Difference between revisions of "2002 AMC 12P Problems/Problem 8"
(→Solution 1) |
m |
||
Line 14: | Line 14: | ||
</math> | </math> | ||
− | |||
− | == Solution | + | == Solution== |
We can solve this with some simple coordinate geometry. Let <math>A</math> be the origin at let <math>AB</math> be located on the positive <math>x-</math>axis. The equation of semi-circle <math>AB</math> is <math>(x-13)^2+y^2=13^2, y \geq 0.</math> Since <math>E</math> and <math>F</math> are both perpendicular to <math>C</math> and <math>D</math> respectively, they must have the same <math>x -</math> coordinate. Plugging in <math>1</math> and <math>8</math> into our semi-circle equation gives us <math>y=5</math> and <math>y=12</math> respectively. The distance formula on <math>(1, 5)</math> and <math>(8, 12)</math> gives us our answer of <math>\sqrt{(1-8)^2 + (5-12)^2}=\sqrt{2(7^2)}=\boxed{\textbf{(D) } 7\sqrt{2}}.</math> | We can solve this with some simple coordinate geometry. Let <math>A</math> be the origin at let <math>AB</math> be located on the positive <math>x-</math>axis. The equation of semi-circle <math>AB</math> is <math>(x-13)^2+y^2=13^2, y \geq 0.</math> Since <math>E</math> and <math>F</math> are both perpendicular to <math>C</math> and <math>D</math> respectively, they must have the same <math>x -</math> coordinate. Plugging in <math>1</math> and <math>8</math> into our semi-circle equation gives us <math>y=5</math> and <math>y=12</math> respectively. The distance formula on <math>(1, 5)</math> and <math>(8, 12)</math> gives us our answer of <math>\sqrt{(1-8)^2 + (5-12)^2}=\sqrt{2(7^2)}=\boxed{\textbf{(D) } 7\sqrt{2}}.</math> | ||
− | |||
− | |||
== See also == | == See also == | ||
{{AMC12 box|year=2002|ab=P|num-b=7|num-a=9}} | {{AMC12 box|year=2002|ab=P|num-b=7|num-a=9}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:24, 19 May 2024
Problem
Let be a segment of length , and let points and be located on such that and . Let and be points on one of the semicircles with diameter for which and are perpendicular to . Find
Solution
We can solve this with some simple coordinate geometry. Let be the origin at let be located on the positive axis. The equation of semi-circle is Since and are both perpendicular to and respectively, they must have the same coordinate. Plugging in and into our semi-circle equation gives us and respectively. The distance formula on and gives us our answer of
See also
2002 AMC 12P (Problems • Answer Key • Resources) | |
Preceded by Problem 7 |
Followed by Problem 9 |
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.