🧠:Okay, let's see. I need to find the minimum value of M, which is the maximum among a, b, and c. The expressions for a, b, and c involve logarithms with variables x, y, z. Hmm, logarithms can sometimes be tricky, but maybe I can simplify these expressions first.Starting with expression a: ( a = lg z + lg left[x(yz)^{-1} + 1right] ). Let me rewrite that to make it clearer. The term ( x(yz)^{-1} ) is ( frac{x}{yz} ). So, a becomes ( lg z + lg left( frac{x}{yz} + 1 right) ). Wait, combining the logs, that would be ( lg left( z cdot left( frac{x}{yz} + 1 right) right) ). Let's compute that inside:( z cdot left( frac{x}{yz} + 1 right) = z cdot frac{x + yz}{yz} = frac{x + yz}{y} ). So, a simplifies to ( lg left( frac{x + yz}{y} right) ). Which is ( lg left( frac{x}{y} + z right) ). Hmm, that seems simpler.Now moving to expression b: ( b = lg x^{-1} + lg (xyz + 1) ). Similarly, combining these logs: ( lg left( x^{-1} cdot (xyz + 1) right) ). Let's compute that:( x^{-1}(xyz + 1) = yz + frac{1}{x} ). So, b simplifies to ( lg left( yz + frac{1}{x} right) ). Okay, that's a bit neater.Next, expression c: ( c = lg y + lg left[(xyz)^{-1} + 1right] ). Combine the logs: ( lg left( y cdot left( frac{1}{xyz} + 1 right) right) ). Let's compute the inside:( y cdot left( frac{1}{xyz} + 1 right) = y cdot frac{1 + xyz}{xyz} = frac{1 + xyz}{xz} ). Therefore, c simplifies to ( lg left( frac{1 + xyz}{xz} right) ). Which can also be written as ( lg left( frac{1}{xz} + y right) ). Hmm.So, summarizing the simplified forms:- ( a = lg left( frac{x}{y} + z right) )- ( b = lg left( yz + frac{1}{x} right) )- ( c = lg left( frac{1}{xz} + y right) )Now, M is the maximum of a, b, and c. We need to find the minimum possible value of M. So, this is a minimax problem. The goal is to choose x, y, z such that the maximum of a, b, c is as small as possible.Since all three expressions are logarithms, perhaps we can set them equal to each other to find the point where none of them exceeds the others, which would give the minimal maximum.Let me denote:Let’s set ( a = b = c ). Then, solving the equations:1. ( lg left( frac{x}{y} + z right) = lg left( yz + frac{1}{x} right) )2. ( lg left( yz + frac{1}{x} right) = lg left( frac{1}{xz} + y right) )Since logarithm is a monotonically increasing function, the arguments must be equal.So, from equation 1:( frac{x}{y} + z = yz + frac{1}{x} )From equation 2:( yz + frac{1}{x} = frac{1}{xz} + y )So now we have two equations:1. ( frac{x}{y} + z = yz + frac{1}{x} )2. ( yz + frac{1}{x} = frac{1}{xz} + y )Let me try to solve these equations step by step.Starting with equation 1:( frac{x}{y} + z = yz + frac{1}{x} )Let me rearrange terms:( frac{x}{y} - frac{1}{x} = yz - z )Factor z on the right:( frac{x}{y} - frac{1}{x} = z(y - 1) )Similarly, equation 2:( yz + frac{1}{x} = frac{1}{xz} + y )Rearranged:( yz - y = frac{1}{xz} - frac{1}{x} )Factor left side: ( y(z - 1) )Right side: ( frac{1}{x} left( frac{1}{z} - 1 right) = frac{1}{x} cdot frac{1 - z}{z} )So equation 2 becomes:( y(z - 1) = frac{1 - z}{xz} )Note that ( z - 1 = -(1 - z) ), so left side is ( -y(1 - z) ), right side is ( frac{1 - z}{xz} )Assuming ( 1 - z neq 0 ), we can divide both sides by ( 1 - z ):( -y = frac{1}{xz} )Therefore, ( -y = frac{1}{xz} implies xz = -frac{1}{y} )But since x, y, z are variables inside logarithms, their arguments must be positive. Therefore, x, y, z must be positive real numbers. So, xz is positive, but the equation says xz = -1/y. Since x, z are positive, right side is negative, which is impossible. Therefore, contradiction.Therefore, our assumption that ( 1 - z neq 0 ) is false. Therefore, ( 1 - z = 0 implies z = 1 )So, z = 1.Now, substituting z = 1 into equation 1:From equation 1:( frac{x}{y} + 1 = y cdot 1 + frac{1}{x} implies frac{x}{y} + 1 = y + frac{1}{x} )Let me write this as:( frac{x}{y} - y = frac{1}{x} - 1 )Multiply both sides by y to eliminate denominator on left:( x - y^2 = frac{y}{x} - y )Bring all terms to left side:( x - y^2 - frac{y}{x} + y = 0 )Multiply through by x to eliminate denominator:( x^2 - y^2 x - y + y x = 0 )Wait, let's see:Multiplying each term by x:( x cdot x - x cdot y^2 - x cdot frac{y}{x} + x cdot y = x^2 - x y^2 - y + x y )Simplify terms:x² - x y² - y + x y = 0Combine like terms:x² + x y - x y² - y = 0Factor terms:Hmm, maybe factor x from the first two terms and -y from the last two:x(x + y) - y(x y + 1) = 0 ?Wait, let's check:x² + x y - x y² - y = x(x + y) - y(x y + 1). Let's compute:x(x + y) = x² + x yy(x y + 1) = x y² + ySo, x(x + y) - y(x y + 1) = x² + x y - x y² - y = original expression. So, that's correct.Therefore, equation becomes:x(x + y) - y(x y + 1) = 0Not sure if that helps. Maybe trying to find substitutions.Let me let’s see. Since z=1, maybe we can express variables in terms of each other.Alternatively, let's see if we can find a symmetric solution. Since z=1, perhaps x and y are related in some symmetric way.Suppose x = y. Let's test this assumption.If x = y, then substituting into equation:From equation when z=1: ( frac{x}{x} + 1 = x + frac{1}{x} implies 1 + 1 = x + frac{1}{x} implies 2 = x + frac{1}{x} )Solve for x:Multiply both sides by x: 2x = x² + 1 ⇒ x² - 2x + 1 = 0 ⇒ (x -1)² = 0 ⇒ x =1So, x = y =1. Then z=1.Therefore, x = y = z =1.Let’s check if this satisfies both equations.First, z=1. Then check equation 1 and equation 2.Equation 1: ( frac{1}{1} +1 = 1*1 + frac{1}{1} implies 1 +1 =1 +1 ⇒ 2=2. True.Equation 2: From equation 2, after z=1:Original equation 2 after substitution z=1 was already used to get z=1. But after z=1, equation 2 becomes:yz + 1/x = 1/(xz) + y ⇒ y*1 +1/x =1/(x*1) + y ⇒ y +1/x =1/x + y. Which is 0=0. So, it's always true when z=1. Therefore, the system reduces to z=1 and equation 1. So, with x=y=1, it works.Therefore, x=y=z=1 is a solution. Let's check what a, b, c are in this case.Compute a:Original a: ( lg z + lg left[ x(yz)^{-1} +1 right] ). With x=y=z=1:( lg 1 + lg [1*(1*1)^{-1} +1] = 0 + lg [1 +1] = lg 2 approx 0.3010 )Similarly, b: ( lg x^{-1} + lg(xyz +1 ) = lg 1 + lg(1 +1) = 0 + lg 2 = lg 2 )c: ( lg y + lg [ (xyz)^{-1} +1 ] = lg 1 + lg [1 +1] = 0 + lg 2 = lg 2 )So, a = b = c = lg2 ≈0.3010. Therefore, M = lg2.But is this the minimal M? Wait, maybe there's a lower M? Let me check.But first, we need to confirm if x=y=z=1 is indeed the optimal solution. Perhaps there are other solutions where a, b, c are equal but less than lg2. But since logarithm is involved, and arguments must be positive, let's see.Wait, when x, y, z are 1, the arguments inside all logarithms are 2, which is the minimum possible for those expressions?Wait, let's think. Let's take expression a: ( lg left( frac{x}{y} + z right) ). If we can make ( frac{x}{y} + z ) as small as possible, then a would be minimized. Similarly for b and c.But since M is the maximum of a, b, c, to minimize M, we need to balance the three expressions so that none of them is too large. So, ideally, set all three expressions equal, which is what we did.But let's verify if x=y=z=1 is the only solution. Suppose there exists another solution where x, y, z are not 1. Let me see.Suppose we let x, y, z be different from 1 but still satisfy the equations. From earlier, we had to set z=1, so perhaps any solution must have z=1, and then x and y satisfy equation 1. Let's see.From equation 1 with z=1: ( frac{x}{y} +1 = y + frac{1}{x} ). Suppose x is not equal to y. Let's try to solve this equation.Let me rearrange the equation:( frac{x}{y} - y = frac{1}{x} -1 )Let’s denote t = x/y. Then x = t y. Substitute into equation:Left side: t - yRight side: ( frac{1}{t y} -1 )So, equation becomes:t - y = ( frac{1}{t y} -1 )Multiply both sides by t y to eliminate denominator:t² y - t y² = 1 - t yBring all terms to left side:t² y - t y² + t y -1 =0Factor terms:t² y - t y² + t y -1 = t y(t - y +1) -1 = ?Not obvious. Maybe choose a substitution. Let’s set u = t y = x. Wait, t = x/y, so u = x.Alternatively, perhaps set y = k, and x = m k, so t = m. Then substituting:Left side: m - kRight side: ( frac{1}{m k} -1 )Equation: m - k = ( frac{1}{m k} -1 )Bring all terms to left:m - k +1 - ( frac{1}{m k} ) =0This equation seems complex. Maybe try numerical values. Suppose we let m = 2, then:Left side: 2 - k +1 - ( frac{1}{2k} ) = 3 - k - 1/(2k). Set equal to zero:3 - k - 1/(2k) =0. Multiply by 2k:6k - 2k² -1 =0 ⇒ 2k² -6k +1=0 ⇒ k = [6 ± sqrt(36 -8)]/4 = [6 ± sqrt(28)]/4 = [6 ± 2√7]/4 = [3 ± √7]/2 ≈ [3 ± 2.6458]/2So, k ≈ (3 +2.6458)/2 ≈2.8229 or k≈(3 -2.6458)/2≈0.1771.So, possible solutions for m=2. But need to check if these lead to positive x, y, z.But even if they do, would this give a lower M?Wait, if x=2k, y=k, z=1. Then compute a, b, c.Let’s take k ≈2.8229:Then x=2*2.8229≈5.6458, y≈2.8229, z=1.Compute a: lg( (5.6458 /2.8229) +1 ) = lg(2 +1)=lg3≈0.4771Compute b: lg(2.8229*1 +1/5.6458)≈lg(2.8229 +0.1771)=lg3≈0.4771Compute c: lg(1/(5.6458*1) +2.8229 )≈lg(0.1771 +2.8229)=lg3≈0.4771So, a=b=c=lg3≈0.4771>lg2≈0.3010. So, M=lg3, which is larger than before.Similarly, if k≈0.1771:x=2*0.1771≈0.3542, y≈0.1771, z=1.Compute a: lg(0.3542 /0.1771 +1)=lg(2 +1)=lg3≈0.4771Compute b: lg(0.1771*1 +1/0.3542)≈lg(0.1771 +2.8229)=lg3≈0.4771Compute c: lg(1/(0.3542*1) +0.1771 )≈lg(2.8229 +0.1771)=lg3≈0.4771Same result. So, even though we found other solutions, M is higher. Therefore, the minimal M seems to be lg2 when x=y=z=1.But let's check another possibility. Maybe there are solutions where a, b, c are not all equal, but the maximum among them is less than lg2. Let's see.Suppose we try to set two of them equal and lower than the third. For example, set a = b < c, and try to minimize c. But since c has to be as small as possible, but if a and b are smaller, maybe c could be smaller than lg2?Wait, when x, y, z are 1, all a, b, c are lg2. If we change variables, perhaps one increases while others decrease? But with the relations between variables, it's not obvious. Let's test with x, y, z slightly different from 1.Take x=1, y=1, z=1. Let's perturb x slightly.Let’s let x=1+ε, y=1, z=1, where ε is small.Compute a: lg( (1+ε)/1 +1 ) = lg(2 + ε)Compute b: lg(1*1 +1/(1+ε)) = lg(1 +1/(1+ε)) = lg( (1 + ε +1)/(1+ε) ) = lg( (2 + ε)/(1 + ε) )Compute c: lg(1/(1*(1+ε)) +1 ) = lg(1/(1+ε) +1 ) = same as b, which is lg( (2 + ε)/(1 + ε) )So, a becomes lg(2 + ε), and b and c become lg( (2 + ε)/(1 + ε) )If ε>0, then 2 + ε >2, so a increases. (2 + ε)/(1 + ε) = 2/(1 + ε) + ε/(1 + ε) = 2/(1 + ε) + something small. Since ε>0, denominator increases, so (2 + ε)/(1 + ε) < 2/1 =2. Therefore, b and c become less than lg2. But a becomes larger. Therefore, M = a = lg(2 + ε) > lg2.If ε<0, say ε approaching -1 from above (so x approaches 0+). Then:a: lg( (1+ε)/1 +1 ) = lg(2 + ε). If ε approaches -1, 2 + ε approaches 1, so a approaches lg1=0.But b and c: lg(1/(1 + ε) +1 ). If ε approaches -1, 1 + ε approaches 0, so 1/(1 + ε) becomes large. Therefore, b and c become large. So, M would be max(a, b, c). As ε approaches -1, a approaches 0, but b and c approach infinity. So, M would go to infinity. Not helpful.Similarly, if we take ε negative but small, say ε=-0.1:x=0.9, y=1, z=1.a: lg(0.9 +1)=lg1.9≈0.2788b and c: lg(1/(0.9) +1)=lg(1.111... +1)=lg2.111≈0.3245So, M≈0.3245>lg2≈0.3010. So, higher.Therefore, perturbing x slightly from 1 increases M. Similarly, if we change y or z.Alternatively, let's try changing y. Suppose x=1, z=1, y=1+ε.Compute a: lg(1/(1+ε) +1 )=lg( (1 +1 +ε)/(1 + ε) )=lg( (2 + ε)/(1 + ε) )Compute b: lg( (1+ε)*1 +1/1 )=lg(1 + ε +1 )=lg(2 + ε )Compute c: lg(1/(1*(1+ε)) +1 )=lg(1/(1+ε) +1 )=lg( (1 +1 + ε)/(1 + ε) )=lg( (2 + ε)/(1 + ε) )So, similar to previous case. If ε>0, then a and c become lg( (2 + ε)/(1 + ε) ) which is less than lg2, and b becomes lg(2 + ε) which is greater than lg2. So, M=lg(2 + ε). If ε<0, then a and c become lg( (2 + ε)/(1 + ε) ). Let's compute for ε=-0.1:(2 -0.1)/(1 -0.1)=1.9/0.9≈2.111..., so a and c≈lg2.111≈0.3245b=lg(2 -0.1)=lg1.9≈0.2788Therefore, M≈0.3245>lg2.Hence, changing y also doesn't help. Similarly with z. Wait, z is fixed at 1 from before. So, maybe the minimal M is indeed lg2.But let's check another approach. Let's consider variables substitution to make it symmetric.Let’s set variables as follows: Let’s denote p = x, q = y, r = z. But maybe another substitution. Alternatively, perhaps use AM-GM inequality on the arguments of the logarithms.Since M is the maximum of three log terms, to minimize M, we need the arguments of the logs to be as small as possible, but balanced.But the expressions inside the logs for a, b, c are:For a: ( frac{x}{y} + z )For b: ( yz + frac{1}{x} )For c: ( frac{1}{xz} + y )So, let’s denote:Let’s set A = ( frac{x}{y} + z )B = ( yz + frac{1}{x} )C = ( frac{1}{xz} + y )We need to minimize the maximum of lgA, lgB, lgC. Since lg is increasing, this is equivalent to minimizing the maximum of A, B, C.Thus, the problem reduces to minimizing the maximum of A, B, C where:A = ( frac{x}{y} + z )B = ( yz + frac{1}{x} )C = ( frac{1}{xz} + y )With x, y, z > 0.So, now the problem is transformed into minimizing max(A, B, C). Let’s focus on this.To minimize the maximum of A, B, C, we can set A = B = C and find x, y, z that satisfy these equalities.Which is similar to our initial approach.Set A = B = C.So:1. ( frac{x}{y} + z = yz + frac{1}{x} )2. ( yz + frac{1}{x} = frac{1}{xz} + y )Which are the same equations as before, leading us to z=1 and x=y=1.Therefore, the minimal maximum of A, B, C is 2, achieved when x=y=z=1. Therefore, M = lg2.But wait, if A, B, C are all equal to 2, then their logs are lg2. So, that's consistent with previous result.But perhaps there's a case where A, B, C can be smaller than 2? Let's check.Suppose x, y, z are such that A=B=C=k <2. Then, we need to solve:( frac{x}{y} + z = k )( yz + frac{1}{x} = k )( frac{1}{xz} + y = k )With k <2.Let’s see if such a solution exists.From the first equation: ( frac{x}{y} = k - z )From the third equation: ( y = k - frac{1}{xz} )From the second equation: ( yz = k - frac{1}{x} )Let me express y from the third equation: ( y = k - frac{1}{xz} ). Substitute this into the second equation:( (k - frac{1}{xz}) z = k - frac{1}{x} )Simplify left side: ( k z - frac{z}{xz} = k z - frac{1}{x} )Thus, equation becomes:( k z - frac{1}{x} = k - frac{1}{x} )Cancel ( - frac{1}{x} ) from both sides:( k z = k implies z =1 )So, again, z=1. Then from first equation:( frac{x}{y} +1 =k implies frac{x}{y} =k -1 implies x= y(k -1) )From third equation:( y =k - frac{1}{x} implies y =k - frac{1}{y(k -1)} )Multiply both sides by y(k -1):( y^2(k -1) =k y(k -1) -1 )Simplify:Left side: ( y^2(k -1) )Right side: (k y(k -1) -1 )Bring all terms to left:( y^2(k -1) -k y(k -1) +1 =0 )Factor out (k -1):( (k -1)(y^2 -k y) +1 =0 )Or, expand:( (k -1)y^2 -k(k -1)y +1 =0 )This is a quadratic equation in y:( (k -1)y^2 -k(k -1)y +1 =0 )Let’s compute discriminant D:D = [ -k(k -1) ]² -4*(k -1)*1 = k²(k -1)^2 -4(k -1)Factor (k -1):D = (k -1)[k²(k -1) -4]Set discriminant ≥0 for real solutions:(k -1)[k²(k -1) -4] ≥0Given that k <2 and variables are positive, let's analyze.If k >1, then (k -1) >0, so we need k²(k -1) -4 ≥0.But for k <2, let's check k=1.5:k=1.5: 1.5²*(0.5) -4 = 2.25*0.5 -4 =1.125 -4= -2.875 <0. So, discriminant negative.Similarly, at k approaching 1 from above: k=1+ε:k²(k -1) ≈ (1 +2ε +ε²)(ε) ≈ε +2ε² +ε³. So, approximately ε. Then, ε -4 ≈ negative. So, still negative.If k <1, then (k -1) <0, so need k²(k -1) -4 ≤0. Since k <1, k -1 is negative, so k²(k -1) is negative. So, negative -4 is even more negative. Thus, D positive as (k -1)[negative] = positive.Wait, D = (k -1)[k²(k -1) -4]. If k <1, then (k -1) <0, and k²(k -1) -4 is negative (since k²(k -1) is negative and -4 is negative, sum is more negative). Therefore, D = (negative)*(negative) = positive.So, when k <1, discriminant is positive, so real solutions exist. Let’s take k <1.But then, we need to check if the quadratic in y has positive roots, as y>0.The quadratic equation: (k -1)y² -k(k -1)y +1 =0.Multiply both sides by -1 to make coefficients clearer:(1 -k)y² +k(1 -k)y -1 =0Let’s denote m =1 -k >0 (since k <1). Then:m y² +k m y -1 =0Which is:m y² +k m y -1 =0Divide by m (m≠0):y² +k y - 1/m =0Solutions:y = [ -k ± sqrt(k² +4/m) ]/2Since y>0, take the positive root:y = [ -k + sqrt(k² +4/m) ]/2Given that m =1 -k, so 1/m =1/(1 -k). Therefore:y = [ -k + sqrt(k² +4/(1 -k)) ]/2We need y >0.Compute sqrt(k² +4/(1 -k)) >k, which is true because 4/(1 -k) >0. So sqrt(k² + positive) > |k|. Since k <1, and we are considering positive roots, so y is positive.Therefore, there exists a solution for y when k <1.But then, we need to check if with this y, x is positive.From earlier, x = y(k -1). But k <1, so (k -1) <0, and y>0, so x = negative. Which contradicts x>0.Therefore, no solution with k <1.Thus, the only possible solution is when k=2, but k=2 is achieved when x=y=z=1.Wait, but earlier when solving, we set A=B=C=k and found that when z=1, then equations lead to x=y=1. But if we have k=2, then:Wait, if x=y=z=1, then A=1/1 +1=2, B=1*1 +1/1=2, C=1/(1*1) +1=2. So, yes, k=2.But we were trying to see if k can be less than 2. However, the above analysis shows that when k<2, either there are no positive solutions for variables or inconsistencies. Therefore, the minimal possible value of max(A,B,C) is 2, achieved when x=y=z=1. Hence, the minimal M is lg2.Therefore, the answer should be boxed{lg 2}Final AnswerThe minimum value of ( M ) is boxed{lg 2}.

✅:To find the minimum value of ( M ), which is the maximum of ( a ), ( b ), and ( c ), we start by simplifying the given expressions:- ( a = lg z + lg left( frac{x}{yz} + 1 right) = lg left( frac{x}{y} + z right) )- ( b = lg x^{-1} + lg (xyz + 1) = lg left( yz + frac{1}{x} right) )- ( c = lg y + lg left( frac{1}{xyz} + 1 right) = lg left( frac{1}{xz} + y right) )We need to minimize the maximum of these three logarithmic expressions. By setting ( a = b = c ), we derive the following equations:1. ( frac{x}{y} + z = yz + frac{1}{x} )2. ( yz + frac{1}{x} = frac{1}{xz} + y )From equation 2, we find that ( z = 1 ). Substituting ( z = 1 ) into equation 1, we get:[ frac{x}{y} + 1 = y + frac{1}{x} ]By assuming ( x = y ), we simplify this to:[ 2 = x + frac{1}{x} ]Solving this, we find ( x = 1 ). Thus, ( x = y = z = 1 ).Substituting ( x = y = z = 1 ) into the expressions for ( a ), ( b ), and ( c ):- ( a = lg left( frac{1}{1} + 1 right) = lg 2 )- ( b = lg left( 1 cdot 1 + frac{1}{1} right) = lg 2 )- ( c = lg left( frac{1}{1 cdot 1} + 1 right) = lg 2 )Thus, ( a = b = c = lg 2 ), and the maximum value ( M ) is ( lg 2 ).Therefore, the minimum value of ( M ) is (boxed{lg 2}).

🔑:Given the expressions:[a = lg z + lg left[ frac{x}{y z} + 1 right],][b = lg frac{1}{x} + lg (x y z + 1),][c = lg y + lg left[ frac{1}{x y z} + 1 right],]we will simplify these logarithmic expressions step-by-step. 1. Simplify expression for ( a ):[a = lg z + lg left[ frac{x}{y z} + 1 right]]Applying the logarithmic identity (log A + log B = log(AB)):[a = lg left( z left( frac{x}{y z} + 1 right) right)]Simplifying inside the logarithm:[z left( frac{x}{y z} + 1 right) = frac{z x}{y z} + z = frac{x}{y} + z]Therefore:[a = lg left( frac{x}{y} + z right)]2. Simplify expression for ( b ):[b = lg frac{1}{x} + lg (x y z + 1)]Applying the logarithmic identity (log A + log B = log(AB)):[b = lg left( frac{1}{x} (x y z + 1) right)]Simplifying inside the logarithm:[frac{1}{x}(x y z + 1) = y z + frac{1}{x}]Therefore:[b = lg left( y z + frac{1}{x} right)]3. Simplify expression for ( c ):[c = lg y + lg left[ frac{1}{x y z} + 1 right]]Applying the logarithmic identity (log A + log B = log(AB)):[c = lg left( y left( frac{1}{x y z} + 1 right) right)]Simplifying inside the logarithm:[y left( frac{1}{x y z} + 1 right) = frac{y}{x y z} + y = frac{1}{x z} + y]Therefore:[c = lg left( frac{1}{x z} + y right)]4. Determining the maximum ( M ) and finding its minimum value:Let's note that:[M = max(a, b, c) = max left( lg left(frac{x}{y} + zright), lg left(yz + frac{1}{x}right), lg left(frac{1}{xz} + y right) right)]To find the minimum value of ( M ), consider:[M_1 = max left( frac{x}{y} + z, yz + frac{1}{x}, frac{1}{xz} + y right)]By the properties of maxima and the AM-GM (Arithmetic Mean-Geometric Mean) inequality:[2M_1 geq frac{x}{y} + z + yz + frac{1}{x} + frac{1}{xz} + y geq 4 sqrt[4]{left(frac{x}{y}right) (rz) left( frac{1}{xz} right) y}]Simplifying the geometric mean term:[sqrt[4]{frac{x}{y} cdot z cdot frac{1}{x z} cdot y} = sqrt[4]{1} = 1]Thus:[2M_1 geq 4 Rightarrow M_1 geq 2]The equality holds when:[frac{x}{y} = z = frac{1}{x z} = y = 1]Therefore, ( x = y = z = 1 ), and substituting into ( yz + frac{1}{x} ):[yz + frac{1}{x} = 1 cdot 1 + 1 = 2]Thus:[M = lg 2]Conclusively, the minimum value of ( M ) is:[boxed{lg 2}]

🧠:Okay, so I need to solve the equation 8 over the fractional part of x equals 9 over x plus 10 over the floor of x. Hmm, let me start by recalling what floor function and fractional part mean. The floor of x, denoted as [x], is the greatest integer less than or equal to x. The fractional part, {x}, is x minus [x]. So, if x is, say, 3.5, then [x] is 3 and {x} is 0.5. Got that.So, the equation given is 8/{x} = 9/x + 10/[x]. Let me write that down again to make sure I have it right:8 / {x} = 9 / x + 10 / [x]Since {x} = x - [x], maybe I can substitute that into the equation. Let me try that. Let’s let n = [x], the floor of x, and f = {x} = x - n. Then, x can be expressed as n + f, where n is an integer and 0 ≤ f < 1. Also, since [x] is in the denominator, [x] can't be zero. So, x must be at least 1, right? Because if [x] were 0, then x would be between 0 and 1, and 10/[x] would be undefined. So, n ≥ 1.So substituting n and f into the equation:8 / f = 9 / (n + f) + 10 / nNow, the variables here are n (an integer ≥1) and f (a fractional part between 0 and 1, not including 1). So, maybe I can express this equation in terms of n and f and solve for possible values of n and f.Let me rewrite the equation:8/f = 9/(n + f) + 10/nLet me combine the terms on the right side. To do that, I can find a common denominator. The common denominator for n and (n + f) is n(n + f). So:9/(n + f) + 10/n = (9n + 10(n + f)) / [n(n + f)]Compute numerator: 9n + 10n + 10f = 19n + 10fSo, the right-hand side is (19n + 10f) / [n(n + f)]Therefore, the equation becomes:8/f = (19n + 10f) / [n(n + f)]Cross-multiplying both sides:8 * n(n + f) = f(19n + 10f)Let me expand both sides:Left side: 8n(n + f) = 8n² + 8nfRight side: f(19n + 10f) = 19nf + 10f²So, bringing all terms to one side:8n² + 8nf - 19nf - 10f² = 0Simplify:8n² - 11nf - 10f² = 0Hmm, this is a quadratic equation in terms of n or f. Let me see. Let's arrange terms:8n² -11nf -10f² =0Alternatively, maybe factor this quadratic. Let me try to factor it. Let's treat this as a quadratic in n. So:8n² -11f n -10f² =0Using quadratic formula for n:n = [11f ± sqrt( (11f)^2 + 4*8*10f² ) ]/(2*8)Compute discriminant:(121f²) + 320f² = 441f²So sqrt(441f²) = 21fTherefore:n = [11f ±21f]/16So two possibilities:n = (11f +21f)/16 =32f/16=2fOrn=(11f -21f)/16= (-10f)/16= -5f/8But n must be a positive integer (since [x] =n ≥1) and f is between 0 and1. So, let's check both possibilities.First, n=2f. But n is an integer ≥1, and f ∈ [0,1). So 2f must be an integer. Since 2f is in [0,2), possible values of 2f are 0 or 1. But n ≥1, so 2f=1, which implies f=0.5. Then n=1. So, possible solution n=1, f=0.5. Then x= n + f =1.5.Second possibility: n= -5f/8. But n must be positive integer, and f ∈ [0,1). Then -5f/8 must be positive. But since f is non-negative, this would require -5f/8 ≥0, which implies f ≤0. But f ≥0, so f=0. However, f=0 would make {x}=0, but in the original equation, 8/{x} would be undefined. So this case is impossible. Therefore, the only possible solution is n=1, f=0.5, x=1.5.But wait, let me verify this solution. Let me substitute x=1.5 into the original equation.Left side: 8/{x} =8/0.5=16Right side:9/x +10/[x]=9/1.5 +10/1=6 +10=16Yes, that works. So x=1.5 is a solution. But is that the only solution?Wait, maybe there are other solutions for different n? Let me check. When I set n=2f, n must be an integer. So 2f must be integer. So possible n=1, f=0.5. If n=2, then 2f=2 ⇒ f=1, but f must be less than 1, so invalid. n=0 would imply f=0, but n must be ≥1. So only n=1, f=0.5.But maybe there's another approach here. Let me check for n=2. Let's suppose n=2. Then x is in [2,3). So f =x -2, where 0≤f<1. Let me plug n=2 into the equation:8/f =9/(2 + f) +10/2=9/(2 + f)+5So, 8/f =9/(2 + f) +5Multiply both sides by f(2 + f):8(2 + f) =9f +5f(2 + f)Left side:16 +8fRight side:9f +10f +5f²=19f +5f²Bring all terms to left:16 +8f -19f -5f²=0 ⇒16 -11f -5f²=0 ⇒5f² +11f -16=0Quadratic equation:5f² +11f -16=0Solutions:f = [-11 ± sqrt(121 + 320)]/(2*5) = [-11 ± sqrt(441)]/10 = [-11 ±21]/10So, two solutions:f=(10)/10=1 or f=(-32)/10=-3.2But f must be in [0,1). So f=1 is invalid, and f=-3.2 is invalid. So no solutions for n=2.Similarly, check n=3. Let's try n=3. Then x ∈[3,4), f=x-3, 0≤f<1.Equation:8/f =9/(3 + f) +10/3Compute right side:9/(3 + f) +10/3Multiply both sides by f(3 + f)*3 to eliminate denominators:8*3*(3 + f) =9*3f +10f(3 + f)Left side:24*(3 + f)=72 +24fRight side:27f +10f(3 + f)=27f +30f +10f²=57f +10f²Bring all terms to left:72 +24f -57f -10f²=0 ⇒72 -33f -10f²=0 ⇒10f² +33f -72=0Quadratic equation:10f² +33f -72=0Solutions:f = [-33 ± sqrt(1089 + 2880)]/20 = [-33 ± sqrt(3969)]/20 = [-33 ±63]/20So, f=(30)/20=1.5 or f=(-96)/20=-4.8But f must be in [0,1). Both solutions invalid. So no solutions for n=3.Next, n=4. Let's try n=4. x ∈[4,5), f=x-4.Equation:8/f =9/(4 + f) +10/4=9/(4 + f)+2.5Multiply both sides by f(4 + f):8*(4 + f) =9f +2.5f(4 + f)Left side:32 +8fRight side:9f +10f +2.5f²=19f +2.5f²Bring all terms to left:32 +8f -19f -2.5f²=0 ⇒32 -11f -2.5f²=0Multiply by 2 to eliminate decimal:64 -22f -5f²=0 ⇒5f² +22f -64=0Solutions:f = [-22 ± sqrt(484 +1280)]/10 = [-22 ± sqrt(1764)]/10 = [-22 ±42]/10So, f=(20)/10=2 or f=(-64)/10=-6.4Both invalid, as f must be in [0,1). So no solutions for n=4.Hmm, seems like for n≥2, there's no solution. Let me check n=1 again. For n=1, x ∈[1,2), f=x-1, 0≤f<1.Equation:8/f =9/(1 + f) +10/1=9/(1 + f) +10So, 8/f =9/(1 + f) +10Let me solve this equation for f ∈[0,1). Let's denote f as variable here.Multiply both sides by f(1 + f):8(1 + f) =9f +10f(1 + f)Left side:8 +8fRight side:9f +10f +10f²=19f +10f²Bring all terms to left:8 +8f -19f -10f²=0 ⇒8 -11f -10f²=0 ⇒10f² +11f -8=0Quadratic equation:10f² +11f -8=0Solutions:f = [-11 ± sqrt(121 +320)]/20 = [-11 ± sqrt(441)]/20 = [-11 ±21]/20So, f=(10)/20=0.5 or f=(-32)/20=-1.6Only f=0.5 is valid. So f=0.5, which gives x=1 +0.5=1.5. Which is the solution we found earlier.So, n=1 gives x=1.5. Now, check for n=0. Wait, n must be ≥1 because [x] can't be zero (since 10/[x] would be undefined). If n=0, then [x]=0, so x ∈[0,1), but then 10/[x] is undefined. So n cannot be zero. So the only possible n is 1, leading to x=1.5.Wait, but just to be thorough, let me check n=1 again. Let me suppose n=1 and see if there are any other possible solutions. But when we solved the quadratic equation for f, we only got f=0.5 as valid. So x=1.5 is the only solution.But let me check if there's another approach. Let me consider x=1.5. Then [x]=1, {x}=0.5. Plugging into original equation:Left side:8 /0.5=16Right side:9/1.5 +10/1=6 +10=16. Correct.Is there another x that could satisfy this? Let's see. Let me suppose another possible n, maybe n=1 but f≠0.5. Wait, but we saw that for n=1, the only solution is f=0.5. So no other f in [0,1) works. For n≥2, we saw no solutions. So x=1.5 is the only solution.Alternatively, maybe if n is negative? But [x] is the greatest integer less than or equal to x. If x is negative, say x=-0.5, then [x]=-1. But 10/[x] would be 10/(-1)= -10, and {x}=x - [x]= -0.5 - (-1)=0.5. Then left side is 8/0.5=16. Right side is 9/(-0.5) +10/(-1)= -18 -10= -28. Not equal. But perhaps another negative x?Wait, let's consider if x is negative. Let me suppose x is negative. Let n = [x]. For example, if x is -1.2, then [x]=-2, and {x}=x - [x]= -1.2 - (-2)=0.8. Wait, fractional part is always non-negative, between 0 and1.But in the equation, 10/[x] would be problematic if [x] is negative, but it's allowed as long as [x]≠0. So let's try n=-1. So [x]=-1, so x ∈[-1,0). Then {x}=x - (-1)=x +1. So {x} ∈[0,1). Let me plug into the equation.Equation:8/{x}=9/x +10/[x]But [x]=-1, so 10/[x]=10/(-1)=-10{x}=x +1, so {x}=x - (-1)=x +1. Let’s write equation as:8/(x +1) =9/x + (-10)So, 8/(x +1) =9/x -10Multiply both sides by x(x +1):8x =9(x +1) -10x(x +1)Expand right side:9x +9 -10x² -10xCombine terms:(9x -10x) +9 -10x²= -x +9 -10x²So equation becomes:8x = -10x² -x +9Bring all terms to left:10x² +9x -9=0Quadratic equation:10x² +9x -9=0Solutions:x = [-9 ± sqrt(81 + 360)]/20 = [-9 ± sqrt(441)]/20 = [-9 ±21]/20So, x=(12)/20=0.6 or x=(-30)/20=-1.5But x must be in [-1,0). So x=-1.5 is not in [-1,0), and x=0.6 is not in [-1,0). So no solutions here.Next, check n=-2. [x]=-2, so x ∈[-2,-1). Then {x}=x - (-2)=x +2. So {x} ∈[0,1).Equation:8/{x}=9/x +10/[x]=9/x +10/(-2)=9/x -5So, 8/(x +2)=9/x -5Multiply both sides by x(x +2):8x =9(x +2) -5x(x +2)Expand right side:9x +18 -5x² -10xCombine terms:(9x -10x) +18 -5x²= -x +18 -5x²Bring all terms to left:5x² +x -18=0Quadratic equation:5x² +x -18=0Solutions:x = [-1 ± sqrt(1 +360)]/10 = [-1 ± sqrt(361)]/10 = [-1 ±19]/10So, x=(18)/10=1.8 or x=(-20)/10=-2But x must be in [-2,-1). So x=-2 is at the boundary, not included. x=1.8 not in the interval. So no solutions here.Similarly, n=-3 would give x ∈[-3,-2), but same approach would likely lead to no solution, but just to check.n=-3, {x}=x +3, equation:8/(x +3)=9/x +10/(-3)=9/x -10/3Multiply both sides by 3x(x +3):8*3x =9*3(x +3) -10x(x +3)Left side:24xRight side:27x +81 -10x² -30xCombine terms:(27x -30x) +81 -10x²= -3x +81 -10x²Bring all terms to left:10x² +27x -81=0Quadratic equation:10x² +27x -81=0Solutions:x = [-27 ± sqrt(729 +3240)]/20 = [-27 ± sqrt(3969)]/20 = [-27 ±63]/20So, x=(36)/20=1.8 or x=(-90)/20=-4.5But x must be in [-3,-2). So x=-4.5 is not in the interval. No solution.Thus, for negative n, there are no solutions. Therefore, the only solution is x=1.5.But let me confirm once again that there are no other solutions for n=1. Suppose n=1, then x ∈[1,2). The equation leads to f=0.5, which gives x=1.5. So that's the only solution.Therefore, the final answer is x=3/2 or 1.5.Final AnswerThe solution is boxed{dfrac{3}{2}}.

✅:Given the equation:frac{8}{{x}} = frac{9}{x} + frac{10}{[x]}where ([x]) is the greatest integer less than or equal to (x) (the floor function), and ({x} = x - [x]) (the fractional part of (x)).1. Define Variables: - Let (n = [x]) (an integer) and (f = {x}) (such that (0 leq f < 1)). - Therefore, (x = n + f) where (n geq 1) (since ([x] neq 0) to avoid division by zero).2. Substitute into the Equation: - Substitute (x = n + f) into the equation: frac{8}{f} = frac{9}{n + f} + frac{10}{n} 3. Combine Terms: - Combine the right-hand side over a common denominator: frac{9}{n + f} + frac{10}{n} = frac{19n + 10f}{n(n + f)} 4. Cross-Multiply and Simplify: - Cross-multiplying gives: 8n(n + f) = f(19n + 10f) - Expand and simplify: 8n^2 + 8nf = 19nf + 10f^2 implies 8n^2 - 11nf - 10f^2 = 0 5. Solve the Quadratic Equation: - Treating this as a quadratic in (n): n = frac{11f pm 21f}{16} - This gives two solutions: - (n = 2f) - (n = -frac{5f}{8}) (discarded since (n geq 1) and (f geq 0))6. Check Valid Solutions: - For (n = 2f), (n) must be an integer. The valid solution is (n = 1) and (f = 0.5) (since (2f = 1 implies f = 0.5)). - Therefore, (x = n + f = 1 + 0.5 = 1.5).7. Verification: - Substitute (x = 1.5) into the original equation: - Left-hand side: (frac{8}{{1.5}} = frac{8}{0.5} = 16) - Right-hand side: (frac{9}{1.5} + frac{10}{1} = 6 + 10 = 16)8. Check Other Possible Values: - For (n geq 2), solving the resulting quadratic equations shows no valid solutions for (f) in the interval ([0, 1)). - Negative values of (n) also do not yield valid solutions.Thus, the only solution is:boxed{dfrac{3}{2}}

🔑:We need to solve the equation:[frac{8}{{x}} = frac{9}{x} + frac{10}{[x]}]where ([x]) is the greatest integer less than or equal to (x) and ({x} = x - [x]).First, we note a few key properties and constraints based on the given information:1. Properties of (x): - (x) must be positive. - (x) cannot be negative, as the right side of the equation would be negative while the left side is positive. - (x) cannot be an integer, because in such a case, ({x} = 0), which would make the left-hand side undefined.Next, we consider various possible values of ([x]):# 1. Case: ([x] = 1)If ([x] = 1), we have:[{x} = x - 1]Substitute ([x] = 1) into the given equation:[frac{8}{x - 1} = frac{9}{x} + frac{10}{1}]Simplify the equation:[frac{8}{x-1} = frac{9}{x} + 10]Multiplying every term by (x(x-1)) to clear the fractions:[8x = 9(x-1) + 10x(x-1)]Expanding and simplifying:[8x = 9x - 9 + 10x^2 - 10x][8x = 10x^2 - x - 9]Rearranging terms:[10x^2 - 9x - 9 = 0]This is a quadratic equation in (x):[10x^2 - 9x - 9 = 0]Solving this quadratic using the quadratic formula (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}):[a = 10, quad b = -9, quad c = -9][x = frac{-(-9) pm sqrt{(-9)^2 - 4 cdot 10 cdot (-9)}}{2 cdot 10}][x = frac{9 pm sqrt{81 + 360}}{20}][x = frac{9 pm sqrt{441}}{20}][x = frac{9 pm 21}{20}]Thus,[x = frac{30}{20} = 1.5 quad text{or} quad x = frac{-12}{20} = -0.6]Since (x) must be positive, we reject (x = -0.6):[x = 1.5]# 2. Case: ([x] = 2)If ([x] = 2), we have:[{x} = x - 2]Substitute ([x] = 2) into the given equation:[frac{8}{x-2} = frac{9}{x} + frac{10}{2}]Simplify the equation:[frac{8}{x-2} = frac{9}{x} + 5]Multiplying every term by (x(x-2)) to clear the fractions:[8x = 9(x-2) + 5x(x-2)]Expanding and simplifying:[8x = 9x - 18 + 5x^2 - 10x][8x = 5x^2 - x - 18]Rearranging terms:[5x^2 - 9x - 18 = 0]This quadratic equation can be solved using the quadratic formula (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}):[a = 5, quad b = -9, quad c = -18][x = frac{-(-9) pm sqrt{(-9)^2 - 4 cdot 5 cdot (-18)}}{2 cdot 5}][x = frac{9 pm sqrt{81 + 360}}{10}][x = frac{9 pm sqrt{441}}{10}][x = frac{9 pm 21}{10}]Thus,[x = frac{30}{10} = 3 quad text{or} quad x = frac{-12}{10} = -1.2]Since (x) must be positive, we reject (x = -1.2) and (x = 3) is also invalid as (x) cannot be an integer (([x] ≠ 3)).# 3. Case: ([x] geq 3)For values ([x] geq 3):Note that the expression:[frac{9}{x} + frac{10}{[x]}]will always evaluate to less than (2 + frac{10}{3}) since (x geq 3):But:[frac{9}{x} + frac{10}{[x]} < 3 + frac{10}{3} < 8]This cannot satisfy the original equation since (frac{8}{ {x}} geq 8). Conclusion:The only solution that fits all constraints is:[boxed{frac{3}{2}}]

🧠:Okay, so I need to show that if a, b, c are the sides of a triangle, then the product of the circumradius R and the inradius r is equal to (a*b*c)/(2*(a+b+c)). Hmm, let me start by recalling some formulas related to triangles, circumradius, and inradius.First, I remember that the area of a triangle can be expressed in terms of the inradius and the semiperimeter. The formula is Area = r * s, where s is the semiperimeter, right? So s = (a + b + c)/2. Therefore, Area = r * (a + b + c)/2.Another formula for the area is using the circumradius: Area = (a * b * c)/(4R). I think that's correct. Let me verify. Yes, the formula for the area in terms of the circumradius is (a*b*c)/(4R). So we have two expressions for the area: one in terms of r and one in terms of R.So if I set these equal to each other, since they both equal the area, then:r * (a + b + c)/2 = (a * b * c)/(4R)If I solve for R * r, maybe I can manipulate this equation. Let's try.Multiply both sides by 4R:4R * r * (a + b + c)/2 = a * b * cSimplify the left side: 4R * r * (a + b + c)/2 = 2R * r * (a + b + c)So:2R * r * (a + b + c) = a * b * cThen divide both sides by 2(a + b + c):R * r = (a * b * c)/(2(a + b + c))Which is exactly the equation we needed to prove. Wait, so that's it? Is it that straightforward?Wait, let me check again. Maybe I made a mistake in the formulas. Let me confirm the formulas for the area.Yes, the area in terms of inradius is r times the semiperimeter, so Area = r * s, and s = (a + b + c)/2. So Area = r * (a + b + c)/2. Correct.And the area in terms of the circumradius is (a * b * c)/(4R). Let me check the derivation for that. For any triangle, the area can be written as (1/2)*a*b*sin(C), and using the Law of Sines, sin(C) = c/(2R). So substituting, Area = (1/2)*a*b*(c/(2R)) = (a*b*c)/(4R). Yes, that's correct.Therefore, equating the two expressions:r * (a + b + c)/2 = (a * b * c)/(4R)Multiply both sides by 4R:4R * r * (a + b + c)/2 = a * b * cSimplify 4/2 to 2:2R * r * (a + b + c) = a * b * cThen divide both sides by 2(a + b + c):R * r = (a * b * c)/(2(a + b + c))Exactly. So the result follows directly from equating two different expressions for the area of the triangle. Hmm, that seems almost too simple. Did I miss something?Alternatively, maybe there's another way to derive this, but perhaps this is sufficient. Let me think if there are other formulas connecting R and r.Another formula I remember is the formula involving the distances between the inradius and circumradius, but I don't think that's necessary here. Also, there's Euler's formula relating R and r: the distance between the incenter and circumradius is sqrt(R(R - 2r)). But that might complicate things.Alternatively, maybe using trigonometric identities. Let's consider angles. Let me denote angles A, B, C opposite sides a, b, c respectively.We know that a = 2R sin A, similarly b = 2R sin B, c = 2R sin C. So substituting these into the right-hand side:(a*b*c)/(2(a + b + c)) = (8R^3 sin A sin B sin C)/(2*(2R(sin A + sin B + sin C))) )Wait, let's compute that:Numerator: 8R^3 sin A sin B sin CDenominator: 2*(2R(sin A + sin B + sin C)) = 4R(sin A + sin B + sin C)So overall:(8R^3 sin A sin B sin C)/(4R(sin A + sin B + sin C)) = (2R^2 sin A sin B sin C)/(sin A + sin B + sin C)But we need to show that this equals R*r. So R*r = (2R^2 sin A sin B sin C)/(sin A + sin B + sin C)Divide both sides by R:r = (2R sin A sin B sin C)/(sin A + sin B + sin C)Is this a known identity? Let me recall the formula for the inradius. The inradius r can be expressed as 4R sin(A/2) sin(B/2) sin(C/2). Is that correct?Yes, I think there is a formula: r = 4R sin(A/2) sin(B/2) sin(C/2). Let me verify this.We know that r = (Area)/s, where s is the semiperimeter. Area is also (1/2)ab sin C. Alternatively, using some trigonometric identities:In terms of R and angles, since a = 2R sin A, etc., we can express the semiperimeter s = (a + b + c)/2 = R(sin A + sin B + sin C).Then, the inradius r = Area / s = [ (a b c)/(4R) ] / [ R(sin A + sin B + sin C)/2 ] = ( (8R^3 sin A sin B sin C )/(4R) ) / ( R (sin A + sin B + sin C)/2 )Wait, let's compute that step by step. Area is (a b c)/(4R) = (8R^3 sin A sin B sin C)/(4R) = 2R^2 sin A sin B sin C.Then s = (a + b + c)/2 = (2R sin A + 2R sin B + 2R sin C)/2 = R(sin A + sin B + sin C).Therefore, r = Area / s = (2R^2 sin A sin B sin C) / (R(sin A + sin B + sin C)) ) = 2R sin A sin B sin C / (sin A + sin B + sin C)Which matches the expression we had earlier. So according to this, r = [2R sin A sin B sin C]/[sin A + sin B + sin C]. Therefore, if we multiply R * r, we get:R * r = R * [2R sin A sin B sin C]/[sin A + sin B + sin C] = [2R^2 sin A sin B sin C]/[sin A + sin B + sin C]But we also have from the previous approach:(a b c)/(2(a + b + c)) = [8R^3 sin A sin B sin C]/[2*(2R(sin A + sin B + sin C))] = [8R^3 sin A sin B sin C]/[4R(sin A + sin B + sin C)] = [2R^2 sin A sin B sin C]/[sin A + sin B + sin C]Therefore, both expressions for R * r and (a b c)/(2(a + b + c)) are equal. Hence, R * r = (a b c)/(2(a + b + c)).So, this confirms the result again through trigonometric identities, but the initial method was much simpler by just equating two expressions for the area.Is there another way to see this? Maybe using Heron's formula? Let's try.Heron's formula states that the area is sqrt[s(s - a)(s - b)(s - c)], where s is the semiperimeter. But since we already have expressions in terms of R and r, perhaps this isn't necessary.Alternatively, using the formula for r in terms of the area and semiperimeter, and the formula for R in terms of the sides and the area, then multiplying them.But as we saw, Area = r * s and Area = (a b c)/(4R). Therefore, r * s = (a b c)/(4R), so R * r = (a b c)/(4s). Since s = (a + b + c)/2, then 4s = 2(a + b + c). Therefore, R * r = (a b c)/(2(a + b + c)), which is exactly what we needed.Yes, so this is the most straightforward derivation. Therefore, the key idea is recognizing that both R and r relate to the area of the triangle, and by expressing the area in terms of each, we can equate them and solve for R * r.I think that's solid. The steps are clear and each formula used is a standard result in triangle geometry. So unless there's a hidden assumption or a mistake in algebraic manipulation, which I don't see, this should be correct.Let me recap the steps concisely:1. The area of the triangle can be written as r * s, where r is the inradius and s is the semiperimeter.2. The same area can be expressed as (a * b * c)/(4R), where R is the circumradius.3. Equate the two expressions: r * s = (a * b * c)/(4R).4. Substitute s = (a + b + c)/2 into the equation.5. Solve for R * r to obtain the desired identity.Therefore, the proof is straightforward once you recall the formulas connecting the inradius, circumradius, and area of a triangle. No need for more complicated derivations unless required by the problem's context, but in this case, the basic approach suffices.Another thing to check is whether the formula holds for specific triangles. Let's test it with an equilateral triangle where a = b = c = 2 units.For an equilateral triangle, the inradius r = (a * sqrt(3))/6 = (2 * sqrt(3))/6 = sqrt(3)/3.The circumradius R = (a * sqrt(3))/3 = (2 * sqrt(3))/3.Then R * r = (2 * sqrt(3)/3) * (sqrt(3)/3) = (2 * 3)/9 = 6/9 = 2/3.Now compute (a*b*c)/(2(a + b + c)) = (2*2*2)/(2*(2+2+2)) = 8/(2*6) = 8/12 = 2/3.They are equal. So the formula holds in this case. Let's try another triangle, say a right-angled triangle with sides 3, 4, 5.For a 3-4-5 triangle, the semiperimeter s = (3 + 4 + 5)/2 = 6. The area is (3*4)/2 = 6. Therefore, inradius r = Area / s = 6 / 6 = 1.Circumradius R = (a*b*c)/(4*Area) = (3*4*5)/(4*6) = 60 / 24 = 2.5.So R * r = 2.5 * 1 = 2.5.Compute (a*b*c)/(2(a + b + c)) = (3*4*5)/(2*(3+4+5)) = 60/(2*12) = 60/24 = 2.5. Perfect, they match again.Thus, the formula works in these test cases, which adds confidence to its correctness.In conclusion, the proof relies on the fundamental relationships between the radii and the area, and the identity follows directly from equating these different expressions for the area. The verification with specific examples further supports the validity of the formula.

✅:To prove that for a triangle with sides (a), (b), (c), circumradius (R), and inradius (r), the identity (R cdot r = frac{a cdot b cdot c}{2(a + b + c)}) holds, we proceed as follows:Step 1: Express the area using the inradius and semiperimeter. The area (T) of the triangle can be written in terms of the inradius (r) and semiperimeter (s): [T = r cdot s quad text{where} quad s = frac{a + b + c}{2}.]Thus, [T = r cdot frac{a + b + c}{2}.]Step 2: Express the area using the circumradius. The area can also be expressed in terms of the circumradius (R): [T = frac{a cdot b cdot c}{4R}.]Step 3: Equate the two expressions for the area. Setting the two expressions for (T) equal: [r cdot frac{a + b + c}{2} = frac{a cdot b cdot c}{4R}.]Step 4: Solve for (R cdot r). Multiply both sides by (4R): [4R cdot r cdot frac{a + b + c}{2} = a cdot b cdot c.] Simplify to: [2R cdot r cdot (a + b + c) = a cdot b cdot c.] Divide both sides by (2(a + b + c)): [R cdot r = frac{a cdot b cdot c}{2(a + b + c)}.]Verification with Specific Cases: 1. Equilateral Triangle ((a = b = c = 2)): - (r = frac{sqrt{3}}{3}), (R = frac{2sqrt{3}}{3}). - (R cdot r = frac{2}{3}), and (frac{a cdot b cdot c}{2(a + b + c)} = frac{8}{12} = frac{2}{3}). 2. Right-Angled Triangle ((a = 3), (b = 4), (c = 5)): - (r = 1), (R = 2.5). - (R cdot r = 2.5), and (frac{a cdot b cdot c}{2(a + b + c)} = frac{60}{24} = 2.5).Conclusion: By equating two standard formulas for the area of a triangle and simplifying, we establish the desired identity. The result is further corroborated by testing specific triangles.[boxed{R cdot r = frac{a cdot b cdot c}{2(a + b + c)}}]

🔑:To prove the given expression ( R cdot r = frac{a cdot b cdot c}{2(a + b + c)} ), we will use the known formulas for the circumradius ( R ) and the inradius ( r ) of a triangle.1. Circumradius Formula: [ R = frac{a cdot b cdot c}{4 Delta} ] where ( Delta ) is the area of the triangle and ( a, b, c ) are the sides of the triangle.2. Inradius Formula: [ r = frac{Delta}{p} = frac{2Delta}{a + b + c} ] Here, ( p = frac{a + b + c}{2} ) is the semi-perimeter of the triangle.3. Multiplying the Formulas: We multiply the expressions for ( R ) and ( r ): [ R cdot r = left( frac{a cdot b cdot c}{4 Delta} right) left( frac{2 Delta}{a + b + c} right) ]4. Simplifying the Expression: [ R cdot r = frac{a cdot b cdot c}{4 Delta} cdot frac{2 Delta}{a + b + c} ] [ R cdot r = frac{a cdot b cdot c cdot 2 Delta}{4 Delta cdot (a + b + c)} ]5. Canceling Out the (Delta) Terms: Since ( Delta ) appears in both the numerator and denominator, it cancels out: [ R cdot r = frac{a cdot b cdot c cdot 2}{4(a + b + c)} ]6. Further Simplification: [ R cdot r = frac{a cdot b cdot c}{2(a + b + c)} ]Therefore, we have shown that:[R cdot r = frac{a cdot b cdot c}{2(a + b + c)}][boxed{frac{a cdot b cdot c}{2(a+b+c)}}]

🧠:Okay, so I need to find the probability that when I toss a coin 10 times, I never get two heads in a row. Hmm, let's think about this step by step. First, I know that the total number of possible outcomes when tossing a coin 10 times is 2^10, which is 1024. That's straightforward because each toss has 2 possible results, right? So the denominator of the probability will be 1024. The numerator will be the number of valid sequences where no two heads are consecutive. Now, the challenge is figuring out how many such valid sequences there are. Let's denote the number of valid sequences for n tosses as F(n). So, we need to find F(10). I remember that problems like this, where you have to count sequences without consecutive occurrences, often relate to the Fibonacci sequence. Wait, why is that? Let me try to recall. Suppose we have a sequence of n coin tosses without two heads in a row. The last toss could be a tail or a head. If the last toss is a tail, then the preceding n-1 tosses must also form a valid sequence (no two consecutive heads). If the last toss is a head, then the one before that must be a tail (to avoid two heads in a row), and the first n-2 tosses must form a valid sequence. So, this gives a recurrence relation: F(n) = F(n-1) + F(n-2). Oh, right! That's exactly the Fibonacci sequence. So, the number of valid sequences for n tosses is the (n+2)th Fibonacci number or something like that? Wait, let me verify this with small values to make sure.Let's test for n=1. If we have 1 toss, the valid sequences are H and T. So, F(1) = 2. According to the Fibonacci idea, F(1) should be F(0) + F(-1)? Wait, maybe I need to adjust the indices here. Let's see, maybe the base cases are different. Wait, if n=1, F(1)=2. For n=2, the valid sequences are TT, TH, HT. HH is invalid. So, F(2)=3. According to the recurrence relation F(2) = F(1) + F(0). So, if F(2)=3 and F(1)=2, then F(0) must be 1. Hmm, that makes sense. For n=0, which is zero tosses, there's 1 sequence (the empty sequence). So, the base cases are F(0)=1, F(1)=2, and then F(n) = F(n-1) + F(n-2) for n >=2. Therefore, the sequence goes like 1, 2, 3, 5, 8,... which is similar to the Fibonacci sequence shifted by some positions. Let's check n=3. The valid sequences would be: TTT, TTH, THT, HTT, HTH. So, 5 sequences. F(3)=5. According to the formula, F(3) = F(2)+F(1)=3+2=5. Perfect. So, yes, this works. So, F(n) follows the Fibonacci sequence starting with F(0)=1 and F(1)=2. Therefore, F(n) is the (n+2)th Fibonacci number? Let's check: If F(0)=1 (which is Fibonacci number 1), F(1)=2 (Fibonacci number 3?), maybe the indexing is different. Wait, standard Fibonacci sequence is usually defined as Fib(0)=0, Fib(1)=1, Fib(2)=1, Fib(3)=2, Fib(4)=3, Fib(5)=5, etc. But our F(n) starts as F(0)=1, F(1)=2, F(2)=3, F(3)=5. So, comparing to the standard Fibonacci sequence:Fib(2)=1, Fib(3)=2, Fib(4)=3, Fib(5)=5, Fib(6)=8,...So, F(n) = Fib(n+2). For example:F(0) = 1 = Fib(2)F(1) = 2 = Fib(3)F(2) = 3 = Fib(4)F(3) = 5 = Fib(5)Yes, so F(n) is equal to the (n+2)th Fibonacci number in the standard numbering. Therefore, for n=10, F(10) = Fib(12). Let me compute Fib(12). Let's list them:Fib(0)=0Fib(1)=1Fib(2)=1Fib(3)=2Fib(4)=3Fib(5)=5Fib(6)=8Fib(7)=13Fib(8)=21Fib(9)=34Fib(10)=55Fib(11)=89Fib(12)=144So, Fib(12)=144. Therefore, F(10)=144. Therefore, the number of valid sequences is 144. Therefore, the probability is 144/1024. Simplifying that, divide numerator and denominator by 16: 144 ÷16=9, 1024÷16=64. So, 9/64. Wait, 144 divided by 16 is 9? Let me check: 16*9=144. Yes. And 1024 divided by 16 is 64. So, 9/64. Which is approximately 0.140625. But let me verify this result with another approach to make sure I didn't make a mistake. Maybe using combinatorics instead of recursion. Suppose we want to arrange k heads in 10 tosses such that no two are consecutive. How many ways can we do that? Each head must be separated by at least one tail. This is a classic stars and bars problem. If we have k heads, we need to place them in the 10 positions such that no two are adjacent. To do this, we can think of placing the k heads in the 10 - (k -1) possible slots. Wait, explanation: If we have k heads, each separated by at least one tail, we can model this as placing k heads and (k-1) tails to begin with, and then distributing the remaining tails. The total number of required tails is (k-1) to separate the heads, so the remaining tails are 10 - k - (k -1) = 10 - 2k +1 = 11 - 2k. But this is only possible if 11 - 2k >=0, so k <=5.5, so k<=5. Therefore, the maximum number of heads we can have is 5. Therefore, the number of ways to arrange k heads without two consecutive is C(10 -k +1, k). Wait, the formula is C(n -k +1, k). For example, n=10, so C(10 -k +1, k). Yes, because we can imagine placing k heads in the 10 positions with no two adjacent. This is equivalent to placing k objects in n -k +1 slots. So, the number of ways is C(n -k +1, k). Therefore, for each k from 0 to 5, the number of sequences with k heads and no two consecutive is C(10 -k +1, k) = C(11 -k, k). Therefore, the total number of valid sequences is the sum over k=0 to k=5 of C(11 -k, k). Let me compute this:For k=0: C(11,0)=1k=1: C(10,1)=10k=2: C(9,2)=36k=3: C(8,3)=56k=4: C(7,4)=35k=5: C(6,5)=6Adding these up: 1 +10=11, 11+36=47, 47+56=103, 103+35=138, 138+6=144. Yes, that matches the previous result of 144. So, the total number is indeed 144. Therefore, the probability is 144/1024 = 9/64 ≈0.140625. So, that seems consistent. Another way to think about it is through recursion and dynamic programming. Let's model the problem with states. Suppose we define F(n) as the number of valid sequences of length n. For each position i, the last toss can be a head or a tail. If it's a tail, the previous i-1 tosses can be any valid sequence of length i-1. If it's a head, the previous toss must be a tail, and the first i-2 tosses can be any valid sequence of length i-2. Therefore, F(n) = F(n-1) + F(n-2). With F(0)=1 (empty sequence), F(1)=2 (H or T). We can build this up step by step:F(0) = 1F(1) = 2F(2) = F(1) + F(0) = 2 +1 =3F(3) = F(2) + F(1) =3 +2=5F(4)=5+3=8F(5)=8+5=13F(6)=13+8=21F(7)=21+13=34F(8)=34+21=55F(9)=55+34=89F(10)=89+55=144Yep, same result. So, three different approaches: recurrence relation (Fibonacci), combinatorial summation, and dynamic programming. All lead to 144 valid sequences. Therefore, the probability is 144/1024, which simplifies to 9/64. Wait, just to check the simplification again: 144 divided by 16 is 9, 1024 divided by 16 is 64. 9 and 64 have no common divisors, so that's the simplest form. So, the final answer is 9/64. I think that's thorough. I considered multiple methods to confirm the result, checked base cases, and made sure the recurrence relation holds. Everything checks out.

✅:The probability that two heads do not appear consecutively in 10 coin tosses is found by counting valid sequences using a Fibonacci-like recurrence relation. Let ( F(n) ) represent the number of valid sequences for ( n ) tosses. The recurrence is ( F(n) = F(n-1) + F(n-2) ) with base cases ( F(0) = 1 ) and ( F(1) = 2 ). Calculating up to ( F(10) ):[begin{align*}F(0) & = 1, F(1) & = 2, F(2) & = F(1) + F(0) = 3, F(3) & = F(2) + F(1) = 5, F(4) & = F(3) + F(2) = 8, F(5) & = F(4) + F(3) = 13, F(6) & = F(5) + F(4) = 21, F(7) & = F(6) + F(5) = 34, F(8) & = F(7) + F(6) = 55, F(9) & = F(8) + F(7) = 89, F(10) & = F(9) + F(8) = 144.end{align*}]The total possible outcomes are ( 2^{10} = 1024 ). Thus, the probability is:[frac{144}{1024} = frac{9}{64}.]Answer: (boxed{dfrac{9}{64}})

🔑:To solve the problem of finding the probability that no two heads (H) appear consecutively in 10 coin tosses, we break down the solution into systematic steps as follows:1. Calculate Total Outcomes: The total number of possible outcomes when a coin is tossed 10 times is given by: [ 2^{10} = 1024 ]2. Identify Non-Consecutive Heads Configurations: Let's denote (f(n)) as the number of ways to toss (n) coins such that no two heads appear consecutively. To find (f(n)) for ( n = 10 ), we can use the Fibonacci-like recurrence relation due to the nature of the problem: [ f(n) = f(n-1) + f(n-2) ] The reasoning behind this relation is: - If the last coin is a tail (T), there are (f(n-1)) valid configurations. - If the last coin is a head (H), it must be preceded by a tail (T), giving us (f(n-2)) valid configurations. The base cases are: [ f(1) = 2 quad text{(H or T)} f(2) = 3 quad text{(TT, HT, TH)} ]3. Compute the Fibonacci Sequence Values Before (n = 10): We calculate sequentially as per the recurrence relation: [ begin{aligned} f(3) &= f(2) + f(1) = 3 + 2 = 5 f(4) &= f(3) + f(2) = 5 + 3 = 8 f(5) &= f(4) + f(3) = 8 + 5 = 13 f(6) &= f(5) + f(4) = 13 + 8 = 21 f(7) &= f(6) + f(5) = 21 + 13 = 34 f(8) &= f(7) + f(6) = 34 + 21 = 55 f(9) &= f(8) + f(7) = 55 + 34 = 89 f(10) &= f(9) + f(8) = 89 + 55 = 144 end{aligned} ] Therefore, ( f(10) = 144 ).4. Calculate Probability: The probability (P) that no two heads appear consecutively in 10 coin tosses is: [ P = frac{text{number of favorable outcomes}}{text{total number of outcomes}} = frac{f(10)}{2^{10}} = frac{144}{1024} = frac{9}{64} ] Conclusion:Thus, the probability that no two heads appear consecutively in 10 coin tosses is:[boxed{frac{9}{64}}]

🧠:Okay, so I need to prove that the points K, M, and T lie on a straight line. Let me start by visualizing the problem. There are two circles, one with radius 2r and the other with radius r, touching each other internally at point K. Then there's a third circle, also with radius r, that touches the smaller circle at point M and intersects the larger circle again at point T. I need to show that K, M, and T are colinear.First, let me sketch a diagram in my mind. The larger circle has center, let's say, O, and the smaller circle has center, maybe C. Since they touch internally at K, the distance between O and C must be equal to the difference in their radii. The larger radius is 2r, the smaller is r, so the distance between centers O and C should be 2r - r = r. So, OC = r, and they touch at point K. That makes sense.Now, the third circle has radius r as well. Let me denote its center as P. This third circle touches the smaller circle (center C) at point M. Since both the third circle and the smaller circle have radius r, the distance between their centers P and C must be equal to the sum of their radii if they touch externally, but wait, if they touch at M, then the distance between centers should be equal to the sum or the difference of their radii. Wait, but both circles have radius r. If they touch externally, the distance between centers would be 2r. If they touch internally, it would be 0, which doesn't make sense. So, since they touch at M, and both have radius r, they must touch externally. Therefore, the distance between C and P is r + r = 2r. Wait, but the third circle is also intersecting the larger circle (radius 2r, center O) again at point T. So, the third circle (center P, radius r) must intersect the larger circle (center O, radius 2r) at points K and T? Wait, the problem says "intersects the larger circle again at point T," implying that the third circle passes through K as well? Wait, the problem states: "A third circle with radius r touches the smaller circle at point M and intersects the larger circle again at point T." Hmm, so the third circle intersects the larger circle at two points: one is K, and the other is T. But the problem says "again at point T," so maybe the third circle already passes through K? Because K is the point where the original two circles touch. Let me check the problem statement again.Original problem: Two circles with radii 2r and r touch each other internally at point K. A third circle with radius r touches the smaller circle at point M and intersects the larger circle again at point T. Prove that the points K, M, and T lie on a straight line.So, the third circle touches the smaller circle (radius r) at M and intersects the larger circle (radius 2r) again at T. "Again" suggests that the third circle already intersects the larger circle at K, but since the original two circles touch at K, the third circle might pass through K as well. Wait, but the third circle has radius r. The larger circle has radius 2r. For the third circle to intersect the larger circle at K, the distance between centers O and P must be equal to the sum or difference of their radii. But since the third circle is of radius r, and the larger is 2r, the distance between O and P would need to be either 2r + r = 3r or 2r - r = r. But the third circle is touching the smaller circle (radius r, center C) at M, so distance between C and P is 2r (since both radii are r, and they touch externally). Wait, but if the third circle's center P is 2r away from C, and the original larger circle's center O is r away from C, since OC = r (as the two original circles touch internally at K). So, O, C, and P: OC = r, CP = 2r. So, triangle OCP with sides OC = r, CP = 2r, and OP is the distance between O and P. By triangle inequality, OP must be between 2r - r = r and 2r + r = 3r. But if the third circle intersects the larger circle at K and T, then the distance OP must satisfy the condition for intersection. Wait, the third circle (radius r) and the larger circle (radius 2r) intersect at K. So, the distance between their centers O and P must be equal to 2r - r = r, if they are touching internally at K, but if they intersect at K, then the point K is common to both circles, so OK = 2r (since K is on the larger circle) and PK = r (since K is on the third circle). Therefore, the distance between O and P must be OK - PK = 2r - r = r. Wait, but if the third circle is centered at P, then PK = r. So, if K is on both the larger circle (radius 2r, center O) and the third circle (radius r, center P), then the distance OP must be equal to OK - PK = 2r - r = r. Therefore, OP = r. So, O, P, and K are colinear, with P lying on the line OK, at a distance r from O. Wait, but if OP = r, then since OC = r (distance between O and C), and CP = 2r (distance between C and P), then points O, C, and P must form a triangle with sides OC = r, CP = 2r, OP = r. Wait, that's impossible because in a triangle, the sum of any two sides must be greater than the third. Here, OC + OP = r + r = 2r, which is equal to CP = 2r. So, the triangle inequality is barely satisfied, but it would mean that points O, C, and P are colinear, with C lying between O and P, and OC = r, CP = r, so OP = OC + CP = 2r. But wait, we thought OP should be r? There's a contradiction here.Wait, maybe my assumption is wrong. Let me go back. If the third circle passes through K, then K is on both the larger circle (radius 2r, center O) and the third circle (radius r, center P). So, the distance from P to K must be equal to the radius of the third circle, which is r. Therefore, PK = r. Also, since K is on the larger circle, OK = 2r. Therefore, the distance between centers O and P must satisfy OP = OK ± PK. Depending on whether P is inside or outside the larger circle. Since the third circle intersects the larger circle again at T, which is another point besides K, so they must intersect at two points, so OP must be between 2r - r = r and 2r + r = 3r. But since the third circle has radius r, and OP is the distance between centers, for two circles to intersect at two points, the distance between centers must be less than the sum and more than the difference of the radii. The larger circle has radius 2r, the third circle has radius r, so OP must be between (2r - r) = r and (2r + r) = 3r. However, since the third circle is touching the smaller circle (radius r) at M, the distance between their centers, C and P, must be 2r (since they touch externally, both radius r). So, we have two equations: CP = 2r and OP is between r and 3r.But OC is the distance between O and C. Since the original two circles (O, radius 2r and C, radius r) touch internally at K, the distance between O and C is 2r - r = r. So, OC = r. Therefore, we have triangle OCP, with OC = r, CP = 2r, and OP = ?By the triangle inequality, OP + OC >= CP, so OP + r >= 2r => OP >= r. Similarly, OC + CP >= OP => r + 2r >= OP => OP <= 3r. Which matches the earlier condition. So, OP is between r and 3r. But since we also have that the third circle passes through K, which gives us that OP = OK ± PK = 2r - r = r (if P is between O and K) or 2r + r = 3r (if P is on the other side). But if OP = 3r, then the third circle (radius r) centered at P would be too far from O to intersect the larger circle (radius 2r). The maximum distance for intersection is 3r, but in that case, the circles would touch externally at one point. Wait, no. If two circles have radii 2r and r and centers separated by 3r, they would touch externally at one point. So, in that case, the third circle would touch the larger circle at one point, but the problem says it intersects again at T, implying two intersection points. Therefore, OP must be less than 3r. So, the only possible value is OP = r. But if OP = r, then since OC = r and CP = 2r, points O, C, P must be colinear with C between O and P, and OC = r, CP = 2r, so OP = OC + CP = 3r. But that contradicts OP = r. Wait, this is confusing.Perhaps my initial assumption that the third circle passes through K is wrong. The problem states that the third circle intersects the larger circle again at T. The word "again" might imply that the third circle intersects the larger circle at two points: K and another point T. But if the original two circles (O and C) touch at K, then K is on both O's circle and C's circle. The third circle touches C's circle at M and intersects O's circle again at T. So, does the third circle pass through K? The problem doesn't explicitly say that. It says "intersects the larger circle again at point T." Maybe "again" refers to another intersection apart from some other point, but perhaps the third circle only intersects the larger circle at T and another point. Wait, the problem says "intersects the larger circle again at point T," which might mean that the third circle and the larger circle intersect at two points, one of which is T, and the other is not mentioned. But since the original two circles touch at K, which is a point on both the larger and the smaller circle. If the third circle touches the smaller circle at M, maybe M is different from K. Then, does the third circle pass through K? Not necessarily. Hmm. The problem is a bit ambiguous here. Let me parse the problem again:"Two circles with radii 2r and r touch each other internally at point K. A third circle with radius r touches the smaller circle at point M and intersects the larger circle again at point T. Prove that the points K, M, and T lie on a straight line."So, the third circle touches the smaller circle (radius r) at M and intersects the larger circle (radius 2r) again at T. The word "again" might be implying that T is another intersection point besides the original contact point K. But if the third circle does not pass through K, then "again" might just mean another intersection point apart from wherever else. But this is unclear. Wait, maybe "intersects the larger circle again at point T" means that apart from touching the smaller circle at M, it intersects the larger circle at T. But the problem didn't mention the third circle passing through K. Hmm. So, perhaps the third circle intersects the larger circle only at T, but the wording is "intersects the larger circle again at point T," which might mean that the third circle intersects the larger circle at two points, one of which is T, and the other is... perhaps K? But why would the third circle pass through K?Alternatively, maybe the third circle is tangent to the smaller circle at M and intersects the larger circle at two points, one of which is T, and the other is not mentioned. The problem wants us to prove that K, M, T are colinear. Maybe the line KMT is the radical axis of the third circle and the larger circle? But the radical axis of two circles is the set of points with equal power with respect to both circles, which is a straight line perpendicular to the line connecting their centers. Alternatively, if we can show that K, M, T lie on the radical axis, then they must be colinear. Alternatively, use homothety or inversion.Alternatively, coordinate geometry. Let me try setting up coordinates. Let me place the center of the larger circle, O, at the origin (0,0). The smaller circle touches the larger circle internally at point K. Since the distance between their centers is 2r - r = r, the center C of the smaller circle is at a distance of r from O. Let's assume that the point K is at (2r, 0). Then, since the two circles touch at K, the center C of the smaller circle must be along the line OK, which is the x-axis. The distance from O to K is 2r (radius of the larger circle). The center C of the smaller circle is r away from O, so since K is the point where they touch, the center C is located at (r, 0). Because the smaller circle has radius r, so from C (r,0) to K (2r,0) is distance r, which matches the radius of the smaller circle.Now, the third circle has radius r, touches the smaller circle at M, and intersects the larger circle again at T. Let me denote the center of the third circle as P. Since it touches the smaller circle (center C at (r,0), radius r) at M, the distance between C and P must be equal to the sum of their radii (since they touch externally). Therefore, CP = r + r = 2r. So, the center P lies on a circle of radius 2r centered at C (r,0). Let me parameterize point P. Let's suppose that the coordinates of P are (r + 2r cos θ, 0 + 2r sin θ) for some angle θ. That is, P is located at a point 2r away from C, at an angle θ from the x-axis.But wait, the third circle also intersects the larger circle (center O at (0,0), radius 2r) at point T. The third circle's center is P, radius r, so the equation of the third circle is (x - (r + 2r cos θ))² + (y - 2r sin θ)² = r². The equation of the larger circle is x² + y² = (2r)² = 4r². The intersection points of these two circles are solutions to both equations. Let me subtract the two equations to find the radical axis:(x - (r + 2r cos θ))² + (y - 2r sin θ)² - r² = x² + y² - 4r².Expanding the left side:(x² - 2x(r + 2r cos θ) + (r + 2r cos θ)²) + (y² - 4r y sin θ + (2r sin θ)²) - r² =x² - 2x(r + 2r cos θ) + r² + 4r² cos θ + 4r² cos² θ + y² - 4r y sin θ + 4r² sin² θ - r².Simplify:x² + y² - 2x(r + 2r cos θ) - 4r y sin θ + 4r² cos θ + 4r² cos² θ + 4r² sin² θ.The terms x² + y² cancel with those on the right side (x² + y² - 4r²). So, moving to the right side equation:Left side expanded - (x² + y² - 4r²) = 0 becomes:[-2x(r + 2r cos θ) - 4r y sin θ + 4r² cos θ + 4r² cos² θ + 4r² sin² θ] - (-4r²) = 0.Simplify:-2x(r + 2r cos θ) - 4r y sin θ + 4r² cos θ + 4r² (cos² θ + sin² θ) + 4r² = 0.Since cos² θ + sin² θ = 1:-2x(r + 2r cos θ) - 4r y sin θ + 4r² cos θ + 4r²(1) + 4r² = 0.Simplify further:-2r x(1 + 2 cos θ) - 4r y sin θ + 4r² cos θ + 4r² + 4r² = 0.Combine constants:-2r x(1 + 2 cos θ) - 4r y sin θ + 4r² cos θ + 8r² = 0.Divide both sides by 2r to simplify:- x(1 + 2 cos θ) - 2 y sin θ + 2r cos θ + 4r = 0.This is the equation of the radical axis between the third circle and the larger circle. Therefore, points K, M, T lie on this radical axis if they are colinear. But point K is (2r, 0). Let's check if K lies on this radical axis.Substitute x = 2r, y = 0 into the equation:-2r(1 + 2 cos θ) - 0 + 2r cos θ + 4r = 0.Simplify:-2r - 4r cos θ + 2r cos θ + 4r = (-2r + 4r) + (-4r cos θ + 2r cos θ) = 2r - 2r cos θ.This equals 2r(1 - cos θ). For this to be zero, 1 - cos θ must be zero, i.e., cos θ = 1. But θ = 0 would mean the center P is at (r + 2r*1, 0 + 2r*0) = (3r, 0). Then, the third circle is centered at (3r, 0) with radius r. But the distance from O (0,0) to P (3r, 0) is 3r, which is equal to the sum of the radii (2r + r), so the two circles touch externally at one point. So, in that case, they would intersect at only one point, which would be T. But the problem says "intersects the larger circle again at point T," implying two intersection points, so θ cannot be 0. Therefore, unless θ is 0, point K is not on the radical axis. Therefore, K is not on the radical axis unless the third circle is tangent to the larger circle, which it's not. Therefore, my initial approach may be flawed.Alternatively, perhaps points K, M, T lie on the line connecting centers C and P? Let's see. Since the third circle touches the smaller circle at M, the point M lies on the line connecting their centers C and P. So, M is on CP. Similarly, if the line KMT is the same as line CP, then K must also lie on CP. But the centers O, C are on the line OK, which is the x-axis in my coordinate system. If CP is another line, then K would not be on CP unless CP coincides with OK. Let's see.In my coordinate system, O is at (0,0), C is at (r,0), K is at (2r,0). The center P of the third circle is at (r + 2r cos θ, 2r sin θ). The line CP connects C (r,0) to P (r + 2r cos θ, 2r sin θ). So, the direction vector of CP is (2r cos θ, 2r sin θ), which is the same as 2r (cos θ, sin θ). So, the line CP is in the direction of angle θ from the x-axis.Point M is where the third circle touches the smaller circle, so it's along CP at a distance of r from C (since the smaller circle has radius r and the third circle has radius r, and they touch externally). Therefore, M is located at C plus the unit vector in the direction of CP times r. Since CP has length 2r, the unit vector is (cos θ, sin θ). Therefore, M = (r,0) + r*(cos θ, sin θ) = (r + r cos θ, r sin θ).Similarly, point T is the other intersection point of the third circle and the larger circle. Let's find coordinates for T. The radical axis we found earlier is the line - x(1 + 2 cos θ) - 2 y sin θ + 2r cos θ + 4r = 0. Since radical axis is perpendicular to the line joining the centers O and P. The line joining O (0,0) and P (r + 2r cos θ, 2r sin θ) has slope (2r sin θ)/(r + 2r cos θ) = (2 sin θ)/(1 + 2 cos θ). Therefore, the radical axis, being perpendicular, has slope - (1 + 2 cos θ)/(2 sin θ). But we have the equation of the radical axis as above. Alternatively, since points T and K (if it were on the radical axis) lie on this line. Wait, but earlier substitution showed that K is on the radical axis only if θ = 0, which is the case when the third circle is tangent to the larger circle. But in our problem, the third circle intersects the larger circle again at T, implying two distinct points, so θ ≠ 0.Therefore, perhaps points K, M, T are colinear on another line. Let's see.We need to prove that K, M, T are colinear. Let me compute coordinates for T.The third circle and the larger circle intersect at points T and... another point? If the third circle is not passing through K, then the two intersection points are T and another point. But the problem statement mentions only T. Alternatively, maybe when θ is chosen such that the third circle passes through K, but earlier analysis suggests that would require θ = 0, leading to a single intersection point. Hmm.Wait, maybe in the problem statement, the third circle passes through K as one intersection point, and T is the other. Then, the radical axis would be the line KT, and since M is on the radical axis, then K, M, T would be colinear. But earlier, substituting K into the radical axis equation gave 2r(1 - cos θ) = 0, which requires θ = 0, but θ = 0 would make the third circle centered at (3r, 0), radius r, which touches the larger circle (radius 2r, center O) at (3r, 0), but the distance between O and P is 3r, which is equal to 2r + r, so they touch externally. Therefore, only one intersection point. Therefore, contradiction. Therefore, the third circle does not pass through K. Therefore, the radical axis does not include K, so K is not on the radical axis. Therefore, another approach is needed.Alternatively, perhaps inversion could help. Inversion with respect to point K. Or homothety. Since the two original circles touch at K, maybe there is a homothety centered at K mapping one circle to the other. The smaller circle (radius r) and the larger circle (radius 2r) are tangent at K, so the homothety centered at K with ratio 2 maps the smaller circle to the larger one. Let's see. Homothety with center K, ratio 2: any point P on the smaller circle is mapped to a point P' such that KP' = 2 KP. Since the smaller circle has radius r and the larger has radius 2r, this homothety would map the smaller circle to the larger one.Now, the third circle has radius r and touches the smaller circle at M. Let's apply the homothety centered at K with ratio 2. The image of the smaller circle (center C) is the larger circle (center O). The image of the third circle (center P) would be a circle with radius 2r (since homothety ratio is 2) touching the image of the smaller circle, which is the larger circle. Wait, no: the third circle touches the smaller circle at M, so under homothety, the image of the third circle would be a circle touching the image of the smaller circle (the larger circle) at the image of M, which is M'. The image of the third circle would have radius 2r (since original radius is r, ratio 2) and center at P', where KP' = 2 KP. So, if the original third circle touches the smaller circle at M, then the image circle under homothety touches the larger circle at M', and since homothety preserves tangency, the image circle is tangent to the larger circle at M'.But the image circle (radius 2r) touches the larger circle (radius 2r) at M', so the distance between centers O and P' must be 2r + 2r = 4r, but since O is the image of C under the homothety (ratio 2, center K), then OC = r, so KO = 2r (since K is on the larger circle). Wait, maybe this is getting too complicated.Alternatively, since homothety maps the smaller circle to the larger one, and the third circle (radius r) tangent to the smaller circle at M, then the image of the third circle under homothety is a circle tangent to the larger circle at M', where M' is the image of M. The image circle would have radius 2r. If we can show that this image circle passes through T, then T would be the image of some point on the original third circle. Then, since homothety preserves colinearity, points K, M, M', T would lie on the line through the homothety center K. But this is vague.Alternatively, consider that line KMT should be the common tangent to the smaller circle and the third circle at point M. Since both the smaller circle and the third circle have radius r and touch at M, the line tangent at M to both circles is the same line, which is perpendicular to the line joining their centers C and P. Therefore, the tangent line at M is perpendicular to CP. If K, M, T are colinear, then line KMT must be this tangent line. But we need to show that T is also on this tangent line. Alternatively, since T is on both the larger circle and the third circle, perhaps the tangent line at M passes through T and K.Alternatively, consider power of point T with respect to the smaller circle. The power of T with respect to the smaller circle (center C) is TC² - r². Since T lies on the third circle (center P, radius r), TP = r. If we can relate TC² - r² to something, maybe we can find that TM is tangent to the smaller circle, implying that TC² - r² = TM². But since M is the point of tangency, TM would be tangent, so indeed TC² - r² = TM². Therefore, if we can show that power of T with respect to the smaller circle is equal to TM², then TM is tangent, which it is by problem statement. Wait, but T is on the third circle and the larger circle. Let me compute TC² - r².TC is the distance from T to C. C is at (r,0) in my coordinate system. Let T be a point (x,y) on both the larger circle x² + y² = 4r² and the third circle (x - (r + 2r cos θ))² + (y - 2r sin θ)² = r². Then, TC² = (x - r)^2 + y^2. Compute TC² - r² = (x - r)^2 + y^2 - r² = x² - 2rx + r² + y² - r² = x² + y² - 2rx. Since T is on the larger circle, x² + y² = 4r². Therefore, TC² - r² = 4r² - 2rx - r² = 3r² - 2rx.On the other hand, TM is the distance from T to M. M is (r + r cos θ, r sin θ). So, TM² = (x - (r + r cos θ))² + (y - r sin θ)^2. Let's expand this:= (x - r - r cos θ)^2 + (y - r sin θ)^2= x² - 2rx - 2rx cos θ + r² + 2r² cos θ + r² cos² θ + y² - 2ry sin θ + r² sin² θSimplify:= x² + y² - 2rx(1 + cos θ) - 2ry sin θ + r²(1 + 2 cos θ + cos² θ + sin² θ)Since cos² θ + sin² θ = 1:= x² + y² - 2rx(1 + cos θ) - 2ry sin θ + r²(2 + 2 cos θ)But x² + y² = 4r²:= 4r² - 2rx(1 + cos θ) - 2ry sin θ + 2r² + 2r² cos θ= 6r² + 2r² cos θ - 2rx(1 + cos θ) - 2ry sin θCompare this to TC² - r² = 3r² - 2rx. For TM² to equal TC² - r², we need:6r² + 2r² cos θ - 2rx(1 + cos θ) - 2ry sin θ = 3r² - 2rxSimplify:6r² + 2r² cos θ - 2rx - 2rx cos θ - 2ry sin θ = 3r² - 2rxSubtract 3r² - 2rx from both sides:3r² + 2r² cos θ - 2rx cos θ - 2ry sin θ = 0Divide by r:3r + 2r cos θ - 2x cos θ - 2y sin θ = 0This is the condition that must be satisfied for TM² = TC² - r². However, since T lies on the third circle and the larger circle, perhaps this equation holds true. Let me check using the equations of the circles.We know T is on the third circle:(x - (r + 2r cos θ))² + (y - 2r sin θ)² = r²Expand this:x² - 2x(r + 2r cos θ) + (r + 2r cos θ)² + y² - 4ry sin θ + (2r sin θ)² = r²= x² + y² - 2rx - 4rx cos θ + r² + 4r² cos θ + 4r² cos² θ - 4ry sin θ + 4r² sin² θ = r²Simplify using x² + y² = 4r²:4r² - 2rx - 4rx cos θ + r² + 4r² cos θ + 4r² cos² θ - 4ry sin θ + 4r² sin² θ = r²Combine like terms:4r² + r² + 4r² cos θ + 4r² (cos² θ + sin² θ) - 2rx - 4rx cos θ - 4ry sin θ = r²Since cos² θ + sin² θ = 1:4r² + r² + 4r² cos θ + 4r² - 2rx - 4rx cos θ - 4ry sin θ = r²Total left side:(4r² + r² + 4r²) + 4r² cos θ - 2rx - 4rx cos θ - 4ry sin θ = 9r² + 4r² cos θ - 2rx - 4rx cos θ - 4ry sin θSet equal to r²:9r² + 4r² cos θ - 2rx - 4rx cos θ - 4ry sin θ = r²Subtract r²:8r² + 4r² cos θ - 2rx - 4rx cos θ - 4ry sin θ = 0Divide by 2r:4r + 2r cos θ - x - 2x cos θ - 2y sin θ = 0Compare this with the earlier condition from TM² = TC² - r²:3r + 2r cos θ - 2x cos θ - 2y sin θ = 0These two equations:1. 4r + 2r cos θ - x - 2x cos θ - 2y sin θ = 02. 3r + 2r cos θ - 2x cos θ - 2y sin θ = 0Subtract equation 2 from equation 1:(4r + 2r cos θ - x - 2x cos θ - 2y sin θ) - (3r + 2r cos θ - 2x cos θ - 2y sin θ) = 0 - 0Simplifies to:r - x = 0 => x = rSo, unless x = r, the equations aren't compatible. Therefore, the only way both conditions hold is if x = r. Therefore, when x = r, TM² = TC² - r². But x = r would mean that T is at (r, y), lying on the larger circle x² + y² = 4r², so y² = 3r², y = ±√3 r. Also, from the third circle's equation:(x - (r + 2r cos θ))² + (y - 2r sin θ)² = r²If x = r, then:(r - r - 2r cos θ)^2 + (y - 2r sin θ)^2 = r²=> ( -2r cos θ )² + (y - 2r sin θ)^2 = r²=> 4r² cos² θ + y² - 4r y sin θ + 4r² sin² θ = r²=> 4r² (cos² θ + sin² θ) + y² - 4r y sin θ = r²=> 4r² + y² - 4r y sin θ = r²=> y² - 4r y sin θ + 3r² = 0Solve for y:y = [4r sin θ ± √(16r² sin² θ - 12r²)] / 2= [4r sin θ ± 2r√(4 sin² θ - 3)] / 2= 2r sin θ ± r√(4 sin² θ - 3)For real solutions, 4 sin² θ - 3 ≥ 0 => sin² θ ≥ 3/4 => |sin θ| ≥ √3/2. So, θ in [60°, 120°] or [240°, 300°]. This suggests that for certain angles θ, point T could be at x = r. However, this is a specific case and doesn't necessarily solve the general problem.This seems too complicated. Maybe there's a simpler geometric approach.Let me consider the line KM. Since M is the point of contact between the third circle and the smaller circle, the line KM should pass through the center of the third circle, P? Wait, no. The line joining centers C and P passes through M, since they are tangent at M. So, C, M, P are colinear. So, line CP passes through M. Also, K is the point where the original two circles touch, so line OK passes through K and C, since OC = r, and OK is the line connecting centers O and C, with K located at distance 2r from O and r from C.Wait, in my coordinate system, O is at (0,0), C is at (r,0), K is at (2r,0). So, line OK is the x-axis. The third circle's center P is at (r + 2r cos θ, 2r sin θ). So, line CP is from (r,0) to (r + 2r cos θ, 2r sin θ). Point M is on CP at (r + r cos θ, r sin θ). So, coordinates of M are (r + r cos θ, r sin θ).Now, we need to show that K, M, T are colinear. Let's parametrize line KM and check if T lies on it.Line KM goes from K (2r, 0) to M (r + r cos θ, r sin θ). The direction vector is (r + r cos θ - 2r, r sin θ - 0) = (-r + r cos θ, r sin θ). So, the parametric equations for line KM are:x = 2r + t(-r + r cos θ)y = 0 + t(r sin θ)We need to see if point T lies on this line for some t.Point T is the other intersection point of the third circle and the larger circle. We can find coordinates for T by solving the two circle equations:1. x² + y² = 4r² (larger circle)2. (x - (r + 2r cos θ))² + (y - 2r sin θ)² = r² (third circle)Expanding equation 2:x² - 2x(r + 2r cos θ) + (r + 2r cos θ)² + y² - 4r y sin θ + (2r sin θ)² = r²Simplify:x² + y² - 2rx - 4rx cos θ + r² + 4r² cos θ + 4r² cos² θ - 4r y sin θ + 4r² sin² θ = r²Using equation 1, replace x² + y² with 4r²:4r² - 2rx - 4rx cos θ + r² + 4r² cos θ + 4r² (cos² θ + sin² θ) - 4r y sin θ = r²Simplify cos² θ + sin² θ = 1:4r² - 2rx - 4rx cos θ + r² + 4r² cos θ + 4r² - 4r y sin θ = r²Combine terms:4r² + r² + 4r² = 9r²-2rx - 4rx cos θ + 4r² cos θ - 4r y sin θ = r²Thus:9r² -2rx -4rx cos θ +4r² cos θ -4r y sin θ = r²Subtract r² from both sides:8r² -2rx -4rx cos θ +4r² cos θ -4r y sin θ = 0Divide all terms by 2r:4r - x -2x cos θ + 2r cos θ -2 y sin θ = 0Rearrange:- x(1 + 2 cos θ) - 2 y sin θ + 4r + 2r cos θ = 0This is the equation we derived earlier for the radical axis.Now, let's assume that T lies on line KM. The parametric equations for KM are:x = 2r - t r (1 - cos θ)y = t r sin θSubstitute x and y into the radical axis equation:- [2r - t r (1 - cos θ)] (1 + 2 cos θ) - 2 [t r sin θ] sin θ + 4r + 2r cos θ = 0Let's expand this:-2r(1 + 2 cos θ) + t r (1 - cos θ)(1 + 2 cos θ) - 2 t r sin² θ + 4r + 2r cos θ = 0Factor out the terms:[-2r(1 + 2 cos θ) + 4r + 2r cos θ] + t r [ (1 - cos θ)(1 + 2 cos θ) - 2 sin² θ ] = 0Simplify the first bracket:-2r -4r cos θ +4r +2r cos θ = ( -2r +4r ) + ( -4r cos θ + 2r cos θ ) = 2r - 2r cos θNow, the second bracket:(1 - cos θ)(1 + 2 cos θ) - 2 sin² θFirst, expand (1 - cos θ)(1 + 2 cos θ):=1*(1) +1*(2 cos θ) - cos θ*(1) - cos θ*(2 cos θ)=1 + 2 cos θ - cos θ - 2 cos² θ=1 + cos θ - 2 cos² θNow subtract 2 sin² θ:=1 + cos θ - 2 cos² θ - 2 sin² θ=1 + cos θ - 2 (cos² θ + sin² θ )=1 + cos θ - 2*1= cos θ -1Therefore, the equation becomes:2r(1 - cos θ) + t r (cos θ -1 ) = 0Factor out (1 - cos θ):(1 - cos θ)(2r - t r ) = 0Therefore, either 1 - cos θ = 0 or 2r - t r = 0.Case 1: 1 - cos θ = 0 => cos θ = 1 => θ = 0. But as before, this leads to P at (3r, 0), where the third circle touches the larger circle externally at (3r, 0), which is T. However, in this case, M would be at (r + r*1, 0) = (2r, 0), which is point K. But the problem states that the third circle touches the smaller circle at M, which would coincide with K only if M=K. But in the problem statement, M and K are distinct points because the third circle touches the smaller circle at M and intersects the larger circle again at T. If θ=0, then M=K and T=K as well, which contradicts the problem's "again at point T." Therefore, θ ≠ 0, so cos θ ≠ 1.Case 2: 2r - t r = 0 => t = 2.Therefore, when t = 2, the equation is satisfied regardless of θ. Therefore, substituting t = 2 into the parametric equations for line KM:x = 2r - 2r(1 - cos θ) = 2r - 2r + 2r cos θ = 2r cos θy = 2r sin θSo, point (2r cos θ, 2r sin θ) lies on both the line KM and the radical axis. Therefore, this must be the other intersection point T. But wait, the larger circle has equation x² + y² = (2r)^2, so plugging in x = 2r cos θ, y = 2r sin θ, we get (2r cos θ)^2 + (2r sin θ)^2 = 4r²(cos² θ + sin² θ) = 4r², which satisfies the equation. Therefore, T is at (2r cos θ, 2r sin θ), which is indeed on the larger circle.Moreover, checking if T is on the third circle:Third circle's equation is (x - (r + 2r cos θ))² + (y - 2r sin θ)² = r²Substitute x = 2r cos θ, y = 2r sin θ:(2r cos θ - r - 2r cos θ)^2 + (2r sin θ - 2r sin θ)^2 = (-r)^2 + 0 = r², which satisfies the equation.Therefore, T is indeed the point (2r cos θ, 2r sin θ) on both circles, and this point lies on line KM when t=2. Therefore, points K, M, T are colinear.Thus, the proof is complete.

✅:To prove that points ( K ), ( M ), and ( T ) lie on a straight line, we analyze the geometric relationships and use coordinate geometry.Setup:- Let the larger circle (radius ( 2r )) have center ( O ) at the origin ((0,0)).- The smaller circle (radius ( r )) touches the larger circle internally at ( K ), so its center ( C ) is at ((r,0)), and ( K ) is at ((2r,0)).- The third circle (radius ( r )) touches the smaller circle at ( M ) and intersects the larger circle at ( T ). Let its center be ( P ).Key Observations:1. Position of ( P ): Since the third circle touches the smaller circle externally, ( CP = 2r ). Parametrize ( P ) as ( (r + 2rcostheta, 2rsintheta) ).2. Point ( M ): Lies on the line ( CP ), one-third from ( C ) to ( P ): ( M = (r + rcostheta, rsintheta) ).3. Intersection ( T ): Found by solving the equations of the larger circle and the third circle. The radical axis (line where their intersections lie) is derived and shown to contain ( T ).Parametric Line ( KM ):- Line ( KM ) is parametrized from ( K(2r,0) ) to ( M(r + rcostheta, rsintheta) ).- Substituting the parametric coordinates into the radical axis equation confirms that ( T ) lies on this line when ( t = 2 ).Coordinates of ( T ):- Solving the intersection, ( T ) is found to be ( (2rcostheta, 2rsintheta) ), which lies on both circles and satisfies the parametric equation of line ( KM ).Conclusion:Since ( T ) lies on the line ( KM ) for all valid ( theta ), points ( K ), ( M ), and ( T ) are collinear.(boxed{K, M, T text{ are collinear.}})

🔑:1. Understanding the Geometry and Coordinates: - We have two circles, the larger one with radius (2r) and the smaller one with radius (r). - These circles touch internally at point (K). - A third circle with the same radius (r) is tangent to the smaller circle at point (M) and intersects the larger circle again at point (T). We denote the centers of the large, medium, and small circles as (O_1, O_2,) and (O_3) respectively.2. Identifying Key Distances: - Since the larger circle has radius (2r) and the smaller circle has radius (r), the center-to-center distance (O_1O_2 = 2r - r = r). - The distance from the large circle’s center to the point where the third circle intersects it again (point (T)) is (2r).3. Utilizing the Properties of Tangency and Distance: - (O_2O_3 = O_1T = r), where (O_2) is the center of the smaller circle and (O_3) is the center of the third circle. - Triangle (TO_1O_2) is an isosceles triangle with (TO_1 = 2r) and (O_1O_2 = r), this implies (TO_3 = r) due to symmetry and the property of the tangents being equal from a central point.4. Considering Parallelogram Formation: - Since (O_2O_3 = r) and (O_1T = 2r), the lengths of both diagonals in the parallelogram (TO_1O_2O_3) also match the necessary criteria for parallelogram. - This implies (TO_3) is parallel to (O_1K), meaning (TO_3 parallel O_1 K).5. Applying Archimedes' Lemma: - Lemma (general form): If two circles touch each other internally or externally, the line through their points of tangency is collinear. - Since (TO_3 | O_1K) and by the direct conclusion of Archimedes' Lemma, point (M) (tangency point on the small circle) must be collinear with points (K) and (T).Therefore, points (K), (M), and (T) must lie on the same straight line.To conclude:[boxed{text{K, M, and T are collinear}}]

🧠:Okay, let's see. So Louise first had her car when 50 litres of petrol cost £40. Then recently, she noticed that 40 litres cost £50. The question is asking by approximately what percentage the cost of petrol has increased over this time. The options are from 50% up to 80%, so I need to figure out the percentage increase.First, I think I need to find the price per litre at both times and then calculate the percentage increase between them. That makes sense because percentage increase usually involves comparing the old and new values.So, initially, 50 litres cost £40. To find the price per litre back then, I divide £40 by 50 litres. Let me do that calculation: 40 divided by 50. Hmm, 40 divided by 50 is 0.8. So that's £0.80 per litre.Now, recently, 40 litres cost £50. So the new price per litre is £50 divided by 40 litres. Let me calculate that. 50 divided by 40 is 1.25. So that's £1.25 per litre now.Okay, so the old price was £0.80 per litre, and the new price is £1.25 per litre. To find the percentage increase, I need to compute the difference between the new and old prices, divide that by the old price, and then multiply by 100 to get the percentage.The difference in price is £1.25 minus £0.80, which is £0.45. Then, divide that by the original price of £0.80. So, 0.45 divided by 0.80. Let me calculate that. Hmm, 0.45 divided by 0.8. Well, 0.45 divided by 0.8 is the same as 45 divided by 80, right? Because multiplying numerator and denominator by 100 to eliminate the decimals. So 45 ÷ 80.Let me compute that. 80 goes into 45 zero times. Then 80 goes into 450 five times (5*80=400), subtract 400 from 450, you get 50. Bring down a zero: 500. 80 goes into 500 six times (6*80=480), remainder 20. Bring down another zero: 200. 80 goes into 200 two times (2*80=160), remainder 40. Bring down another zero: 400. 80 goes into 400 exactly five times. So putting it all together, 45 ÷ 80 = 0.5625.So 0.5625 as a decimal. To get the percentage, multiply by 100, so 56.25%. The question asks for the approximate percentage increase, so 56.25% rounds to 56%, which is option B.Wait, let me double-check. Sometimes percentage increase can be tricky. The formula is ((New - Old)/Old) * 100. So yes, that's exactly what I did. The old price was £0.80, new is £1.25. Difference is £0.45. 0.45/0.80 = 0.5625, so 56.25%. That's correct. So the answer is B) 56%.But let me just confirm once more. Maybe I mixed up the prices? Like, maybe I reversed old and new. Let me check the problem statement again. When she had her first car, 50 litres cost £40. Then, the other day, 40 litres cost £50. So yes, first time: 50L = £40, so £0.80 per litre. Second time: 40L = £50, so £1.25 per litre. So the increase is from 0.80 to 1.25. So yes, the calculation is right. So the percentage increase is 56.25%, so 56% approximately. So B is correct.I don't think there's any other way around this. If someone thought the percentage was calculated based on total cost or something else, maybe they could get a different answer, but the question specifically asks about the cost of petrol, which would be per litre. So yes, per litre price increased from £0.80 to £1.25, which is a 56.25% increase. So the answer is B.

✅:To determine the percentage increase in the cost of petrol, we first calculate the price per litre at both times:1. Initial price: ( frac{£40}{50 text{ litres}} = £0.80 text{ per litre} ).2. Recent price: ( frac{£50}{40 text{ litres}} = £1.25 text{ per litre} ).Next, compute the absolute increase in price: ( £1.25 - £0.80 = £0.45 ).The percentage increase is then calculated as: [frac{£0.45}{£0.80} times 100 = 0.5625 times 100 = 56.25%.]Rounding to the nearest whole number, the approximate percentage increase is 56%.Answer: B) 56%

🔑:1. Determine the initial cost per litre in pence: - Initially, 50 litres of petrol cost £40. - First, convert £40 to pence: £40 is equivalent to (40 times 100 = 4000) pence. - Calculate the cost per litre in pence: [ frac{4000 text{ pence}}{50 text{ litres}} = 80 text{ pence/litre}. ]2. Determine the current cost per litre in pence: - Currently, 40 litres of petrol cost £50. - Convert £50 to pence: £50 is equivalent to (50 times 100 = 5000) pence. - Calculate the cost per litre in pence: [ frac{5000 text{ pence}}{40 text{ litres}} = 125 text{ pence/litre}. ]3. Calculate the increase in cost per litre: - The initial cost per litre was 80 pence, and the current cost per litre is 125 pence. - The increase in cost per litre is: [ 125 text{ pence/litre} - 80 text{ pence/litre} = 45 text{ pence/litre}. ]4. Calculate the percentage increase in the cost: - To find the percentage increase, use the formula: [ text{Percentage Increase} = left(frac{text{Increase}}{text{Original Amount}}right) times 100 ] - Plug in the values: [ text{Percentage Increase} = left(frac{45 text{ pence/litre}}{80 text{ pence/litre}}right) times 100 ] - Perform the division: [ frac{45}{80} = 0.5625 ] - Convert to percentage: [ 0.5625 times 100 = 56.25% ]5. Approximate the percentage: - The increase can be approximated to the nearest whole number: [ 56.25 % approx 56 % ]Conclusion:The increase in the cost of petrol over the given period is approximately:[boxed{B : 56%}]