Source: 2017 Canadian Open Math Challenge, Problem B4 Numbers $a$, $b$ and $c$ form an arithmetic sequence if $b - a = c - b$. Let $a$, $b$, $c$ be positive integers forming an arithmetic sequence with $a < b < c$. Let $f(x) = ax2 + bx + c$. Two distinct real numbers $r$ and $s$ satisfy $f(r) = s$ and $f(s) = r$. If $rs = 2017$, determine the smallest possible value of $a$.
Problem
Source:
Tags: Comc, 2017 COMC
Immanuel Bonfils
14.10.2018 05:14
I'm not sure I got the question, but seems to be $a=1$ since the AP are positive?
the common difference should be
$2018$
cooljoseph
14.10.2018 05:37
Note that $f(x)+x-(r+s)$ has two solutions at $r$ and $s$. As this is a quadratic, those are the only solutions it can have. So $f(x)=k(x-r)(x-s)-x+(r+s)$ for some $k$. As $rs=2017$, this simplifies to $f(x)=kx^2-k(r+s)x+2017k+(r+s)\longrightarrow 2018k+r+s = 2k(r+s)\longrightarrow k = \frac{r+s}{2(r+s-2018)}$ This means we must have $2(r+s-2018)|r+s \longrightarrow (r+s-2018)|2018$ To minimize $a=k$, we want to maximize $r+s$, so the answer is when $r+s=4036\longrightarrow a=k=\boxed{1}$ We then can double check to ensure that this yields real values for $r$ and $s$ (which it does).
toN_yhW
30.09.2019 09:34
cooljoseph wrote: Note that $f(x)+x-(r+s)$ has two solutions at $r$ and $s$. As this is a quadratic, those are the only solutions it can have. So $f(x)=k(x-r)(x-s)-x+(r+s)$ for some $k$. As $rs=2017$, this simplifies to $f(x)=kx^2-k(r+s)x+2017k+(r+s)\longrightarrow 2018k+r+s = 2k(r+s)\longrightarrow k = \frac{r+s}{2(r+s-2018)}$ This means we must have $2(r+s-2018)|r+s \longrightarrow (r+s-2018)|2018$ To minimize $a=k$, we want to maximize $r+s$, so the answer is when $r+s=4036\longrightarrow a=k=\boxed{1}$ We then can double check to ensure that this yields real values for $r$ and $s$ (which it does). I checked the official solution and this is wrong, a=9.
gamerrk1004
30.09.2019 10:49
toN_yhW wrote: cooljoseph wrote: Note that $f(x)+x-(r+s)$ has two solutions at $r$ and $s$. As this is a quadratic, those are the only solutions it can have. So $f(x)=k(x-r)(x-s)-x+(r+s)$ for some $k$. As $rs=2017$, this simplifies to $f(x)=kx^2-k(r+s)x+2017k+(r+s)\longrightarrow 2018k+r+s = 2k(r+s)\longrightarrow k = \frac{r+s}{2(r+s-2018)}$ This means we must have $2(r+s-2018)|r+s \longrightarrow (r+s-2018)|2018$ To minimize $a=k$, we want to maximize $r+s$, so the answer is when $r+s=4036\longrightarrow a=k=\boxed{1}$ We then can double check to ensure that this yields real values for $r$ and $s$ (which it does). I checked the official solution and this is wrong, a=9. can you please share the link for the official solutions ??