AMC 10 · 2023 · #17

Grade 8 geometry-2d
pythagorean-theoreminteger-pythagorean-triplesarea-rectangles caseworksystematic-enumerationidentify-subproblems ↑ Prerequisites: pythagorean-theoremarea-rectangles
📏 Medium solution 💡 3 insights
Problem
ABCD is a rectangle with AB = 30 and BC = 28. Point P lies on BC and point Q lies on CD so that all three right triangles △ ABP, △ PCQ, and △ QDA have integer side lengths. Find the perimeter of △ APQ.

Pick an answer.

(A)
84
(B)
86
(C)
88
(D)
90
(E)
92

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Three right triangles glued inside a rectangle is a classic call for Tool #1 (Draw a Diagram) — labelling sides BP, PC, DQ, CQ on a sketch makes the constraints visible at a glance. Tool #7 (Identify Subproblems) splits the puzzle into the three triangles: each one is an independent "find an integer hypotenuse" problem. Tool #2 (Systematic List) lists all Pythagorean triples with a leg of 30 and a leg of 28 — short lists once you remember the (8,15,17) and (3,4,5) families. The third triangle △ PCQ is then forced, and we only need its sides to be integers too.

1STEP 1

Sketch rectangle ABCD with P on BC, Q on CD. Three right triangles sit at corners B, C, D; △ APQ inside is what we measure.

AB = CD = 30, BC = AD = 28
2STEP 2

Subproblem 1: leg 30. Among triples, only (16, 30, 34) keeps the other leg ≤ 28, so BP = 16 and AP = 34.

30² + 16² = 900 + 256 = 1156 = 34² → BP = 16, AP = 34
3STEP 3

Subproblem 2: leg 28. Only (21, 28, 35) keeps DQ under 30, so DQ = 21 and AQ = 35.

28² + 21² = 784 + 441 = 1225 = 35² → DQ = 21, AQ = 35
4STEP 4

Subproblem 3: PC = 12, CQ = 9 are forced, and 9² + 12² = 15², so PQ = 15 — the third triangle is integer too.

PC = 12, CQ = 9, PQ = √(12² + 9²) = 15
5STEP 5

Add the three hypotenuses: AP + AQ + PQ = 34 + 35 + 15.

34 + 35 + 15 = 84 → (A)
Answer
84
Diagram sanity: BP = 16 ＜ 28 = BC and DQ = 21 ＜ 30 = CD, so P and Q sit on their intended segments. Triangle-inequality on △ APQ with sides (15, 34, 35): 15 + 34 = 49 ＞ 35, 15 + 35 = 50 ＞ 34, 34 + 35 = 69 ＞ 15 — a valid triangle. All three corner triangles are integer Pythagorean triples: (16, 30, 34), (21, 28, 35), (9, 12, 15). The perimeter 84 matches answer (A), and the other choices 86, 88, 90, 92 correspond to no consistent assignment of triples on this rectangle.
💡Key takeaway

Three right triangles around the rectangle's corners are just three Pythagorean triples in disguise. Scan the short list of triples with a leg matching 30 and a leg matching 28: only (16, 30, 34) and (21, 28, 35) fit, and the third triangle (9, 12, 15) falls out for free. Sum the three hypotenuses to get (A) 84.