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2008 AMC 10B Problem 7

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2008 AMC 10B Problem 7AMC 10 Step by Step
Name:                             Date: September 13, 2026
Similar & Congruent Triangles · difficulty 1 of 5 · about 1 min · position P1-10

An equilateral triangle of side length 1010 is completely filled in by non-overlapping equilateral triangles of side length 11 . How many small triangles are required?

  1. A)

    1010

  2. B)

    2525

  3. C)

    100100

  4. D)

    250250

  5. E)

    10001000

Answer

C

Key insight

The big triangle is a scaled copy of the small one with ratio 10, so its area is 10^2 = 100 times as large.

Solution

All equilateral triangles are similar, and the side ratio here is 10:110 : 1. Areas of similar figures scale by the square of the side ratio, so the big triangle has 102=10010^2 = 100 times the area of a unit triangle. Since the small triangles tile it without gaps or overlaps, exactly 100100 of them are needed.

The answer is (C) 100\boxed{\textbf{(C)}\ 100}.

Why this works

"How many unit copies fill a figure" is an area question, and area scales quadratically with length. This is the same reason a 10×1010 \times 10 square holds 100100 unit squares. Whenever a figure is tiled by scaled copies of itself, count by area ratio rather than by drawing.

Alternative approach

Count by rows: the top row holds 11 small triangle, the next 33, then 55, and so on up to 1919 in the bottom row. The sum of the first ten odd numbers is 102=10010^2 = 100.

The trap

Scaling the area by 10 instead of 10^2, giving 10, or guessing 25 from the bottom row of 19 triangles.

Common mistakes

  • Scaling the area by 10 instead of 10^2, giving 10, or guessing 25 from the bottom row of 19 triangles.
  • Counting only the upward-pointing triangles (1+2+⋯+10=551 + 2 + \dots + 10 = 55) and forgetting the inverted ones between them.

Techniques

Set up the equation/formula and compute; no special trick needed