Difference between revisions of "2015 AIME I Problems/Problem 6"
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Let <math>O</math> be the center of the circle with <math>ABCDE</math> on it. | Let <math>O</math> be the center of the circle with <math>ABCDE</math> on it. | ||
− | 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>. | + | 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 <cmath>\frac{360-4x}{2}=180-2x</cmath> by way of circle <math>O</math>. | <math>\angle ECA</math> is, therefore, <math>5y</math> by way of circle <math>C</math> and <cmath>\frac{360-4x}{2}=180-2x</cmath> by way of circle <math>O</math>. | ||
<math>\angle ABD</math> is <math>180 - \frac{3x}{2}</math> by way of circle <math>O</math>, and <cmath>\angle AHG = 180 - \frac{3y}{2}</cmath> 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 <cmath>\angle AHG = 180 - \frac{3y}{2}</cmath> by way of circle <math>C</math>. | ||
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Let <math>m</math> be the degree measurement of <math>\angle GCH</math>. Since <math>G,H</math> lie on a circle with center <math>C</math>, <math>\angle GHC=\frac{180-m}{2}=90-\frac{m}{2}</math>. | Let <math>m</math> be the degree measurement of <math>\angle GCH</math>. Since <math>G,H</math> lie on a circle with center <math>C</math>, <math>\angle GHC=\frac{180-m}{2}=90-\frac{m}{2}</math>. | ||
− | Since <math>\angle ACH=2 \angle GCH=2m</math>, <math>\angle AHC=\frac{180-2m}{2}=90-m</math>. Adding <math>\angle GHC</math> and <math>\angle AHC</math> gives <math>\angle AHG=180-\frac{3m}{2}</math>, and <math>\angle ABD=\angle AHG+12=192-\frac{3m}{2}</math>. Since <math>AE</math> is parallel to <math>BD</math>, <math>\angle DBA=180-\angle ABD=\frac{3m}{2}-12=\overarc{BE}</math>. | + | Since <math>\angle ACH=2 \angle GCH=2m</math>, <math>\angle AHC=\frac{180-2m}{2}=90-m</math>. Adding <math>\angle GHC</math> and <math>\angle AHC</math> gives <math>\angle AHG=180-\frac{3m}{2}</math>, and <math>\angle ABD=\angle AHG+12=192-\frac{3m}{2}</math>. Since <math>AE</math> is parallel to <math>BD</math>, <math>\angle DBA=180-\angle ABD=\frac{3m}{2}-12=</math><math>\overarc{BE}</math>. |
We are given that <math>A,B,C,D,E</math> are evenly distributed on a circle. Hence, | We are given that <math>A,B,C,D,E</math> are evenly distributed on a circle. Hence, | ||
− | <math>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}=\frac{\angle DBA}{3}=\frac{m}{2}-4</math> | + | <math>\overarc{ED}=\overarc{DC}=\overarc{CB}=\overarc{BA}</math><math>=\frac{\angle DBA}{3}=\frac{m}{2}-4</math> |
+ | |||
+ | Here comes the key: Draw a line through <math>C</math> parallel to <math>AE</math>, and select a point <math>X</math> to the right of point <math>C</math>. | ||
+ | |||
+ | <math>\angle ACX</math> = <math>\overarc{AB}</math> + <math>\overarc{BC}</math> = <math>m-8</math>. | ||
− | + | Let the midpoint of <math>\overline{HG}</math> be <math>Y</math>, then <math>\angle YCX=\angle ACX+\angle ACY=(m-8)+\frac{5m}{2}=90</math>. Solving gives <math>m=28</math> | |
The rest of the solution proceeds as in solution 1, which gives <math>\boxed{058}</math> | The rest of the solution proceeds as in solution 1, which gives <math>\boxed{058}</math> |
Revision as of 17:57, 7 February 2022
Contents
[hide]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 1
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 by way of circle .
This means that:
which when simplified yields or Since: and So: is equal to + , which equates to . Plugging in yields , or .
Solution 2
Let be the degree measurement of . Since lie on a circle with center , .
Since , . Adding and gives , and . Since is parallel to , .
We are given that are evenly distributed on a circle. Hence,
Here comes the key: Draw a line through parallel to , and select a point to the right of point .
= + = .
Let the midpoint of be , then . Solving gives
The rest of the solution proceeds as in solution 1, which gives
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.