Difference between revisions of "2007 AMC 12B Problems/Problem 22"
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==Solution== | ==Solution== | ||
First, notice that each of the midpoints of <math>AB</math>,<math>BC</math>, and <math>CA</math> are on the locus. | First, notice that each of the midpoints of <math>AB</math>,<math>BC</math>, and <math>CA</math> are on the locus. | ||
| − | Suppose after some time the particles have each been displaced by a short distance <math>x</math>, to new positions <math>A'</math> and <math>M'</math> respectively. Consider <math>\triangle ABM</math> and drop a perpendicular from <math>M'</math> to hit <math>AB</math> at <math>Y</math>. Then, <math>BA'=1-x</math> and <math>BM'=1/2+x</math>. From here, we can use properties of a <math>30-60-90</math> triangle to determine the lengths <math>YA'</math> and <math>YM'</math> as monomials in <math>x</math>. Thus, the locus of the midpoint will be linear between each of the three special points mentioned above. It follows that the locus consists of the only triangle with those three points as vertices. | + | Suppose after some time the particles have each been displaced by a short distance <math>x</math>, to new positions <math>A'</math> and <math>M'</math> respectively. Consider <math>\triangle ABM</math> and drop a perpendicular from <math>M'</math> to hit <math>AB</math> at <math>Y</math>. Then, <math>BA'=1-x</math> and <math>BM'=1/2+x</math>. From here, we can use properties of a <math>30-60-90</math> triangle to determine the lengths <math>YA'</math> and <math>YM'</math> as monomials in <math>x</math>. Thus, the locus of the midpoint will be linear between each of the three special points mentioned above. It follows that the locus consists of the only triangle with those three points as vertices. (A cheaper way to find the shape of the region is to look at the answer choices: if it were any sort of conic section then the ratio would not generally be rational.) |
Comparing inradii between this "midpoint" triangle and the original triangle, the area contained by <math>R</math> must be <math>\textbf{(A)} \frac{1}{16}</math> of the total area. | Comparing inradii between this "midpoint" triangle and the original triangle, the area contained by <math>R</math> must be <math>\textbf{(A)} \frac{1}{16}</math> of the total area. | ||
Revision as of 02:06, 23 November 2017
Problem 22
Two particles move along the edges of equilateral 
in the direction ![\[A\Rightarrow B\Rightarrow C\Rightarrow A,\]](http://latex.artofproblemsolving.com/c/f/e/cfe840ebf4592b958ba4c17b5f42b74b2d93403b.png)
starting simultaneously and moving at the same speed. One starts at 
, and the other starts at the midpoint of 
. The midpoint of the line segment joining the two particles traces out a path that encloses a region 
. What is the ratio of the area of to the area of ?
Solution
First, notice that each of the midpoints of 
,
, and 
are on the locus.
Suppose after some time the particles have each been displaced by a short distance 
, to new positions 
and 
respectively. Consider 
and drop a perpendicular from to hit at 
. Then, 
and 
. From here, we can use properties of a 
triangle to determine the lengths 
and 
as monomials in . Thus, the locus of the midpoint will be linear between each of the three special points mentioned above. It follows that the locus consists of the only triangle with those three points as vertices. (A cheaper way to find the shape of the region is to look at the answer choices: if it were any sort of conic section then the ratio would not generally be rational.)
Comparing inradii between this "midpoint" triangle and the original triangle, the area contained by must be 
of the total area.
See Also
| 2007 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 21 |
Followed by Problem 23 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
