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Difference between revisions of "2025 SSMO Speed Round Problems/Problem 10"

(Created page with "==Problem== Let <math>p</math> be a quadratic with a positive leading coefficient, and let <math>r</math> be a real number satisfying <math>r < 1 < \tfrac{5}{2r} < 5</math>....")
 
(→‎Solution)
 
Line 4: Line 4:
  
 
==Solution==
 
==Solution==
 +
 +
Note that any fixed point of <math>p(x)</math> is also a fixed point of <math>p(p(x))</math>. From the condition that <math>p(p(x)) = x</math> for <math>x\in \{r,1,\tfrac{5}{2r},5\}</math>, we can conclude two things:
 +
<UL>
 +
<LI>The fixed points of <math>p</math> are a subset of <math>\{r,1,\tfrac{5}{2r},5\}</math>;</LI>
 +
<LI><math>p</math> has <math>0</math>, <math>2</math>, or <math>4</math> fixed points.</LI>
 +
</UL>
 +
The main claim is that <math>p</math> has exactly <math>2</math> fixed points, and that these fixed points are <math>1</math> and <math>5</math>.
 +
 +
First, we prove that <math>p</math> has exactly two fixed points. It is impossible that <math>p</math> has <math>4</math> fixed points because that would imply <math>\deg p \ge 4</math>. It is also impossible that <math>p</math> has <math>0</math> fixed points. For the sake of contradiction, suppose that this is the case. Then, the parabola <math>y=p(x)</math> does not intersect the line <math>y=x</math>, and since <math>p</math> has a positive leading coefficient, the parabola <math>y=p(x)</math> lies entirely above the line <math>y=x</math>. Since <math>p</math> has no fixed points, the points <math>(1, p(1))</math> and <math>(p(1), p(p(1))) = (p(1),1)</math> are distinct and lie on opposite sides of the line <math>y=x</math>. However, both of these points lie on the parabola <math>y=p(x)</math>, which is a contradiction. Thus, <math>p</math> has exactly <math>2</math> fixed points.
 +
 +
<asy>
 +
unitsize(0.27cm);
 +
 +
import contour;
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real f(real x, real y) { return 0.3x^2-y; }
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guide[][] thegraphs = contour(f,
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a=(-20,-20), b=(20,20), new real[] {0});
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draw(thegraphs[0],blue);
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draw((-9,-5)--(12,16),Arrows);
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pair A,B,C,D;
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B=(-2.3472,1.6528);
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D=(5.68053,9.68053);
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A=(-3.92744,4.62744);
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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);
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dot(B);
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dot(C);
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dot(D);
 +
 +
label("$y=x$",(-9,-2.5));
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label("$y=p(x)$",(-11,18));
 +
 +
label("$(c,d)$",(-5.82744,4.62744));
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label("$(a,a)$",(-4.4472,1.7528));
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label("$(d,c)$",(0.59411,-1.10589));
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label("$(b,b)$",(7.28053,8.78053));
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 +
label("$x=a$",(-4.8472,23));
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label("$x=b$",(3.28053,23.2));
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</asy>
 +
 +
Now, we show that the fixed points of <math>p</math> are <math>1</math> and <math>5</math>. Let <math>(a,b,c,d)</math> be a permutation of <math>(r,1,\tfrac{5}{2r},5)</math>, and suppose that <math>p(a)=a</math>, <math>p(b)=b</math>, <math>p(c)=d</math>, and <math>p(d)=c</math>. WLOG, suppose <math>a<b</math>. We first note that a point on <math>y=p(x)</math> lies under the line <math>y=x</math> if and only if it lies between the lines <math>x = a</math> and <math>x=b</math>. WLOG, suppose <math>a<d<b</math>. Since <math>(d,c)</math> lies under the line <math>y=x</math>, we have <math>c<d</math>. Notice that <math>(c,d)</math> cannot lie between <math>x=a</math> and <math>x=b</math> because exactly one of the points <math>(c,d)</math> and <math>(d,c)</math> lies below <math>y=x</math>. Since <math>c<d</math>, it follows that <math>c<a<d<b</math>, so <math>(a,b,c,d) = (1,5,r,\tfrac{5}{2r})</math>. Thus, <math>p(1)=1</math> and <math>p(5)=5</math>, as desired.
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All that is left to do is to determine the leading coefficient of <math>p</math>, which we call <math>m</math>. We now know that <math>p(x) = m(x-1)(x-5) + x</math>. For convenience, replace <math>r</math> and <math>\tfrac{5}{2r}</math> by <math>s</math> and <math>t</math>, respectively, satisfying <math>st=\tfrac{5}{2}</math>. 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 <math>6m(s+t) = m(s^2+t^2) + 10m</math>. Dividing through by <math>m</math> and using <math>s^2+t^2 = (s+t)^2 - 2st = (s+t)^2 - 5</math>, we find that <math>s+t=1</math> or <math>s+t=5</math>. Since <math>t>1</math> and <math>s>0</math>, we must have <math>s+t=5</math>. Then, subtracting the second centered equation from the first and rearranging gives us <math>t-s + m(t^2-s^2) = (6m-1)(t-s)</math>. As <math>s\ne t</math>, we may divide through by <math>t-s</math> to obtain <math>1 + m(s+t) = 1+5m = 6m-1</math>. Finally, we have <math>m=2</math>, which gives <math>p(x) = 2x^2 - 11x + 10</math>. The answer is <math>2(12)^2 - 11(12)+10 = \boxed{166}</math>.
 +
 +
~Sedro

Latest revision as of 19:29, 12 September 2025

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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