1983 AIME Problem 2

Attempt Problem 2 of the 1983 AIME below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1983 AIME solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

Let

f(x)=|x-p|+|x-15|

{}+|x-p-15|,

where

0<p<15.

Determine the minimum value taken by

f(x)

for

x

in the interval

p\leq x\leq15.

Answer

Determine the sign of each expression inside an absolute value on the given interval

p\leq x\leq15,

simplify

f(x)

to a decreasing linear function

Since

p\leq x\leq15,

we have

\begin{aligned} |x-p|&=x-p,\\ |x-15|&=15-x,\\ |x-p-15|&=p+15-x. \end{aligned}

Hence

f(x)=30-x.

This is minimized at the right endpoint

x=15,

where its value is

15.

Problem 2 in other years