That's because the polynomial is continuous, so if it is 1 for every s in the neighbourhood of p, q, r, then it follows that it is 1 also at p, q, r. Of course the fractions are not defined at p, q, r, but if you were to take the limit as s approaches p, q, r respectively from both sides, you would find that that limit exists.
First of all, the question said that s can't be p, q, or r because it had to prevent dividing by zero. Stewart's explanation is correct. But to be more simple, you can think that the two expressions are independent (at 02:00, the original one and the handwritten one). As we've cleared the denominators, we only have the polynomial with variable 's'. So from this point, we don't have to care about dividing by zero - and that means everything can be the value of s! The polynomial itself is a identity so we can try various methods like unfolding them all, or just substituting beautiful values (you know, results of these are the same. Always). It does not matter because we are only to find A, B, and C. If you are reluctant, try thinking that the variable is x, instead of s.
It's harsh because it forces you to use continuity, which isn't something most people know about at this level. So if you're quite a rigorous and precise mathematician you might struggle with this problem thinking that you can't plug in p,q,r.
This technique for 'partial fraction decomposition' by substituting in the forbidden values is actually a completely standard technique, and in a calculus class, this would just be an exercise. The only reason it becomes an AMC problem is because at this level most people won't have seen this technique before, but for those who had seen it before, it was basically a free 6 points.
As for the "coefficient" thing, if you mean the relationships between the roots and coefficients of the cubic (or any polynomial), which are formally known as Vieta's Formulae, these are expected knowledge for any mathematical competition at this level.
However, once we've cleared the denominators, what we have is a polynomial identity. The left-hand side is identically equal to the right-hand side for all values of s except p, q, and r, but that's still infinitely many possible values of s, and two polynomials are equal for infinitely many values of the variable only when they are identically equal for all possible values of the variable. Thus, the two polynomials must be equal for all values of s, including s = p, q, and r. (This technique for 'partial fraction decomposition' by substituting in the forbidden values is a standard method, and so is taught in calculus classes - but I guess you're not expected to have seen it before as an AMC 10 student).
You can think about it like this: This will be true for any value but p, q, r, so you can substitute in a number that gets closer and closer to p, q, r, like p+0.000001. This is the concept of limits in calculus. Since this equation is true as s gets really close to p, doesn't it make sense it holds for p too?
No it doesn't. Only for the reciprocal of the polynomial to stop us from dividing by zero, try doing the same for a normal quadratic you will see the same
But the question says that s can't be p , Q and r , I dont understand how he substituted s=p
That's because the polynomial is continuous, so if it is 1 for every s in the neighbourhood of p, q, r, then it follows that it is 1 also at p, q, r. Of course the fractions are not defined at p, q, r, but if you were to take the limit as s approaches p, q, r respectively from both sides, you would find that that limit exists.
First of all, the question said that s can't be p, q, or r because it had to prevent dividing by zero.
Stewart's explanation is correct. But to be more simple, you can think that the two expressions are independent (at 02:00, the original one and the handwritten one). As we've cleared the denominators, we only have the polynomial with variable 's'. So from this point, we don't have to care about dividing by zero - and that means everything can be the value of s!
The polynomial itself is a identity so we can try various methods like unfolding them all, or just substituting beautiful values (you know, results of these are the same. Always). It does not matter because we are only to find A, B, and C. If you are reluctant, try thinking that the variable is x, instead of s.
the function is continuous, so as s approaches p, it's 1, but it's continuous, so s=p also gives 1
i took basically the same soln but i didnt substitute s=p bc i thought it was restricted, had to monkey around with the variables a lot
Woah... that's such a cool prob :) It has a nice, clean sol too.
real aops user jndd? or an imposta???????
@@name-np4gr lol yea, the real aops user. not an impasta :)
beautiful solution
Intermediate algebra by aops makes this problem very free
This is a legit routine AP calculus BC problem
I think this is a harsh one because I couldn't think of the coefficient thing TT
It's harsh because it forces you to use continuity, which isn't something most people know about at this level. So if you're quite a rigorous and precise mathematician you might struggle with this problem thinking that you can't plug in p,q,r.
This technique for 'partial fraction decomposition' by substituting in the forbidden values is actually a completely standard technique, and in a calculus class, this would just be an exercise. The only reason it becomes an AMC problem is because at this level most people won't have seen this technique before, but for those who had seen it before, it was basically a free 6 points.
As for the "coefficient" thing, if you mean the relationships between the roots and coefficients of the cubic (or any polynomial), which are formally known as Vieta's Formulae, these are expected knowledge for any mathematical competition at this level.
@@amritlohia8240 Thanks for your explanation!
@@alephnull4044 Exactly what I was concerning, thanks!
This questions put to our semifinal test! Uh!
What web can I find out these problem
Aops wiki
Why p=s, s=r and s=q ?
It gets rid of some terms and makes it more simple
But it says s cannot be equal to p, q, r
However, once we've cleared the denominators, what we have is a polynomial identity. The left-hand side is identically equal to the right-hand side for all values of s except p, q, and r, but that's still infinitely many possible values of s, and two polynomials are equal for infinitely many values of the variable only when they are identically equal for all possible values of the variable. Thus, the two polynomials must be equal for all values of s, including s = p, q, and r. (This technique for 'partial fraction decomposition' by substituting in the forbidden values is a standard method, and so is taught in calculus classes - but I guess you're not expected to have seen it before as an AMC 10 student).
You can think about it like this: This will be true for any value but p, q, r, so you can substitute in a number that gets closer and closer to p, q, r, like p+0.000001. This is the concept of limits in calculus. Since this equation is true as s gets really close to p, doesn't it make sense it holds for p too?
No it doesn't. Only for the reciprocal of the polynomial to stop us from dividing by zero, try doing the same for a normal quadratic you will see the same
the best math problems are always explained by quirkiest ppl 😂
What's your expression
Bro nice board as well the question.
srr but i dont know why s^3-22s^2+80s-67=(s-p)*(s-q)*(s-r)
Since p,q, and r are the roots of x^3-22x^2+80x-67, you get (x-p)(x-q)(x-r)=x^3-22x^2+80x-67. Now just replace x with s.
Check out Vieta's formulas. That's where this idea comes in. :D
alg 1 review for u :D
first!
nuh uh lil bro
🤡