SAT1000 - P153

Algebra Level 3

f ( x ) f(x) is a function defined at [ 0 , 1 ] [0,1] such that:

  • f ( 0 ) = f ( 1 ) = 0 f(0)=f(1)=0 .

  • ∀ x , y ∈ [ 0 , 1 ] ( x ≠ y ) , ∣ f ( x ) − f ( y ) ∣ < 1 2 ∣ x − y ∣ \forall x,y \in [0,1]\ (x \neq y), |f(x)-f(y)| < \dfrac{1}{2} |x-y| .

If ∀ x , y ∈ [ 0 , 1 ] , ∣ f ( x ) − f ( y ) ∣ < k \forall x,y \in [0,1], |f(x)-f(y)|<k , find the minimum value of k k .

Let K K be the minimum value. Submit ⌊ 1000 K ⌋ \lfloor 1000K \rfloor .

Have a look at my problem set: SAT 1000 problems


The answer is 250.

Alice Smith by Alice Smith , 20, China

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1 solution

Abhishek Sinha
Jul 1, 2020

Without any loss of generality, assume 1 ≥ y ≥ x ≥ 0. 1\geq y \geq x \geq 0. . We have ∣ f ( x ) − f ( y ) ∣ ≤ y − x 2 . |f(x)-f(y)| \leq \frac{y-x}{2}. also ∣ f ( x ) − f ( y ) ∣ = ∣ f ( x ) − f ( 0 ) + f ( 1 ) − f ( y ) ∣ ≤ ∣ f ( x ) − f ( 0 ) ∣ + ∣ f ( 1 ) − f ( y ) ∣ ≤ x 2 + 1 − y 2 = 1 2 − y − x 2 . |f(x)- f(y)| = |f(x)-f(0)+ f(1)-f(y)| \leq |f(x)-f(0)| + |f(1)-f(y)| \leq \frac{x}{2}+ \frac{1-y}{2}= \frac{1}{2}-\frac{y-x}{2}. Adding up the above two inequalities, we have ∣ f ( x ) − f ( y ) ∣ ≤ 1 4 . |f(x)-f(y)| \leq \frac{1}{4}. The above bound is met with equality for a function f ( x ) f(x) which consists of two triangles located side by side, each with base 1 2 \frac{1}{2} and height 1 4 \frac{1}{4} .

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