Difference between revisions of "2003 AMC 12A Problems/Problem 17"
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==Solution 8== | ==Solution 8== | ||
+ | |||
+ | <asy> | ||
+ | size(8cm, 8cm); | ||
+ | pair A,B,C,D,M,P,Q,R; | ||
+ | D = (0,0); | ||
+ | C = (10,0); | ||
+ | B = (10,10); | ||
+ | A = (0,10); | ||
+ | M = (5,0); | ||
+ | P = (8,4); | ||
+ | Q = (D+P)/2; | ||
+ | R = (0,4); | ||
+ | dot(M); | ||
+ | dot(P); | ||
+ | draw(A--B--C--D--cycle,linewidth(0.7)); | ||
+ | draw((5,5)..D--C..cycle,linewidth(0.7)); | ||
+ | draw((7.07,2.93)..B--A--D..cycle,linewidth(0.7)); | ||
+ | draw(A--M,linewidth(0.7)); | ||
+ | draw(A--P,linewidth(0.7)); | ||
+ | draw(D--P,linewidth(0.7)); | ||
+ | draw(R--P,linewidth(0.7)); | ||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,NE); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,SW); | ||
+ | label("$M$",M,S); | ||
+ | label("$P$",P,N); | ||
+ | label("$Q$",Q,W); | ||
+ | label("$R$",R,W); | ||
+ | </asy> | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2003|ab=A|num-b=16|num-a=18}} | {{AMC12 box|year=2003|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 01:08, 27 August 2022
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 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 the Law of Cosines:
Solution 7
Let be the foot of the perpendicular from to , and let . Then we have , and . Since , we have , or . Solving gives .
Solution 8
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.