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Difference between revisions of "2007 AMC 12A Problems/Problem 8"

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We look at the angle between 12, 5, and 10. It subtends <math>\frac 16</math> of the circle, or <math>60</math> degrees (or you can see that the [[arc]] is <math>\frac 23</math> of the [[right angle]]). Thus, the angle at each vertex is an [[inscribed angle]] subtending <math>60</math> degrees, making the answer <math>\frac 1260 = 30^{\circ} \Longrightarrow \mathrm{(C)}</math>
 
We look at the angle between 12, 5, and 10. It subtends <math>\frac 16</math> of the circle, or <math>60</math> degrees (or you can see that the [[arc]] is <math>\frac 23</math> of the [[right angle]]). Thus, the angle at each vertex is an [[inscribed angle]] subtending <math>60</math> degrees, making the answer <math>\frac 1260 = 30^{\circ} \Longrightarrow \mathrm{(C)}</math>
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==Solution 2==
 
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Consider one chord from 12 to 5 on the clockface. Since this chord subtends 5 of the 12 given hours, it subtends an angle at the center corresponding to  <math>\frac{5}{12}</math>  x <math>360</math> degrees. Given every triangle has a unique circumcircle, we can construct a quadrilateral from that triangle within that circumcircle. Hence the required angle, being in a cyclic quadrilateral, is supplementary to <math>150</math> degrees. Hence <math>\mathrm{(C)}</math>
 
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==See also==
 
==See also==
 
{{AMC12 box|year=2007|ab=A|num-b=7|num-a=9}}
 
{{AMC12 box|year=2007|ab=A|num-b=7|num-a=9}}

Latest revision as of 02:00, 20 July 2016

Problem

A star-polygon is drawn on a clock face by drawing a chord from each number to the fifth number counted clockwise from that number. That is, chords are drawn from 12 to 5, from 5 to 10, from 10 to 3, and so on, ending back at 12. What is the degree measure of the angle at each vertex in the star polygon?

$\mathrm{(A)}\ 20\qquad \mathrm{(B)}\ 24\qquad \mathrm{(C)}\ 30\qquad \mathrm{(D)}\ 36\qquad \mathrm{(E)}\ 60$

Solution

2007 AMC12A-8.png

We look at the angle between 12, 5, and 10. It subtends $\frac 16$ of the circle, or $60$ degrees (or you can see that the arc is $\frac 23$ of the right angle). Thus, the angle at each vertex is an inscribed angle subtending $60$ degrees, making the answer $\frac 1260 = 30^{\circ} \Longrightarrow \mathrm{(C)}$

See also

2007 AMC 12A (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
All AMC 12 Problems and Solutions

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