2025 SSMO Speed Round Problems/Problem 10

Problem

Let $p$ be a quadratic with a positive leading coefficient, and let $r$ be a real number satisfying $r < 1 < \tfrac{5}{2r} < 5$ . Given that $p(p(x)) = x$ for $x \in \{ r,1, \tfrac{5}{2r} , 5\}$ , find $p(12)$ .

Solution

Note that any fixed point of $p(x)$ is also a fixed point of $p(p(x))$ . From the condition that $p(p(x)) = x$ for $x\in \{r,1,\tfrac{5}{2r},5\}$ , we can conclude two things:

The fixed points of $p$ are a subset of $\{r,1,\tfrac{5}{2r},5\}$ ;
$p$ has $0$ , $2$ , or $4$ fixed points.

The main claim is that $p$ has exactly $2$ fixed points, and that these fixed points are $1$ and $5$ .

First, we prove that $p$ has exactly two fixed points. It is impossible that $p$ has $4$ fixed points because that would imply $\deg p \ge 4$ . It is also impossible that $p$ has $0$ fixed points. For the sake of contradiction, suppose that this is the case. Then, the parabola $y=p(x)$ does not intersect the line $y=x$ , and since $p$ has a positive leading coefficient, the parabola $y=p(x)$ lies entirely above the line $y=x$ . Since $p$ has no fixed points, the points $(1, p(1))$ and $(p(1), p(p(1))) = (p(1),1)$ are distinct and lie on opposite sides of the line $y=x$ . However, both of these points lie on the parabola $y=p(x)$ , which is a contradiction. Thus, $p$ has exactly $2$ fixed points.

$[asy] unitsize(0.27cm); import contour; real f(real x, real y) { return 0.3x^2-y; } guide[][] thegraphs = contour(f, a=(-20,-20), b=(20,20), new real[] {0}); draw(thegraphs[0],blue); draw((-9,-5)--(12,16),Arrows); pair A,B,C,D; B=(-2.3472,1.6528); D=(5.68053,9.68053); A=(-3.92744,4.62744); C=(0.59411,0.10589); draw((-2.3472,-7)--(-2.3472,25),red+dashed); draw((5.68053,-7)--(5.68053,25),red+dashed); dot(A); dot(B); dot(C); dot(D); label("$y=x$",(-9,-2.5)); label("$y=p(x)$",(-11,18)); label("$(c,d)$",(-5.82744,4.62744)); label("$(a,a)$",(-4.4472,1.7528)); label("$(d,c)$",(0.59411,-1.10589)); label("$(b,b)$",(7.28053,8.78053)); label("$x=a$",(-4.8472,23)); label("$x=b$",(3.28053,23.2)); [/asy]$

Now, we show that the fixed points of $p$ are $1$ and $5$ . Let $(a,b,c,d)$ be a permutation of $(r,1,\tfrac{5}{2r},5)$ , and suppose that $p(a)=a$ , $p(b)=b$ , $p(c)=d$ , and $p(d)=c$ . WLOG, suppose $a<b$ . We first note that a point on $y=p(x)$ lies under the line $y=x$ if and only if it lies between the lines $x = a$ and $x=b$ . WLOG, suppose $a<d<b$ . Since $(d,c)$ lies under the line $y=x$ , we have $c<d$ . Notice that $(c,d)$ cannot lie between $x=a$ and $x=b$ because exactly one of the points $(c,d)$ and $(d,c)$ lies below $y=x$ . Since $c<d$ , it follows that $c<a<d<b$ , so $(a,b,c,d) = (1,5,r,\tfrac{5}{2r})$ . Thus, $p(1)=1$ and $p(5)=5$ , as desired.

All that is left to do is to determine the leading coefficient of $p$ , which we call $m$ . We now know that $p(x) = m(x-1)(x-5) + x$ . For convenience, replace $r$ and $\tfrac{5}{2r}$ by $s$ and $t$ , respectively, satisfying $st=\tfrac{5}{2}$ . We have\begin{align*} t &= ms^2 - (6m-1)s + 5m \\ s &= mt^2 - (6m-1)t + 5m.\end{align*}Adding these two equations gives us $6m(s+t) = m(s^2+t^2) + 10m$ . Dividing through by $m$ and using $s^2+t^2 = (s+t)^2 - 2st = (s+t)^2 - 5$ , we find that $s+t=1$ or $s+t=5$ . Since $t>1$ and $s>0$ , we must have $s+t=5$ . Then, subtracting the second centered equation from the first and rearranging gives us $t-s + m(t^2-s^2) = (6m-1)(t-s)$ . As $s\ne t$ , we may divide through by $t-s$ to obtain $1 + m(s+t) = 1+5m = 6m-1$ . Finally, we have $m=2$ , which gives $p(x) = 2x^2 - 11x + 10$ . The answer is $2(12)^2 - 11(12)+10 = \boxed{166}$ .

~Sedro

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2025 SSMO Speed Round Problems/Problem 10

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