Difference between revisions of "2015 AIME I Problems/Problem 6"
(→Problem) |
(→Problem) |
||
| Line 26: | Line 26: | ||
label("$H$",H,dir(280)); | label("$H$",H,dir(280)); | ||
label("$I$",I,dir(315));</asy> | label("$I$",I,dir(315));</asy> | ||
| + | |||
| + | ==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. | ||
| + | <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/2 by way of circle O, and <math>\angle AHG</math> is 180-3y/2 by way of circle C. | ||
| + | This means that 180-3x/2=180-3y/2+12, which when simplified yields 3x/2+12=3y/2, or x+8=y. | ||
| + | Since 5y=180-2x, | ||
| + | 5x+40=180-2x | ||
| + | 7x=140 | ||
| + | x=20 | ||
| + | y=28. | ||
| + | <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 | ||
==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 22:09, 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 x=ED=DC=CB=BA and y=EF=FG=GH=HI=IA.
is therefore 5y by way of circle C and 180-2x by way of circle O.
is 180-3x/2 by way of circle O, and is 180-3y/2 by way of circle C.
This means that 180-3x/2=180-3y/2+12, which when simplified yields 3x/2+12=3y/2, or x+8=y.
Since 5y=180-2x,
5x+40=180-2x
7x=140
x=20
y=28.
is equal to 
+ 
, which equates to 3x/2+y.
Plugging in yields 30+28, or 058
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. 
