1964 AHSME Problems/Problem 27

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Problem

If $x$ is a real number and $|x-4|+|x-3|<a$ where $a>0$, then:

$\textbf{(A)}\ 0<a<.01\qquad \textbf{(B)}\ .01<a<1 \qquad \textbf{(C)}\ 0<a<1\qquad \\ \textbf{(D)}\ 0<a \le 1\qquad \textbf{(E)}\ a>1$

Solution

Let $A$ be point $3$ on a number line, and let $B$ be point $4$. Let $X$ be a mobile point at $x$. Geometrically, $|x-4| + |x-3|$ represents $AX + BX$. If $X$ is between $A$ and $B$, then $AX + BX = AB = 1$. Otherwise, if $X$ is to the left of $A$, then $AX + BX = AX + (AX + AB) = 2AX + 1$, which is greater than $1$. If $X$ is to the right of $B$, we have $AX + BX = (AB + BX) + BX = 2BX + 1$.

In all $3$ cases, the minimum value of $AX + BX$ is $1$. Thus, $|x-4|+|x-3| < a$ will always be true if $a >1$. If $a=1$, it can be false for $3 \le x \le 4$. If $a < 1$, then $|x-4|+|x-3| < a$ is always false because the LHS is too big.

Thus, the answer is $\boxed{\textbf{(E)}}$.

See Also

1964 AHSC (ProblemsAnswer KeyResources)
Preceded by
Problem 26
Followed by
Problem 28
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