Question

Considera funciones $f$ que satisfacen $|f(x)-f(y)|\leq \dfrac{1}{2}|x-y|$ para todos los números reales $x$ y $y.$ De todas esas funciones que además satisfacen la ecuación $f(300) = f(900),$ ¿cuál es el mayor valor posible de $f(f(800))-f(f(400))$ ?

Consider functions $f$ that satisfy $|f(x)-f(y)|\leq \dfrac{1}{2}|x-y|$ for all real numbers $x$ and $y.$ Of all such functions that also satisfy the equation $f(300) = f(900),$ what is the greatest possible value of $f(f(800))-f(f(400))?$

$25$

$50$

$100$

$150$

$200$

Solución:

Nota que $|f(f(400))-f(f(300))|$ $\leq \dfrac 12 | f(400) - f(300)|$ $\leq \dfrac 14 |400-300| = 25,$ y $|f(f(900))-f(f(800))|$ $\leq \dfrac 12 | f(900) - f(800)|$ $\leq \dfrac 14 |900-800| = 25.$

Como $f(900) = f(300),$ por la desigualdad triangular, sabemos que $|f(f(800)) - f(f(400))| =$ $|(f(f(800)) - f(f(900))) -$ $(f(f(400))-f(f(300)))| \leq$ $|(f(f(800)) - f(f(900)))| +$ $|(f(f(400))-f(f(300)))|$ $\leq 50.$

Para alcanzar la cota, define $f$ por interpolación lineal a través de los puntos $\begin{gathered}(300,600),(400,550),(550,575),\\ (650,625),(800,650),(900,600).\end{gathered}$ y define $f(x)=600$ para $x\leq300$ o $x\geq900$ . Cada segmento tiene pendiente con valor absoluto a lo sumo $\dfrac12$ , así que la condición de contracción se cumple. En particular, $f(300)=f(900)=600$ , $f(400)=550$ y $f(800)=650$ . Por lo tanto $f(f(400))=f(550)=575$ , mientras que $f(f(800))=f(650)=625$ , dando la diferencia $50$ .

Así, la respuesta es B.

Note that $|f(f(400))-f(f(300))|$ $\leq \dfrac 12 | f(400) - f(300)|$ $\leq \dfrac 14 |400-300| = 25,$ and $|f(f(900))-f(f(800))|$ $\leq \dfrac 12 | f(900) - f(800)|$ $\leq \dfrac 14 |900-800| = 25.$

Since $f(900) = f(300),$ by the triangle inequality, we know $|f(f(800)) - f(f(400))| =$ $|(f(f(800)) - f(f(900))) -$ $(f(f(400))-f(f(300)))| \leq$ $|(f(f(800)) - f(f(900)))| +$ $|(f(f(400))-f(f(300)))|$ $\leq 50.$

To attain the bound, define $f$ by linear interpolation through the points $\begin{gathered}(300,600),(400,550),(550,575),\\ (650,625),(800,650),(900,600).\end{gathered}$ and set $f(x)=600$ for $x\leq300$ or $x\geq900$ . Every segment has slope with absolute value at most $\dfrac12$ , so the contraction condition holds. In particular, $f(300)=f(900)=600$ , $f(400)=550$ , and $f(800)=650$ . Hence $f(f(400))=f(550)=575$ , while $f(f(800))=f(650)=625$ , giving the difference $50$ .

Thus, the answer is B .

2022 AMC 10B Problema 24

Solución:

El Problema 24 en otros años