AMC 10 · 2015 · #19

Grade 8 geometry-2d
pythagorean-theoremdifference-of-squaresperfect-squares bound-inequality-then-enumerate ↑ Prerequisites: pythagorean-theorem
📏 Long solution 💡 4 insights
Problem
A quadrilateral with two right angles has whole-number sides and two of them equal. Count the perimeters below a bound.

Pick an answer.

(A)
30
(B)
31
(C)
61
(D)
62
(E)
63
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

A grid makes both right angles explicit.

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

Name the unknown sides

Two letters name the unknown sides.

BC=y, CD=AD=x, AB=2
3STEP 3

Apply the Pythagorean theorem

One right triangle links them.

y²+(x-2)²=x²
4STEP 4

Simplify to a square condition

Whole numbers force a perfect square.

y²=4(x-1) → x=k²+1, y=2k
5STEP 5

Write the perimeter as a formula

The perimeter becomes one clean formula.

p=2+2k+2(k²+1)=2k²+2k+4
6STEP 6

Find the boundary under 2015

The bound is crossed at 31, choice (B).

k=31: p=1988 < 2015; k=32: p=2116 > 2015
Answer
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 you a Pythagorean equation, and the only shapes that work 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