Problem:
Let ABCD be a rectangle with AB=30 and BC=28. Points P and Q lie on BC and CD, respectively, so that all sides of β³ABP,β³PCQ, and β³QDA have integer lengths. What is the perimeter of β³APQ?
Answer Choices:
A. 84
B. 86
C. 88
D. 90
E. 92
Solution:
The figure below shows the given information.

The Pythagorean triples with smallest total lengths are 3β4β5,5β12β13,7β24β25,8β15β17, and multiples thereof. Note that AD=28=4β
7. Then triangle β³ADQ must be a multiple of 3β 4β5 (with multiplier 7) or a multiple of 7β24β25 (with multiplier 4). But the latter option implies DQ=4β
24>30, so the first option is correct. This gives AQ=5β
7=35.\
Because DQ=3β
7=21, it follows that QC=30β21=9. Thus triangle β³QCP is also a multiple of 3β4β5 (with multiplier 3). This gives PC=4β
3=12 and PQ=5β
3=15.\
Finally, because PC=12, it follows that PB=BCβPC=28β12=16. Then triangle β³ABQ must be a multiple of 8β15β17 (with multiplier 2), so AP=17β
2=34. The perimeter of β³APQ is 34+15+35=(A)84β.
The problems on this page are the property of the MAA's American Mathematics Competitions