1969 AMC 12 Problem 8

Attempt Problem 8 of the 1969 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 1969 AMC 12 solutions, or check the answer key.

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

Triangle $ABC$ is inscribed in a circle. The measures of the non-overlapping minor arcs $AB,$ $BC,$ and $CA$ are, respectively, $x+75^\circ,$ $2x+25^\circ,$ $3x-22^\circ.$ Then one interior angle of the triangle, in degrees, is:

$57\dfrac12$

$59$

$60$

$61$

$122$

Answer: D

Concepts:arc inscribed angle angle sum

Difficulty rating: 1500

Small Hint:

The three non-overlapping arcs sum to $360^\circ$

Big Hint:

Each inscribed angle is half the measure of its intercepted arc

Solution:

The arc sum gives $\begin{aligned} 6x+78&=360,\\ x&=47. \end{aligned}$ The arcs then measure $122^\circ,$ $119^\circ,$ and $119^\circ.$ The angle intercepting the $122^\circ$ arc measures $61^\circ.$

Therefore, the correct answer is D.

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