 5 = 6- 2 + 4 = 6- 3 + 3 = 6, but since the cards are numbered from 1 to 9 and each is unique, she can't have two 3s. So, only the first two pairs are possible: (1,5) and (2,4).Next, Bart says the difference between his numbers is 5. Let's find all pairs where the difference is 5. The larger number minus the smaller one equals 5.Possible pairs for Bart (difference 5):- 6 - 1 = 5 → (1,6)- 7 - 2 = 5 → (2,7)- 8 - 3 = 5 → (3,8)- 9 - 4 = 5 → (4,9)So, Bart could have any of these four pairs: (1,6), (2,7), (3,8), (4,9).Clara states that the product of her numbers is 18. Let's find all pairs of distinct numbers that multiply to 18.Possible pairs for Clara (product 18):- 2 × 9 = 18 → (2,9)- 3 × 6 = 18 → (3,6)- 1 × 18 = 18, but 18 isn't in the cards, so this is invalid.So Clara's possible pairs are (2,9) and (3,6).Deindra says one of her numbers is twice the other. Let's list possible pairs where one is double the other. Note that both numbers must be between 1 and 9 and distinct.Possible pairs for Deindra (one is twice the other):- 1 and 2 (2 is twice 1)- 2 and 4 (4 is twice 2)- 3 and 6 (6 is twice 3)- 4 and 8 (8 is twice 4)So Deindra's possible pairs are (1,2), (2,4), (3,6), (4,8).Now, the key here is that all these pairs must use distinct numbers from 1 to 9, with each number used at most once. The remaining number is the one left on the table.We need to figure out which combination of pairs for each person uses 8 distinct numbers, leaving one number unused. Let's try to find overlapping numbers and see if we can eliminate possibilities.Let's start by considering possible overlaps. For example, if Aleksa has (1,5), then Bart can't have 1 or 5. Similarly, if Clara has (3,6), then Bart can't have 3 or 6, etc.This might require trying different combinations. Let's make a table of possibilities.First, let's note the possible pairs:Aleksa: (1,5) or (2,4)Bart: (1,6), (2,7), (3,8), (4,9)Clara: (2,9), (3,6)Deindra: (1,2), (2,4), (3,6), (4,8)Let's start by considering each possibility for Aleksa and see which options for the others remain.Case 1: Aleksa has (1,5)Then, the remaining numbers are 2,3,4,6,7,8,9.Now, Bart's possible pairs: (2,7), (3,8), (4,9), since (1,6) is invalid now.Clara's possible pairs: (2,9), (3,6). But 6 is still available here.Deindra's possible pairs: (2,4), (3,6), (4,8), (1,2) is invalid as 1 is taken.Let's try to assign Bart, Clara, Deindra.Suppose Bart takes (2,7). Then remaining numbers: 3,4,6,8,9.Clara's options: (3,6) or (2,9). But 2 is taken by Bart, so Clara must take (3,6). Then Deindra's options: remaining numbers are 4,8,9.Deindra's possible pairs: (4,8). Because (3,6) is taken by Clara, (2,4) is taken by Bart (2 is taken), (4,8) is possible. So Deindra takes (4,8). Then the remaining number is 9. Wait, but Clara took (3,6), Bart took (2,7), Aleksa (1,5), Deindra (4,8). So numbers used: 1,5,2,7,3,6,4,8. Leftover is 9. But let's check Clara's product: (3,6) gives 18, correct. Deindra's pair (4,8): 8 is twice 4, correct. Bart's difference (7-2=5). Correct. So this seems possible. The leftover is 9. But 9 is one of the options (E). But let's check other possibilities to see if this is the only solution.Alternatively, what if in this case, Bart takes (4,9). So Bart has (4,9). Then remaining numbers: 2,3,6,7,8.Clara's possible pairs: (2,9) but 9 is taken by Bart, so Clara must take (3,6). Then Deindra's options: remaining numbers are 2,7,8.Deindra's possible pairs: (2,4) but 4 is taken, (3,6) taken, (4,8) taken, (1,2) no. So possible pairs? Deindra needs a pair where one is twice the other. From 2,7,8. The possible pairs here are (2,4) but 4 is taken, (7,14) invalid, (8,4) taken. So no valid pairs. Therefore, this path is invalid. So Bart cannot take (4,9) in this scenario.Next, Bart could take (3,8). Let's try that. Bart has (3,8). Remaining numbers: 2,4,6,7,9.Clara's options: (2,9) or (3,6). But 3 is taken, so Clara takes (2,9). Then Deindra's options: remaining numbers 4,6,7.Deindra needs a pair where one is twice the other. Possible pairs: (4,8) but 8 is taken, (3,6) taken, (2,4) 2 is taken. So from 4,6,7, can we form such a pair? 6 and 3, but 3 is taken. 4 and 2, taken. 7 and... nothing. So no valid pairs. So this path is invalid.Therefore, in Case 1 where Aleksa has (1,5), the only valid possibility is Bart takes (2,7), Clara takes (3,6), Deindra takes (4,8), leaving 9. But let's check if Deindra can take (4,8). Yes, because 8 is twice 4. That works. So the leftover is 9. So answer E. But wait, let's check other cases to ensure there's no other possibility.Case 2: Aleksa has (2,4). Then remaining numbers are 1,3,5,6,7,8,9.Bart's possible pairs: (1,6), (2,7) but 2 is taken, (3,8), (4,9) but 4 is taken. So possible Bart pairs: (1,6), (3,8), (4,9). Wait, 4 is taken by Aleksa, so Bart can't have (4,9). So Bart's options are (1,6) and (3,8).Clara's possible pairs: (2,9) is possible? 2 is taken by Aleksa, so (2,9) is invalid. So Clara must take (3,6).Deindra's possible pairs: (1,2) is invalid (2 taken), (2,4) taken by Aleksa, (3,6) if available, (4,8) 4 is taken. So Deindra's possible pairs: (3,6) if available, or (4,8) taken, or maybe others. Wait, remaining numbers after Aleksa (2,4) are 1,3,5,6,7,8,9.Let's proceed step by step.Subcase 2a: Bart takes (1,6). Then remaining numbers: 3,5,7,8,9.Clara must take (3,6), but 6 is taken by Bart. So Clara can't take (3,6). Wait, Clara's only possible pair was (3,6) because (2,9) is invalid. But if Bart took (1,6), then Clara cannot take (3,6) because 6 is taken. This is a problem. Therefore, this subcase is invalid.Subcase 2b: Bart takes (3,8). Then remaining numbers: 1,5,6,7,9.Clara must take (3,6), but 3 is taken by Bart, so she can't. Again, Clara's only possible pair is (3,6), which is now impossible. Therefore, this subcase is invalid as well.Wait, this is a problem. In Case 2 where Aleksa has (2,4), both of Bart's possible pairs lead to Clara not having any valid pairs. That means Case 2 is impossible. Therefore, the only possible case is Case 1, where Aleksa has (1,5), leading to the leftover number 9.But wait, the options given include E) 9, but the answer choices also have other numbers. Let me double-check my reasoning.In Case 1, the leftover number is 9. But let's verify all pairs again:Aleksa: 1 and 5 (sum 6)Bart: 2 and 7 (difference 5)Clara: 3 and 6 (product 18)Deindra: 4 and 8 (one is twice the other, 8=2×4)Used numbers: 1,5,2,7,3,6,4,8. Leftover is 9. Correct.But wait, another possible combination: Let's see if there's another way. For example, in Case 1, if Bart takes (3,8), then Clara might take (2,9), but then Deindra would need to take (4, something). Let's check again.Wait, in Case 1: Aleksa (1,5), remaining numbers 2,3,4,6,7,8,9.If Bart takes (3,8), then remaining numbers: 2,4,6,7,9.Clara can take (2,9) (since 3 is taken by Bart). Then Deindra has remaining numbers 4,6,7.Deindra needs a pair where one is twice the other. From 4,6,7: 4 and 2 (but 2 is taken by Clara), 6 and 3 (taken), 7 and... nothing. So no valid pairs. So this doesn't work. Hence, the only valid combination in Case 1 is Bart taking (2,7), Clara taking (3,6), Deindra taking (4,8), leaving 9. Therefore, the answer should be E) 9. But let's check the options again. The choices are A)1, B)3, C)6, D)8, E)9. So E is an option.Wait, but wait another thing. Let me check if there's another possible combination where the leftover is different. Maybe I missed something.Alternatively, let's consider if Aleksa has (2,4), and even though in my previous analysis it leads to a contradiction, maybe there's a way.Case 2: Aleksa has (2,4). Remaining numbers: 1,3,5,6,7,8,9.Bart's options: (1,6) and (3,8).If Bart takes (1,6), then Clara can't take (3,6) because 6 is taken. Clara's only possible pair was (3,6), so this is invalid.If Bart takes (3,8), then Clara needs to take (3,6), but 3 is taken, so again invalid. So yes, Case 2 is impossible. Thus, the only possible leftover is 9.But wait, the problem states that each person picked two cards, so all four have two each, totaling 8 cards, leaving one. So the answer must be among the options. However, I recall that in some similar puzzles, sometimes there's an alternative solution. Let me check again.Wait, in Deindra's possible pairs, when Aleksa is (1,5), Bart (2,7), Clara (3,6), Deindra must take (4,8). So numbers used: 1,5,2,7,3,6,4,8. Leftover is 9. That's correct. All statements are satisfied.But let's check if there's another possible combination where the leftover is different. Suppose, for example, that Aleksa has (2,4), but as we saw, this leads to a dead end. Another way: let's check if Clara could have (2,9) instead of (3,6) in some scenario.Wait, in Case 1 where Aleksa has (1,5), Bart takes (4,9). Let's see:Aleksa (1,5), Bart (4,9). Remaining numbers: 2,3,6,7,8.Clara could take (2,9), but 9 is taken by Bart. So she must take (3,6). Deindra has remaining numbers: 2,7,8.Deindra needs a pair where one is twice the other. 2 and 4 (4 is taken by Aleksa?), no. 8 is twice 4 (4 is taken). 7 and... nothing. So no valid pairs. Thus, this path is invalid.Alternatively, if Clara takes (2,9) when possible. But if Bart took (1,6), then Clara could take (2,9) only if 2 and 9 are available. Let's see:Case 1: Aleksa (1,5), Bart (1,6) is invalid because 1 is taken by Aleksa. Wait, Bart's options in Case 1 are (2,7), (3,8), (4,9). So Bart can't take (1,6) in Case 1 because Aleksa has 1. So no, that's not possible.Hmm, so all possibilities lead to the leftover being 9. Therefore, the answer should be E) 9. But wait, the options given include B)3, which I might have overlooked. Let me check again.Wait, maybe there's a different combination. Let me think differently. Suppose Clara takes (2,9). Then:Clara has (2,9), product 18. Then, if Aleksa has (1,5) sum 6. Then Bart needs a pair with difference 5. Possible pairs: (3,8), (4,9). But 9 is taken by Clara, so Bart must take (3,8). Then Deindra needs to have numbers where one is twice the other. Remaining numbers would be 4,6,7. Wait, let's check:Aleksa: 1,5Clara: 2,9Bart: 3,8Used numbers: 1,5,2,9,3,8. Remaining: 4,6,7.Deindra needs two of these. Possible pairs: 4 and 8 (taken), 6 and 3 (taken), 7 and... no. Wait, remaining numbers are 4,6,7. So possible pairs for Deindra: 4 and 2 (taken), 6 and 3 (taken), 4 and 8 (taken), 6 and 3 (taken). So no possible pairs. Thus, this is invalid.Alternatively, if Clara takes (3,6), then:Aleksa: 1,5Clara:3,6Bart: let's say (2,7). Then Deindra has 4,8,9. Deindra can take (4,8), leaving 9. That's the same as before.Alternatively, Bart takes (4,9), but then Deindra has 2,7,8. No valid pairs.So no, the only valid solution is leftover 9. Therefore, the answer is E)9.But wait, the options include B)3. Could there be a mistake? Let me check again.Wait, perhaps I made a mistake in assuming that in Case 1, when Aleksa has (1,5), Bart takes (2,7), Clara takes (3,6), and Deindra takes (4,8). Then the leftover is 9. Correct. All numbers used except 9.But let's check if Deindra's pair is valid: 4 and 8. Yes, 8 is twice 4. Correct.Clara's product:3×6=18. Correct.Bart's difference:7-2=5. Correct.Aleksa's sum:1+5=6. Correct.All statements are satisfied. So the leftover is 9, which is option E. Therefore, the answer should be E.But the options given in the problem are A)1, B)3, C)6, D)8, E)9. So E is the correct answer. But wait, the initial problem says "nine cards numbered from 1 to 9" and four people each took two, so 8 cards taken, leaving one. Yes.Wait, but I'm a bit confused because sometimes these puzzles can have alternative solutions. Let me check one more time.Is there any other combination where the leftover is different? For example, suppose Aleksa has (2,4), but we saw that leads to a contradiction. Let's try to see if there's a way to make it work.Aleksa (2,4). Remaining numbers:1,3,5,6,7,8,9.Bart must take (1,6) or (3,8).If Bart takes (1,6), then remaining numbers:3,5,7,8,9.Clara must take (3,6), but 6 is taken. So invalid.If Bart takes (3,8), remaining numbers:1,5,6,7,9.Clara must take (3,6), but 3 is taken. Invalid.So no, Case 2 is impossible. Therefore, only Case 1 works, leaving 9. So the answer is E.

✅:The remaining card is determined by analyzing the possible pairs for each person based on their statements and ensuring all numbers are unique. 1. Aleksa's pairs (sum 6): (1,5) or (2,4). 2. Bart's pairs (difference 5): (1,6), (2,7), (3,8), (4,9). 3. Clara's pairs (product 18): (2,9) or (3,6). 4. Deindra's pairs (one number twice the other): (1,2), (2,4), (3,6), (4,8). Testing combinations where all pairs use distinct numbers: - If Aleksa takes (1,5), remaining numbers are 2,3,4,6,7,8,9. - Bart must take (2,7) (difference 5). Remaining: 3,4,6,8,9. - Clara takes (3,6) (product 18). Remaining: 4,8,9. - Deindra takes (4,8) (8 is twice 4). Used numbers: 1,5,2,7,3,6,4,8. Leftover: 9. This configuration satisfies all conditions, leaving 9 as the remaining card. Answer: E) 9

🔑:Consider the following guidelines for the distribution of cards among Aleksa, Bart, Clara, and Deindra. Each statement they made is true, and we must identify the single number left on the table.1. Aleksa's Cards: - Aleksa's numbers add up to 6. The possible pairs are (1, 5) and (2, 4).2. Bart's Cards: - Bart's numbers have a difference of 5. The possible pairs are (1, 6), (2, 7), (3, 8), and (4, 9).3. Clara's Cards: - Clara's numbers have a product of 18. The possible pairs are (2, 9) and (3, 6).4. Deindra's Cards: - Deindra's numbers include one number twice the other. The possible pairs are (1, 2), (2, 4), (3, 6), and (4, 8).We'll examine feasible distributions of cards: Step-by-step Analysis:Step 1: Determine Aleksa's cards.- Suppose Aleksa has (2, 4). Potential pairs for others: - Clara's pairs: (3, 6) (remaining since (2, 9) does not work for others). - Attempt Deindra: Check pairs 1, 2; 2, 4; 3, 6; and 4, 8. However, all pairs using 2 , 4 are invalid (other uses of 2 and 4). Hence, Aleksa's cards must be (1, 5).Step 2: Determine Clara's cards.Given (1, 5) for Aleksa, check remaining pairs for Bart and Clara:- For Clara, the potential pairs with product 18 are (2, 9) and (3, 6). - Suppose Clara's pair is (2, 9). - Bart's pair possibilities then (3, 8) and (4, 9) but this conflicts with already used numbers. - Hence, Clara's Cards must be (3, 6).Step 3: Determine Bart's cards. - Remaining feasible pairs for Bart exist: (1, 6), (2, 7), (3, 8), (4, 9). - With (3, 6) set, examine Bart's `(difference 5)` pairs: - (1, 6) consumes leftover, hence invalid. - (2, 7) only valid (otherwise (4, 9) empty for Deindra).Therefore, Bart's cards are (2, 7).Step 4: Determine Deindra's cards.Last check on remaining pairs of (3, 6) already used: - For pairs with one twice the other: - Remaining numbers are 4, 8.Finally, verifying if: - Aleksa: (1, 5) - Bart: (2, 7) - Clara: (3, 6) - Deindra: (4, 8)We find number left on the table is 9, boxed. Conclusion:boxed{9}

🧠:Okay, let me try to figure out how to solve this problem. So, we have a line segment AB with length a. We need to find a point P on AB such that if we take AP as the edge of a regular tetrahedron and PB as the edge of a regular hexahedron (which is a cube), the sum of their surface areas is minimized. Hmm, okay.First, let me recall the formulas for the surface areas of a regular tetrahedron and a cube. A regular tetrahedron has four equilateral triangular faces. So, the surface area should be 4 times the area of one of those triangles. The area of an equilateral triangle with side length s is (√3/4)s². Therefore, the surface area of the tetrahedron would be 4*(√3/4)s² = √3 s². So, if AP is x, then the surface area of the tetrahedron is √3 x².Then, for the regular hexahedron, which is a cube, each face is a square. A cube has 6 faces, each with area s² where s is the edge length. So, the surface area is 6 s². Since PB is the edge here, if PB is (a - x), then the surface area of the cube is 6(a - x)².Therefore, the total surface area S is the sum of these two: S = √3 x² + 6(a - x)². Our goal is to find the value of x in [0, a] that minimizes S. That seems like a calculus problem where we can take the derivative of S with respect to x, set it to zero, and solve for x. Let me check that.So, S(x) = √3 x² + 6(a - x)². To find the minimum, take the derivative S’(x) = 2√3 x + 6*2*(a - x)*(-1). Wait, hold on. Let me compute that correctly. The derivative of √3 x² is 2√3 x. The derivative of 6(a - x)² is 6*2*(a - x)*(-1) = -12(a - x). So, S’(x) = 2√3 x - 12(a - x).Set this derivative equal to zero for critical points: 2√3 x - 12(a - x) = 0. Let me solve for x.2√3 x - 12a + 12x = 0Combine like terms: (2√3 x + 12x) = 12aFactor out x: x(2√3 + 12) = 12aTherefore, x = 12a / (2√3 + 12)Simplify numerator and denominator by dividing numerator and denominator by 2: x = 6a / (√3 + 6)Hmm, so x = 6a / (6 + √3). To rationalize the denominator, maybe multiply numerator and denominator by (6 - √3):x = 6a (6 - √3) / [(6 + √3)(6 - √3)] = 6a(6 - √3)/(36 - 3) = 6a(6 - √3)/33Simplify further: 6/33 = 2/11, so x = 2a(6 - √3)/11. Wait, let me check:Wait, 6a(6 - √3)/33 = (6/33)a(6 - √3) = (2/11)a(6 - √3). Yes, correct. So x = (2a/11)(6 - √3). Let me compute that expression:6 - √3 ≈ 6 - 1.732 ≈ 4.268, so 2/11*4.268 ≈ (8.536)/11 ≈ 0.776. So x ≈ 0.776a. But maybe we can leave it in exact form.So the critical point is at x = [6/(6 + √3)]a. But we rationalized it to x = [2(6 - √3)/11]a. Either form is acceptable, but perhaps the first form is simpler: x = (6a)/(6 + √3). Either way, since it's a minimum, we need to confirm that this critical point is indeed a minimum. Since the function S(x) is a quadratic-like function (sum of two quadratics), and the coefficients of x² are positive (√3 + 6), the function is convex, so the critical point is indeed a minimum.Therefore, the point P should be located at a distance of x = 6a/(6 + √3) from point A along AB. To construct this point, given that AB = a, we can use a compass and straightedge construction. Since the ratio is 6/(6 + √3), which involves √3, which is constructible, we can find this point.But how exactly to construct it? Let's think. The length x is 6a/(6 + √3). To construct this, maybe we can construct a right triangle with legs 6 and √3, then use similar triangles. Alternatively, perhaps using the harmonic mean or something. Let me see.Alternatively, since √3 is approximately 1.732, but in exact terms, we can represent it as the height of an equilateral triangle with side length 2, for example. But perhaps a better approach is to set up a line segment of length 6 + √3 and then divide it into 6 parts and so on. Hmm.Alternatively, note that 6/(6 + √3) = 1/(1 + √3/6). So, if we can construct √3/6, which is (√3)/6, that's the same as (1/6)√3. But constructing such a ratio might involve creating similar triangles.Alternatively, here's a method:1. Draw segment AB with length a.2. Construct a right triangle where one leg is 6 units and the other leg is √3 units. The hypotenuse would then be √(6² + (√3)²) = √(36 + 3) = √39. But this might not be directly helpful.Wait, perhaps we can use algebra to express x/a = 6/(6 + √3). Let me rewrite that as x/a = 1/(1 + √3/6). So, if we let k = √3/6, then x/a = 1/(1 + k). To construct 1/(1 + k), we can use similar triangles.Here's a way to do it:- Let’s consider a line segment of length 1 + k. If we can construct k = √3/6, then 1 + k is the denominator. Then, x/a is the reciprocal segment.But maybe a better approach is to use the concept of dividing a segment in a given ratio. The ratio here is 6 : √3. So, if we can construct a segment with ratio 6 to √3, then we can divide AB into that ratio.Given that, the point P divides AB in the ratio AP : PB = 6 : √3. Wait, is that correct?Wait, the critical point was x = [6/(6 + √3)]a, so AP = x = 6a/(6 + √3), and PB = a - x = a - 6a/(6 + √3) = [ (6 + √3)a - 6a ]/(6 + √3) = (√3 a)/(6 + √3). Therefore, AP/PB = [6a/(6 + √3)] / [√3 a/(6 + √3)] = 6/√3 = 2√3. So the ratio AP : PB is 2√3 : 1.Wait, that's interesting. So AP is 2√3 times PB. So if we can divide AB in the ratio 2√3 : 1, that will give the position of P.But how to construct a ratio involving √3? Since √3 is constructible, we can do this by constructing a right triangle with legs 1 and √3, hypotenuse 2, which is a 30-60-90 triangle. For example, if we construct an equilateral triangle, then the height is √3/2 times the side length.Alternatively, here's a step-by-step construction method:1. Draw segment AB with length a.2. At point A, construct a perpendicular line.3. On this perpendicular, mark a point D such that AD = 1 unit (using some arbitrary unit length).4. From D, construct a line perpendicular to AD (i.e., parallel to AB). The length from D along this line should be √3 units. But since we can't measure √3 directly, perhaps construct an equilateral triangle on AD, so each side is 1, then the height would be √3/2. Hmm, maybe another way.Alternatively, construct a 30-60-90 triangle where the sides are 1, √3, 2. For example:- Draw a line segment AE perpendicular to AB at A, of length 1.- From E, draw a circle with radius 2. The intersection with AB extended would be at a point F such that AF = 2. Then, triangle AEF is a right triangle with AE =1, AF=2, so EF = √3. But maybe that's complicating.Alternatively, use similar triangles. To get the ratio AP : PB = 2√3 : 1, we can use the following:- Let’s consider that we need to divide AB into parts AP and PB such that AP/PB = 2√3. So AP = 2√3 * PB.Since AP + PB = a, substituting gives 2√3 PB + PB = a => PB(2√3 + 1) = a => PB = a/(2√3 + 1). Then AP = 2√3 a/(2√3 + 1). But this seems different from earlier, but wait, maybe I made a miscalculation.Wait, earlier we had AP = 6a/(6 + √3) and PB = √3 a/(6 + √3). Let's compute AP/PB:AP/PB = (6a/(6 + √3)) / (√3 a/(6 + √3)) ) = 6/√3 = 2√3. So yes, AP : PB = 2√3 : 1. Therefore, to divide AB in the ratio 2√3 :1.To construct this ratio, since 2√3 is approximately 3.464, but in exact terms, we can use geometric construction.A standard method to divide a segment into a given ratio is using similar triangles or the intercept theorem (Thales' theorem).Here's how we can do it:1. Draw segment AB of length a.2. At point A, draw a ray making an angle with AB.3. On this ray, mark off a distance equal to 2√3 units (let’s say using some scale) from A to a point C.4. From point C, mark off another distance of 1 unit in the same direction to a point D, so that AD = 2√3 + 1 units.5. Connect point D to point B.6. Construct a line through point C parallel to DB, intersecting AB at point P.By the intercept theorem, AP/PB = AC/CD = 2√3 /1, so P divides AB in the ratio 2√3 :1, which is the required ratio.But the challenge is how to mark off 2√3 units. Since √3 is constructible, we can construct a right triangle with legs 1 and √3, but how?Alternatively, use an equilateral triangle to get √3. For example:- Construct an equilateral triangle on AB; each side is a, so the height is (√3/2)a. But we need 2√3 units. Wait, perhaps we can use a different approach.Alternatively, set up a coordinate system where A is at (0,0) and B is at (a,0). Then, the point P is at x = 6a/(6 + √3). But construction-wise, how to get this?Alternatively, use the following steps:1. Draw AB with length a.2. Construct a square on AB, so each side is a.3. Construct an equilateral triangle on AB, so the height is (√3/2)a.4. Using the height, create a segment of length √3 a.But this might not directly help. Alternatively, use a unit construction:Let’s use a unit length. Let’s say we take a unit length u. Then:1. Construct a right triangle with legs 1u and √3 u, hypotenuse 2u. This is a 30-60-90 triangle.2. Use this triangle to measure 2√3 u.But since we need to apply this to segment AB of length a, we need to scale accordingly.Alternatively, use the intercept theorem with a constructed triangle:1. From point A, draw a line at an angle.2. On this line, construct a segment AC = 2√3 units and CB' = 1 unit (total AB' = 2√3 +1 units).3. Connect B' to B.4. Draw a line through C parallel to B'B, intersecting AB at P.But again, the problem is constructing 2√3 units. To do this:1. Construct an equilateral triangle with side length 2u. Its height is √3 u.2. So, if we take a segment of 2u, the height is √3 u, so we can use that to get √3.Alternatively, here's a practical construction:1. Let’s extend AB beyond B.2. Construct an equilateral triangle on AB, say triangle ABC.3. The height of this triangle is (√3/2)a.4. Extend this height beyond C by some amount. Wait, maybe not.Alternatively, use similar triangles by constructing a triangle outside AB that has sides in the ratio 2√3 :1.But honestly, this is getting complicated. Maybe there's a simpler way.Alternatively, use the fact that the ratio is 6/(6 + √3). Let's write this as 6/(6 + √3) = 1/(1 + √3/6). So, if we can construct √3/6 of AB, then we can subtract that from 1 and invert it. But constructing √3/6 of AB would require some steps.Alternatively, perhaps use the following:1. Divide AB into 6 equal parts.2. Construct a segment of length equal to one part (a/6).3. Construct an equilateral triangle on this segment (a/6), so its height is (√3/2)(a/6) = √3 a /12.4. Use this height to measure √3 a /6 by doubling it.But I'm not sure.Alternatively, here's a step-by-step construction:1. Draw AB with length a.2. At point A, construct a perpendicular line.3. On this perpendicular, mark a point C such that AC = a/6.4. Construct an equilateral triangle on AC, so each side is a/6. The height of this triangle is (√3/2)(a/6) = √3 a /12.5. From point C, measure a distance of √3 a /12 upwards to point D.6. Now, the segment AD is a/6 in the horizontal direction and √3 a /12 in the vertical, but not sure how this helps.Alternatively, create a right triangle where one leg is 6 units and the other is √3 units, then use that to set the ratio.Wait, maybe use coordinates:Let’s place point A at (0,0) and B at (a,0). We need to find point P at (x,0) where x = 6a/(6 + √3). To construct this x, perhaps:1. Construct a coordinate system with A at (0,0), B at (a,0).2. Construct a line segment from A at an angle, say upwards.3. On this line, mark a point C at (6,0) if we scale the coordinate system, but need to relate to length a.Alternatively, use similar triangles outside of AB.Let me think differently. The ratio we need is 6 : √3. So if we can create a triangle with sides 6 and √3, then the hypotenuse would be √(6² + (√3)²) = √39, but maybe not helpful.Wait, perhaps using the intercept theorem:To divide AB into the ratio 6:√3.1. From point A, draw a ray at an angle.2. On this ray, mark off 6 units to point C and then another √3 units to point D, so that AC = 6 units and CD = √3 units, making AD = 6 + √3 units.3. Connect D to B.4. Draw a line through C parallel to DB, intersecting AB at P.Then, by the intercept theorem, AP / PB = AC / CD = 6 / √3 = 2√3, which is the required ratio.Yes, this seems feasible. However, we need to ensure that the units used on the ray from A are in the same scale as AB. Since AB is length a, we need to adjust the construction accordingly.But in classical geometric construction, the units are arbitrary. So if we construct AC and CD with lengths 6 and √3 respectively, regardless of the scale of AB, the ratio will be maintained.However, since AB is length a, we need to map this ratio onto AB. To do that, we can use the intercept theorem (Thales' theorem) which states that if two lines are cut by parallel lines, the segments are proportional.Here's a precise construction method:1. Draw segment AB with length a.2. From point A, draw a ray AX that is not colinear with AB.3. On ray AX, using any convenient unit length, mark off a point C such that AC = 6 units.4. From point C, mark off another point D on AX such that CD = √3 units. Now, AD = AC + CD = 6 + √3 units.5. Connect point D to point B.6. Construct a line through point C parallel to DB, intersecting AB at point P.By the intercept theorem, the ratio AP / PB = AC / CD = 6 / √3 = 2√3, which is the desired ratio. Therefore, point P is the required point that minimizes the total surface area.To verify, since the intercept theorem gives the proportion based on the parallel lines, this construction ensures that AP/PB = 6/√3, hence AP = 6a/(6 + √3) and PB = √3 a/(6 + √3), which is the critical point we found earlier.Therefore, the construction is valid, and point P is located at AP = 6a/(6 + √3) from point A along AB.

✅:To solve the problem of finding point ( P ) on segment ( AB ) such that the sum of the surface areas of a regular tetrahedron (with edge ( AP )) and a regular hexahedron (cube with edge ( PB )) is minimized, follow these steps: Mathematical Analysis:1. Surface Areas: - Regular tetrahedron surface area: ( sqrt{3} cdot AP^2 ). - Cube surface area: ( 6 cdot PB^2 ). - Total surface area: ( S = sqrt{3}x^2 + 6(a - x)^2 ), where ( x = AP ).2. Optimization: - Differentiate ( S ) with respect to ( x ): ( S' = 2sqrt{3}x - 12(a - x) ). - Set ( S' = 0 ): ( 2sqrt{3}x = 12(a - x) ). - Solve for ( x ): ( x = frac{6a}{6 + sqrt{3}} ). Geometric Construction:1. Divide Segment ( AB ) in the ratio ( AP : PB = 6 : sqrt{3} ): - Steps: 1. Draw segment ( AB ) of length ( a ). 2. From ( A ), draw a ray ( AX ) not collinear with ( AB ). 3. On ( AX ), mark points ( C ) and ( D ) such that ( AC = 6 ) units and ( CD = sqrt{3} ) units (using a chosen unit length). 4. Connect ( D ) to ( B ). 5. Draw a line through ( C ) parallel to ( DB ), intersecting ( AB ) at ( P ). Final Answer:The point ( P ) that minimizes the total surface area is located at a distance of ( boxed{dfrac{6a}{6 + sqrt{3}}} ) from point ( A ) along ( AB ). This is geometrically constructed using the intercept theorem to divide ( AB ) in the ratio ( 6 : sqrt{3} ).

🔑: Problem:Adva van egy ( AB = a ) távolság. Szerkesszük meg e távolságon a ( P ) pontot úgy, hogy az ( AP ), illetőleg ( PB ) távolságokkal, mint élekkel alkotott szabályos tetraéder, illetőleg szabályos hexaéder felületeinek összege a lehető legkisebb legyen.1. Legyen ( AP = x ). A tetraéder éle ( x ), így a tetraéder felülete: [ 4 cdot frac{x^2}{4} sqrt{3} = x^2 sqrt{3} ]2. A hexaéder éle ( a - x ), így a hexaéder felülete: [ 6(a - x)^2 ]3. Az összesített felület ( y ) kifejezése: [ y = x^2 sqrt{3} + 6(a - x)^2 ]4. Fejtsük ki a kifejezést: [ y = x^2 sqrt{3} + 6[a^2 - 2ax + x^2] ] [ y = x^2 sqrt{3} + 6a^2 - 12ax + 6x^2 ] [ y = (6 + sqrt{3})x^2 - 12ax + 6a^2 ]5. Ez egy másodfokú függvény az ( x ) változóval. Mivel a kifejezés ( x^2 ) együtthatója pozitív (( 6 + sqrt{3} )), a függvénynek van minimuma.6. A másodfokú függvény általános alakja: ( y = ax^2 + bx + c ). Az ( x_0 ) szélső érték (minimum) helye: [ x_0 = -frac{b}{2a} ]7. Alkalmazzuk ezt a jelen esetben: [ a = 6 + sqrt{3}, quad b = -12a ] [ x_0 = frac{12a}{2(6 + sqrt{3})} = frac{12a}{12 + 2sqrt{3}} = frac{6a}{6 + sqrt{3}} ]8. Tehát: [ AP = frac{6a}{6 + sqrt{3}} ] [ PB = a - x = a - frac{6a}{6 + sqrt{3}} = frac{asqrt{3}}{6 + sqrt{3}} ]9. Vizsgáljuk meg az arányt ( frac{AP}{PB} ): [ frac{AP}{PB} = frac{frac{6a}{6 + sqrt{3}}}{frac{asqrt{3}}{6 + sqrt{3}}} = frac{6}{sqrt{3}} = frac{2sqrt{3}}{1} = frac{sqrt{12}}{1} ]Következtetésünk:()boxed{AP = frac{6a}{6 + sqrt{3}}, quad PB = a - frac{6a}{6 + sqrt{3}} = frac{asqrt{3}}{6 + sqrt{3}}}(boxed{frac{AP}{PB} = frac{sqrt{12}}{1}}

🧠:Okay, let me try to figure out whether every triangle where the incenter is equidistant from the midpoints of two sides must be isosceles. Hmm, so first, I need to recall some properties of triangles, incenter, midpoints, and maybe coordinate geometry. Let's break this down step by step.First, the incenter is the point where the angle bisectors of the triangle meet, and it's equidistant from all three sides. The midpoints of the sides are connected to the concept of the midsegments of a triangle. If the incenter is equidistant from two midpoints, does that force the triangle to have two sides equal, making it isosceles?Let me start by setting up a coordinate system. Maybe placing the triangle in a coordinate plane will make it easier to compute distances. Let's assume vertices A, B, C with coordinates. Let me assign coordinates such that the triangle is positioned conveniently. For example, place vertex A at (0, 0), vertex B at (b, 0), and vertex C at (c, d). Then, the midpoints of the sides can be calculated. The midpoints of AB, BC, and AC would be ((b/2, 0), ((b + c)/2, d/2), and (c/2, d/2) respectively. Wait, maybe that's a bit messy. Alternatively, maybe place the triangle symmetrically if we suspect it's isosceles. But since we don't know yet, maybe a general triangle is better.Alternatively, maybe using coordinate geometry with side lengths. Let's denote the triangle as ABC with sides opposite to vertices A, B, C as a, b, c respectively. The incenter's coordinates can be found using the formula ( (aA_x + bB_x + cC_x)/(a + b + c), (aA_y + bB_y + cC_y)/(a + b + c) ). But maybe that's getting complicated. Alternatively, if I place the triangle with side BC on the x-axis for simplicity. Let's try that.Let me position vertex B at (0, 0), vertex C at (c, 0), and vertex A at (a, b). Then, the midpoints of sides AB and AC would be at ((a/2, b/2) and ((a + c)/2, b/2) respectively. The midpoint of BC is (c/2, 0). The incenter coordinates can be calculated using the formula: ( (a_A * x_A + a_B * x_B + a_C * x_C ) / perimeter, similarly for y-coordinates ), where a_A, a_B, a_C are the lengths of the sides opposite vertices A, B, C. Wait, actually, the formula for the incenter is ( (a x_A + b x_B + c x_C ) / (a + b + c ), (a y_A + b y_B + c y_C ) / (a + b + c ) ), where a, b, c are the lengths of the sides opposite to A, B, C respectively. Wait, no, actually, the incenter coordinates are given by weighted average of the vertices, with weights equal to the lengths of the sides opposite those vertices. So if the triangle has vertices A (x_A, y_A), B (x_B, y_B), C (x_C, y_C), then incenter (I_x, I_y) is:I_x = (a x_A + b x_B + c x_C) / (a + b + c)I_y = (a y_A + b y_B + c y_C) / (a + b + c)where a is the length of BC, b the length of AC, c the length of AB.Wait, let me confirm. Yes, the incenter coordinates can be calculated using the formula:I_x = (a x_A + b x_B + c x_C)/(a + b + c),I_y = (a y_A + b y_B + c y_C)/(a + b + c),where a, b, c are the lengths of the sides opposite to angles A, B, C respectively.So in my coordinate system, let's assign:- Let’s set vertex B at (0, 0), vertex C at (c, 0), and vertex A at (d, e). Then:- The length of BC is a = c.- The length of AC is b = sqrt( (d - c)^2 + e^2 )- The length of AB is c_side = sqrt( d^2 + e^2 )Wait, maybe I should use standard notation. Let's denote:- Let’s have triangle ABC with coordinates: - B at (0, 0) - C at (c, 0) - A at (d, e)Then:- Length BC = a = c- Length AC = b = sqrt( (d - c)^2 + e^2 )- Length AB = c_side = sqrt( d^2 + e^2 )Therefore, the incenter I has coordinates:I_x = (a*d + b*0 + c_side*c) / (a + b + c_side)Wait, hold on. Wait, the formula is (a x_A + b x_B + c x_C)/(a + b + c). Wait, but here, a is the length opposite vertex A, which is BC, so a = BC = c. Then b is the length opposite vertex B, which is AC = sqrt( (d - c)^2 + e^2 ). c_side is the length opposite vertex C, which is AB = sqrt( d^2 + e^2 ). Therefore, incenter coordinates are:I_x = (a*d + b*0 + c_side*c) / (a + b + c_side)Similarly,I_y = (a*e + b*0 + c_side*0) / (a + b + c_side) = (a e)/(a + b + c_side)So, I_x = (a d + c_side c)/(a + b + c_side), I_y = (a e)/(a + b + c_side)Now, the midpoints:Midpoint of AB: Let's compute coordinates. AB is from (0, 0) to (d, e), so midpoint M_AB is (d/2, e/2)Midpoint of AC: AC is from (d, e) to (c, 0), so midpoint M_AC is ((d + c)/2, e/2)Midpoint of BC: (c/2, 0)But the problem states that the incenter is equidistant from the midpoints of two sides. Let's say, for instance, the incenter is equidistant from midpoints of AB and AC. Then, the distance from I to M_AB equals the distance from I to M_AC.Alternatively, maybe equidistant to midpoints of two sides, not necessarily AB and AC. But since the triangle is general, we need to consider which two midpoints. But maybe the problem states "the midpoints of two sides", so it could be any two. But to check if the triangle must be isosceles, perhaps regardless of which two midpoints, but in the problem statement, it's "the center of the inscribed circle is equidistant from the midpoints of two sides". So, for a given triangle, if the incenter is equidistant to two midpoints, is that triangle necessarily isosceles?Alternatively, maybe the problem is that for a triangle, if the incenter is equidistant from midpoints of two sides, then it's isosceles. So perhaps the answer is yes, such a triangle must be isosceles.Alternatively, maybe not necessarily. Let's check.Let me consider two cases. Suppose the incenter is equidistant to midpoints of AB and AC. Then set up the equation:Distance from I to M_AB = Distance from I to M_ACCompute both distances.Coordinates of I: ( (a d + c_side c)/(a + b + c_side), (a e)/(a + b + c_side) )Coordinates of M_AB: (d/2, e/2)Coordinates of M_AC: ((d + c)/2, e/2)Compute distance squared from I to M_AB:[ ( (a d + c_side c)/(a + b + c_side) - d/2 )^2 + ( (a e)/(a + b + c_side) - e/2 )^2 ]Similarly, distance squared from I to M_AC:[ ( (a d + c_side c)/(a + b + c_side) - (d + c)/2 )^2 + ( (a e)/(a + b + c_side) - e/2 )^2 ]Set these equal:First component squared plus second component squared equal for both distances.But notice that the y-coordinates of M_AB and M_AC are both e/2. The y-coordinate of I is (a e)/(a + b + c_side). So the vertical distance from I to both M_AB and M_AC is | (a e)/(a + b + c_side) - e/2 |. Therefore, the vertical component is the same for both distances. Therefore, the difference in distances comes only from the horizontal component.Therefore, setting the distances equal reduces to setting the horizontal components equal.Therefore:| ( (a d + c_side c)/(a + b + c_side) - d/2 ) | = | ( (a d + c_side c)/(a + b + c_side) - (d + c)/2 ) |.Since distances are non-negative, we can square both sides to remove the absolute value:[ (a d + c_side c)/(a + b + c_side) - d/2 ]^2 = [ (a d + c_side c)/(a + b + c_side) - (d + c)/2 ]^2Let me denote S = a + b + c_side.Then the equation becomes:[ (a d + c_side c)/S - d/2 ]^2 = [ (a d + c_side c)/S - (d + c)/2 ]^2Let me compute the left-hand side (LHS) and right-hand side (RHS):Compute LHS:( (a d + c_side c)/S - d/2 ) = (2(a d + c_side c) - S d ) / (2 S )Similarly, RHS:( (a d + c_side c)/S - (d + c)/2 ) = (2(a d + c_side c) - S (d + c) ) / (2 S )So squaring both:[2(a d + c_side c) - S d]^2 / (4 S^2 ) = [2(a d + c_side c) - S (d + c)]^2 / (4 S^2 )Multiply both sides by 4 S^2 to eliminate denominators:[2(a d + c_side c) - S d]^2 = [2(a d + c_side c) - S (d + c)]^2Let me expand both sides.Let’s denote T = 2(a d + c_side c)Left side: (T - S d)^2Right side: (T - S (d + c))^2Expand both:Left: T^2 - 2 T S d + S^2 d^2Right: T^2 - 2 T S (d + c) + S^2 (d + c)^2Set equal:T^2 - 2 T S d + S^2 d^2 = T^2 - 2 T S (d + c) + S^2 (d + c)^2Cancel T^2 on both sides:-2 T S d + S^2 d^2 = -2 T S (d + c) + S^2 (d + c)^2Bring all terms to left side:-2 T S d + S^2 d^2 + 2 T S (d + c) - S^2 (d + c)^2 = 0Factor terms:-2 T S d + 2 T S (d + c) + S^2 d^2 - S^2 (d + c)^2 = 0Factor 2 T S:2 T S [ -d + (d + c) ] + S^2 [ d^2 - (d + c)^2 ] = 0Simplify inside the brackets:For the first term: -d + d + c = cFor the second term: d^2 - (d^2 + 2 d c + c^2) = -2 d c - c^2So:2 T S * c + S^2 (-2 d c - c^2 ) = 0Factor S c:S c [ 2 T - S (2 d + c ) ] = 0Since S is the perimeter, which is positive, and c is the length BC, which is positive (assuming the triangle is non-degenerate), so S c ≠ 0. Therefore, the remaining factor must be zero:2 T - S (2 d + c ) = 0Recall that T = 2(a d + c_side c )So:2 * 2(a d + c_side c ) - S (2 d + c ) = 0=> 4(a d + c_side c ) - S (2 d + c ) = 0But S = a + b + c_sideThus:4(a d + c_side c ) - (a + b + c_side)(2 d + c ) = 0Let me expand the second term:(a + b + c_side)(2 d + c ) = a (2 d + c ) + b (2 d + c ) + c_side (2 d + c )So:4(a d + c_side c ) - [ 2 a d + a c + 2 b d + b c + 2 c_side d + c_side c ] = 0Now distribute the 4:4 a d + 4 c_side c - 2 a d - a c - 2 b d - b c - 2 c_side d - c_side c = 0Combine like terms:(4 a d - 2 a d) + (4 c_side c - c_side c) + (-a c) + (-2 b d) + (-b c) + (-2 c_side d ) = 0Simplify:2 a d + 3 c_side c - a c - 2 b d - b c - 2 c_side d = 0Let me group terms by variables:Terms with d:2 a d - 2 b d - 2 c_side d = d (2 a - 2 b - 2 c_side )Terms with c:3 c_side c - a c - b c = c (3 c_side - a - b )So:d (2 a - 2 b - 2 c_side ) + c (3 c_side - a - b ) = 0Hmm, this seems complicated. Let me recall what a, b, c_side are in terms of the coordinates.Earlier, we defined:a = BC = c (distance from B (0,0) to C (c,0))b = AC = sqrt( (d - c)^2 + e^2 )c_side = AB = sqrt( d^2 + e^2 )So, a = c, b = sqrt( (d - c)^2 + e^2 ), c_side = sqrt( d^2 + e^2 )Therefore, substituting into the equation:d (2 c - 2 sqrt( (d - c)^2 + e^2 ) - 2 sqrt( d^2 + e^2 )) + c (3 sqrt( d^2 + e^2 ) - c - sqrt( (d - c)^2 + e^2 )) = 0This equation looks very complicated. Maybe there's a simpler approach.Alternatively, maybe consider specific cases. Let's assume the triangle is isosceles and check if the incenter is equidistant to the midpoints of two sides.Suppose triangle ABC is isosceles with AB = AC. Then, midpoints of AB and AC would be symmetric with respect to the axis of symmetry. The incenter lies on the axis of symmetry (the altitude from A), so the distances from the incenter to the midpoints of AB and AC should be equal. Thus, in this case, the incenter is equidistant to those midpoints. So isosceles triangles do satisfy the condition.But the question is whether only isosceles triangles satisfy this condition. Suppose we have a triangle that is not isosceles, but the incenter is equidistant to two midpoints. If we can find such a triangle, then the answer is no. If not, then yes.Alternatively, maybe through properties rather than coordinates. Let's think about the inradius and the midpoints.The midpoint of a side is part of the midline of the triangle. The distance from the incenter to the midpoint might be related to the sides and angles.Alternatively, use vectors. Let me try vector approach.Let the triangle be ABC, with position vectors A, B, C. The incenter I is given by:I = (a A + b B + c C ) / (a + b + c )where a, b, c are lengths of sides opposite to A, B, C.Midpoint of AB is (A + B)/2, midpoint of BC is (B + C)/2, midpoint of AC is (A + C)/2.The distance from I to midpoint of AB is |I - (A + B)/2|.Similarly for other midpoints.Set |I - M1| = |I - M2|, square both sides:(I - M1) · (I - M1) = (I - M2) · (I - M2)Expanding both:|I|² - 2 I · M1 + |M1|² = |I|² - 2 I · M2 + |M2|²Cancel |I|²:-2 I · M1 + |M1|² = -2 I · M2 + |M2|²Rearranged:2 I · (M2 - M1) = |M2|² - |M1|²Let me compute this for midpoints of AB and AC.Let M1 = (A + B)/2, M2 = (A + C)/2.Then, M2 - M1 = (A + C)/2 - (A + B)/2 = (C - B)/2|M2|² - |M1|² = |(A + C)/2|² - |(A + B)/2|²Compute this:= ( |A + C|² - |A + B|² ) / 4= [ (A + C) · (A + C) - (A + B) · (A + B) ] / 4= [ |A|² + 2 A · C + |C|² - |A|² - 2 A · B - |B|² ] / 4= [ 2 A · (C - B) + |C|² - |B|² ] / 4Therefore, the equation becomes:2 I · (C - B)/2 = [ 2 A · (C - B) + |C|² - |B|² ] / 4Simplify left side:I · (C - B) = [ 2 A · (C - B) + |C|² - |B|² ] / 4Multiply both sides by 4:4 I · (C - B) = 2 A · (C - B) + |C|² - |B|²Now substitute I = (a A + b B + c C ) / (a + b + c )Thus:4 [ (a A + b B + c C ) / (a + b + c ) ] · (C - B ) = 2 A · (C - B ) + |C|² - |B|²Multiply left side:4/(a + b + c ) [ a A · (C - B ) + b B · (C - B ) + c C · (C - B ) ] = 2 A · (C - B ) + |C|² - |B|²Compute each dot product:A · (C - B ) = A · C - A · BB · (C - B ) = B · C - |B|²C · (C - B ) = |C|² - C · BTherefore, left side becomes:4/(a + b + c ) [ a (A · C - A · B ) + b (B · C - |B|² ) + c (|C|² - C · B ) ]Let me rearrange terms:= 4/(a + b + c ) [ a (A · C - A · B ) + (b B · C - b |B|² ) + c |C|² - c C · B )= 4/(a + b + c ) [ a (A · (C - B )) + B · C (b - c ) - b |B|² + c |C|² ]Now, this seems complicated. Let me see if I can relate these terms to the sides of the triangle.Recall that in a triangle, the lengths are related to the dot products. For example, |B - C|² = |B|² + |C|² - 2 B · C = a² (since a is BC). Therefore, B · C = ( |B|² + |C|² - a² ) / 2Similarly, A · B = ( |A|² + |B|² - c_side² ) / 2, where c_side is AB.And A · C = ( |A|² + |C|² - b² ) / 2, where b is AC.Let me substitute these into the equation.First, let's note that:A · (C - B ) = A · C - A · B = [ ( |A|² + |C|² - b² ) / 2 ] - [ ( |A|² + |B|² - c_side² ) / 2 ] = ( |C|² - |B|² - b² + c_side² ) / 2Similarly, B · C = ( |B|² + |C|² - a² ) / 2Therefore, substitute back:Left side:4/(a + b + c ) [ a ( ( |C|² - |B|² - b² + c_side² ) / 2 ) + (b - c ) ( ( |B|² + |C|² - a² ) / 2 ) - b |B|² + c |C|² ]This is getting extremely algebra-heavy. Maybe I need to consider specific cases or use another approach.Alternatively, maybe consider the converse: if a triangle is not isosceles, can the incenter be equidistant to two midpoints?Suppose triangle is scalene, all sides different. Let's assume in such a triangle, the incenter is equidistant to midpoints of two sides. Is this possible?Alternatively, maybe in a scalene triangle, the incenter is closer to the midpoint of the longer side? Not sure.Alternatively, maybe use barycentric coordinates. The incenter in barycentric coordinates is (a : b : c). Midpoints of sides have coordinates (0, 1/2, 1/2), (1/2, 0, 1/2), (1/2, 1/2, 0). The distance from incenter to midpoint can be computed using barycentric distance formulas.But barycentric coordinates might not be straightforward here. Alternatively, recall that in barycentric coordinates, the distance between two points (u1, v1, w1) and (u2, v2, w2) is given by sqrt( -a² (v1 - v2)(w1 - w2) - b² (u1 - u2)(w1 - w2) - c² (u1 - u2)(v1 - v2) ) ) / (sum of coordinates). But this is complicated.Alternatively, since midpoints are the centers of the edges, maybe the distance from the incenter to the midpoint can be expressed in terms of the triangle's parameters.Alternatively, think about the formula for the distance from the incenter to a midpoint.Alternatively, use the formula for the distance between two points. Let's consider triangle ABC with sides a, b, c, inradius r, semiperimeter s = (a + b + c)/2. The coordinates of the incenter are known in trilinear coordinates as 1:1:1, but maybe this isn't helpful.Wait, maybe using vectors again. Let's define vectors with B at origin, C at vector c, and A at vector a. Then, the incenter is (a a + b B + c C ) / (a + b + c ). Wait, no, better to set coordinates.Wait, perhaps taking specific values. Let me take a specific triangle and check.Suppose we take an isosceles triangle, say AB = AC = 5, BC = 6. Then, the inradius can be calculated, and midpoints.But maybe better to take coordinates. Let’s place B at (-3, 0), C at (3, 0), A at (0, 4). This makes BC = 6, AB = AC = 5.Compute incenter: the incenter in an isosceles triangle lies on the altitude, which is the y-axis. The coordinates would be (0, r), where r is the inradius.Semiperimeter s = (5 + 5 + 6)/2 = 8Area = (base * height)/2 = (6 * 4)/2 = 12Inradius r = area / s = 12 / 8 = 1.5Therefore, incenter is at (0, 1.5)Midpoints of AB and AC:Midpoint of AB: between (-3, 0) and (0, 4): (-1.5, 2)Midpoint of AC: between (0, 4) and (3, 0): (1.5, 2)Distance from incenter (0, 1.5) to (-1.5, 2):sqrt( (-1.5 - 0)^2 + (2 - 1.5)^2 ) = sqrt(2.25 + 0.25) = sqrt(2.5) ≈ 1.581Distance to (1.5, 2): same value. So yes, equidistant. As expected.Now, take a scalene triangle and see if we can have the incenter equidistant to two midpoints.Let’s take triangle with sides 3, 4, 5. Right-angled triangle. Coordinates: B(0,0), C(3,0), A(0,4). Midpoints:AB: (0,2), BC: (1.5, 0), AC: (1.5, 2)Incenter: coordinates formula. The inradius r = (3 + 4 - 5)/2 = 1? Wait, no. For a right-angled triangle, inradius is (a + b - c)/2 where c is the hypotenuse. So (3 + 4 - 5)/2 = 2/2 = 1. So inradius is 1. Coordinates of incenter: distance 1 from each side. So in a right-angled triangle at B(0,0), the incenter is at (r, r) = (1,1).Distance from incenter (1,1) to midpoint of AB (0,2):sqrt( (1 - 0)^2 + (1 - 2)^2 ) = sqrt(1 + 1) = sqrt(2) ≈ 1.414Distance to midpoint of BC (1.5, 0):sqrt( (1.5 - 1)^2 + (0 - 1)^2 ) = sqrt(0.25 + 1) = sqrt(1.25) ≈ 1.118Distance to midpoint of AC (1.5, 2):sqrt( (1.5 - 1)^2 + (2 - 1)^2 ) = sqrt(0.25 + 1) = sqrt(1.25) ≈ 1.118So in the 3-4-5 triangle, incenter is equidistant to midpoints of BC and AC, which are both at distance ≈1.118, but not to the midpoint of AB. So in this scalene triangle, the incenter is equidistant to midpoints of two sides. Wait, but the 3-4-5 triangle is scalene, but here the incenter is equidistant to midpoints of BC and AC. But this contradicts the original question's implication. But according to this example, a scalene triangle can have the incenter equidistant to two midpoints.Wait, but in the 3-4-5 triangle, the midpoints of BC and AC are equidistant to the incenter. So does that mean the answer is no, not necessarily isosceles?But wait, let me confirm the calculation.Incenter coordinates: (1,1)Midpoint of BC: (1.5, 0)Distance: sqrt( (0.5)^2 + (-1)^2 ) = sqrt(0.25 + 1) = sqrt(1.25)Midpoint of AC: (1.5, 2)Distance: sqrt( (0.5)^2 + (1)^2 ) = same sqrt(1.25)So yes, equidistant.Therefore, in this scalene triangle, the incenter is equidistant to midpoints of BC and AC. Therefore, the answer is no, such a triangle does not have to be isosceles.But wait, the 3-4-5 triangle is scalene, yet satisfies the condition. Therefore, the statement "every triangle with incenter equidistant to two midpoints is isosceles" is false.But wait, the problem is stated as:"Is every triangle in which the center of the inscribed circle is equidistant from the midpoints of two sides an isosceles triangle?"So the answer would be no, since we found a scalene triangle (3-4-5) where incenter is equidistant to midpoints of two sides.But wait, in the 3-4-5 example, which two midpoints are equidistant? Midpoints of BC and AC. In this case, BC = 3, AC = 5, so the sides opposite to these are different. But the midpoints are at (1.5, 0) and (1.5, 2). The distances to incenter (1,1) are both sqrt(1.25). So yes, it works.Therefore, the answer is no, there exist non-isosceles triangles where the incenter is equidistant from midpoints of two sides.But wait, the initial analysis using coordinates led to a complicated equation, but the example with 3-4-5 triangle shows that it's possible. So maybe the answer is no.Wait, but I need to verify once again. Let me check another scalene triangle.Take triangle with sides 5, 6, 7. Compute inradius and coordinates.Semiperimeter s = (5 + 6 + 7)/2 = 9Area = sqrt(9*(9-5)*(9-6)*(9-7)) = sqrt(9*4*3*2) = sqrt(216) = 6*sqrt(6)Inradius r = area / s = 6√6 / 9 = 2√6 / 3 ≈ 1.632Coordinates: Let's place the triangle with side BC = 5 on x-axis, B at (0,0), C at (5,0), and A somewhere in the plane.Lengths: AB = 7, AC = 6.Coordinates of A: let's find using distance formulas.Let A = (x, y)Distance from A to B: sqrt(x² + y²) = 7 → x² + y² = 49Distance from A to C: sqrt( (x - 5)^2 + y² ) = 6 → (x -5)^2 + y² = 36Subtract first equation from second:(x -5)^2 + y² - x² - y² = 36 - 49 → x² -10x +25 -x² = -13 → -10x +25 = -13 → -10x = -38 → x = 3.8Then from x² + y² = 49:(3.8)^2 + y² = 49 → 14.44 + y² = 49 → y² = 34.56 → y = √34.56 ≈ 5.88So A is at (3.8, ≈5.88)Incenter coordinates:I_x = (a x_A + b x_B + c x_C)/(a + b + c )Here, a = BC = 5, b = AC = 6, c = AB =7Wait, no: in the formula, a, b, c are lengths opposite to A, B, C. If we placed B at (0,0), C at (5,0), A at (3.8,5.88), then:Length opposite A is BC = 5Length opposite B is AC = 6Length opposite C is AB =7Therefore, incenter coordinates:I_x = (a x_A + b x_B + c x_C ) / (a + b + c ) = (5*3.8 + 6*0 +7*5)/(5+6+7) = (19 + 0 +35)/18 = 54/18 = 3I_y = (a y_A + b y_B + c y_C ) / (a + b + c ) = (5*5.88 +6*0 +7*0)/18 = 29.4 /18 ≈1.633Which matches the inradius ≈1.632, since the y-coordinate is the inradius (as the triangle is on the x-axis with height y).Midpoints:Midpoint of AB: between B(0,0) and A(3.8,5.88): (1.9, 2.94)Midpoint of BC: (2.5, 0)Midpoint of AC: between A(3.8,5.88) and C(5,0): (4.4, 2.94)Now compute distance from incenter (3, ≈1.633) to each midpoint.Distance to midpoint of AB (1.9, 2.94):sqrt( (3 - 1.9)^2 + (1.633 - 2.94)^2 ) ≈ sqrt(1.1² + (-1.307)^2 ) ≈ sqrt(1.21 + 1.708) ≈ sqrt(2.918) ≈1.708Distance to midpoint of BC (2.5, 0):sqrt( (3 -2.5)^2 + (1.633 -0)^2 ) ≈ sqrt(0.5² +1.633² )≈ sqrt(0.25 +2.666)≈sqrt(2.916)≈1.708Distance to midpoint of AC (4.4, 2.94):sqrt( (3 -4.4)^2 + (1.633 -2.94)^2 )≈sqrt( (-1.4)^2 + (-1.307)^2 )≈sqrt(1.96 +1.708 )≈sqrt(3.668 )≈1.915So in this 5-6-7 triangle, the incenter is equidistant to midpoints of AB and BC, both approximately 1.708. Therefore, this scalene triangle also has the incenter equidistant to two midpoints.Therefore, this confirms that there exist scalene triangles where the incenter is equidistant to two midpoints. Therefore, the answer to the original question is no, such triangles are not necessarily isosceles.But wait, this contradicts the initial thought. How come?In both the 3-4-5 triangle and the 5-6-7 triangle, which are scalene, the incenter is equidistant to two midpoints. Therefore, the answer is no, not every triangle with this property is isosceles.But let me check if this is a general property. Maybe in any triangle, the incenter is equidistant to midpoints of two sides if and only if the triangle is either isosceles or has sides in a particular ratio.But according to the examples, even scalene triangles can have this property. Therefore, the answer is no.But why in the coordinate analysis earlier, the equation led to a complicated relation? Perhaps because it's not necessarily leading to a condition that enforces the triangle to be isosceles, but rather allows for other solutions.Therefore, the conclusion is that the statement is false; there exist scalene triangles where the incenter is equidistant from midpoints of two sides. Hence, not every such triangle is isosceles.

✅:The answer is no, not every triangle with the incenter equidistant from the midpoints of two sides is necessarily isosceles. Here's the detailed reasoning:1. Counterexample with a Scalene Triangle: Consider a 3-4-5 right-angled triangle (scalene). Placing vertices at ( B(0,0) ), ( C(3,0) ), and ( A(0,4) ): - Incenter: Located at ( (1,1) ), calculated using the formula for the inradius ( r = frac{a + b - c}{2} ) (where ( c ) is the hypotenuse) and coordinates. - Midpoints: - Midpoint of ( BC ): ( (1.5, 0) ) - Midpoint of ( AC ): ( (1.5, 2) ) - Distances: - Distance from incenter to midpoint of ( BC ): ( sqrt{(1.5 - 1)^2 + (0 - 1)^2} = sqrt{1.25} ) - Distance from incenter to midpoint of ( AC ): ( sqrt{(1.5 - 1)^2 + (2 - 1)^2} = sqrt{1.25} ) - Conclusion: The incenter is equidistant to midpoints of ( BC ) and ( AC ), yet the triangle is scalene.2. General Verification: Similar calculations for other scalene triangles (e.g., 5-6-7 triangle) confirm that the incenter can indeed be equidistant to two midpoints without the triangle being isosceles. This occurs due to the symmetry in distances arising from the incenter's position relative to midpoints, even in asymmetric triangles.3. Key Insight: While isosceles triangles inherently satisfy this property due to their symmetry, scalene triangles can also fulfill the condition depending on the specific side lengths and the resulting position of the incenter.Final Answer: boxed{text{No}}

🔑:To determine whether every triangle in which the incenter is equidistant from the midpoints of two sides must be isosceles, we analyze the given properties and consider counterexamples.# First Solution:1. Define T Property: - A triangle is said to have the T property if the incenter (center of the incircle) is equidistant from (at least) two midpoints of the triangle's sides.2. Trivially Isosceles Triangles: - Any isosceles triangle inherently has the T property. For an isosceles triangle with equal sides, the incenter is equidistant from the midpoints of these sides due to symmetry.3. Counterexample with Non-Isosceles Triangle: - Consider the well-known right triangle called the "Egyptian triangle," with sides 3, 4, and 5 units. This triangle is not isosceles but can still have the T property. - Calculation of the inradius varrho (radius of the incircle): [ varrho = frac{A}{s} = frac{6}{6} = 1 text{ unit}, ] where A is the area of the triangle and s is the semiperimeter. - The incenter is one unit away from the longer leg (4 units) and therefore lies on the vertical axis of symmetry that bisects the segment joining the midpoints of the shorter leg (3 units) and the hypotenuse (5 units). - As a result, the midpoints of the 3-unit and 5-unit sides are equidistant from the incenter.4. Conclusion: - This counterexample shows that a triangle having the T property need not necessarily be isosceles. - Therefore, it is not true that every triangle with the T property must be isosceles.5. [text{boxed{text{Thus, the given property does not imply the triangle is isosceles.}}}]# Second Solution:1. Assumptions: - Consider a triangle ABC with incenter O. Assume that O is equidistant from the midpoints A_1 and B_1 of sides BC and AC, respectively. - Let the points where the incircle touches the sides be A_0, B_0, and C_0.2. Right Triangles: - Triangles OA_1A_0 and OB_1B_0 are right triangles with OA_1 = OB_1 and OA_0 = OB_0. Therefore, these triangles are congruent by the RHS (Right angle-Hypotenuse-Side) congruence criterion.3. Calculate distances: - The distances A_0C = B_0C are equal and denote them as s-c, where s is the semiperimeter and c is the side opposite to C. - Distances from C to the midpoints: CA_1 = frac{a}{2} and CB_1 = frac{b}{2}. 4. Express Segment Lengths: - Calculate segments A_0A_1 and B_0B_1: [ A_0 A_1 = left{ begin{array}{ll} frac{a}{2} - (s - c) & text{if } CA_1 geq CA_0 (s - c) - frac{a}{2} & text{if } CA_0 > CA_1 end{array} right. ] [ B_0 B_1 = left{ begin{array}{ll} frac{b}{2} - (s - c) & text{if } CB_1 geq CB_0 (s - c) - frac{b}{2} & text{if } CB_0 > CB_1 end{array} right. ]5. Equidstance Equality: - When the corresponding segments are equal, we get two possible equations: [ frac{a}{2} - (s - c) = frac{b}{2} - (s - c), ] resulting in a = b (triangle is isosceles). - Or, if: [ frac{a}{2} - (s - c) = (s - c) - frac{b}{2}, ] leading to: [ c = frac{a + b}{2}. ]6. Conclusions from Cases: - If a = b, the triangle is isosceles. - If c = frac{a + b}{2}, the side c is the arithmetic mean of the other two sides, and the triangle is not necessarily isosceles.7. General Insight: - Thus, from both approaches, we conclude: [ text{A triangle having the T property need not be isosceles.} ] - The hypothesis only implies either isosceles or that one side is the arithmetic mean of the other two sides.8. [text{boxed{text{Hence, not every triangle with the incenter equidistant from two side midpoints is isosceles.}}}]