Difference between revisions of "2003 AMC 12A Problems/Problem 17"
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Square <math>ABCD</math> has sides of length <math>4</math>, and <math>M</math> is the midpoint of <math>\overline{CD}</math>. A circle with radius <math>2</math> and center <math>M</math> intersects a circle with radius <math>4</math> and center <math>A</math> at points <math>P</math> and <math>D</math>. What is the distance from <math>P</math> to <math>\overline{AD}</math>? | Square <math>ABCD</math> has sides of length <math>4</math>, and <math>M</math> is the midpoint of <math>\overline{CD}</math>. A circle with radius <math>2</math> and center <math>M</math> intersects a circle with radius <math>4</math> and center <math>A</math> at points <math>P</math> and <math>D</math>. What is the distance from <math>P</math> to <math>\overline{AD}</math>? | ||
− | + | <asy> | |
+ | pair A,B,C,D,M,P; | ||
+ | D=(0,0); | ||
+ | C=(10,0); | ||
+ | B=(10,10); | ||
+ | A=(0,10); | ||
+ | M=(5,0); | ||
+ | P=(8,4); | ||
+ | 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)); | ||
+ | label("$A$",A,NW); | ||
+ | label("$B$",B,NE); | ||
+ | label("$C$",C,SE); | ||
+ | label("$D$",D,SW); | ||
+ | label("$M$",M,S); | ||
+ | label("$P$",P,N); | ||
+ | </asy> | ||
<math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}</math> | <math>\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}</math> |
Revision as of 21:08, 12 February 2017
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
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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