1999 AMC 8 Problem 25
Attempt Problem 25 of the 1999 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1999 AMC 8 solutions, or check the answer key.
All problems are used with official legal permission of the Mathematical Association of America (MAA).
25.
Points and are midpoints of the sides of right triangle Points are midpoints of the sides of triangle etc. If the dividing and shading process is done times (the first three are shown) and then the total area of the shaded triangles is nearest
Answer: A
Small hint:
The shaded triangles form a geometric area pattern
Big hint:
Each new shaded triangle has one-fourth the previous shaded area
Solution:
The first shaded triangle has legs and , so its area is .
At each later step, the relevant triangle has half the leg length, so the shaded area is multiplied by .
The total shaded area after many steps is therefore very close to the geometric sum
After steps the omitted tail is tiny, so the total area is nearest .
Thus, A is the correct answer.
Problem 25 in other years
- 1985 AMC 8
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