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Difference between revisions of "1969 AHSME Problems/Problem 8"

(Created page with "== Problem == Triangle <math>ABC</math> is inscribed in a circle. The measure of the non-overlapping minor arcs <math>AB</math>, <math>BC</math> and <math>CA</math> are, respect...")
 
(Solution to Problem 8)
 
(One intermediate revision by one other user not shown)
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
−
<math>\fbox{D}</math>
+
<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 <math>360^\circ</math>.  Thus,
 +
<cmath>x+75+2x+25+3x-22=360</cmath>
 +
<cmath>6x+78=360</cmath>
 +
Solving this yields <math>x = 47</math>, so the inscribed angles are <math>122^\circ</math>, <math>99^\circ</math>, and <math>119^\circ</math>.  Noting that an angle of <math>\triangle ABC</math> is half of its corresponding inscribed angle, so the angles of <math>\triangle ABC</math> are <math>59.5^\circ</math>, <math>49.5^\circ</math>, and <math>\boxed{\textbf{(D) } 61^\circ}</math>.
  
 
== See also ==
 
== See also ==
−
{{AHSME box|year=1969|num-b=7|num-a=9}}   
+
{{AHSME 35p box|year=1969|num-b=7|num-a=9}}   
  
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 02:45, 7 June 2018

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 $\boxed{\textbf{(D) } 61^\circ}$.

See also

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

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