Difference between revisions of "2003 AMC 12A Problems/Problem 17"
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We use LoC: | We use LoC: | ||
− | <math>32-32 \cos \theta = 8 + 8 \cos \theta \ | + | <math>32-32 \cos \theta = 8 + 8 \cos \theta \linebreak</math> |
− | <math>\frac{3}{5} = \cos \theta \ | + | <math>\frac{3}{5} = \cos \theta \linebreak</math> |
<math>2 + 2*\frac{3}{5} = \frac{16}{5}</math> | <math>2 + 2*\frac{3}{5} = \frac{16}{5}</math> | ||
Revision as of 14:04, 12 July 2020
Contents
[hide]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
obviously forms a kite. Let the intersection of the diagonals be . Let . Then, .
By Pythagorean Theorem, . Thus, . Simplifying, . By Pythagoras again, . Then, the area of is .
Using instead as the base, we can drop a altitude from P. . . Thus, the horizontal distance is
~BJHHar
Solution 3
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 4
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 5
Using the double-angle formula for sine, what we need to find is .
Solution 6(LoC)
We use LoC:
See Also
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 16 |
Followed by Problem 18 |
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 |
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