answer：First, I need to determine the cost of the barbell. The problem states that the barbell costs 1/10 of the squat rack's price, which is 2500.To find the cost of the barbell, I calculate 1/10 of 2500:2500 × (1/10) = 250.Next, I add the cost of the squat rack and the barbell to find the total amount Jack paid:2500 + 250 = 2750.Therefore, Jack paid a total of 2750 for everything.

answer：First, I recognize that Lisa is investing 1,000 in a savings account that earns 1% annual interest, compounded yearly.To find the total amount after five years, I'll use the compound interest formula: A = P × (1 + r)^t, where P is the principal amount, r is the annual interest rate, and t is the time in years.Plugging in the values, A = 1000 × (1 + 0.01)^5. Calculating this gives A ≈ 1,051.01.Finally, to determine the interest earned, I'll subtract the principal from the total amount: 1,051.01 - 1,000 = 51.01. Rounding to the nearest dollar, Lisa will earn 51 in interest after five years.

answer：I need to compute the binomial coefficient (dbinom{1293}{1}).The binomial coefficient (dbinom{n}{k}) is calculated as (frac{n!}{k!(n - k)!}).Substituting (n = 1293) and (k = 1), the formula becomes (frac{1293!}{1!(1293 - 1)!}).Simplifying the denominator, (1293 - 1 = 1292), so the expression is (frac{1293!}{1! times 1292!}).Since (1293! = 1293 times 1292!), the expression simplifies to (frac{1293 times 1292!}{1 times 1292!}).The (1292!) terms cancel out, leaving (frac{1293}{1} = 1293).Therefore, (dbinom{1293}{1} = 1293).

answer：Okay, so I have this limit problem to solve: [ lim _{x rightarrow frac{pi}{3}} frac{1-2 cos x}{sin (pi-3 x)} ]Hmm, let me see. I remember that when dealing with limits, especially as x approaches a certain value, I should first try plugging in the value to see if it gives me an indeterminate form. If it does, then I might need to use some techniques like L'Hospital's Rule or maybe simplify the expression using trigonometric identities.Alright, let's try plugging in x = π/3 into the function. First, the numerator: 1 - 2 cos(x). So, cos(π/3) is 0.5, right? So, 1 - 2*(0.5) = 1 - 1 = 0. Now the denominator: sin(π - 3x). Let's substitute x = π/3 here. So, sin(π - 3*(π/3)) = sin(π - π) = sin(0) = 0. So, we have 0/0, which is an indeterminate form. That means I can't just evaluate it directly; I need to manipulate it somehow. Maybe L'Hospital's Rule is the way to go here since both numerator and denominator approach 0.L'Hospital's Rule says that if the limit of f(x)/g(x) as x approaches a is of the form 0/0 or ∞/∞, then it's equal to the limit of f'(x)/g'(x) as x approaches a, provided that this latter limit exists.So, let's find the derivatives of the numerator and the denominator.First, the numerator is 1 - 2 cos x. The derivative of that with respect to x is 0 - 2*(-sin x) = 2 sin x.The denominator is sin(π - 3x). Let's find its derivative. The derivative of sin(u) is cos(u)*u', so here u = π - 3x, so u' = -3. Therefore, the derivative is cos(π - 3x)*(-3) = -3 cos(π - 3x).So, applying L'Hospital's Rule, the limit becomes:[ lim _{x rightarrow frac{pi}{3}} frac{2 sin x}{-3 cos (pi - 3x)} ]Hmm, let's see if this simplifies. Maybe we can plug in x = π/3 now.First, the numerator: 2 sin(π/3). Sin(π/3) is √3/2, so 2*(√3/2) = √3.Denominator: -3 cos(π - 3*(π/3)) = -3 cos(π - π) = -3 cos(0) = -3*1 = -3.So, the limit becomes √3 / (-3) = -√3 / 3.Wait, but let me double-check. Maybe I made a mistake in the derivative or in simplifying.Wait, the derivative of the denominator was -3 cos(π - 3x). So, when x approaches π/3, π - 3x approaches 0, so cos(0) is 1. So, denominator becomes -3*1 = -3. Numerator is 2 sin(π/3) = 2*(√3/2) = √3. So, yeah, √3 / (-3) is -√3/3.Alternatively, maybe I can approach this without using L'Hospital's Rule, just to verify.Let me try to manipulate the original expression:[ frac{1 - 2 cos x}{sin (pi - 3x)} ]I know that sin(π - θ) = sin θ, so sin(π - 3x) = sin(3x). So, the denominator can be rewritten as sin(3x). So, the expression becomes:[ frac{1 - 2 cos x}{sin 3x} ]Hmm, maybe I can express sin 3x in terms of sin x or something else. I remember that sin 3x can be expanded as 3 sin x - 4 sin^3 x. Let me write that down:sin 3x = 3 sin x - 4 sin^3 x.So, the denominator is 3 sin x - 4 sin^3 x.So, the expression is:[ frac{1 - 2 cos x}{3 sin x - 4 sin^3 x} ]I wonder if the numerator can be related to the denominator somehow. Let's see.Alternatively, maybe I can factor the denominator or numerator.Wait, the numerator is 1 - 2 cos x. I don't immediately see a trigonometric identity for that, but maybe I can express it in terms of sin or something else.Alternatively, let's consider multiplying numerator and denominator by something to make it easier.Wait, another approach: since x approaches π/3, maybe we can make a substitution t = x - π/3, so that as x approaches π/3, t approaches 0. Let's try that.Let t = x - π/3, so x = t + π/3. Then, as x approaches π/3, t approaches 0.So, rewrite the original limit in terms of t:Numerator: 1 - 2 cos(t + π/3)Denominator: sin(π - 3(t + π/3)) = sin(π - 3t - π) = sin(-3t) = -sin(3t)So, the expression becomes:[ frac{1 - 2 cos(t + pi/3)}{ - sin 3t } ]Let me compute numerator and denominator separately.First, numerator: 1 - 2 cos(t + π/3). Let's expand cos(t + π/3) using the cosine addition formula:cos(t + π/3) = cos t cos π/3 - sin t sin π/3 = (0.5) cos t - (√3/2) sin t.So, 1 - 2 cos(t + π/3) = 1 - 2*(0.5 cos t - (√3/2) sin t) = 1 - cos t + √3 sin t.So, numerator is 1 - cos t + √3 sin t.Denominator: - sin 3t.So, the expression is:[ frac{1 - cos t + sqrt{3} sin t}{ - sin 3t } ]Hmm, maybe I can factor numerator or denominator.Alternatively, let's use the small angle approximations as t approaches 0. Since t approaches 0, sin t ≈ t, cos t ≈ 1 - t²/2, and sin 3t ≈ 3t.But before that, let me see if I can express the numerator in terms of sin and cos t.Wait, 1 - cos t is 2 sin²(t/2), and √3 sin t is 2√3 sin(t/2) cos(t/2). So, maybe I can factor something out.Wait, numerator: 1 - cos t + √3 sin t = 2 sin²(t/2) + 2√3 sin(t/2) cos(t/2). Hmm, that's 2 sin(t/2) [ sin(t/2) + √3 cos(t/2) ].Let me write that:Numerator: 2 sin(t/2) [ sin(t/2) + √3 cos(t/2) ]Denominator: - sin 3t = -3 sin t + higher order terms, but maybe we can write sin 3t as 3 sin t - 4 sin³ t.Wait, but maybe using the approximation for small t.So, as t approaches 0, sin t ≈ t, cos t ≈ 1 - t²/2, so let's approximate numerator and denominator.Numerator: 1 - cos t + √3 sin t ≈ 1 - (1 - t²/2) + √3 t = t²/2 + √3 t.Denominator: - sin 3t ≈ -3t.So, the expression becomes approximately:(t²/2 + √3 t) / (-3t) = [ t(t/2 + √3) ] / (-3t) = (t/2 + √3)/(-3) as t approaches 0.But as t approaches 0, t/2 approaches 0, so the limit becomes (√3)/(-3) = -√3/3.So, same result as before. So, that's consistent.Alternatively, maybe I can factor the numerator and denominator more cleverly.Wait, let's consider the numerator: 1 - 2 cos x. Maybe I can write this as something squared or something else.Alternatively, maybe I can use the identity for sin 3x in terms of sin x and cos x.Wait, another thought: maybe express the numerator in terms of sin or cos of multiple angles.Wait, 1 - 2 cos x. Hmm, 1 can be written as 2*(1/2), so 2*(1/2) - 2 cos x = 2*(1/2 - cos x). Not sure if that helps.Alternatively, maybe 1 - 2 cos x can be related to some trigonometric identity.Wait, I recall that 1 - 2 cos x is similar to the expression for cos(2x) or something else. Let me check:cos(2x) = 2 cos²x - 1, so 1 - 2 cos²x = -cos(2x). Hmm, not quite.Alternatively, 1 - 2 cos x doesn't directly correspond to a standard identity, but maybe if I consider it as part of a product or something else.Alternatively, maybe I can write 1 - 2 cos x as a multiple of sin or cos of some angle.Wait, another approach: since we have sin(3x) in the denominator, and 3x, maybe we can relate the numerator to sin(3x) somehow.Wait, let me think about the original expression again:[ frac{1 - 2 cos x}{sin (pi - 3x)} = frac{1 - 2 cos x}{sin 3x} ]So, maybe I can write 1 - 2 cos x in terms of sin(3x) or something.Wait, let me recall that sin(3x) = 3 sin x - 4 sin³x. So, if I can express 1 - 2 cos x in terms of sin x, maybe I can factor something out.Alternatively, maybe I can write 1 - 2 cos x as a combination of sin and cos terms.Wait, let me consider that 1 - 2 cos x can be written as something like A sin x + B cos x, but that might not be helpful.Alternatively, maybe I can use the identity for sin(A - B) or something like that.Wait, another idea: maybe factor the numerator and denominator.Wait, the numerator is 1 - 2 cos x. Let me see if I can factor this as something.Alternatively, maybe I can write 1 - 2 cos x as 2*(1/2 - cos x). Hmm, not sure.Wait, another thought: perhaps use the substitution x = π/3 - t, so that as x approaches π/3, t approaches 0. Let me try that.Let x = π/3 - t, so t = π/3 - x, and as x approaches π/3, t approaches 0.So, substituting into the original expression:Numerator: 1 - 2 cos(π/3 - t)Denominator: sin(π - 3*(π/3 - t)) = sin(π - π + 3t) = sin(3t)So, the expression becomes:[ frac{1 - 2 cos(pi/3 - t)}{sin 3t} ]Let me expand cos(π/3 - t) using the cosine subtraction formula:cos(π/3 - t) = cos π/3 cos t + sin π/3 sin t = 0.5 cos t + (√3/2) sin t.So, numerator: 1 - 2*(0.5 cos t + (√3/2) sin t) = 1 - cos t - √3 sin t.So, numerator is 1 - cos t - √3 sin t.Denominator is sin 3t.So, the expression is:[ frac{1 - cos t - sqrt{3} sin t}{sin 3t} ]Hmm, maybe we can factor numerator or denominator.Alternatively, let's use small angle approximations as t approaches 0.So, as t approaches 0, cos t ≈ 1 - t²/2, sin t ≈ t, sin 3t ≈ 3t.So, numerator: 1 - (1 - t²/2) - √3 t ≈ t²/2 - √3 t.Denominator: 3t.So, the expression becomes approximately:(t²/2 - √3 t) / (3t) = [ t(t/2 - √3) ] / (3t) = (t/2 - √3)/3 as t approaches 0.As t approaches 0, t/2 approaches 0, so the limit becomes (-√3)/3 = -√3/3.Same result again. So, that's consistent.Alternatively, maybe I can factor the numerator.Wait, numerator: 1 - cos t - √3 sin t. Let me see if I can factor something out.Hmm, 1 - cos t - √3 sin t. Maybe factor out a 2?Wait, 1 - cos t - √3 sin t = 1 - (cos t + √3 sin t). Hmm, cos t + √3 sin t is similar to 2 cos(t - π/3), because cos A cos B + sin A sin B = cos(A - B). Wait, let me check:cos(t - π/3) = cos t cos π/3 + sin t sin π/3 = 0.5 cos t + (√3/2) sin t. So, 2 cos(t - π/3) = cos t + √3 sin t.So, cos t + √3 sin t = 2 cos(t - π/3). Therefore, 1 - (cos t + √3 sin t) = 1 - 2 cos(t - π/3).Wait, but that's similar to the original numerator, which was 1 - 2 cos x. Hmm, maybe this isn't helpful.Alternatively, maybe I can write the numerator as 1 - cos t - √3 sin t = something else.Wait, another idea: maybe express 1 - cos t as 2 sin²(t/2) and √3 sin t as 2√3 sin(t/2) cos(t/2). So, let's try that.Numerator: 1 - cos t - √3 sin t = 2 sin²(t/2) - 2√3 sin(t/2) cos(t/2).Factor out 2 sin(t/2):= 2 sin(t/2) [ sin(t/2) - √3 cos(t/2) ]Hmm, so numerator is 2 sin(t/2) [ sin(t/2) - √3 cos(t/2) ]Denominator is sin 3t = 3 sin t - 4 sin³ t, but maybe we can write it as 3 sin t for small t.Alternatively, let's use the identity sin 3t = 3 sin t - 4 sin³ t.But maybe it's better to write sin 3t in terms of sin(t/2) or something else.Alternatively, maybe we can factor the denominator as well.Wait, sin 3t = 3 sin t - 4 sin³ t. Hmm, not sure if that helps.Alternatively, maybe express sin 3t in terms of sin(t/2):Wait, 3t = 6*(t/2), so sin 3t = sin(6*(t/2)) = 2 sin(3*(t/2)) cos(3*(t/2)). Hmm, not sure.Alternatively, maybe use multiple-angle identities.Wait, perhaps it's getting too complicated. Since all the methods I've tried so far give me the same result, maybe I can just stick with that.So, using L'Hospital's Rule, I got -√3/3, and using substitution and small angle approximations, I also got -√3/3. So, that seems consistent.Alternatively, maybe I can use another substitution or identity.Wait, another thought: since the denominator is sin(3x), and the numerator is 1 - 2 cos x, maybe I can relate this to some known limit.Wait, I recall that sometimes, when you have limits involving trigonometric functions, you can use the identity that sin(a) ≈ a when a is small, but in this case, we already used that approach.Alternatively, maybe I can write the numerator as a combination of sin and cos terms and then see if it can be expressed as a multiple of sin(3x) or something else.Wait, let me try to write 1 - 2 cos x in terms of sin(3x). Hmm, not sure.Wait, another approach: maybe use the identity for sin(A) - sin(B) or something like that.Wait, 1 - 2 cos x. Hmm, 1 can be written as 2*(1/2), so 2*(1/2) - 2 cos x = 2*(1/2 - cos x). Hmm, not sure.Alternatively, maybe I can write 1 - 2 cos x as -2 cos x + 1. Hmm, not helpful.Wait, another idea: maybe use the identity for cos(2x) or something else.Wait, cos(2x) = 2 cos²x - 1, so 1 - 2 cos²x = -cos(2x). Hmm, but we have 1 - 2 cos x, not 1 - 2 cos²x.Wait, maybe I can write 1 - 2 cos x as 1 - 2 cos x = 2*(1/2 - cos x). Hmm, not helpful.Alternatively, maybe I can write 1 - 2 cos x as a linear combination of sin and cos terms.Wait, 1 - 2 cos x = A sin x + B cos x + C. Hmm, maybe not.Alternatively, maybe I can write it as something like 2 sin(x + φ) or something. Let me try.Wait, 1 - 2 cos x. Let me consider expressing this as R sin(x + φ) + something.Wait, R sin(x + φ) = R sin x cos φ + R cos x sin φ. So, if I set this equal to 1 - 2 cos x, then:R cos φ sin x + R sin φ cos x = 1 - 2 cos x.Comparing coefficients:Coefficient of sin x: R cos φ = 0.Coefficient of cos x: R sin φ = -2.Constant term: 1. Hmm, but the left side doesn't have a constant term, so this approach might not work.Alternatively, maybe I can write 1 - 2 cos x as a combination of sin and cos terms plus a constant.Wait, maybe 1 - 2 cos x = A sin x + B cos x + C.But then, to match 1 - 2 cos x, we have:A sin x + B cos x + C = 1 - 2 cos x.So, equating coefficients:A = 0 (since there's no sin x on the right side),B = -2,C = 1.So, 1 - 2 cos x = 0*sin x - 2 cos x + 1. Well, that's trivial, so it doesn't help.Hmm, maybe this isn't the right approach.Alternatively, maybe I can use the identity for sin(3x) in terms of sin x and cos x, and see if I can relate it to the numerator.Wait, sin(3x) = 3 sin x - 4 sin³x. So, maybe I can factor sin x out:sin(3x) = sin x (3 - 4 sin²x).So, the denominator is sin x (3 - 4 sin²x).So, the expression becomes:[ frac{1 - 2 cos x}{sin x (3 - 4 sin^2 x)} ]Hmm, maybe I can split the fraction:= [ frac{1 - 2 cos x}{sin x} times frac{1}{3 - 4 sin^2 x} ]Now, let's evaluate each part as x approaches π/3.First, let's compute [ frac{1 - 2 cos x}{sin x} ] as x approaches π/3.We can write this as:[ frac{1 - 2 cos x}{sin x} ]Let me compute this at x = π/3:Numerator: 1 - 2*(0.5) = 0.Denominator: sin(π/3) = √3/2.So, 0 divided by √3/2 is 0. Hmm, but that's just the value at x = π/3, but we need the limit as x approaches π/3.Wait, maybe we can use L'Hospital's Rule here as well, since numerator approaches 0 and denominator approaches √3/2, which is not zero. Wait, no, because the denominator doesn't approach zero, so L'Hospital's Rule doesn't apply here.Wait, but in the original expression, the denominator was sin(3x), which does approach 0 as x approaches π/3. So, maybe this approach isn't helpful.Alternatively, maybe I can write 1 - 2 cos x as something involving sin x.Wait, another idea: maybe express 1 - 2 cos x in terms of sin(x - π/3) or something like that.Wait, let me try to write 1 - 2 cos x as a function of (x - π/3).Let me set t = x - π/3, so x = t + π/3. Then, as x approaches π/3, t approaches 0.So, 1 - 2 cos x = 1 - 2 cos(t + π/3) = 1 - 2*(cos t cos π/3 - sin t sin π/3) = 1 - 2*(0.5 cos t - (√3/2) sin t) = 1 - cos t + √3 sin t.Which is what I had earlier. So, same as before.So, 1 - 2 cos x = 1 - cos t + √3 sin t, where t = x - π/3.So, the expression becomes:[ frac{1 - cos t + sqrt{3} sin t}{sin 3x} ]But sin 3x = sin(3(t + π/3)) = sin(3t + π) = -sin(3t), because sin(θ + π) = -sin θ.So, sin 3x = -sin 3t.So, the expression is:[ frac{1 - cos t + sqrt{3} sin t}{ - sin 3t } ]Which is the same as before.So, as t approaches 0, we can approximate numerator and denominator.Numerator: 1 - cos t + √3 sin t ≈ t²/2 + √3 t.Denominator: - sin 3t ≈ -3t.So, the expression ≈ (t²/2 + √3 t)/(-3t) = (t/2 + √3)/(-3) as t approaches 0.So, as t approaches 0, t/2 approaches 0, so the limit is √3/(-3) = -√3/3.Same result again.So, all methods point to the limit being -√3/3.Alternatively, maybe I can use the Taylor series expansion of the numerator and denominator around x = π/3.Let me try that.Let me set x = π/3 + h, where h approaches 0.So, numerator: 1 - 2 cos(π/3 + h).Denominator: sin(π - 3(π/3 + h)) = sin(π - π - 3h) = sin(-3h) = -sin(3h).So, numerator: 1 - 2 cos(π/3 + h).Let me expand cos(π/3 + h) using the Taylor series around h = 0.cos(π/3 + h) = cos π/3 cos h - sin π/3 sin h ≈ 0.5*(1 - h²/2) - (√3/2)*(h - h³/6).So, cos(π/3 + h) ≈ 0.5 - 0.5 h²/2 - (√3/2) h + (√3/2)*(h³/6).Simplify:≈ 0.5 - (h²)/4 - (√3 h)/2 + (√3 h³)/12.So, 1 - 2 cos(π/3 + h) ≈ 1 - 2*(0.5 - (h²)/4 - (√3 h)/2 + (√3 h³)/12).Compute this:= 1 - [1 - (h²)/2 - √3 h + (√3 h³)/6]= 1 - 1 + (h²)/2 + √3 h - (√3 h³)/6= (h²)/2 + √3 h - (√3 h³)/6.So, numerator ≈ (h²)/2 + √3 h.Denominator: -sin(3h) ≈ -3h + ( (3h)^3 )/6 = -3h + (27 h³)/6 = -3h + (9 h³)/2.So, the expression becomes:[ (h²)/2 + √3 h ] / [ -3h + (9 h³)/2 ]Factor h from numerator and denominator:= h [ h/2 + √3 ] / [ h (-3 + (9 h²)/2 ) ]Cancel h (since h ≠ 0):= [ h/2 + √3 ] / [ -3 + (9 h²)/2 ]Now, as h approaches 0, the terms with h in the numerator and denominator become negligible, so we have:[ 0 + √3 ] / [ -3 + 0 ] = √3 / (-3) = -√3/3.Same result again.So, all these methods—L'Hospital's Rule, substitution with small angle approximations, Taylor series expansion—lead me to the same limit: -√3/3.Therefore, I can be confident that the limit is -√3/3.**Final Answer**The limit is boxed{-dfrac{sqrt{3}}{3}}.

answer：Okay, so I have this problem where Gavrila wants to maximize the distance his car can travel by swapping the front and rear tires at some point. The front tires last for 42,000 kilometers and the rear tires last for 56,000 kilometers. Hmm, interesting. I need to figure out when to swap them so that the total distance is maximized. Let me think. So, the front tires wear out faster than the rear ones. That makes sense because front tires usually handle more of the steering and braking, so they wear out quicker. But if we swap them, maybe we can make use of the rear tires' longer lifespan on the front, and vice versa. I guess the key here is to balance the wear on all four tires. If we don't swap them, the front tires will only last 42,000 km, and the rear ones would still have some life left. But by swapping, we can distribute the wear more evenly. Let me denote the total distance the car can travel as D. We need to find D such that both the front and rear tires are used optimally. Let's say we swap the tires after some distance x. So, for the first x kilometers, the front tires are on the front, and the rear tires are on the rear. Then, after swapping, the front tires move to the rear, and the rear tires move to the front for the remaining distance (D - x). So, each tire will have been used on both the front and the rear. The wear on each tire will depend on how much distance it was used on each position. The wear on a tire can be thought of as the distance driven divided by the lifespan of that tire in that position. So, for the original front tires, they will have been used on the front for x km and on the rear for (D - x) km. Similarly, the original rear tires will have been used on the rear for x km and on the front for (D - x) km. Since all tires must wear out completely by the time the car has traveled D km, the total wear on each tire should equal 1 (representing 100% wear). So, for the original front tires:- Wear on front: x / 42,000- Wear on rear: (D - x) / 56,000Total wear: x / 42,000 + (D - x) / 56,000 = 1Similarly, for the original rear tires:- Wear on rear: x / 56,000- Wear on front: (D - x) / 42,000Total wear: x / 56,000 + (D - x) / 42,000 = 1Wait, so both equations are the same. That makes sense because swapping is symmetric. So, I can just solve one equation for D in terms of x, but I have two variables here. Hmm, maybe I need another equation or a way to relate x and D.Wait, actually, since both equations are the same, I can just solve one of them. Let me write it out:x / 42,000 + (D - x) / 56,000 = 1Let me solve for D. First, let's find a common denominator for the fractions. 42,000 and 56,000. Let's see, 42,000 is 42 * 1000, and 56,000 is 56 * 1000. The least common multiple of 42 and 56 is... let's see, 42 is 6*7, 56 is 8*7, so LCM is 8*6*7 = 336. So, LCM of 42,000 and 56,000 is 336,000.So, converting both fractions to have denominator 336,000:x / 42,000 = (8x) / 336,000(D - x) / 56,000 = (6(D - x)) / 336,000So, adding them together:(8x + 6(D - x)) / 336,000 = 1Simplify numerator:8x + 6D - 6x = (2x + 6D)So, (2x + 6D) / 336,000 = 1Multiply both sides by 336,000:2x + 6D = 336,000Divide both sides by 2:x + 3D = 168,000So, x = 168,000 - 3DWait, that seems a bit odd. Let me check my steps.Wait, when I converted x / 42,000 to 8x / 336,000, that's correct because 42,000 * 8 = 336,000. Similarly, (D - x)/56,000 is 6(D - x)/336,000 because 56,000 * 6 = 336,000. So, that part is correct.Adding them: 8x + 6(D - x) = 8x + 6D - 6x = 2x + 6D. Correct.Then, 2x + 6D = 336,000. Dividing by 2: x + 3D = 168,000. So, x = 168,000 - 3D.Wait, but x is the distance before swapping, so x must be positive. Similarly, D must be greater than x. So, 168,000 - 3D > 0 => 3D < 168,000 => D < 56,000. But that can't be, because if we don't swap, the front tires would last 42,000 km, which is less than 56,000. So, swapping should allow us to go beyond 42,000 km, but according to this, D is less than 56,000. Hmm, maybe I made a mistake in setting up the equation.Wait, let's think again. The total wear on each tire should be 1. So, for the front tires, which are used on the front for x km and on the rear for (D - x) km, their total wear is x / 42,000 + (D - x) / 56,000 = 1.Similarly, for the rear tires, which are used on the rear for x km and on the front for (D - x) km, their total wear is x / 56,000 + (D - x) / 42,000 = 1.So, both equations are the same, so we can just solve one of them.But when I tried solving, I got x = 168,000 - 3D, which seems to give D < 56,000, but that contradicts the idea that swapping should allow us to go beyond 42,000 km.Wait, maybe I need to set up the equation differently. Let me think about the wear on each tire.Each tire is used on two positions: front and rear. So, for each tire, the total wear is the sum of wear in both positions.But actually, each tire is either front or rear. Wait, no, when we swap, the original front tires go to the rear, and the original rear tires go to the front.So, each tire is used on both front and rear positions, but for different distances.So, for each original front tire, it's used on front for x km and on rear for (D - x) km. Similarly, each original rear tire is used on rear for x km and on front for (D - x) km.So, the wear on each original front tire is x / 42,000 + (D - x) / 56,000 = 1.Same for each original rear tire: x / 56,000 + (D - x) / 42,000 = 1.So, both equations are the same, so we can just solve one.So, let's write the equation again:x / 42,000 + (D - x) / 56,000 = 1Let me solve this equation for D.First, multiply both sides by 336,000 (the LCM of 42,000 and 56,000) to eliminate denominators:336,000 * [x / 42,000 + (D - x) / 56,000] = 336,000 * 1Simplify each term:336,000 / 42,000 = 8, so 8x336,000 / 56,000 = 6, so 6(D - x)So, 8x + 6(D - x) = 336,000Expand:8x + 6D - 6x = 336,000Combine like terms:2x + 6D = 336,000Divide both sides by 2:x + 3D = 168,000So, x = 168,000 - 3DWait, that's the same result as before. Hmm, but if I plug this back into the equation, I can solve for D.Wait, but x is the distance before swapping, so x must be positive. So, 168,000 - 3D > 0 => 3D < 168,000 => D < 56,000.But that doesn't make sense because without swapping, the front tires would limit the distance to 42,000 km. Swapping should allow us to go beyond that, but according to this, D is less than 56,000, which is the rear tire lifespan. Hmm, maybe I need to think differently.Wait, perhaps I need to consider that each tire is being used on both front and rear, so the total wear on each tire is the sum of wear in both positions. So, each tire is used for some distance on front and some on rear, and the sum should equal 1.But in this case, the original front tires are used on front for x km and on rear for (D - x) km. So, their total wear is x / 42,000 + (D - x) / 56,000 = 1.Similarly, the original rear tires are used on rear for x km and on front for (D - x) km, so their total wear is x / 56,000 + (D - x) / 42,000 = 1.So, both equations are the same, so we can just solve one.But when I solve, I get x = 168,000 - 3D, which seems problematic because D is the total distance, which should be greater than 42,000 km.Wait, maybe I need to express D in terms of x instead.From x + 3D = 168,000, we can write D = (168,000 - x) / 3But D must be greater than x, so (168,000 - x)/3 > xMultiply both sides by 3: 168,000 - x > 3x168,000 > 4xx < 42,000So, x must be less than 42,000 km. That makes sense because if we swap before the front tires wear out, which would be at 42,000 km.So, x is the distance before swapping, which is less than 42,000 km.So, D = (168,000 - x)/3But we need to find D, so we need another equation or a way to find x.Wait, but we have only one equation and two variables. Maybe I need to think differently.Wait, perhaps I can consider that all four tires are being used equally. Since each tire is used on both front and rear, the total wear on each tire is 1.But each tire is used for x km on one position and (D - x) km on another.So, for each tire, the wear is x / lifespan1 + (D - x) / lifespan2 = 1But for front tires, lifespan1 is 42,000, and lifespan2 is 56,000.Similarly, for rear tires, lifespan1 is 56,000, lifespan2 is 42,000.But since we have four tires, two front and two rear, but when we swap, each front tire becomes a rear tire and vice versa.Wait, maybe I need to consider the total wear on all tires.Wait, the total wear on all four tires should be equal to 4, since each tire is worn out completely.But each tire is used for x km on one position and (D - x) km on another.So, for each front tire: x / 42,000 + (D - x) / 56,000 = 1Similarly, for each rear tire: x / 56,000 + (D - x) / 42,000 = 1Since there are two front tires and two rear tires, the total wear is 2*(x / 42,000 + (D - x)/56,000) + 2*(x / 56,000 + (D - x)/42,000) = 4But since each equation is the same, it's just 4*(x / 42,000 + (D - x)/56,000) = 4So, x / 42,000 + (D - x)/56,000 = 1Which is the same equation as before.So, we have x / 42,000 + (D - x)/56,000 = 1And from this, we get x = 168,000 - 3DBut we need another relation. Wait, maybe I can express D in terms of x.From x + 3D = 168,000, we have D = (168,000 - x)/3But we also know that x must be less than 42,000, as established earlier.So, let's think about the total wear on the front tires. Each front tire is used on front for x km and on rear for (D - x) km.So, the wear on each front tire is x / 42,000 + (D - x)/56,000 = 1Similarly, for rear tires.But since we have two front tires and two rear tires, the total wear is 2*(x / 42,000 + (D - x)/56,000) + 2*(x / 56,000 + (D - x)/42,000) = 4But simplifying, it's 4*(x / 42,000 + (D - x)/56,000) = 4So, x / 42,000 + (D - x)/56,000 = 1Which is the same equation.So, perhaps I need to find D such that both front and rear tires are fully worn out after D km, considering the swap.Wait, maybe I can think of it as each tire is used for a certain amount of time on front and rear, and the total wear is 1.But since we have four tires, each tire is used for some distance on front and some on rear.Wait, but actually, each tire is either front or rear, but after swapping, they switch positions.So, each tire is used on front for x km and on rear for (D - x) km, or vice versa.Wait, no, actually, the original front tires are used on front for x km and on rear for (D - x) km. Similarly, the original rear tires are used on rear for x km and on front for (D - x) km.So, each tire is used on both positions, but for different distances.So, the wear on each tire is x / lifespan1 + (D - x)/lifespan2 = 1So, for front tires: x / 42,000 + (D - x)/56,000 = 1For rear tires: x / 56,000 + (D - x)/42,000 = 1Which are the same equations.So, solving for D:x / 42,000 + (D - x)/56,000 = 1Multiply both sides by 336,000:8x + 6(D - x) = 336,0008x + 6D - 6x = 336,0002x + 6D = 336,000Divide by 2:x + 3D = 168,000So, x = 168,000 - 3DBut we also know that x must be positive, so 168,000 - 3D > 0 => D < 56,000But as I thought earlier, D should be more than 42,000 km.Wait, but 56,000 is the rear tire lifespan. So, if we don't swap, the rear tires would last longer, but the front tires would limit us to 42,000 km.By swapping, we can make the front tires last longer by using them on the rear, and the rear tires can be used on the front, which wears them out faster, but allows the car to go further.Wait, maybe I need to think about the total wear on all tires.Each front tire is used on front for x km and on rear for (D - x) km.Each rear tire is used on rear for x km and on front for (D - x) km.So, the total wear on all front tires is 2*(x / 42,000 + (D - x)/56,000) = 2Similarly, the total wear on all rear tires is 2*(x / 56,000 + (D - x)/42,000) = 2So, adding them together, total wear is 4, which makes sense because all four tires are worn out.But this doesn't help us find D.Wait, maybe I can think of it as the harmonic mean of the two tire lifespans.Wait, harmonic mean is used when rates are involved. So, if we have two different rates, the total distance can be found using harmonic mean.But in this case, we have two different lifespans for front and rear tires, and swapping them.Wait, maybe the maximum distance is the harmonic mean of 42,000 and 56,000.But harmonic mean of two numbers a and b is 2ab / (a + b). So, 2*42,000*56,000 / (42,000 + 56,000)Let me compute that.First, 42,000 + 56,000 = 98,00042,000 * 56,000 = 2,352,000,000So, 2 * 2,352,000,000 = 4,704,000,000Divide by 98,000: 4,704,000,000 / 98,000Let me compute that.4,704,000,000 divided by 98,000.First, 4,704,000,000 / 98,000 = (4,704,000,000 / 100,000) * (100,000 / 98,000) = 47,040 * (100/98) ≈ 47,040 * 1.0204 ≈ 47,040 + (47,040 * 0.0204) ≈ 47,040 + 959 ≈ 48,000 km.Wait, so the harmonic mean is approximately 48,000 km.But is that the maximum distance?Wait, let me check.If we don't swap, the distance is limited by the front tires at 42,000 km.If we swap at some point, we can make the total distance longer.But according to the harmonic mean, it's 48,000 km.But let's see if that makes sense.If we swap at x km, then the front tires have been used for x km on front and (D - x) km on rear.Similarly, rear tires have been used for x km on rear and (D - x) km on front.So, the wear on each front tire is x / 42,000 + (D - x)/56,000 = 1Similarly for rear tires.So, solving for D, we have:x / 42,000 + (D - x)/56,000 = 1Multiply both sides by 336,000:8x + 6(D - x) = 336,0008x + 6D - 6x = 336,0002x + 6D = 336,000Divide by 2:x + 3D = 168,000So, x = 168,000 - 3DBut we also know that x must be positive, so 168,000 - 3D > 0 => D < 56,000But we can also think about the wear on the rear tires when they are on the front.Each rear tire is used on front for (D - x) km, and their lifespan on front is 42,000 km.So, the wear on rear tires when on front is (D - x)/42,000Similarly, when on rear, it's x / 56,000So, total wear is (D - x)/42,000 + x / 56,000 = 1Which is the same equation as before.So, we have x = 168,000 - 3DBut we can also express D in terms of x.From x + 3D = 168,000, we have D = (168,000 - x)/3But we need another equation to solve for D.Wait, maybe I can think about the fact that the total wear on all tires is 4.But each tire is worn out completely, so total wear is 4.But each tire's wear is 1, so total wear is 4.But that doesn't help because we already have that.Wait, maybe I can consider that the wear on front tires and rear tires must be equal.Wait, no, because front and rear tires have different wear rates.Wait, perhaps I can think of it as the total wear on front tires is 2*(x / 42,000 + (D - x)/56,000) = 2Similarly, total wear on rear tires is 2*(x / 56,000 + (D - x)/42,000) = 2So, both total wears are 2, which is correct because each set of two tires is worn out.But again, that doesn't help us find D.Wait, maybe I can set up the equation as follows:The total distance D is such that each tire is used on both front and rear, and the sum of their wear is 1.So, for each tire, whether it was originally front or rear, it's used on front for some distance and on rear for some distance.So, for each tire, the wear is distance_on_front / 42,000 + distance_on_rear / 56,000 = 1But for the original front tires, distance_on_front = x, distance_on_rear = D - xFor the original rear tires, distance_on_rear = x, distance_on_front = D - xSo, both equations are the same.So, x / 42,000 + (D - x)/56,000 = 1Which is the same equation.So, solving for D, we have:x / 42,000 + (D - x)/56,000 = 1Multiply both sides by 336,000:8x + 6(D - x) = 336,0008x + 6D - 6x = 336,0002x + 6D = 336,000Divide by 2:x + 3D = 168,000So, x = 168,000 - 3DBut we need another equation to solve for D.Wait, perhaps I can think about the fact that the wear on the front tires when they are on the front is x / 42,000, and when they are on the rear is (D - x)/56,000.Similarly, the wear on the rear tires when they are on the rear is x / 56,000, and when they are on the front is (D - x)/42,000.But since all tires must be worn out completely, the total wear on each tire is 1.So, the wear on each tire is the same, regardless of whether it was originally front or rear.So, for each tire, whether it was front or rear, the wear is x / 42,000 + (D - x)/56,000 = 1Wait, but that's the same equation.So, I think we need to accept that we have one equation with two variables, but we can express D in terms of x, and since x must be positive and less than D, we can find the maximum D.Wait, but how?Wait, maybe I can think about the fact that the wear on the front tires when they are on the front is x / 42,000, and when they are on the rear is (D - x)/56,000.Similarly, the wear on the rear tires when they are on the rear is x / 56,000, and when they are on the front is (D - x)/42,000.But since all tires are worn out completely, the total wear on each tire is 1.So, for each tire, the wear is x / 42,000 + (D - x)/56,000 = 1Which is the same equation.So, solving for D, we have:x / 42,000 + (D - x)/56,000 = 1Multiply both sides by 336,000:8x + 6(D - x) = 336,0008x + 6D - 6x = 336,0002x + 6D = 336,000Divide by 2:x + 3D = 168,000So, x = 168,000 - 3DBut we also know that x must be positive, so 168,000 - 3D > 0 => D < 56,000But we can also think about the fact that when we swap, the rear tires are now on the front, which wear out faster.So, the distance after swapping, which is (D - x), must be such that the rear tires, now on the front, don't wear out before the front tires, now on the rear.Wait, so the wear on the rear tires when on front is (D - x)/42,000, and the wear on the front tires when on rear is (D - x)/56,000.But since both must be equal to 1 when added to their respective x terms.Wait, maybe I can set up another equation based on the fact that the wear on the rear tires when on front must equal the wear on the front tires when on rear.Wait, no, that's not necessarily true.Wait, perhaps I can think about the fact that the wear on the rear tires when on front is (D - x)/42,000, and the wear on the front tires when on rear is (D - x)/56,000.But since both sets of tires are worn out completely, the total wear on each set is 2.Wait, no, each tire is worn out completely, so each tire's wear is 1.Wait, I'm getting confused.Let me try plugging in D = 48,000 km, which is the harmonic mean.So, if D = 48,000, then x = 168,000 - 3*48,000 = 168,000 - 144,000 = 24,000 km.So, x = 24,000 km.So, the car is driven 24,000 km on the original front and rear tires, then swapped, and driven another 24,000 km.Wait, but 24,000 + 24,000 = 48,000 km.So, let's check the wear on each tire.For the original front tires:- Used on front for 24,000 km: wear = 24,000 / 42,000 = 0.5714- Used on rear for 24,000 km: wear = 24,000 / 56,000 = 0.4286- Total wear = 0.5714 + 0.4286 = 1Similarly, for the original rear tires:- Used on rear for 24,000 km: wear = 24,000 / 56,000 = 0.4286- Used on front for 24,000 km: wear = 24,000 / 42,000 = 0.5714- Total wear = 0.4286 + 0.5714 = 1So, that works. So, D = 48,000 km is the maximum distance.But wait, earlier I thought D must be less than 56,000 km, but 48,000 is less than 56,000, so that's fine.But let me check if D can be more than 48,000 km.Suppose D = 50,000 km.Then, x = 168,000 - 3*50,000 = 168,000 - 150,000 = 18,000 km.So, x = 18,000 km.So, the car is driven 18,000 km on original tires, then swapped, and driven another 32,000 km.Wait, 18,000 + 32,000 = 50,000 km.Let's check the wear on each tire.For the original front tires:- Used on front for 18,000 km: wear = 18,000 / 42,000 ≈ 0.4286- Used on rear for 32,000 km: wear = 32,000 / 56,000 ≈ 0.5714- Total wear ≈ 0.4286 + 0.5714 = 1Similarly, for the original rear tires:- Used on rear for 18,000 km: wear = 18,000 / 56,000 ≈ 0.3214- Used on front for 32,000 km: wear = 32,000 / 42,000 ≈ 0.7619- Total wear ≈ 0.3214 + 0.7619 ≈ 1.0833Wait, that's more than 1, which is not possible because the tires can't be worn out more than 100%.So, that means D = 50,000 km is too much because the rear tires would be overworn.So, D cannot be 50,000 km.Similarly, if I try D = 48,000 km, as before, the wear is exactly 1 on all tires.So, that seems to be the maximum distance.Therefore, the maximum distance is 48,000 km.Wait, but let me confirm with another value.Suppose D = 42,000 km, which is the original front tire lifespan.Then, x = 168,000 - 3*42,000 = 168,000 - 126,000 = 42,000 km.So, x = 42,000 km, which means no swapping occurs because x = D.So, the car is driven 42,000 km without swapping, and the front tires are worn out, while the rear tires have 42,000 km remaining (since 56,000 - 42,000 = 14,000 km remaining).But that's not optimal because we can swap and use the rear tires on the front to go further.So, D = 48,000 km is better.Wait, another way to think about it is that the total wear on all tires is 4 (since each of the four tires is worn out completely).Each tire is used on front for some distance and on rear for some distance.So, the total wear on all tires is:For front tires: 2*(x / 42,000 + (D - x)/56,000) = 2For rear tires: 2*(x / 56,000 + (D - x)/42,000) = 2So, total wear is 4, which is correct.But that doesn't help us find D.Wait, but if I think about the total wear on all tires as 4, and the total distance driven is D, then the total wear can also be expressed as:Total wear = (number of front tire wear) + (number of rear tire wear)But each front tire is used on front for x km and on rear for (D - x) km.Similarly, each rear tire is used on rear for x km and on front for (D - x) km.So, total wear is:2*(x / 42,000 + (D - x)/56,000) + 2*(x / 56,000 + (D - x)/42,000) = 4Which simplifies to:2*(x / 42,000 + (D - x)/56,000 + x / 56,000 + (D - x)/42,000) = 4But that's the same as:2*( (x / 42,000 + x / 56,000) + ((D - x)/56,000 + (D - x)/42,000) ) = 4Which is:2*( x*(1/42,000 + 1/56,000) + (D - x)*(1/56,000 + 1/42,000) ) = 4But 1/42,000 + 1/56,000 is the same as (56,000 + 42,000)/(42,000*56,000) = 98,000 / (42,000*56,000) = 98,000 / 2,352,000,000 = 1 / 24,000Wait, let me compute 1/42,000 + 1/56,000.Find a common denominator, which is 42,000*56,000 / gcd(42,000,56,000)gcd(42,000,56,000) is 14,000.So, common denominator is (42,000*56,000)/14,000 = 42,000*4 = 168,000.So, 1/42,000 = 4 / 168,0001/56,000 = 3 / 168,000So, 1/42,000 + 1/56,000 = 7 / 168,000 = 1 / 24,000Similarly, 1/56,000 + 1/42,000 is also 1 / 24,000So, going back:2*( x*(1/24,000) + (D - x)*(1/24,000) ) = 4Simplify:2*( (x + D - x)/24,000 ) = 4Which is:2*( D / 24,000 ) = 4So, (2D) / 24,000 = 4Multiply both sides by 24,000:2D = 96,000So, D = 48,000 kmAh, there we go. So, that's another way to get D = 48,000 km.So, the maximum distance is 48,000 km.Therefore, the answer is 48,000 kilometers.

answer：Okay, so I have this geometry problem here. Let me read it again to make sure I understand it correctly."The distance from the intersection point of the diameter of a circle with a radius of 11 cm and a chord of length 18 cm to the center of the circle is 7 cm. In what ratio does the intersection point divide the chord?"Hmm, okay. Let me visualize this. There's a circle with radius 11 cm. So, the diameter would be 22 cm. There's a chord in the circle, which is 18 cm long. The diameter intersects this chord at some point, and the distance from this intersection point to the center of the circle is 7 cm. I need to find the ratio in which this intersection point divides the chord.Alright, let me try to draw this mentally. There's a circle, center O. Let's say the diameter is AB, so A and B are endpoints of the diameter. The chord is CD, which is 18 cm long. The diameter AB intersects the chord CD at point E. The distance from E to the center O is 7 cm. So, EO = 7 cm.I need to find the ratio CE:ED.Hmm, okay. So, I think I can use the perpendicular distance from the center to the chord to find some relations. Wait, but in this case, the diameter is intersecting the chord, but is it necessarily perpendicular? I don't think so. Because if the diameter is not perpendicular to the chord, then the distance from the center to the chord isn't necessarily 7 cm. Wait, but the distance from E to O is 7 cm. So, E is a point on the diameter, 7 cm away from the center.Let me recall some theorems. There's the theorem that says the perpendicular from the center to a chord bisects the chord. But in this case, the diameter is not necessarily perpendicular to the chord, so E might not be the midpoint of CD. So, we can't directly say that CE = ED.But maybe we can use the power of a point or something similar. Alternatively, coordinate geometry might help here. Let me try to set up a coordinate system.Let me place the center O at (0,0). Let the diameter AB be along the x-axis, so A is (-11,0) and B is (11,0). The chord CD is somewhere in the circle, intersecting the diameter AB at point E, which is 7 cm away from O. So, E is either at (7,0) or (-7,0). Let's assume it's at (7,0) for simplicity.So, point E is (7,0). The chord CD passes through E and has a length of 18 cm. I need to find the ratio CE:ED.Let me denote point C as (x1, y1) and point D as (x2, y2). Since CD is a chord of the circle, both points lie on the circle, so they satisfy the equation x² + y² = 11² = 121.Also, since E is the intersection point of AB and CD, and E is (7,0), the points C, E, D are colinear, so the line CD passes through (7,0).Let me find the equation of line CD. Since it's a chord passing through (7,0), I can write its equation in the form y = m(x - 7), where m is the slope.But I don't know the slope. Maybe I can find it using the fact that CD has length 18 cm.Alternatively, maybe I can use the distance from the center to the chord. Wait, but since CD is not necessarily perpendicular to AB, the distance from O to CD isn't 7 cm. Wait, no, the distance from E to O is 7 cm, but E is on AB, which is the diameter. So, maybe the distance from O to CD is something else.Wait, perhaps I can use the formula for the length of a chord. The length of a chord is 2√(r² - d²), where d is the distance from the center to the chord. But in this case, I don't know d, but I know that the chord passes through E, which is 7 cm from the center.Hmm, maybe I can relate these two. Let me think.If I denote the distance from O to CD as h, then the length of CD is 2√(121 - h²) = 18. So, √(121 - h²) = 9, so 121 - h² = 81, so h² = 40, so h = √40 = 2√10 cm.So, the distance from the center O to the chord CD is 2√10 cm. But how does that help me? Because E is 7 cm away from O, but E is on AB, which is the diameter. So, perhaps the distance from E to CD is something?Wait, no, E is on CD as well. So, E is a point on CD, and it's 7 cm away from O. The distance from O to CD is 2√10 cm. So, the distance from O to CD is less than the distance from E to O, which is 7 cm. Hmm, that seems contradictory because if the distance from O to CD is 2√10 ≈ 6.32 cm, which is less than 7 cm, so E is outside the region where the perpendicular distance is.Wait, maybe I'm confusing something here. Let me clarify.The distance from the center O to the chord CD is h = 2√10 cm. So, the shortest distance from O to CD is 2√10 cm. But E is a point on CD that's 7 cm away from O. So, E is not the closest point on CD to O, because the closest point is only 2√10 ≈ 6.32 cm away. So, E is somewhere else on CD, 7 cm away from O.So, perhaps we can model this as a right triangle. Let me consider the point F, which is the foot of the perpendicular from O to CD. So, OF = h = 2√10 cm. Then, E is another point on CD, 7 cm away from O. So, triangle OFE is a right triangle, where OF = 2√10, OE = 7, and FE is the distance along CD from F to E.Wait, yes, that makes sense. So, in triangle OFE, we have:OF² + FE² = OE²So, (2√10)² + FE² = 7²Calculating that:(4*10) + FE² = 4940 + FE² = 49FE² = 9FE = 3 cmSo, the distance from F to E along CD is 3 cm.Since F is the midpoint of CD (because the perpendicular from the center to the chord bisects the chord), and CD is 18 cm long, so CF = FD = 9 cm.But E is 3 cm away from F along CD. So, depending on the direction, E could be either 3 cm towards C or 3 cm towards D from F.But since E is on the diameter AB, which is a straight line passing through O, and F is the foot of the perpendicular from O to CD, the position of E relative to F depends on the orientation.Wait, but in our coordinate system, we placed E at (7,0). So, let's see.In our coordinate system, O is at (0,0), E is at (7,0). The chord CD passes through E and has length 18 cm. The distance from O to CD is 2√10 cm.So, the foot of the perpendicular from O to CD is point F, which is 2√10 cm away from O. Since E is on CD, and FE = 3 cm, then F is either 3 cm towards C or 3 cm towards D from E.But since F is the midpoint of CD, which is 18 cm long, so CF = FD = 9 cm.So, if FE = 3 cm, then E is either 3 cm away from F towards C or towards D. So, if E is closer to C, then CE = CF - FE = 9 - 3 = 6 cm, and ED = FD + FE = 9 + 3 = 12 cm. Alternatively, if E is closer to D, then CE = 9 + 3 = 12 cm, and ED = 9 - 3 = 6 cm.But in our case, E is at (7,0), which is on the diameter AB. So, depending on the slope of CD, E could be on one side or the other. But since the problem doesn't specify, I think both possibilities are there, but the ratio would be the same in either case, just inverted.But let's see. Let me try to figure out the exact positions.In our coordinate system, O is (0,0), E is (7,0). The chord CD passes through E and has length 18 cm. The distance from O to CD is 2√10 cm.So, the line CD has the equation y = m(x - 7). The distance from O to CD is |0 - m(0 - 7)| / √(1 + m²) = |7m| / √(1 + m²) = 2√10.So, |7m| / √(1 + m²) = 2√10Squaring both sides:49m² / (1 + m²) = 4049m² = 40(1 + m²)49m² = 40 + 40m²49m² - 40m² = 409m² = 40m² = 40/9m = ±(2√10)/3So, the slope of CD is either (2√10)/3 or -(2√10)/3.So, the equation of CD is y = (2√10)/3 (x - 7) or y = -(2√10)/3 (x - 7).Now, let's find the points C and D where this line intersects the circle x² + y² = 121.Let me substitute y from the line equation into the circle equation.Case 1: y = (2√10)/3 (x - 7)Substitute into x² + y² = 121:x² + [(2√10)/3 (x - 7)]² = 121Compute [(2√10)/3 (x - 7)]²:= (4*10)/9 (x - 7)²= 40/9 (x² - 14x + 49)So, the equation becomes:x² + (40/9)(x² - 14x + 49) = 121Multiply through by 9 to eliminate denominators:9x² + 40(x² - 14x + 49) = 1089Expand:9x² + 40x² - 560x + 1960 = 1089Combine like terms:49x² - 560x + 1960 - 1089 = 049x² - 560x + 871 = 0Hmm, let's check my calculations.Wait, 40*49 is 1960, and 121*9 is 1089. So, 1960 - 1089 is 871. So, yes, 49x² - 560x + 871 = 0.Let me try to solve this quadratic equation.Discriminant D = (-560)^2 - 4*49*871Calculate D:560² = 3136004*49*871 = 4*49*871First, 4*49 = 196196*871: Let's compute that.Compute 200*871 = 174200Subtract 4*871 = 3484So, 174200 - 3484 = 170716So, D = 313600 - 170716 = 142884√D = √142884. Let me see, 378² = 142884 because 400²=160000, 380²=144400, so 378² = (380 - 2)² = 380² - 4*380 + 4 = 144400 - 1520 + 4 = 142884. Yes, so √D = 378.So, solutions are:x = [560 ± 378] / (2*49) = [560 ± 378] / 98Compute 560 + 378 = 938560 - 378 = 182So, x = 938/98 and x = 182/98Simplify:938 ÷ 14 = 67, 98 ÷14=7, so 938/98 = 67/7 ≈ 9.571182 ÷14=13, 98 ÷14=7, so 182/98 =13/7 ≈1.857So, x = 67/7 and x =13/7.So, the x-coordinates of C and D are 67/7 and 13/7.Now, let's find the corresponding y-coordinates.Using y = (2√10)/3 (x - 7)For x = 67/7:y = (2√10)/3 (67/7 - 7) = (2√10)/3 (67/7 - 49/7) = (2√10)/3 (18/7) = (36√10)/21 = (12√10)/7For x =13/7:y = (2√10)/3 (13/7 - 7) = (2√10)/3 (13/7 - 49/7) = (2√10)/3 (-36/7) = (-72√10)/21 = (-24√10)/7So, the points are:C: (67/7, 12√10/7) and D: (13/7, -24√10/7)Wait, hold on. Let me check the calculation for y when x =13/7.Wait, x =13/7, so x -7 =13/7 -49/7= -36/7.So, y = (2√10)/3*(-36/7)= (-72√10)/21= (-24√10)/7. Yes, correct.So, points C and D are (67/7, 12√10/7) and (13/7, -24√10/7).Now, let's compute the distances CE and ED.Point E is at (7,0). So, let's compute the distance from E to C and E to D.First, distance CE:C is (67/7, 12√10/7), E is (7,0).Compute the differences:Δx = 67/7 - 7 = 67/7 - 49/7 = 18/7Δy = 12√10/7 - 0 = 12√10/7So, CE = √[(18/7)² + (12√10/7)²] = √[(324/49) + (144*10)/49] = √[(324 + 1440)/49] = √[1764/49] = √[36] = 6 cm.Similarly, distance ED:D is (13/7, -24√10/7), E is (7,0).Δx =13/7 -7 =13/7 -49/7= -36/7Δy= -24√10/7 -0= -24√10/7So, ED=√[(-36/7)² + (-24√10/7)²] = √[(1296/49) + (576*10)/49] = √[(1296 + 5760)/49] = √[7056/49] = √[144] =12 cm.So, CE is 6 cm and ED is 12 cm. So, the ratio CE:ED is 6:12, which simplifies to 1:2.Wait, so the intersection point E divides the chord CD into segments of 6 cm and 12 cm, so the ratio is 1:2.But let me just double-check my calculations because sometimes when dealing with coordinates, it's easy to make a mistake.Wait, in the calculation for CE, Δx was 18/7 and Δy was 12√10/7. Squaring those:(18/7)² = 324/49(12√10/7)² = (144*10)/49 =1440/49324 +1440=1764, and 1764/49=36, whose square root is 6. Correct.Similarly, for ED:Δx=-36/7, so squared is 1296/49Δy=-24√10/7, squared is (576*10)/49=5760/491296 +5760=7056, 7056/49=144, square root is12. Correct.So, yes, CE=6, ED=12, ratio 1:2.Alternatively, if the slope was negative, would the ratio be different?Let me check. If I take the other case where the slope is negative, y = -(2√10)/3 (x -7). Let's see if that affects the ratio.But actually, in that case, the points C and D would just be mirrored over the x-axis, so the distances CE and ED would remain the same. So, the ratio would still be 1:2.Therefore, regardless of the slope, the ratio is 1:2.So, the intersection point divides the chord in the ratio 1:2.**Final Answer**The intersection point divides the chord in the ratio boxed{dfrac{1}{2}}.Wait, hold on. The ratio is CE:ED = 1:2, which can be written as 1/2. But sometimes ratios are expressed in the order they are divided. Since the problem says "In what ratio does the intersection point divide the chord?" It might be expecting the ratio as parts of the chord, so from one end to the other. So, if CE is 6 and ED is 12, the ratio is 6:12 or 1:2. But depending on the convention, sometimes it's expressed as the segment closer to the center first, but in this case, E is closer to D because E is at (7,0), which is on the right side of the center. Wait, no, in our coordinate system, E is at (7,0), which is to the right of the center. So, point C is at (67/7 ≈9.57, 12√10/7 ≈5.43), which is to the right of E, and D is at (13/7≈1.86, -24√10/7≈-7.64), which is to the left of E.Wait, so CE is from C to E, which is from right to center, and ED is from E to D, which is from center to left. So, in terms of the chord CD, which goes from C to D, passing through E, the ratio is CE:ED =6:12=1:2.So, the ratio is 1:2, which can be written as 1/2. But in the problem statement, it's asking "In what ratio does the intersection point divide the chord?" So, depending on the convention, sometimes it's expressed as the part before the colon is the segment closer to the starting point. But in this case, since the chord is CD, and E is between C and D, so the ratio is CE:ED=1:2.But to write it as a single number ratio, it's 1:2, which can be represented as 1/2. Alternatively, sometimes it's written as 1:2. But in the answer, I think it's better to write it as 1:2 inside a box.Wait, but the initial answer I wrote was boxed{dfrac{1}{2}}, but actually, the ratio is 1:2, so maybe it's better to write it as boxed{1:2}.Let me check the problem again. It says, "In what ratio does the intersection point divide the chord?" So, it's asking for the ratio of the two parts, so 1:2 is appropriate.So, I think the correct final answer is boxed{1:2}.**Final Answer**The intersection point divides the chord in the ratio boxed{1:2}.