💡Alright, so I've got this problem here about stools and chairs in a room. Let me try to wrap my head around it step by step. First off, the problem says there are several stools, each with 3 legs, and chairs, each with 4 legs. Each stool or chair can only seat one person. Then, some people come in for a meeting. If they only chose one type of seating—either all stools or all chairs—there wouldn't be enough seats for everyone. But, since they used both stools and chairs, everyone had a seat, and there were still some empty seats left. Okay, so let's break this down. It seems like the number of people is more than the number of stools alone and more than the number of chairs alone. But when you combine stools and chairs, there are enough seats for everyone, with some seats still empty. That means the total number of seats (stools + chairs) is more than the number of people, but individually, both stools and chairs are fewer than the number of people.Now, the key piece of information is that when all the stools and chairs are occupied, the total number of legs, including people's legs, stools, and chairs, is 32. So, we need to figure out how many people, stools, and chairs there are.Let me assign some variables to make this clearer. Let's say:- ( x ) is the number of people,- ( y ) is the number of stools,- ( z ) is the number of chairs.Each person has 2 legs, each stool has 3 legs, and each chair has 4 legs. So, the total number of legs would be the sum of all these. That gives us the equation:[ 2x + 3y + 4z = 32 ]Now, from the problem, we also know that:1. If everyone tried to sit on stools, there wouldn't be enough. So, ( x > y ).2. Similarly, if everyone tried to sit on chairs, there wouldn't be enough. So, ( x > z ).3. But when using both stools and chairs, everyone can sit, and there are still empty seats. That means the total number of seats is more than the number of people. So, ( x < y + z ).Putting this all together, we have the following system of inequalities and equations:[ begin{align*}2x + 3y + 4z &= 32 x &> y x &> z x &< y + zend{align*}]Alright, so now I need to find integer values for ( x ), ( y ), and ( z ) that satisfy all these conditions.Let me think about possible values for ( x ). Since ( x ) has to be greater than both ( y ) and ( z ), and less than ( y + z ), it's somewhere in the middle. Also, the total number of legs is 32, which includes people's legs, so ( 2x ) is part of that 32.Let me try to find bounds for ( x ). If all 32 legs were just people's legs, that would mean ( 2x = 32 ), so ( x = 16 ). But obviously, there are also stools and chairs, so ( x ) must be less than 16. On the lower end, if all 32 legs were from stools and chairs, and assuming all are chairs (which have more legs), the minimum number of people would be when ( y = 0 ) and ( z = 8 ), giving ( 2x + 0 + 32 = 32 ), so ( x = 0 ). But that doesn't make sense because there are people. So, ( x ) is somewhere between 1 and 15.But given the constraints ( x > y ), ( x > z ), and ( x < y + z ), let's see if we can narrow it down further.Let me consider that ( x ) must be at least 1, but realistically, since there are both stools and chairs, and people, ( x ) is probably a small number. Let me try to find possible values.Let me rearrange the equation:[ 2x = 32 - 3y - 4z ]So, ( 2x ) must be equal to ( 32 - 3y - 4z ). Since ( x ) is an integer, ( 32 - 3y - 4z ) must be even because ( 2x ) is even. So, ( 3y + 4z ) must be even as well because 32 is even.Looking at ( 3y + 4z ), since 4z is always even, 3y must also be even for the sum to be even. That means ( y ) must be even because 3 times an even number is even, and 3 times an odd number is odd. So, ( y ) has to be even.Alright, so ( y ) is even. Let's list possible even values for ( y ): 0, 2, 4, 6, etc., but considering that ( y ) must be less than ( x ), and ( x ) is likely small, let's try small even numbers for ( y ).Let me start with ( y = 2 ). Then, plugging into the equation:[ 2x = 32 - 3(2) - 4z ][ 2x = 32 - 6 - 4z ][ 2x = 26 - 4z ][ x = 13 - 2z ]Now, since ( x > y = 2 ), ( x ) must be greater than 2. Also, ( x < y + z = 2 + z ). So:[ 13 - 2z > 2 ][ 13 - 2z > 2 ][ 11 > 2z ][ z < 5.5 ]Since ( z ) is an integer, ( z leq 5 ).Also, ( x < 2 + z ):[ 13 - 2z < 2 + z ][ 13 - 2z < 2 + z ][ 11 < 3z ][ z > frac{11}{3} ][ z > 3.666 ]So, ( z geq 4 ).Therefore, ( z ) must be 4 or 5.Let's try ( z = 4 ):[ x = 13 - 2(4) = 13 - 8 = 5 ]Check if ( x > y ): 5 > 2, yes.Check if ( x > z ): 5 > 4, yes.Check if ( x < y + z ): 5 < 2 + 4 = 6, yes.So, this works.Let's check ( z = 5 ):[ x = 13 - 2(5) = 13 - 10 = 3 ]Check if ( x > y ): 3 > 2, yes.Check if ( x > z ): 3 > 5, no. So, this doesn't work.So, with ( y = 2 ), the only valid solution is ( z = 4 ) and ( x = 5 ).Let me check if there are other possible values for ( y ). Next even number is ( y = 4 ).Plugging ( y = 4 ) into the equation:[ 2x = 32 - 3(4) - 4z ][ 2x = 32 - 12 - 4z ][ 2x = 20 - 4z ][ x = 10 - 2z ]Now, ( x > y = 4 ), so:[ 10 - 2z > 4 ][ 6 > 2z ][ z < 3 ]Since ( z ) is a positive integer, ( z leq 2 ).Also, ( x < y + z = 4 + z ):[ 10 - 2z < 4 + z ][ 10 - 2z < 4 + z ][ 6 < 3z ][ z > 2 ]So, ( z geq 3 ).But from the previous inequality, ( z < 3 ), so there's no integer ( z ) that satisfies both ( z < 3 ) and ( z geq 3 ). Therefore, no solution for ( y = 4 ).Next, ( y = 6 ):[ 2x = 32 - 3(6) - 4z ][ 2x = 32 - 18 - 4z ][ 2x = 14 - 4z ][ x = 7 - 2z ]Now, ( x > y = 6 ):[ 7 - 2z > 6 ][ 1 > 2z ][ z < 0.5 ]But ( z ) can't be negative, so no solution here.So, ( y = 6 ) doesn't work. Let's try ( y = 0 ), just to check:[ 2x = 32 - 0 - 4z ][ 2x = 32 - 4z ][ x = 16 - 2z ]But ( x > y = 0 ), which is always true. Also, ( x < y + z = 0 + z = z ):[ 16 - 2z < z ][ 16 < 3z ][ z > frac{16}{3} ][ z > 5.333 ]So, ( z geq 6 ).But let's see if ( x ) is positive:[ x = 16 - 2z > 0 ][ 16 > 2z ][ z < 8 ]So, ( z ) must be between 6 and 7.Let's try ( z = 6 ):[ x = 16 - 12 = 4 ]Check ( x > y = 0 ): yes.Check ( x > z = 6 ): 4 > 6? No. So, invalid.( z = 7 ):[ x = 16 - 14 = 2 ]Check ( x > y = 0 ): yes.Check ( x > z = 7 ): 2 > 7? No. Invalid.So, ( y = 0 ) doesn't give a valid solution either.Therefore, the only valid solution is when ( y = 2 ), ( z = 4 ), and ( x = 5 ).Let me double-check:Total legs:- People: 5 people * 2 legs = 10- Stools: 2 stools * 3 legs = 6- Chairs: 4 chairs * 4 legs = 16Total: 10 + 6 + 16 = 32 legs. Perfect.Also, checking the seating:- If everyone tried to sit on stools: 2 stools < 5 people. Not enough.- If everyone tried to sit on chairs: 4 chairs < 5 people. Not enough.- But together: 2 + 4 = 6 seats > 5 people. Everyone can sit, and there's 1 empty seat.Everything checks out!

💡Alright, so I've got this trigonometric expression to prove equals 1. It's a bit intimidating at first glance, but let's break it down step by step. The expression is:[ 4.12 frac{left(sin^{2} alpha + operatorname{tg}^{2} alpha + 1right)left(cos^{2} alpha - operatorname{ctg}^{2} alpha + 1right)}{left(cos^{2} alpha + operatorname{ctg}^{2} alpha + 1right)left(sin^{2} alpha + operatorname{tg}^{2} alpha - 1right)} = 1. ]First off, I notice that the expression involves sine, cosine, tangent, and cotangent squared terms. I remember that tangent is sine over cosine, and cotangent is cosine over sine. Maybe I can express everything in terms of sine and cosine to simplify.Let me rewrite the tangent and cotangent terms:- (operatorname{tg} alpha = frac{sin alpha}{cos alpha}), so (operatorname{tg}^2 alpha = frac{sin^2 alpha}{cos^2 alpha})- (operatorname{ctg} alpha = frac{cos alpha}{sin alpha}), so (operatorname{ctg}^2 alpha = frac{cos^2 alpha}{sin^2 alpha})Substituting these into the original expression:Numerator:[left(sin^{2} alpha + frac{sin^2 alpha}{cos^2 alpha} + 1right)left(cos^{2} alpha - frac{cos^2 alpha}{sin^2 alpha} + 1right)]Denominator:[left(cos^{2} alpha + frac{cos^2 alpha}{sin^2 alpha} + 1right)left(sin^{2} alpha + frac{sin^2 alpha}{cos^2 alpha} - 1right)]Hmm, this looks messy. Maybe I can factor out some terms or find common denominators. Let's tackle the numerator first.For the first part of the numerator:[sin^{2} alpha + frac{sin^2 alpha}{cos^2 alpha} + 1]I can factor out (sin^2 alpha):[sin^2 alpha left(1 + frac{1}{cos^2 alpha}right) + 1]But (1 + frac{1}{cos^2 alpha} = frac{cos^2 alpha + 1}{cos^2 alpha}), so:[sin^2 alpha cdot frac{cos^2 alpha + 1}{cos^2 alpha} + 1 = frac{sin^2 alpha (cos^2 alpha + 1)}{cos^2 alpha} + 1]This seems complicated. Maybe I should combine all terms over a common denominator.Let me rewrite the entire numerator as:[left(sin^{2} alpha + frac{sin^2 alpha}{cos^2 alpha} + 1right) = frac{sin^2 alpha cos^2 alpha + sin^2 alpha + cos^2 alpha}{cos^2 alpha}]Wait, that might not be correct. Let me check:Actually, to combine them, I need a common denominator, which is (cos^2 alpha):[sin^{2} alpha + frac{sin^2 alpha}{cos^2 alpha} + 1 = frac{sin^2 alpha cos^2 alpha + sin^2 alpha + cos^2 alpha}{cos^2 alpha}]But (sin^2 alpha cos^2 alpha + sin^2 alpha + cos^2 alpha) doesn't seem to simplify easily. Maybe I'm overcomplicating things.Let's try another approach. Maybe multiply numerator and denominator by (cos^2 alpha sin^2 alpha) to eliminate the fractions. That might make things clearer.Multiplying the entire expression by (cos^2 alpha sin^2 alpha):Numerator becomes:[(sin^2 alpha + operatorname{tg}^2 alpha + 1)(cos^2 alpha - operatorname{ctg}^2 alpha + 1) times cos^2 alpha sin^2 alpha]Denominator becomes:[(cos^2 alpha + operatorname{ctg}^2 alpha + 1)(sin^2 alpha + operatorname{tg}^2 alpha - 1) times cos^2 alpha sin^2 alpha]But this seems like it might not help much. Maybe I should look for symmetries or patterns in the expression.Looking at the numerator and denominator, I notice that the terms are similar but with some sign changes. Maybe if I factor or rearrange terms, something cancels out.Let me consider the first part of the numerator:[sin^{2} alpha + operatorname{tg}^2 alpha + 1 = sin^2 alpha + frac{sin^2 alpha}{cos^2 alpha} + 1]And the first part of the denominator:[cos^{2} alpha + operatorname{ctg}^2 alpha + 1 = cos^2 alpha + frac{cos^2 alpha}{sin^2 alpha} + 1]These look similar, just with sine and cosine swapped.Similarly, the second part of the numerator:[cos^{2} alpha - operatorname{ctg}^2 alpha + 1 = cos^2 alpha - frac{cos^2 alpha}{sin^2 alpha} + 1]And the second part of the denominator:[sin^{2} alpha + operatorname{tg}^2 alpha - 1 = sin^2 alpha + frac{sin^2 alpha}{cos^2 alpha} - 1]Again, similar structure but with signs and terms swapped.Maybe if I factor out terms or use identities, I can find a relationship.I recall that (sin^2 alpha + cos^2 alpha = 1), which might be useful. Let's see if I can express some terms in terms of this identity.Looking back at the numerator's first part:[sin^2 alpha + frac{sin^2 alpha}{cos^2 alpha} + 1 = sin^2 alpha left(1 + frac{1}{cos^2 alpha}right) + 1]But (1 + frac{1}{cos^2 alpha} = frac{cos^2 alpha + 1}{cos^2 alpha}), so:[sin^2 alpha cdot frac{cos^2 alpha + 1}{cos^2 alpha} + 1]This still seems complicated. Maybe I should consider multiplying out the terms.Let me expand the numerator:[(sin^2 alpha + operatorname{tg}^2 alpha + 1)(cos^2 alpha - operatorname{ctg}^2 alpha + 1)]Expanding this:[sin^2 alpha cos^2 alpha - sin^2 alpha operatorname{ctg}^2 alpha + sin^2 alpha + operatorname{tg}^2 alpha cos^2 alpha - operatorname{tg}^2 alpha operatorname{ctg}^2 alpha + operatorname{tg}^2 alpha + cos^2 alpha - operatorname{ctg}^2 alpha + 1]This is getting really messy. Maybe there's a smarter way.Wait, I remember that (operatorname{tg} alpha cdot operatorname{ctg} alpha = 1), so (operatorname{tg}^2 alpha cdot operatorname{ctg}^2 alpha = 1). That might help simplify some terms.Let me see:- (operatorname{tg}^2 alpha cdot operatorname{ctg}^2 alpha = 1)- (sin^2 alpha operatorname{ctg}^2 alpha = sin^2 alpha cdot frac{cos^2 alpha}{sin^2 alpha} = cos^2 alpha)- (operatorname{tg}^2 alpha cos^2 alpha = frac{sin^2 alpha}{cos^2 alpha} cdot cos^2 alpha = sin^2 alpha)So substituting these back into the expanded numerator:[sin^2 alpha cos^2 alpha - cos^2 alpha + sin^2 alpha + sin^2 alpha - 1 + operatorname{tg}^2 alpha + cos^2 alpha - operatorname{ctg}^2 alpha + 1]Simplify term by term:- (sin^2 alpha cos^2 alpha)- (- cos^2 alpha)- (+ sin^2 alpha)- (+ sin^2 alpha)- (- 1)- (+ operatorname{tg}^2 alpha)- (+ cos^2 alpha)- (- operatorname{ctg}^2 alpha)- (+ 1)Combine like terms:- (sin^2 alpha cos^2 alpha)- (- cos^2 alpha + cos^2 alpha = 0)- (+ sin^2 alpha + sin^2 alpha = 2sin^2 alpha)- (- 1 + 1 = 0)- (+ operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha)So the numerator simplifies to:[sin^2 alpha cos^2 alpha + 2sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha]Now, let's look at the denominator:[(cos^{2} alpha + operatorname{ctg}^2 alpha + 1)(sin^{2} alpha + operatorname{tg}^2 alpha - 1)]Expanding this:[cos^2 alpha sin^2 alpha + cos^2 alpha operatorname{tg}^2 alpha - cos^2 alpha + operatorname{ctg}^2 alpha sin^2 alpha + operatorname{ctg}^2 alpha operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha + sin^2 alpha + operatorname{tg}^2 alpha - 1]Again, using (operatorname{tg}^2 alpha cdot operatorname{ctg}^2 alpha = 1) and simplifying similar terms:- (cos^2 alpha sin^2 alpha)- (cos^2 alpha operatorname{tg}^2 alpha = cos^2 alpha cdot frac{sin^2 alpha}{cos^2 alpha} = sin^2 alpha)- (- cos^2 alpha)- (operatorname{ctg}^2 alpha sin^2 alpha = frac{cos^2 alpha}{sin^2 alpha} cdot sin^2 alpha = cos^2 alpha)- (operatorname{ctg}^2 alpha operatorname{tg}^2 alpha = 1)- (- operatorname{ctg}^2 alpha)- (+ sin^2 alpha)- (+ operatorname{tg}^2 alpha)- (- 1)Combine like terms:- (cos^2 alpha sin^2 alpha)- (+ sin^2 alpha - cos^2 alpha + cos^2 alpha = sin^2 alpha)- (+ 1 - 1 = 0)- (- operatorname{ctg}^2 alpha + operatorname{tg}^2 alpha)So the denominator simplifies to:[cos^2 alpha sin^2 alpha + sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha]Now, comparing the simplified numerator and denominator:- Numerator: (sin^2 alpha cos^2 alpha + 2sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha)- Denominator: (cos^2 alpha sin^2 alpha + sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha)Notice that the numerator has an extra (2sin^2 alpha) compared to the denominator. So, the fraction becomes:[frac{sin^2 alpha cos^2 alpha + 2sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha}{sin^2 alpha cos^2 alpha + sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha} = frac{text{Denominator} + sin^2 alpha}{text{Denominator}} = 1 + frac{sin^2 alpha}{text{Denominator}}]But this doesn't seem to simplify to 1 unless (sin^2 alpha = 0), which isn't generally true. I must have made a mistake somewhere.Let me go back and check my steps. Maybe I messed up the expansion or simplification.Looking back at the numerator expansion:[sin^2 alpha cos^2 alpha - cos^2 alpha + sin^2 alpha + sin^2 alpha - 1 + operatorname{tg}^2 alpha + cos^2 alpha - operatorname{ctg}^2 alpha + 1]Simplifying:- (sin^2 alpha cos^2 alpha)- (- cos^2 alpha + cos^2 alpha = 0)- (+ sin^2 alpha + sin^2 alpha = 2sin^2 alpha)- (- 1 + 1 = 0)- (+ operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha)That seems correct.Denominator expansion:[cos^2 alpha sin^2 alpha + sin^2 alpha - cos^2 alpha + cos^2 alpha + 1 - operatorname{ctg}^2 alpha + sin^2 alpha + operatorname{tg}^2 alpha - 1]Wait, I think I made a mistake here. Let me re-express the denominator expansion carefully.Denominator:[(cos^{2} alpha + operatorname{ctg}^2 alpha + 1)(sin^{2} alpha + operatorname{tg}^2 alpha - 1)]Expanding term by term:- (cos^2 alpha cdot sin^2 alpha)- (cos^2 alpha cdot operatorname{tg}^2 alpha)- (cos^2 alpha cdot (-1))- (operatorname{ctg}^2 alpha cdot sin^2 alpha)- (operatorname{ctg}^2 alpha cdot operatorname{tg}^2 alpha)- (operatorname{ctg}^2 alpha cdot (-1))- (1 cdot sin^2 alpha)- (1 cdot operatorname{tg}^2 alpha)- (1 cdot (-1))Now, simplifying each term:- (cos^2 alpha sin^2 alpha)- (cos^2 alpha cdot frac{sin^2 alpha}{cos^2 alpha} = sin^2 alpha)- (- cos^2 alpha)- (frac{cos^2 alpha}{sin^2 alpha} cdot sin^2 alpha = cos^2 alpha)- (frac{cos^2 alpha}{sin^2 alpha} cdot frac{sin^2 alpha}{cos^2 alpha} = 1)- (- frac{cos^2 alpha}{sin^2 alpha})- (+ sin^2 alpha)- (+ frac{sin^2 alpha}{cos^2 alpha})- (- 1)Now, combine like terms:- (cos^2 alpha sin^2 alpha)- (sin^2 alpha - cos^2 alpha + cos^2 alpha = sin^2 alpha)- (1 - 1 = 0)- (- frac{cos^2 alpha}{sin^2 alpha} + frac{sin^2 alpha}{cos^2 alpha})So the denominator simplifies to:[cos^2 alpha sin^2 alpha + sin^2 alpha - frac{cos^2 alpha}{sin^2 alpha} + frac{sin^2 alpha}{cos^2 alpha}]Wait, this is different from what I had before. So the denominator is:[cos^2 alpha sin^2 alpha + sin^2 alpha - operatorname{ctg}^2 alpha + operatorname{tg}^2 alpha]Which is the same as the numerator except the numerator has (2sin^2 alpha) instead of (sin^2 alpha).So the fraction becomes:[frac{cos^2 alpha sin^2 alpha + 2sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha}{cos^2 alpha sin^2 alpha + sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha}]This simplifies to:[frac{text{Denominator} + sin^2 alpha}{text{Denominator}} = 1 + frac{sin^2 alpha}{text{Denominator}}]But this doesn't equal 1 unless (sin^2 alpha = 0), which isn't generally true. So there must be an error in my approach.Maybe I should try a different strategy. Let's consider using substitution. Let me set (x = sin^2 alpha) and (y = cos^2 alpha). Then, since (x + y = 1), I can express everything in terms of (x) and (y).Given that:- (operatorname{tg}^2 alpha = frac{x}{y})- (operatorname{ctg}^2 alpha = frac{y}{x})Substituting into the original expression:Numerator:[(x + frac{x}{y} + 1)(y - frac{y}{x} + 1)]Denominator:[(y + frac{y}{x} + 1)(x + frac{x}{y} - 1)]Let's simplify each part.First part of numerator:[x + frac{x}{y} + 1 = x + frac{x}{y} + 1 = xleft(1 + frac{1}{y}right) + 1 = xleft(frac{y + 1}{y}right) + 1]But (x + y = 1), so (y = 1 - x). Substitute:[xleft(frac{1 - x + 1}{1 - x}right) + 1 = xleft(frac{2 - x}{1 - x}right) + 1]This seems complicated. Maybe instead, multiply numerator and denominator by (xy) to eliminate fractions.Multiply numerator and denominator by (xy):Numerator becomes:[(x + frac{x}{y} + 1)(y - frac{y}{x} + 1) times xy = (x y + x^2 + y)(y x - y^2 + x) times xy]Wait, this might not be the right approach. Let me try expanding without substitution.Alternatively, maybe I can factor the terms. Let's look at the numerator and denominator again.Numerator:[(x + frac{x}{y} + 1)(y - frac{y}{x} + 1)]Let me factor out (x) and (y) where possible:[x(1 + frac{1}{y} + frac{1}{x}) times y(1 - frac{1}{x} + frac{1}{y})]But this doesn't seem helpful.Wait, maybe I can write (x + frac{x}{y} + 1 = x(1 + frac{1}{y}) + 1 = x cdot frac{y + 1}{y} + 1). Since (x + y = 1), (y = 1 - x), so:[x cdot frac{1 - x + 1}{1 - x} + 1 = x cdot frac{2 - x}{1 - x} + 1]This is getting too convoluted. Maybe I should try specific values for (alpha) to test if the expression equals 1. If it does, then maybe it's always true.Let's choose (alpha = 45^circ), where (sin alpha = cos alpha = frac{sqrt{2}}{2}), (operatorname{tg} alpha = 1), (operatorname{ctg} alpha = 1).Substitute into the expression:Numerator:[left(left(frac{sqrt{2}}{2}right)^2 + 1^2 + 1right)left(left(frac{sqrt{2}}{2}right)^2 - 1^2 + 1right) = left(frac{1}{2} + 1 + 1right)left(frac{1}{2} - 1 + 1right) = left(frac{5}{2}right)left(frac{1}{2}right) = frac{5}{4}]Denominator:[left(left(frac{sqrt{2}}{2}right)^2 + 1^2 + 1right)left(left(frac{sqrt{2}}{2}right)^2 + 1^2 - 1right) = left(frac{1}{2} + 1 + 1right)left(frac{1}{2} + 1 - 1right) = left(frac{5}{2}right)left(frac{1}{2}right) = frac{5}{4}]So the fraction is (frac{5/4}{5/4} = 1). That works.Let's try another angle, say (alpha = 30^circ), where (sin alpha = frac{1}{2}), (cos alpha = frac{sqrt{3}}{2}), (operatorname{tg} alpha = frac{1}{sqrt{3}}), (operatorname{ctg} alpha = sqrt{3}).Numerator:[left(left(frac{1}{2}right)^2 + left(frac{1}{sqrt{3}}right)^2 + 1right)left(left(frac{sqrt{3}}{2}right)^2 - (sqrt{3})^2 + 1right)]Calculate each part:- (left(frac{1}{4} + frac{1}{3} + 1right) = frac{3}{12} + frac{4}{12} + frac{12}{12} = frac{19}{12})- (left(frac{3}{4} - 3 + 1right) = frac{3}{4} - 2 = -frac{5}{4})So numerator: (frac{19}{12} times -frac{5}{4} = -frac{95}{48})Denominator:[left(left(frac{sqrt{3}}{2}right)^2 + (sqrt{3})^2 + 1right)left(left(frac{1}{2}right)^2 + left(frac{1}{sqrt{3}}right)^2 - 1right)]Calculate each part:- (left(frac{3}{4} + 3 + 1right) = frac{3}{4} + 4 = frac{19}{4})- (left(frac{1}{4} + frac{1}{3} - 1right) = frac{3}{12} + frac{4}{12} - frac{12}{12} = -frac{5}{12})So denominator: (frac{19}{4} times -frac{5}{12} = -frac{95}{48})Thus, the fraction is (frac{-95/48}{-95/48} = 1). It works again.Let's try one more, (alpha = 60^circ), where (sin alpha = frac{sqrt{3}}{2}), (cos alpha = frac{1}{2}), (operatorname{tg} alpha = sqrt{3}), (operatorname{ctg} alpha = frac{1}{sqrt{3}}).Numerator:[left(left(frac{sqrt{3}}{2}right)^2 + (sqrt{3})^2 + 1right)left(left(frac{1}{2}right)^2 - left(frac{1}{sqrt{3}}right)^2 + 1right)]Calculate each part:- (left(frac{3}{4} + 3 + 1right) = frac{3}{4} + 4 = frac{19}{4})- (left(frac{1}{4} - frac{1}{3} + 1right) = frac{3}{12} - frac{4}{12} + frac{12}{12} = frac{11}{12})So numerator: (frac{19}{4} times frac{11}{12} = frac{209}{48})Denominator:[left(left(frac{1}{2}right)^2 + left(frac{1}{sqrt{3}}right)^2 + 1right)left(left(frac{sqrt{3}}{2}right)^2 + (sqrt{3})^2 - 1right)]Calculate each part:- (left(frac{1}{4} + frac{1}{3} + 1right) = frac{3}{12} + frac{4}{12} + frac{12}{12} = frac{19}{12})- (left(frac{3}{4} + 3 - 1right) = frac{3}{4} + 2 = frac{11}{4})So denominator: (frac{19}{12} times frac{11}{4} = frac{209}{48})Thus, the fraction is (frac{209/48}{209/48} = 1). It works again.Since testing specific angles works, it suggests that the expression is indeed equal to 1 for these values. However, to prove it generally, I need a more algebraic approach.Let me consider the original expression again:[frac{(sin^2 alpha + operatorname{tg}^2 alpha + 1)(cos^2 alpha - operatorname{ctg}^2 alpha + 1)}{(cos^2 alpha + operatorname{ctg}^2 alpha + 1)(sin^2 alpha + operatorname{tg}^2 alpha - 1)} = 1]Cross-multiplying gives:[(sin^2 alpha + operatorname{tg}^2 alpha + 1)(cos^2 alpha - operatorname{ctg}^2 alpha + 1) = (cos^2 alpha + operatorname{ctg}^2 alpha + 1)(sin^2 alpha + operatorname{tg}^2 alpha - 1)]Expanding both sides:Left side:[sin^2 alpha cos^2 alpha - sin^2 alpha operatorname{ctg}^2 alpha + sin^2 alpha + operatorname{tg}^2 alpha cos^2 alpha - operatorname{tg}^2 alpha operatorname{ctg}^2 alpha + operatorname{tg}^2 alpha + cos^2 alpha - operatorname{ctg}^2 alpha + 1]Right side:[cos^2 alpha sin^2 alpha + cos^2 alpha operatorname{tg}^2 alpha - cos^2 alpha + operatorname{ctg}^2 alpha sin^2 alpha + operatorname{ctg}^2 alpha operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha + sin^2 alpha + operatorname{tg}^2 alpha - 1]Now, let's simplify both sides using (operatorname{tg}^2 alpha cdot operatorname{ctg}^2 alpha = 1) and other identities.Left side simplifies to:[sin^2 alpha cos^2 alpha - cos^2 alpha + sin^2 alpha + sin^2 alpha - 1 + operatorname{tg}^2 alpha + cos^2 alpha - operatorname{ctg}^2 alpha + 1]Which simplifies further to:[sin^2 alpha cos^2 alpha + 2sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha]Right side simplifies to:[cos^2 alpha sin^2 alpha + sin^2 alpha - cos^2 alpha + cos^2 alpha + 1 - operatorname{ctg}^2 alpha + sin^2 alpha + operatorname{tg}^2 alpha - 1]Which simplifies further to:[cos^2 alpha sin^2 alpha + 2sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha]Comparing both sides, they are equal:[sin^2 alpha cos^2 alpha + 2sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha = cos^2 alpha sin^2 alpha + 2sin^2 alpha + operatorname{tg}^2 alpha - operatorname{ctg}^2 alpha]Thus, the original equation holds true, proving that the expression equals 1.

💡Okay, so I have this function f(α) that's given as a big fraction with sine, cosine, and tangent functions in both the numerator and the denominator. The problem has three parts: first, I need to simplify f(α); second, I have to find its value when α is in the third quadrant and given a specific cosine value; and third, I need to compute f(α) when α is -1860 degrees. Let me tackle each part step by step.Starting with part 1: Simplify f(α). The function is:f(α) = [sin(π - α) * cos(2π - α) * tan(-α + 3π/2) * tan(-α - π)] / sin(-π - α)Hmm, okay. I remember that trigonometric functions have various identities that can help simplify expressions. Let me recall some of them.First, for sine: sin(π - α) is equal to sin α. That's a standard identity. Similarly, cos(2π - α) is equal to cos α because cosine is an even function and has a period of 2π. So, cos(2π - α) = cos α.Next, looking at the tangent functions. The first one is tan(-α + 3π/2). Let me rewrite that as tan(3π/2 - α). I remember that tan(π/2 - α) is cot α, but this is 3π/2 - α. Let me think about the tangent function's periodicity and phase shifts. The tangent function has a period of π, so tan(3π/2 - α) is the same as tan(π/2 - α + π). Since tan is periodic with period π, tan(π/2 - α + π) = tan(π/2 - α). But tan(π/2 - α) is cot α. So, tan(3π/2 - α) = cot α.Wait, hold on. Let me verify that. The tangent function is periodic with period π, so tan(x + π) = tan x. So, tan(3π/2 - α) = tan(π/2 - α + π) = tan(π/2 - α). And tan(π/2 - α) is indeed cot α. So, tan(3π/2 - α) = cot α.Now, the second tangent function is tan(-α - π). Let me rewrite that as tan(- (α + π)). Since tangent is an odd function, tan(-x) = -tan x. So, tan(- (α + π)) = -tan(α + π). Again, using the periodicity of tangent, tan(α + π) = tan α. Therefore, tan(-α - π) = -tan α.So, putting it all together, the numerator becomes:sin(π - α) = sin αcos(2π - α) = cos αtan(-α + 3π/2) = cot αtan(-α - π) = -tan αSo, numerator: sin α * cos α * cot α * (-tan α)Denominator: sin(-π - α). Let's simplify that. sin(-π - α) = sin(- (π + α)). Since sine is an odd function, sin(-x) = -sin x. So, sin(- (π + α)) = -sin(π + α). Now, sin(π + α) is equal to -sin α because sine is negative in the third quadrant. So, sin(π + α) = -sin α, which makes sin(-π - α) = -(-sin α) = sin α.Wait, let me double-check that. sin(π + α) = -sin α, so sin(-π - α) = sin(-(π + α)) = -sin(π + α) = -(-sin α) = sin α. Yes, that's correct.So, denominator is sin α.Now, let's write out the entire expression:Numerator: sin α * cos α * cot α * (-tan α)Denominator: sin αSo, f(α) = [sin α * cos α * cot α * (-tan α)] / sin αI can cancel out sin α in numerator and denominator:f(α) = [cos α * cot α * (-tan α)]Now, let's simplify cot α and tan α. Remember that cot α = 1/tan α, so cot α * tan α = 1. So, cot α * (-tan α) = -1.Therefore, f(α) = cos α * (-1) = -cos α.Wait, hold on. Let me verify that step again. So, cot α is 1/tan α, so cot α * tan α is 1. But in the numerator, it's cot α * (-tan α). So, that's - (cot α * tan α) = -1. So, yes, f(α) = cos α * (-1) = -cos α.Wait, but let me make sure I didn't skip any steps. Let me write it out step by step:Numerator after cancellation: cos α * cot α * (-tan α)= cos α * (cot α * (-tan α))= cos α * (- (cot α * tan α))= cos α * (-1)= -cos αYes, that seems correct.So, part 1 simplifies to f(α) = -cos α.Wait, but in the original problem, the function is given as [sin(π - α) * cos(2π - α) * tan(-α + 3π/2) * tan(-α - π)] / sin(-π - α). After simplifying, I got f(α) = -cos α.But let me cross-verify this result. Maybe I made a mistake in simplifying the tangent functions.Let me go back to tan(-α + 3π/2). I said that's equal to cot α. Let me think again. tan(3π/2 - α). Let me consider the angle 3π/2 - α. 3π/2 is 270 degrees, so 270 - α. The tangent of 270 - α is equal to cot α? Wait, tan(270 - α) is actually equal to cot α, but with a sign change because tangent is positive in the third quadrant where 270 - α would be if α is in the first quadrant. Wait, maybe I need to consider the quadrant.Wait, no. Let me use the identity tan(π/2 - α) = cot α. But 3π/2 - α is π + (π/2 - α). So, tan(3π/2 - α) = tan(π + (π/2 - α)) = tan(π/2 - α) because tan has a period of π. But tan(π + x) = tan x. So, tan(3π/2 - α) = tan(π/2 - α) = cot α. So, that part seems correct.Similarly, tan(-α - π) = tan(- (α + π)) = -tan(α + π) = -tan α, since tan is periodic with period π.So, that seems correct.So, the numerator is sin α * cos α * cot α * (-tan α) = sin α * cos α * (1/tan α) * (-tan α) = sin α * cos α * (-1). Then, divided by sin α, which cancels out, leaving -cos α.So, f(α) = -cos α.Wait, but let me check with another approach. Maybe using co-function identities or other simplifications.Alternatively, perhaps I can use the fact that tan(-α + 3π/2) can be rewritten as tan(3π/2 - α). Let me recall that tan(3π/2 - α) = cot α, as I did before.Similarly, tan(-α - π) = tan(- (α + π)) = -tan(α + π) = -tan α.So, the numerator is sin(π - α) * cos(2π - α) * tan(3π/2 - α) * tan(-α - π) = sin α * cos α * cot α * (-tan α).Which is sin α * cos α * (cos α / sin α) * (-tan α) = cos α * cos α * (-tan α) = cos² α * (-tan α).But tan α = sin α / cos α, so cos² α * (-sin α / cos α) = -cos α * sin α.Then, the denominator is sin(-π - α) = sin(- (π + α)) = -sin(π + α) = -(-sin α) = sin α.So, f(α) = (-cos α * sin α) / sin α = -cos α.Yes, same result. So, f(α) simplifies to -cos α.Okay, so part 1 is done. f(α) = -cos α.Moving on to part 2: If α is an angle in the third quadrant, and cos(α - 3π/2) = 1/5, find the value of f(α).Since we have f(α) = -cos α, we just need to find cos α and then take its negative.Given that cos(α - 3π/2) = 1/5.Let me simplify cos(α - 3π/2). Using the cosine of a difference identity:cos(α - 3π/2) = cos α cos(3π/2) + sin α sin(3π/2)We know that cos(3π/2) = 0 and sin(3π/2) = -1.So, cos(α - 3π/2) = cos α * 0 + sin α * (-1) = -sin α.Therefore, -sin α = 1/5, which implies sin α = -1/5.Since α is in the third quadrant, both sine and cosine are negative. So, sin α = -1/5, and we need to find cos α.Using the Pythagorean identity: sin² α + cos² α = 1So, (-1/5)² + cos² α = 1 => 1/25 + cos² α = 1 => cos² α = 1 - 1/25 = 24/25Therefore, cos α = ±√(24/25) = ±(2√6)/5But since α is in the third quadrant, cosine is negative. So, cos α = -2√6/5Therefore, f(α) = -cos α = -(-2√6/5) = 2√6/5Wait, but hold on. Let me double-check.Wait, f(α) = -cos α, and cos α = -2√6/5, so f(α) = -(-2√6/5) = 2√6/5.But wait, in the problem statement, it's given that cos(α - 3π/2) = 1/5, which we found to be equal to -sin α = 1/5, so sin α = -1/5.Yes, that seems correct.But let me think again: cos(α - 3π/2) = 1/5.Alternatively, we can think of α - 3π/2 as an angle, and since cosine is positive 1/5, that angle is in the first or fourth quadrant. But since α is in the third quadrant, let's see where α - 3π/2 would be.If α is in the third quadrant, say between π and 3π/2, then α - 3π/2 would be between π - 3π/2 = -π/2 and 3π/2 - 3π/2 = 0. So, α - 3π/2 is between -π/2 and 0, which is the fourth quadrant. In the fourth quadrant, cosine is positive, which matches the given value of 1/5.So, cos(α - 3π/2) = 1/5, which is in the fourth quadrant, so sin(α - 3π/2) would be negative. But we used the identity to find that cos(α - 3π/2) = -sin α = 1/5, so sin α = -1/5, which is consistent with α being in the third quadrant.Therefore, sin α = -1/5, cos α = -2√6/5, so f(α) = -cos α = 2√6/5.Wait, but in the initial simplification, I got f(α) = -cos α, so plugging in cos α = -2√6/5, f(α) = -(-2√6/5) = 2√6/5.But let me check the calculation again:sin² α + cos² α = 1(-1/5)² + cos² α = 11/25 + cos² α = 1cos² α = 24/25cos α = -2√6/5 (since in third quadrant)Thus, f(α) = -cos α = -(-2√6/5) = 2√6/5.Yes, that seems correct.Wait, but in the problem statement, part 2 says "find the value of f(α)", which is 2√6/5.But let me check if I made a mistake in the sign somewhere.Wait, f(α) = -cos α, and cos α is negative, so f(α) is positive.Yes, that's correct.So, part 2 answer is 2√6/5.Moving on to part 3: If α = -1860 degrees, find the value of f(α).First, let's convert -1860 degrees to a positive angle by adding multiples of 360 degrees until we get an angle between 0 and 360 degrees.-1860 degrees divided by 360 is -5.166..., so adding 6*360 = 2160 degrees to -1860 gives:-1860 + 2160 = 300 degrees.So, -1860 degrees is coterminal with 300 degrees.Therefore, f(-1860°) = f(300°).Since f(α) = -cos α, f(300°) = -cos(300°).Cos(300°) is cos(360° - 60°) = cos(60°) = 0.5, but since 300° is in the fourth quadrant, cosine is positive, so cos(300°) = 0.5.Therefore, f(300°) = -0.5.Wait, but let me double-check.Wait, f(α) = -cos α, so f(300°) = -cos(300°).cos(300°) = cos(360° - 60°) = cos(60°) = 0.5, so f(300°) = -0.5.But wait, in the third quadrant, we had f(α) = 2√6/5, which is positive, but here, f(300°) is negative. That seems okay because 300° is in the fourth quadrant, where cosine is positive, so -cos(300°) is negative.But let me think again: f(α) = -cos α, so for α = 300°, f(α) = -cos(300°) = -0.5.Yes, that seems correct.Alternatively, let's compute f(-1860°) directly without converting to positive angle.f(-1860°) = -cos(-1860°).But cosine is an even function, so cos(-1860°) = cos(1860°).Now, 1860° divided by 360° is 5.166..., so 5*360° = 1800°, so 1860° - 1800° = 60°, so 1860° is coterminal with 60°, but in the positive direction. Wait, no: 1860° is 5*360° + 60°, so it's coterminal with 60°, but since we're dealing with negative angles, -1860° is coterminal with -1860° + 6*360° = 300°, as I did before.So, cos(-1860°) = cos(300°) = 0.5, so f(-1860°) = -0.5.Yes, that's consistent.Therefore, part 3 answer is -0.5, which is -1/2.Wait, but in the initial simplification, I got f(α) = -cos α, so for α = -1860°, f(α) = -cos(-1860°) = -cos(300°) = -0.5.Yes, correct.So, summarizing:1. f(α) simplifies to -cos α.2. When α is in the third quadrant and cos(α - 3π/2) = 1/5, f(α) = 2√6/5.3. When α = -1860°, f(α) = -1/2.Wait, but let me check part 2 again. I got f(α) = 2√6/5, but let me think if that's correct.Given that f(α) = -cos α, and cos α = -2√6/5, so f(α) = -(-2√6/5) = 2√6/5. Yes, that's correct.Alternatively, if I didn't simplify f(α) to -cos α, but instead used the original expression, would I get the same result?Let me try that.Given f(α) = [sin(π - α) * cos(2π - α) * tan(-α + 3π/2) * tan(-α - π)] / sin(-π - α)We can plug in the values we found: sin α = -1/5, cos α = -2√6/5.First, sin(π - α) = sin α = -1/5.cos(2π - α) = cos α = -2√6/5.tan(-α + 3π/2) = tan(3π/2 - α) = cot α = cos α / sin α = (-2√6/5) / (-1/5) = 2√6.tan(-α - π) = tan(- (α + π)) = -tan(α + π) = -tan α. Since tan α = sin α / cos α = (-1/5) / (-2√6/5) = 1/(2√6). So, tan(-α - π) = -1/(2√6).Denominator: sin(-π - α) = sin(- (π + α)) = -sin(π + α) = -(-sin α) = sin α = -1/5.So, putting it all together:Numerator: (-1/5) * (-2√6/5) * 2√6 * (-1/(2√6)).Let's compute step by step:First, multiply (-1/5) * (-2√6/5) = (2√6)/25.Then, multiply by 2√6: (2√6)/25 * 2√6 = (4 * 6)/25 = 24/25.Then, multiply by (-1/(2√6)): 24/25 * (-1/(2√6)) = -24/(50√6) = -12/(25√6).Rationalizing the denominator: -12/(25√6) * (√6/√6) = -12√6/(25*6) = -12√6/150 = -2√6/25.Denominator: -1/5.So, f(α) = (-2√6/25) / (-1/5) = (-2√6/25) * (-5/1) = (10√6)/25 = 2√6/5.Yes, same result as before. So, f(α) = 2√6/5.Therefore, my initial simplification was correct.Similarly, for part 3, using the original expression:f(-1860°) = [sin(π - (-1860°)) * cos(2π - (-1860°)) * tan(-(-1860°) + 3π/2) * tan(-(-1860°) - π)] / sin(-π - (-1860°))But this seems complicated, so it's better to use the simplified form f(α) = -cos α, which gives f(-1860°) = -cos(300°) = -0.5.So, all parts are consistent.Therefore, the answers are:1. f(α) = -cos α2. f(α) = 2√6/53. f(α) = -1/2But wait, in the initial problem statement, part 2 says "find the value of f(α)", and part 3 says "find the value of f(α)". So, I think I have all the answers.But let me just make sure I didn't make any calculation errors.In part 2, sin α = -1/5, so cos α = -2√6/5, so f(α) = -cos α = 2√6/5.Yes.In part 3, α = -1860°, which is coterminal with 300°, so f(α) = -cos(300°) = -0.5.Yes.So, I think I'm confident with these answers.

💡Okay, so I have this geometry problem here about triangle ABC. Let me try to visualize it first. Triangle ABC has sides a, b, and c opposite angles A, B, and C respectively. There's an angle bisector from angle A meeting BC at point D. Then, E is the midpoint of AD. They define x as CD and y as BD. The question is asking for the correct proportion involving x, y, and the side lengths a, b, c.Alright, let's break this down step by step. First, since AD is the angle bisector of angle A, I remember there's a theorem called the Angle Bisector Theorem that might help here. The theorem states that an angle bisector in a triangle divides the opposite side into segments proportional to the adjacent sides. So, in this case, AD bisects angle A, so it should divide BC into segments BD and DC such that BD/DC = AB/AC.Given that AB is opposite angle C, which is side c, and AC is opposite angle B, which is side b. So, BD/DC = c/b. Let me write that down:BD/DC = c/bBut BD is y and DC is x, so substituting those in:y/x = c/bWhich can be rearranged as:y = (c/b)xAlso, since BD + DC = BC, which is side a, we have:y + x = aSubstituting y from the previous equation:(c/b)x + x = aFactor out x:x(c/b + 1) = aCombine the terms:x(c + b)/b = aSo, solving for x:x = (a*b)/(b + c)Similarly, since y = (c/b)x, substituting x:y = (c/b)*(a*b)/(b + c) = (a*c)/(b + c)So, we have expressions for both x and y in terms of a, b, c.Now, the options given are:(A) x/a = b/(2a)(B) y/b = c/(2a)(C) DE/a = 1/2(D) x/c = a/(b + c)(E) y/a = c/(b + c)Let me check each option one by one.Starting with option (A): x/a = b/(2a)From above, x = (a*b)/(b + c). So, x/a = b/(b + c). Comparing to option (A), which is x/a = b/(2a). Wait, that doesn't seem right because x/a is b/(b + c), not b/(2a). So, (A) is incorrect.Moving on to option (B): y/b = c/(2a)We have y = (a*c)/(b + c). So, y/b = (a*c)/(b*(b + c)). Comparing to option (B), which is c/(2a). These are not the same. So, (B) is incorrect.Option (C): DE/a = 1/2Hmm, DE is a segment from D to E, where E is the midpoint of AD. So, DE is half of AD. But the question is about DE in relation to a. I need to find the length of AD first.Wait, maybe I can use the formula for the length of an angle bisector. The formula for the length of angle bisector AD is:AD = (2*b*c*cos(A/2))/(b + c)But I don't know angle A, so maybe that's not helpful. Alternatively, maybe I can use Stewart's theorem to find AD.Stewart's theorem states that in triangle ABC, with a cevian AD, where BD = y, DC = x, and AD = d, then:b^2*y + c^2*x = a*d^2 + x*y*aWe already have expressions for x and y in terms of a, b, c. Let me plug those in.x = (a*b)/(b + c)y = (a*c)/(b + c)So, substituting into Stewart's theorem:b^2*(a*c/(b + c)) + c^2*(a*b/(b + c)) = a*d^2 + (a*b/(b + c))*(a*c/(b + c))*aSimplify the left side:(a*b^2*c + a*c^2*b)/(b + c) = a*d^2 + (a^2*b*c)/(b + c)^2 *aFactor out a*b*c from the numerator on the left:a*b*c*(b + c)/(b + c) = a*d^2 + (a^3*b*c)/(b + c)^2Simplify:a*b*c = a*d^2 + (a^3*b*c)/(b + c)^2Divide both sides by a:b*c = d^2 + (a^2*b*c)/(b + c)^2Now, solve for d^2:d^2 = b*c - (a^2*b*c)/(b + c)^2Factor out b*c:d^2 = b*c*(1 - a^2/(b + c)^2)So,d^2 = b*c*((b + c)^2 - a^2)/(b + c)^2Factor the numerator:(b + c)^2 - a^2 = (b + c - a)(b + c + a)So,d^2 = b*c*(b + c - a)(b + c + a)/(b + c)^2Take square root:d = sqrt[ b*c*(b + c - a)(b + c + a) ] / (b + c)Hmm, this is getting complicated. Maybe there's a simpler way to find DE.Since E is the midpoint of AD, DE is half of AD. So, DE = AD/2.But from the above, AD is expressed in terms of a, b, c. So, DE = (1/2)*AD.But the question is about DE/a. So, DE/a = (AD)/(2a).But unless AD is equal to a, which I don't think it is, DE/a is not necessarily 1/2. So, maybe option (C) is not correct. Let me check with specific values.Suppose triangle ABC is such that a = b = c, an equilateral triangle. Then, angle bisector AD is also a median and altitude. In that case, AD = (sqrt(3)/2)*a. So, DE = AD/2 = (sqrt(3)/4)*a. Then, DE/a = sqrt(3)/4 ≈ 0.433, which is not 1/2. So, option (C) is incorrect in this case.Wait, but in the problem, E is the midpoint of AD, so DE = AD/2. But unless AD = a, DE/a would be 1/2. But in general, AD is not equal to a. So, option (C) is not necessarily correct.Moving on to option (D): x/c = a/(b + c)From earlier, x = (a*b)/(b + c). So, x/c = (a*b)/(c*(b + c)) = (a*b)/(c*(b + c)). Comparing to option (D), which is a/(b + c). These are not the same unless b = c. So, (D) is incorrect.Option (E): y/a = c/(b + c)From earlier, y = (a*c)/(b + c). So, y/a = c/(b + c). That's exactly option (E). So, (E) is correct.Wait, but earlier I thought (C) might be correct, but with the specific example, it wasn't. So, (E) seems correct.Let me double-check. If y = (a*c)/(b + c), then y/a = c/(b + c). Yes, that's exactly option (E). So, (E) is correct.But wait, the initial problem statement says "the correct proportion involving x, y, and side lengths a, b, c." So, (E) is correct, but what about (C)? In my earlier thought process, I thought (C) was incorrect because DE/a is not necessarily 1/2. But maybe I made a mistake.Wait, let's think again. E is the midpoint of AD, so DE = AD/2. So, DE = (1/2)*AD. But unless AD = a, DE/a is not 1/2. So, unless AD = a, which is not generally true, DE/a is not 1/2. So, (C) is incorrect.Therefore, the correct answer is (E).Wait, but in the initial problem, the options are (A) through (E), and (E) is y/a = c/(b + c), which we've shown is correct. So, the answer should be (E).But wait, in the initial problem, the user wrote the same problem twice, and in the second time, they provided a thought process that concluded (C). But in my analysis, (E) is correct. Maybe the initial thought process was incorrect.Alternatively, perhaps I made a mistake. Let me check again.From the Angle Bisector Theorem, BD/DC = AB/AC = c/b. So, y/x = c/b, hence y = (c/b)x. Also, y + x = a, so x = (a*b)/(b + c), y = (a*c)/(b + c). Therefore, y/a = c/(b + c), which is option (E). So, (E) is correct.As for option (C), DE is half of AD, but AD is not necessarily equal to a, so DE/a is not necessarily 1/2. Therefore, (C) is incorrect.So, the correct answer is (E).But wait, in the initial problem, the user wrote the same problem twice, and in the second time, they provided a thought process that concluded (C). But in my analysis, (E) is correct. Maybe the initial thought process was incorrect.Alternatively, perhaps I made a mistake. Let me check again.From the Angle Bisector Theorem, BD/DC = AB/AC = c/b. So, y/x = c/b, hence y = (c/b)x. Also, y + x = a, so x = (a*b)/(b + c), y = (a*c)/(b + c). Therefore, y/a = c/(b + c), which is option (E). So, (E) is correct.As for option (C), DE is half of AD, but AD is not necessarily equal to a, so DE/a is not necessarily 1/2. Therefore, (C) is incorrect.So, the correct answer is (E).

💡Larry starts with the number 4 on his thumb. He applies the function ( f ) repeatedly, writing the result on each subsequent finger. The function ( f ) maps the numbers as follows: ( f(4) = 7 ), ( f(7) = 8 ), ( f(8) = 3 ), ( f(3) = 5 ), and ( f(5) = 4 ). This creates a cycle of 5 numbers: 4, 7, 8, 3, 5.To determine the number on the twelfth finger, we need to see where 12 falls in the cycle. Since the cycle repeats every 5 numbers, we calculate the remainder when 12 is divided by 5. 12 divided by 5 is 2 with a remainder of 2. This means after 2 full cycles (10 numbers), we are at the second position in the cycle. Looking at the cycle:1. 42. 73. 84. 35. 5The second position is 7. Therefore, the number on the twelfth finger is 7.

💡Okay, so I've got this system of linear equations here, and I need to solve it by substitution. The equations are:1. ( y = x - 1 )2. ( x - 2y = 7 )Alright, substitution method. That means I need to plug the expression for ( y ) from the first equation into the second equation. Let me make sure I understand what substitution is. It's a way to solve systems of equations by replacing one variable with an expression involving the other variable. So, since the first equation already gives me ( y ) in terms of ( x ), I can use that to eliminate ( y ) from the second equation.Let me write down the second equation again:( x - 2y = 7 )Now, I need to substitute ( y ) with ( x - 1 ) from the first equation. So, wherever I see ( y ) in the second equation, I'll replace it with ( x - 1 ). Let's do that step by step.Starting with the second equation:( x - 2y = 7 )Substitute ( y = x - 1 ):( x - 2(x - 1) = 7 )Okay, so now I have ( x - 2(x - 1) = 7 ). I need to simplify this equation to solve for ( x ). Let me handle the parentheses first. The term ( -2(x - 1) ) needs to be distributed. That means I'll multiply both ( x ) and ( -1 ) by ( -2 ).So, distributing the ( -2 ):( x - 2x + 2 = 7 )Wait, let me double-check that. Multiplying ( x ) by ( -2 ) gives ( -2x ), and multiplying ( -1 ) by ( -2 ) gives ( +2 ). So, yes, that's correct: ( x - 2x + 2 = 7 ).Now, let's combine like terms. The ( x ) and ( -2x ) can be combined. ( x ) is the same as ( 1x ), so:( 1x - 2x = -1x )So, combining those:( -x + 2 = 7 )Hmm, that seems a bit different from the options given. Let me check the options again:A: ( x - 2x + 1 = 7 )B: ( x - 2x - 2 = 7 )C: ( x + x + 1 = 7 )D: ( x - 2x + 2 = 7 )Wait, in my simplification, I ended up with ( -x + 2 = 7 ), which is equivalent to ( x - 2x + 2 = 7 ) if I rearrange it. Let me see:Starting from ( x - 2(x - 1) = 7 ), distributing gives ( x - 2x + 2 = 7 ). So, that's exactly option D. So, my initial simplification was correct, and it matches option D.But just to be thorough, let me go through each option and see why the others might not be correct.Option A: ( x - 2x + 1 = 7 )If I substitute ( y = x - 1 ) into equation 2, I should have ( x - 2(x - 1) = 7 ). Distributing the -2 gives ( x - 2x + 2 = 7 ), not +1. So, option A is incorrect because the constant term is wrong.Option B: ( x - 2x - 2 = 7 )Again, substituting ( y = x - 1 ) into equation 2 gives ( x - 2(x - 1) = 7 ). Distributing the -2 should give +2, not -2. So, option B is also incorrect.Option C: ( x + x + 1 = 7 )This seems off because when substituting, we have a negative coefficient for y, which should lead to a negative term when distributing. Here, it's adding x and x, which would be ( 2x ), but that doesn't align with the substitution. So, option C is incorrect.Option D: ( x - 2x + 2 = 7 )As I calculated earlier, this is exactly what we get after substituting and distributing. So, option D is correct.Just to make sure, let me solve the equation ( x - 2x + 2 = 7 ) to see if it makes sense.Combine like terms:( -x + 2 = 7 )Subtract 2 from both sides:( -x = 5 )Multiply both sides by -1:( x = -5 )Now, plug this back into equation 1 to find y:( y = x - 1 = -5 - 1 = -6 )So, the solution is ( x = -5 ) and ( y = -6 ). Let me check this in equation 2:( x - 2y = -5 - 2(-6) = -5 + 12 = 7 )Yes, that works. So, the substitution method led us to the correct solution, and the equation we obtained after substitution was indeed option D.I think I've covered all the steps and checked each option carefully. It's important to distribute the negative sign correctly when substituting, as a common mistake could be to forget the negative sign or misapply it, leading to an incorrect equation. But in this case, everything checks out, and option D is the right choice.