Difference between revisions of "2015 AIME I Problems/Problem 6"
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<cmath>180-\frac{3x}{2}=180-\frac{3y}{2}+12</cmath> | <cmath>180-\frac{3x}{2}=180-\frac{3y}{2}+12</cmath> | ||
− | which when simplified yields <cmath>3x | + | which when simplified yields <cmath>\frac{3x}{2}+12=\frac{3y}{2}</cmath> or <cmath>x+8=y</cmath> |
Since: | Since: | ||
<cmath>5y=180-2x</cmath> and <cmath>5x+40=180-2x</cmath> | <cmath>5y=180-2x</cmath> and <cmath>5x+40=180-2x</cmath> |
Revision as of 21:46, 1 September 2021
Problem
Point and are equally spaced on a minor arc of a circle. Points and are equally spaced on a minor arc of a second circle with center as shown in the figure below. The angle exceeds by . Find the degree measure of .
Solution
Let be the center of the circle with on it.
Let be the degree measurement of in circle and be the degree measurement of in circle . is, therefore, by way of circle and by way of circle . is by way of circle , and by way of circle .
This means that:
which when simplified yields or Since: and So: is equal to + , which equates to . Plugging in yields , or .
See Also
2015 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.