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Problem 550

AMC 10/12, early questions

Algebra Difficulty 3.5 Multiple choice CEMC Pascal · Canada · 2026

$P$ , $Q$ , $R$ , and $S$ are four distinct points on a line
segment in the order shown.

If $PR=8$ and $QS=15$ , what is the smallest possible
integer length of $PS$ ?

Pick one

Official solution

Suppose that $QR=x$ . Since
$PR=8$ , then $PQ=PR-QR=8-x$ . Since $QS=15$ , then $RS=QS-QR=15-x$ , and so $PS=PQ+QR+RS=(8-x)+x+(15-x)=23-x$ .

The smallest possible integer length of $PS=23-x$ occurs when $x$ is the largest integer possible. Since
$PQ=8-x$ , then $8-x>0$ and so $x<8$ . The largest integer value of
$x$ that is less than 8 is $x=7$ , and so the smallest possible integer
length of $PS$ is $23-7=16$ .

(We confirm that when $x=7$ , $RS=15-x=8$ and thus $RS$ also has length greater than $0$ .)

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.