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

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==Solution 4== | ==Solution 4== | ||

− | Using the double-angle formula for sine, what we need to find is <math> | + | Using the double-angle formula for sine, what we need to find is <math>AP\cdot \sin(DAP) = AD\cdot 2\sin( DAM) \cos(DAM) = 4\cdot 2\cdot \frac{2}{\sqrt{20}}\cdot\frac{4}{\sqrt{20}} = \frac{16}{5}</math>. |

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

## Revision as of 19:03, 27 April 2016

## 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 1

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:

Subtract the second equation from the first:

Then substitute:

Thus and making and .

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 .

## Solution 3

As in Solution 2, draw in and and denote their intersection point . Next, drop a perpendicular from to and denote the foot as . as they are both radii and similarly so is a kite and by a well-known theorem.

Pythagorean theorem gives us . Clearly by angle-angle and by Hypotenuse Leg. Manipulating similar triangles gives us

## Solution 4

Using the double-angle formula for sine, what we need to find is .

## See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.