Difference between revisions of "2015 AIME I Problems/Problem 6"
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==Solution== | ==Solution== | ||
| − | Let O be the center of the circle with ABCDE on it. Let x=ED=DC=CB=BA and y=EF=FG=GH=HI=IA. | + | Let O be the center of the circle with ABCDE on it. |
| − | <math>\angle ECA</math> is therefore 5y by way of circle C and 180-2x by way of circle O. | + | |
| − | <math>\angle ABD</math> is 180-3x/ | + | Let <math>x=ED=DC=CB=BA</math> and <math>y=EF=FG=GH=HI=IA</math>. |
| − | This means that 180-3x/2=180-3y/2+12, which when simplified yields 3x/2+12=3y/2, or x+8=y. | + | <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>. |
| − | Since 5y=180-2x, | + | <math>\angle ABD</math> is <math>180 - \frac{3x}{2}</math> by way of circle O, and <math>\angle AHG</math> is 180-3y/2 by way of circle <math>C</math>. |
| − | 5x+40=180-2x | + | |
| − | 7x=140 | + | This means that: |
| − | x=20 | + | |
| − | y=28. | + | <math>180-3x/2=180-3y/2+12</math>, |
| + | |||
| + | which when simplified yields <math>3x/2+12=3y/2</math>, or <math>x+8=y</math>. | ||
| + | Since: | ||
| + | <math>5y=180-2x</math>, <math>5x+40=180-2x</math> | ||
| + | So: | ||
| + | <math>7x=140, x=20</math> | ||
| + | <math>y=28.</math> | ||
<math>\angle BAG</math> is equal to <math>\angle BAE</math> + <math>\angle EAG</math>, which equates to 3x/2+y. | <math>\angle BAG</math> is equal to <math>\angle BAE</math> + <math>\angle EAG</math>, which equates to 3x/2+y. | ||
| − | Plugging in yields 30+28, or 058 | + | Plugging in yields <math>30+28</math>, or <math>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}} | ||
Revision as of 21:19, 20 March 2015
Problem
Point 
and 
are equally spaced on a minor arc of a cirle. 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 
.
![[asy] pair A,B,C,D,E,F,G,H,I,O; O=(0,0); C=dir(90); B=dir(70); A=dir(50); D=dir(110); E=dir(130); draw(arc(O,1,50,130)); real x=2*sin(20*pi/180); F=x*dir(228)+C; G=x*dir(256)+C; H=x*dir(284)+C; I=x*dir(312)+C; draw(arc(C,x,200,340)); label("$A$",A,dir(0)); label("$B$",B,dir(75)); label("$C$",C,dir(90)); label("$D$",D,dir(105)); label("$E$",E,dir(180)); label("$F$",F,dir(225)); label("$G$",G,dir(260)); label("$H$",H,dir(280)); label("$I$",I,dir(315));[/asy]](http://latex.artofproblemsolving.com/d/0/9/d09d2462dec1d60eaa6947ab799ce506fb914c99.png)
Solution
Let O be the center of the circle with ABCDE on it.
Let 
and 
.
is therefore 
by way of circle and 
by way of circle 
.
is 
by way of circle O, and is 180-3y/2 by way of circle .
This means that:
,
,
which when simplified yields, or
.
, or 
.
Since:
, 
So:
is equal to 
+ 
, which equates to 3x/2+y.
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. 
