Difference between revisions of "2007 AMC 12A Problems/Problem 8"
(→Solution 2) |
(→Solution 2) |
||
| Line 9: | Line 9: | ||
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> | ||
==Solution 2== | ==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 <math>\frac{5}{12}</math> | + | 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 the given circumcircle. Hence the required angle, being in a cyclic quadrilateral, is supplementary to <math>150</math> degrees. Hence <math>\mathrm{(C)}</math> |
==See also== | ==See also== | ||
Revision as of 01:52, 20 July 2016
Contents
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?
Solution
We look at the angle between 12, 5, and 10. It subtends 
of the circle, or 
degrees (or you can see that the arc is 
of the right angle). Thus, the angle at each vertex is an inscribed angle subtending degrees, making the answer 
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 
x 
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 
degrees. Hence 
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
