AMC 10 · 2010 · #25

Grade 11 countinggeometry-2d

Archetype Casework by Position / Vertex Type └ Sub-archetype Count of Type

cyclic-quadrilateralpolygon-inequalitystars-and-barsprinciple-of-inclusion-exclusionburnsides-lemma complementary-countingsymmetry-argument ↑ Prerequisites: cyclic-quadrilateral

📏 Long solution 💡 4 insights

Problem

Four whole-number sides add to a fixed total and all four corners lie on one circle. Count the different shapes.

Pick an answer.

(A)

560

(B)

564

(C)

568

(D)

1498

(E)

2255

How to solve

Strategy Change Focus / Count the Complement

There are infinitely many circles and infinitely many placements, so nothing can be counted until the shapes are replaced by something finite. Tool #16 (Change Focus) does that: read the four side lengths off in counterclockwise order and a quadrilateral becomes a four-bead necklace (a,b,c,d), taken up to the four cyclic shifts. That trade is only legal if the necklace is a perfect stand-in, and that means proving two separate things — no two different shapes give the same necklace, and every legal necklace is actually achieved by some shape. Both halves are load-bearing and each is easy to skip. Tool #7 (Identify Subproblems) supplies the first: cut along a diagonal and the law of cosines forces one angle, hence the entire shape. Tool #14 (Extreme Principle) supplies the boundary of the second: the shape exists exactly while a certain cosine stays strictly inside [-1,1], which converts into 'every side at most 15'. Tool #1 (Draw a Diagram) then glues two triangles back together to show the shape really is there. Only after that does counting begin, and Tool #16 counts the complement — all 4-tuples minus the ones with an oversized side. The last step is the one that separates three of the answer choices: a necklace count is not 'divide by 4', because a few necklaces are carried onto themselves by a shift, so Tool #2 (Make a Systematic List) sorts the tuples by how much rotational symmetry they have. Tools #3 and #15 close the loop by killing the distractors and recounting a second way.

1STEP 1

Turn each shape into a necklace

The object being counted is a cyclic list.

(a,b,c,d) ∼ (b,c,d,a) ∼ (c,d,a,b) ∼ (d,a,b,c), with a+b+c+d=32 and each of a,b,c,d a positive integer

2STEP 2

The necklace fixes the shape

On a circle the list determines the shape.

AC²=a²+b²-2abcos B=c²+d²+2cdcos B ⟹ cos B=(a²+b²-c²-d²)/(2(ab+cd))

3STEP 3

Which necklaces are legal

A list is legal when no side beats the other three.

|t| ＜ 1⇔ |a-b| ＜ c+d and a+b ＞ |c-d| ⇔ every side ＜ sum of the other three ⇔ 1 ≤ a,b,c,d ≤ 15

4STEP 4

Every legal necklace is really built

Every legal list really is built by some shape.

cos∠ B=t, cos∠ D=-t, ∠ B+∠ D=180^° ⟹ A,B,C,D concyclic in that order

5STEP 5

Count the ordered lists first

Counting ordered lists first gives 2255.

C(31, 3)-4C(16, 3)=4495-2240=2255 legal ordered tuples

6STEP 6

Sort the lists by rotational symmetry

Rotational symmetry sorts them into three groups.

1 list of size 1; 14/2=7 lists of size 2; 2240/4=560 lists of size 4

7STEP 7

Add the three groups

Adding the groups gives 568, choice (C).

560+7+1=568 (C)

8STEP 8

Recount by symmetry type

Recounting by symmetry type confirms 568.

1 · 1+4 · 1+7 · 2+45 · 3+69 · 6=1+4+14+135+414=568

Answer

568

Three independent checks agree. First, the bracket: 2240 of the 2255 legal tuples fall into groups of exactly four, contributing exactly 560 quadrilaterals, and the remaining 15 tuples can contribute at most 15 more and at least 1 more, so the total is forced into [561,575]. That alone eliminates (A) 560, (D) 1498 and (E) 2255, and 568 sits comfortably inside while (B) 564 survives only until the 15 symmetric tuples are actually grouped, which gives 1+7=8 and lands on 568. Second, the orbit-counting (Burnside) identity: average the number of tuples fixed by each of the four shifts — 2255 for the identity, 1 for shift by one, 15 for shift by two, 1 for shift by three — to get (2255+1+15+1)/4=2272/4=568, and note 568 · 4=2272=2255+17 is a genuine arithmetic constraint that a wrong total would fail. Third, a sensitivity check on the wording: if mirror images were also identified, the group would have eight elements instead of four and the count would drop to 316, which is not among the choices. So the problem's 'rotation and a translation' is doing real work, and reading it as plain congruence would be a different problem.

💡Key takeaway

Four side lengths in cyclic order pin a circle-quadrilateral down completely, so counting the shapes is really counting four-bead necklaces — and a necklace count is never just divide by four.

Turn each shape into a necklace
The necklace fixes the shape
Which necklaces are legal
Every legal necklace is really built
Count the ordered lists first
Sort the lists by rotational symmetry
Add the three groups
Recount by symmetry type