AMC 10 · 2015 · #24

Grade 8 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

pythagorean-theoremdifference-of-squaresperfect-squares bound-inequality-then-enumerate ↑ Prerequisites: pythagorean-theorem

📏 Long solution 💡 4 insights

Problem

A quadrilateral ABCD has whole-number side lengths and right angles at B and C. One side AB equals 2, and side CD equals side AD. Count how many whole-number perimeters below 2015 such a shape can have.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 10 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Introduce a Variable

Two right angles let me set the shape on a grid and name the unknown sides with letters. Once a letter stands for BC and a letter stands for CD, the equal sides CD = AD turn into one Pythagorean equation. Solving that equation reveals which side lengths are allowed, and a boundary check finds how many fit under the perimeter limit.

1STEP 1

Set the shape on a grid

Right angles at B and C put the shape on a grid: B=(0,0), C=(y,0), A=(0,2), D=(y,x).

B=(0,0), C=(y,0), A=(0,2), D=(y,x)

2STEP 2

Name the unknown sides

Let y=BC, x=CD; since CD=AD, the slanted side AD is also x: CD = x, DA = x.

BC=y, CD=AD=x, AB=2

3STEP 3

Apply the Pythagorean theorem

AD moves y across and x−2 up, and AD=x, so Pythagoras gives y² + (x−2)² = x².

y²+(x-2)²=x²

4STEP 4

Simplify to a square condition

Expanding gives y²=4(x−1), so x−1 must be a perfect square k²: x = k² + 1 and y = 2k.

y²=4(x-1) → x=k²+1, y=2k

5STEP 5

Write the perimeter as a formula

Summing the four sides and substituting k gives p = 2k² + 2k + 4 for each valid k ≥ 1.

p=2+2k+2(k²+1)=2k²+2k+4

6STEP 6

Find the boundary under 2015

k=31 fits (p=1988) but k=32 doesn't (p=2116), so k runs 1 to 31: the answer is (B).

k=31: p=1988 ＜ 2015; k=32: p=2116 ＞ 2015

Answer

p = 2k² + 2k + 4, 1 ≤ k ≤ 31

Each k from 1 to 31 gives strictly increasing p = 2k² + 2k + 4 (smallest p = 8 at k=1, largest p = 1988 at k=31), so all 31 perimeters are distinct and all stay under 2015. The count 31 matches choice (B), and the near-miss k=32 giving 2116 confirms 31 is exactly the cutoff.

💡Key takeaway

Name the mystery sides with letters, let the right angles give a Pythagorean equation, and valid shapes line up one per whole number k.

Set the shape on a grid
Name the unknown sides
Apply the Pythagorean theorem
Simplify to a square condition
Write the perimeter as a formula
Find the boundary under 2015