Difference between revisions of "2015 AIME I Problems/Problem 6"
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Here comes the key: Draw a line through <math>C</math> parallel to <math>AE</math>, and select a point <math>X</math> to the right of point <math>C</math>. <math>\angle ACX=\overarc{AB}+\overarc{BC}=m-8</math>. Let the midpoint of <math>\overline{HG}</math> be <math>Y</math>, then <math>\angle YCX=\angle ACX+\angle ACY=(m-8)+\frac{5m}{2}=90</math>. Solving gives <math>m=28</math> | Here comes the key: Draw a line through <math>C</math> parallel to <math>AE</math>, and select a point <math>X</math> to the right of point <math>C</math>. <math>\angle ACX=\overarc{AB}+\overarc{BC}=m-8</math>. Let the midpoint of <math>\overline{HG}</math> be <math>Y</math>, then <math>\angle YCX=\angle ACX+\angle ACY=(m-8)+\frac{5m}{2}=90</math>. Solving gives <math>m=28</math> | ||
− | The rest of the solution proceeds as in solution 1 | + | The rest of the solution proceeds as in solution 1, which gives <math>\boxed{058}</math> |
Revision as of 22:22, 22 December 2021
Contents
[hide]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 1
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 .
Solution 2
Let be the degree measurement of . Since lie on a circle with center , .
Since , . Adding and gives , and . Since is parallel to , .
We are given that are evenly distributed on a circle. Hence,
Here comes the key: Draw a line through parallel to , and select a point to the right of point . . Let the midpoint of be , then . Solving gives
The rest of the solution proceeds as in solution 1, which gives
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.