Difference between revisions of "2015 AIME I Problems/Problem 6"
m (→Solution) |
(→Solution) |
||
Line 29: | Line 29: | ||
==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=ED=DC=CB=BA</math> and <math>y=EF=FG=GH=HI=IA</math>. | Let <math>x=ED=DC=CB=BA</math> and <math>y=EF=FG=GH=HI=IA</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>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: |
Revision as of 22:21, 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
.
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 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.