1969 AHSME Problems/Problem 8

Problem

Triangle $ABC$ is inscribed in a circle. The measure 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 is:

$\text{(A) } 57\tfrac{1}{2}^{\circ}\quad \text{(B) } 59^{\circ}\quad \text{(C) } 60^{\circ}\quad \text{(D) } 61^{\circ}\quad \text{(E) } 122^{\circ}$

Solution

$[asy] draw(circle((0,0),65)); draw((25,60)--(39,-52)--(-52,-39)--(25,60)); dot((25,60)); dot((39,-52)); dot((-52,-39)); dot((0,0)); draw((0,0)--(-52,-39)); draw((0,0)--(39,-52)); draw((0,0)--(25,60)); label("A",(-52,-39),SW); label("B",(25,60),NE); label("C",(39,-52),SE); [/asy]$ Because the triangle is inscribed, the sum of the minor arcs equals $360^\circ$ . Thus, $x+75+2x+25+3x-22=360$ $6x+78=360$ Solving this yields $x = 47$ , so the inscribed angles are $122^\circ$ , $99^\circ$ , and $119^\circ$ . Noting that an angle of $\triangle ABC$ is half of its corresponding inscribed angle, so the angles of $\triangle ABC$ are $59.5^\circ$ , $49.5^\circ$ , and $\boxed{\textbf{(D) } 61^\circ}$ .

1969 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

1969 AHSME Problems/Problem 8

Problem

Solution

See also