AMC 10 · 2017 · #10

Grade 7 geometry-2d
Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search
polygon-inequalityinterval-arithmetic bound-inequality-then-enumerate ↑ Prerequisites: polygon-inequality
📏 Medium solution 💡 2 insights
Problem
Three rods of lengths 3, 7, and 15 cm are on a table. From the remaining rods (one of each integer length 1 through 30, except 3, 7, 15), how many single lengths could serve as a fourth rod so the four rods form a quadrilateral with positive area?

Pick an answer.

(A)
16
(B)
17
(C)
18
(D)
19
(E)
20

AMC 10 2017 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Extreme Principle

Four lengths close into a quadrilateral with positive area exactly when no single side is so long that the other three can't reach around it — the same idea as the triangle inequality, but for four sides: each side must be shorter than the sum of the other three. The only side that can break this rule is the longest one, so Tool #14 (Extreme Principle) tells us to test the two boundary cases — the fourth rod being the longest side, and the fixed 15 rod being the longest side. Tool #4 (Introduce a Variable) names the unknown fourth length x so we can write those boundary conditions as inequalities, and Tool #2 (Make a Systematic List) counts the integer lengths that survive, after removing the rods already used.

1STEP 1

Name the rod and state the rule

Let the fourth rod be x. Four sides close into a positive-area quadrilateral only when each side is shorter than the sum of the other three.

sides 3, 7, 15, x → each side ＜ sum of the other three
2STEP 2

Boundary 1: the new rod is the longest

If x is the longest side, x must beat 3+7+15=25; at x=25 the other three lie flat with zero area, so x ＜ 25.

x ＜ 3+7+15 = 25
3STEP 3

Boundary 2: the fixed 15 rod is the longest

If x is short, 15 is the longest and must beat 3+7+x, giving x ＞ 5; at x=5 the sides 3,7,5 sum to 15 and lie flat.

15 ＜ 3 + 7 + x → x ＞ 5
4STEP 4

Count the surviving lengths

Combining gives 5 ＜ x ＜ 25, so x runs 6 to 24 — 19 integers; drop 7 and 15 already in use, leaving 17 → (B).

{6,7,…,24} : 24-6+1=19, 19-2=17 → (B)
Answer
17
The window 5 ＜ x ＜ 25 is symmetric-looking and sensible: it bars rods that are too long to close ( ≥ 25) and rods too short to wrap around the 15 ( ≤ 5). Counting 6 through 24 gives 19 integers, and removing the two already-used rods (7 and 15) leaves 17, matching choice (B). A quick spot check confirms the ends: x=6 works (15 ＜ 3+7+6=16) and x=24 works (24 ＜ 25), while x=5 and x=25 both collapse to zero area.
💡Key takeaway

Four sides only close into a real shape when no side is longer than the other three combined — so find the longest-side cutoffs, then count the lengths in between.

  • Name the rod and state the rule
  • Boundary 1: the new rod is the longest
  • Boundary 2: the fixed 15 rod is the longest
  • Count the surviving lengths

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