1971 AMC 12 Problem 28

Attempt Problem 28 of the 1971 AMC 12 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 1971 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

28.

Nine lines parallel to the base of a triangle divide the other sides each into 1010 equal segments and the area into 1010 distinct parts. If the area of the largest of these parts is 38,38, then the area of the original triangle is:
180180
190190
200200
210210
240240
Answer: C
Concepts:similarityarea ratioarea decomposition
Small hint:
The largest part is the bottom strip
Big hint:
The smaller triangle above that strip has linear scale 910\frac{9}{10}
Solution:
If the whole area is K,K, the triangle above the bottom strip is similar to the original with scale 910,\frac{9}{10}, so its area is 81K100.\frac{81K}{100}. Thus the largest strip has area K−81K100=19K100=38, K-\frac{81K}{100}=\frac{19K}{100}=38, giving K=200.K=200.
Therefore, the correct answer is C.
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