Difference between revisions of "1954 AHSME Problems/Problem 39"
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Note that the midpoint of <math>P</math> to the point <math>Q</math> is the image of <math>Q</math> under a homothety of factor <math>\frac{1}{2}</math> with center <math>P</math>. Since homotheties preserve circles, the image of the midpoint as <math>Q</math> varies over the circle is a circle centered at the midpoint of <math>P</math> and the original center and radius half the original radius. Therefore, our answer is <math>\boxed{\text{(E)}}</math>, and we are done. | Note that the midpoint of <math>P</math> to the point <math>Q</math> is the image of <math>Q</math> under a homothety of factor <math>\frac{1}{2}</math> with center <math>P</math>. Since homotheties preserve circles, the image of the midpoint as <math>Q</math> varies over the circle is a circle centered at the midpoint of <math>P</math> and the original center and radius half the original radius. Therefore, our answer is <math>\boxed{\text{(E)}}</math>, and we are done. | ||
| − | + | ||
| + | <asy> | ||
import graph; | import graph; | ||
unitsize(60); | unitsize(60); | ||
| Line 19: | Line 20: | ||
draw(P--Q, red); | draw(P--Q, red); | ||
draw(c, darkgreen); | draw(c, darkgreen); | ||
| − | label(" | + | label("$P$", P, N); |
| − | label(" | + | label("$Q$", Q, NW); |
pair M; | pair M; | ||
| Line 26: | Line 27: | ||
dot(M, blue); | dot(M, blue); | ||
draw(Circle(P/2, 1/2, 100), blue); | draw(Circle(P/2, 1/2, 100), blue); | ||
| − | label(" | + | label("$M$", M, SE); |
| − | + | </asy> | |
==See Also== | ==See Also== | ||
Latest revision as of 15:18, 5 July 2020
The locus of the midpoint of a line segment that is drawn from a given external point 
to a given circle with center 
and radius 
, is:
Solution
Note that the midpoint of to the point 
is the image of under a homothety of factor 
with center . Since homotheties preserve circles, the image of the midpoint as varies over the circle is a circle centered at the midpoint of and the original center and radius half the original radius. Therefore, our answer is 
, and we are done.
![[asy] import graph; unitsize(60); pair P, Q; path c; P = (2,0); Q = dir(142); c = Circle((0,0), 1, 100); dot(P, red); dot(Q, darkgreen); draw(P--Q, red); draw(c, darkgreen); label("$P$", P, N); label("$Q$", Q, NW); pair M; M = (P+Q)/2; dot(M, blue); draw(Circle(P/2, 1/2, 100), blue); label("$M$", M, SE); [/asy]](http://latex.artofproblemsolving.com/1/a/e/1ae0e163a0098bd0e804fc0942a762d07c7dff00.png)
See Also
| 1954 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 35 |
Followed by Problem 37 | |
| 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 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
