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Algebra Difficulty 3.5 AMC 10/12 Find the answer Canada

$P$ , $Q$ , $R$ , 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

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 $0$ .)

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