AMC 10 · 2017 · #6

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 lengths are fixed and a fourth must complete a real quadrilateral. Count the possible lengths.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

Each side must be beaten by 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

One check caps the new side above.

x ＜ 3+7+15 = 25

3STEP 3

Boundary 2: the fixed 15 rod is the longest

The other caps it from below.

15 ＜ 3 + 7 + x → x ＞ 5

4STEP 4

Count the surviving lengths

Removing the used lengths leaves 17, choice (B).

{6,7,…,24} : 24-6+1=19, 19-2=17 → (B)

Answer

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

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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