Difference between revisions of "2015 AIME I Problems/Problem 6"
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==Problem== | ==Problem== | ||
− | Point <math>A,B,C,D,</math> and <math>E</math> are equally spaced on a minor arc of a | + | Point <math>A,B,C,D,</math> and <math>E</math> are equally spaced on a minor arc of a circle. Points <math>E,F,G,H,I</math> and <math>A</math> are equally spaced on a minor arc of a second circle with center <math>C</math> as shown in the figure below. The angle <math>\angle ABD</math> exceeds <math>\angle AHG</math> by <math>12^\circ</math>. Find the degree measure of <math>\angle BAG</math>. |
<asy> | <asy> | ||
Line 25: | Line 25: | ||
label("$G$",G,dir(260)); | label("$G$",G,dir(260)); | ||
label("$H$",H,dir(280)); | label("$H$",H,dir(280)); | ||
− | label("$I$",I,dir(315));</asy> | + | label("$I$",I,dir(315)); |
+ | </asy> | ||
==Solution== | ==Solution== | ||
− | Let O be the center of the circle with ABCDE on it. | + | Let <math>O</math> be the center of the circle with <math>ABCDE</math> on it. |
− | Let <math>x | + | Let <math>x</math> be the degree measurement of <math>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}</math> in circle <math>O</math> and <math>y</math> be the degree measurement of <math>\overarc{EF}=\overarc{FG}=\overarc{GH}=\overarc{HI}=\overarc{IA}</math> in circle <math>C</math>. |
− | <math>\angle ECA</math> is therefore <math>5y</math> by way of circle <math>C</math> and <math>180-2x</math> by way of circle <math>O</math>. | + | <math>\angle ECA</math> is therefore <math>5y</math> by way of circle <math>C</math> and <math>\frac{360-4x}{2}=180-2x</math> by way of circle <math>O</math>. |
− | <math>\angle ABD</math> is <math>180 - \frac{3x}{2}</math> by way of circle O, and <math>\angle AHG</math> is <math>180 - \frac{3y}{2}</math> by way of circle <math>C</math>. | + | <math>\angle ABD</math> is <math>180 - \frac{3x}{2}</math> by way of circle <math>O</math>, and <math>\angle AHG</math> is <math>180 - \frac{3y}{2}</math> by way of circle <math>C</math>. |
This means that: | This means that: | ||
− | + | <math>180-\frac{3x}{2}=180-\frac{3y}{2}+12</math>, | |
− | + | which when simplified yields <math>3x/2+12=3y/2</math>, or <math>x+8=y</math>. | |
Since: | Since: | ||
<math>5y=180-2x</math>, <math>5x+40=180-2x</math> | <math>5y=180-2x</math>, <math>5x+40=180-2x</math> | ||
Line 45: | Line 46: | ||
<math>7x=140, x=20</math> | <math>7x=140, x=20</math> | ||
<math>y=28.</math> | <math>y=28.</math> | ||
− | <math>\angle BAG</math> is equal to <math>\angle BAE</math> + <math>\angle EAG</math>, which equates to 3x | + | <math>\angle BAG</math> is equal to <math>\angle BAE</math> + <math>\angle EAG</math>, which equates to <math>\frac{3x}{2} + y</math>. |
− | Plugging in yields <math>30+28</math>, or <math>058</math> | + | Plugging in yields <math>30+28</math>, or <math>\boxed{058}</math>. |
==See Also== | ==See Also== | ||
{{AIME box|year=2015|n=I|num-b=5|num-a=7}} | {{AIME box|year=2015|n=I|num-b=5|num-a=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] |
Revision as of 14:05, 23 May 2020
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 is by way of circle .
This means that:
,
which when simplified yields , or . Since: , 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.