Difference between revisions of "2015 AIME I Problems/Problem 6"
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This means that: | This means that: | ||
| − | <math>180-3x | + | <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>. | which when simplified yields <math>3x/2+12=3y/2</math>, or <math>x+8=y</math>. | ||
Revision as of 01:51, 28 November 2017
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 
.
![[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 
be the center of the circle with 
on it.
Let 
and 
.
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. 
