Difference between revisions of "1994 AHSME Problems/Problem 4"
(→Solution) |
Logsobolev (talk | contribs) (→Solution) |
||
Line 7: | Line 7: | ||
--Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | --Solution by [http://www.artofproblemsolving.com/Forum/memberlist.php?mode=viewprofile&u=200685 TheMaskedMagician] | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | The diameter of the circle is <math>25- (-5)=30</math> so the radius is <math>15</math> and the center of the circle is on the <math>x</math>-axis at <math>x=10</math>. The only point with <math>y=15</math> must be the point exactly above the center. | ||
==See Also== | ==See Also== |
Latest revision as of 02:31, 28 May 2021
Contents
Problem
In the -plane, the segment with endpoints and is the diameter of a circle. If the point is on the circle, then
Solution
We see that the center of this circle is at . The radius is . So the equation of this circle is Substituting yields so .
--Solution by TheMaskedMagician
Solution 2
The diameter of the circle is so the radius is and the center of the circle is on the -axis at . The only point with must be the point exactly above the center.
See Also
1994 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 3 |
Followed by Problem 5 | |
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 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.