Difference between revisions of "2015 AIME I Problems/Problem 6"
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The rest of the solution proceeds as in solution 1, which gives <math>\boxed{058}</math> | The rest of the solution proceeds as in solution 1, which gives <math>\boxed{058}</math> | ||
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+ | ==Solution 3== | ||
+ | [[File:2015 AIME I 6.png|500px|right]] | ||
+ | Let <math>\angle GAH = \varphi \implies \overset{\Large\frown} {GH} = 2\varphi \implies</math> | ||
+ | <cmath>\overset{\Large\frown} {EF} = \overset{\Large\frown} {FG} = \overset{\Large\frown} {HI} = \overset{\Large\frown} {IA} = 2\varphi \implies</cmath> | ||
+ | <cmath>\angle AGH = 2\varphi, \angle ACE = 10 \varphi.</cmath> | ||
+ | |||
+ | <cmath>BD||GH \implies \angle AJB = \angle AGH = 2 \varphi.</cmath> | ||
+ | <cmath>\triangle AHG: \hspace{10mm} \angle AHG = \beta = 180^\circ – 3 \varphi.</cmath> | ||
+ | <math>\hspace{10mm} \triangle ABJ: \hspace{10mm} \angle BAG + \angle ABD = \alpha + \gamma = 180^\circ + 2 \varphi. </math> | ||
+ | |||
+ | Let arc <math> \overset{\Large\frown} {AB} = 2\psi \implies</math> | ||
+ | |||
+ | <math>\angle ACE = \frac {360^\circ – 8 \psi}{2}= 180^\circ – 4 \psi, \angle ABD = \gamma =\frac {360^\circ – 6 \psi}{2} =180^\circ – 3 \psi.</math> | ||
+ | <math>\gamma – \beta = 3(\varphi – \psi) = 12^\circ \implies \psi = \varphi – 4^\circ \implies 10 \varphi = 180^\circ – 4(\varphi – 4^\circ) \implies 14 \varphi = 196^\circ \implies \varphi = 14^\circ.</math> | ||
+ | |||
+ | Therefore <math>\gamma = 180^\circ – 3 \cdot (14^\circ – 4^\circ) = 150^\circ \implies \alpha = 180^\circ + 2 \cdot 14^\circ – 150^\circ = \boxed{\textbf{058}}.</math> | ||
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+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/IuwkX2Dv25s | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
Latest revision as of 15:04, 21 July 2023
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
Solution 3
Let
Let arc
Therefore
Video Solution
~MathProblemSolvingSkills.com
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.