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Difference between revisions of "1989 AHSME Problems/Problem 19"

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== Problem ==
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A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle?
 
A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle?
  
 
<math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \frac{18}{\pi^2} } \qquad \mathrm{(C) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(D) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(E) \frac{9}{\pi^2}(\sqrt{3}+3) }  </math>
 
<math> \mathrm{(A) \ 6 } \qquad \mathrm{(B) \frac{18}{\pi^2} } \qquad \mathrm{(C) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(D) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(E) \frac{9}{\pi^2}(\sqrt{3}+3) }  </math>
  
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=== Solution ===
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===Solution===
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The three arcs make up the entire circle, so the circumference of the circle is <math>3+4+5=12</math> and the radius is <math>\frac{12}{2\pi}=\frac{6}{\pi}</math>. Also, the lengths of the arcs are proportional to their corresponding central angles. Thus, we can write the values of the arcs as <math>3\theta</math>, <math>4\theta</math>, and <math>5\theta</math> for some <math>\theta</math>. By Circle Angle Sum, we obtain <math>3\theta+4\theta+5\theta=360</math>. Solving yields <math>\theta=30</math>. Thus, the angles of the triangle are <math>90</math>, <math>120</math>, and <math>150</math>. Using <math>[ABC]=\frac{1}{2}ab\sin{C}</math>, we obtain <math>\frac{r^2}{2}(\sin{90}+\sin{120}+\sin{150})</math>. Substituting <math>\frac{6}{\pi}</math> for <math>r</math> and evaluating yields <math>\frac{9}{\pi^2}(\sqrt{3}+3)\implies{\boxed{E}}</math>.
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-Solution by '''thecmd999'''
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== See also ==
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{{AHSME box|year=1989|num-b=18|num-a=20}} 
  
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The three arcs make up the entire circle, so the circumference of the circle is <math>3+4+5=12</math> and the radius is <math>\frac{12}{2\pi}=\frac{6}{\pi}</math>. Also, the lengths of the arcs are proportional to their corresponding central angles. Thus, we can write the values of the arcs as <math>3\theta</math>, <math>4\theta</math>, and <math>5\theta</math> for some <math>\theta</math>. Circle Angle Sum yields <math>3\theta+4\theta+5\theta=360\implies{\theta=30}</math>. Thus, the angles of the triangle are <math>90</math>, <math>120</math>, and <math>150</math>. Using <math>[ABC]=\frac{1}{2}ab\sin{C}</math>, we obtain <math>\frac{r^2}{2}(\sin{90}+\sin{120}+\sin{150})</math>. Substituting <math>\frac{6}{pi}</math> for <math>r</math> and evaluating yields <math>\frac{9}{\pi}(\sqrt{3}+3)\implies{\boxed{E}}</math>.
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[[Category: Intermediate Geometry Problems]]
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{{MAA Notice}}

Revision as of 07:57, 22 October 2014

Problem

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of lengths 3, 4, and 5. What is the area of the triangle?

$\mathrm{(A) \ 6 } \qquad \mathrm{(B) \frac{18}{\pi^2} } \qquad \mathrm{(C) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(D) \frac{9}{\pi^2}(\sqrt{3}-1) } \qquad \mathrm{(E) \frac{9}{\pi^2}(\sqrt{3}+3) }$

Solution

The three arcs make up the entire circle, so the circumference of the circle is $3+4+5=12$ and the radius is $\frac{12}{2\pi}=\frac{6}{\pi}$. Also, the lengths of the arcs are proportional to their corresponding central angles. Thus, we can write the values of the arcs as $3\theta$, $4\theta$, and $5\theta$ for some $\theta$. By Circle Angle Sum, we obtain $3\theta+4\theta+5\theta=360$. Solving yields $\theta=30$. Thus, the angles of the triangle are $90$, $120$, and $150$. Using $[ABC]=\frac{1}{2}ab\sin{C}$, we obtain $\frac{r^2}{2}(\sin{90}+\sin{120}+\sin{150})$. Substituting $\frac{6}{\pi}$ for $r$ and evaluating yields $\frac{9}{\pi^2}(\sqrt{3}+3)\implies{\boxed{E}}$.


-Solution by thecmd999

See also

1989 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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
All AHSME Problems and Solutions

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