Difference between revisions of "2003 AMC 12A Problems/Problem 17"
(→Solution 2) |
|||
Line 56: | Line 56: | ||
== Solution 2== | == Solution 2== | ||
− | <math>APMD</math> obviously forms a kite. Let the intersection of the diagonals be <math>E</math>. <math>AE+EM=AM=2\sqrt{5}</math> Let <math>AE=x</math>. Then, <math>EM=2\sqrt{5}-x</math>. By Pythagorean Theorem, <math>DE^2=4^2-AE^2=2^2-EM^2</math>. Thus, <math>16-x^2=4-(2\sqrt{5}-x)^2</math>. Simplifying, <math>x=\frac{8}{\sqrt{5}}</math>. By Pythagoras again, <math>DE=\frac{4}{\sqrt{5}}</math>. Then, the area of <math>ADP</math> is <math>DE\cdot AE=\frac{32}{5}</math>. Using <math>4</math> instead as the base, we can drop a altitude from P. <math>\frac{32}{5}=\frac{bh}{2}\implies\frac{32}{5}=\frac{4h}{2}</math>. Thus, the horizontal distance is <math>\frac{16}{5}\implies\boxed{\textbf{(E)\frac{16}{5}}</math> | + | <math>APMD</math> obviously forms a kite. Let the intersection of the diagonals be <math>E</math>. <math>AE+EM=AM=2\sqrt{5}</math> Let <math>AE=x</math>. Then, <math>EM=2\sqrt{5}-x</math>. By Pythagorean Theorem, <math>DE^2=4^2-AE^2=2^2-EM^2</math>. Thus, <math>16-x^2=4-(2\sqrt{5}-x)^2</math>. Simplifying, <math>x=\frac{8}{\sqrt{5}}</math>. By Pythagoras again, <math>DE=\frac{4}{\sqrt{5}}</math>. Then, the area of <math>ADP</math> is <math>DE\cdot AE=\frac{32}{5}</math>. Using <math>4</math> instead as the base, we can drop a altitude from P. <math>\frac{32}{5}=\frac{bh}{2}\implies\frac{32}{5}=\frac{4h}{2}</math>. Thus, the horizontal distance is <math>\frac{16}{5}\implies \boxed{\textbf{(E)\frac{16}{5}}</math> |
− | |||
==Solution 3== | ==Solution 3== |
Revision as of 23:10, 27 September 2019
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 $\frac{16}{5}\implies \boxed{\textbf{(E)\frac{16}{5}}$ (Error compiling LaTeX. Unknown error_msg)
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 .
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.