2014 AMC 8 Problem 15

Attempt Problem 15 of the 2014 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 2014 AMC 8 solutions, or check the answer key.

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

15.

The circumference of the circle with center OO is divided into 1212 equal arcs, marked the letters AA through LL as seen below. What is the number of degrees in the sum of the angles xx and y?y?
75 75
80 80
90 90
120 120
150 150
Answer: C
Concepts:arcisosceles triangleangle chasing
Small hint:
Each adjacent pair of letters cuts off 30∘30^\circ at the center.
Big hint:
The two relevant triangles are isosceles because their sides are radii.
Solution:
Note that each of the 1212 arcs splits the circle evenly, so they each cover 360∘12=30∘.\frac{360^{\circ}}{12} = 30^{\circ}.
∠AOE\angle AOE spans 44 of these arcs, so ∠AOE=4⋅30∘=120∘.\angle AOE = 4 \cdot 30^{\circ} = 120^{\circ}. Similarly, ∠GOI=2⋅30∘=60∘.\angle GOI = 2 \cdot 30^{\circ} = 60^{\circ}. We also know that both triangles are isosceles since two of their sides are radii. Therefore, x=180−1202=30∘ x = \dfrac{180 - 120}{2} = 30^{\circ} and y=180−602=60∘. y = \dfrac{180 - 60}{2} = 60^{\circ}. Therefore, x+y=90∘.x + y = 90^{\circ}.
Thus, C is the correct answer.
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