2007 AMC 12A Problems/Problem 8

Revision as of 01:52, 20 July 2016 by Pahad00 (talk | contribs) (Solution 2)

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 $\frac 1260 = 30^{\circ} \Longrightarrow \mathrm{(C)}$

Solution 2

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 $\frac{5}{12}$ x $360$ degrees. Given every triangle has a unique circumcircle, we can construct a quadrilateral from that triangle within the given circumcircle. Hence the required angle, being in a cyclic quadrilateral, is supplementary to $150$ degrees. Hence $\mathrm{(C)}$

See also

2007 AMC 12A (ProblemsAnswer KeyResources)
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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