Maths Olympiad Prep

Library / CEMC Pascal /566 of 603

CEMC Pascal, 2026

Algebra Difficulty 3.5 AMC 10/12 Find the answer Canada

PP, QQ, RR, and SS are four distinct points on a line
segment in the order shown.

Figure 0

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

Pick one

Solution

Suppose that QR=xQR=x. Since
PR=8PR=8, then PQ=PR−QR=8−xPQ=PR-QR=8-x. Since QS=15QS=15, then RS=QS−QR=15−xRS=QS-QR=15-x, and so PS=PQ+QR+RS=(8−x)+x+(15−x)=23−xPS=PQ+QR+RS=(8-x)+x+(15-x)=23-x.

The smallest possible integer length of PS=23−xPS=23-x occurs when xx is the largest integer possible. Since
PQ=8−xPQ=8-x, then 8-x gt;0\text{8-x gt;0} and so x lt;8\text{x lt;8}. The largest integer value of
xx that is less than 8 is x=7x=7, and so the smallest possible integer
length of PSPS is 23−7=1623-7=16.

(We confirm that when x=7x=7, RS=15−x=8RS=15-x=8 and thus RSRS also has length greater than 00.)

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

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