1973 AMC 12 Problem 16

Attempt Problem 16 of the 1973 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 1973 AMC 12 solutions, or check the answer key.

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

16.

If the sum of all the angles except one of a convex polygon is 2190∘,2190^\circ, then the number of sides of the polygon must be
1313
1515
1717
1919
2121
Answer: B
Concepts:angle suminequalitybounding to limit cases
Small hint:
Let the omitted interior angle be x∘x^\circ and the number of sides be nn
Big hint:
Use 0<x<1800\lt x\lt180 to trap the integer n−2n-2 in an interval of length 11
Solution:
If the omitted angle is x∘,x^\circ, then convexity gives 0<x<180.0\lt x\lt180. Hence 180(n−2)=2190+x, 180(n-2)=2190+x, so 2190180<n−2<2370180. \frac{2190}{180} \lt n-2 \lt\frac{2370}{180}. This is 1216<n−2<1316. 12\frac16\lt n-2\lt13\frac16. The only possible integer is n−2=13,n-2=13, so n=15.n=15.
Therefore, the correct answer is B.
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