# Difference between revisions of "2003 AMC 12A Problems/Problem 17"

m (→See Also) |
(solution 2) |
||

Line 46: | Line 46: | ||

Solving each factor for 0 yields <math>x = 0 , \frac{16}{5}</math>. The first value of <math>0</math> is obviously referring to the x-coordinate of the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x coordinate of <math>P</math>. Since <math>\overline{AD}</math> is the y-axis, the distance to it from <math>P</math> is the same as the x-value of the coordinate of <math>P</math>, so the distance from <math>P</math> to <math>\overline{AD}</math> is <math>\frac{16}{5} \Rightarrow B</math> | Solving each factor for 0 yields <math>x = 0 , \frac{16}{5}</math>. The first value of <math>0</math> is obviously referring to the x-coordinate of the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x coordinate of <math>P</math>. Since <math>\overline{AD}</math> is the y-axis, the distance to it from <math>P</math> is the same as the x-value of the coordinate of <math>P</math>, so the distance from <math>P</math> to <math>\overline{AD}</math> is <math>\frac{16}{5} \Rightarrow B</math> | ||

+ | |||

+ | ==Solution 2== | ||

+ | |||

+ | Note that <math>P</math> is merely a reflection of <math>D</math> over <math>AM</math>. Call the intersection of <math>AM</math> and <math>DP</math> <math>X</math>. Drop perpendiculars from <math>X</math> and <math>P</math> to <math>AD</math>, and denote their respective points of intersection by <math>J</math> and <math>K</math>. We then have <math>\triangle DXJ\sim\triangle DPK</math>, with a scale factor of 2. Thus, we can find <math>XJ</math> and double it to get our answer. With some analytical geometry, we find that <math>XJ=\frac{8}{5}</math>, implying that <math>PK=\frac{16}{5}</math>. | ||

== See Also == | == See Also == |

## Revision as of 17:04, 30 June 2012

## Contents

## Problem

Square has sides of length , and is the midpoint of . A circle with radius and center intersects a circle with radius and center at points and . What is the distance from to ?

## Solution

Let be the origin. is the point and is the point . We are given the radius of the quarter circle and semicircle as and , respectively, so their equations, respectively, are:

Algebraically manipulating the second equation gives:

Substituting this back into the first equation:

Solving each factor for 0 yields . The first value of is obviously referring to the x-coordinate of the point where the circles intersect at the origin, , so the second value must be referring to the x coordinate of . Since is the y-axis, the distance to it from is the same as the x-value of the coordinate of , so the distance from to is

## Solution 2

Note that is merely a reflection of over . Call the intersection of and . Drop perpendiculars from and to , and denote their respective points of intersection by and . We then have , with a scale factor of 2. Thus, we can find and double it to get our answer. With some analytical geometry, we find that , implying that .