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

Revision as of 22:50, 6 December 2012

Problem

Square $ABCD$ has sides of length $4$ , and $M$ is the midpoint of $\overline{CD}$ . A circle with radius $2$ and center $M$ intersects a circle with radius $4$ and center $A$ at points $P$ and $D$ . What is the distance from $P$ to $\overline{AD}$ ?

$\textbf{(A)}\ 3 \qquad \textbf{(B)}\ \frac {16}{5} \qquad \textbf{(C)}\ \frac {13}{4} \qquad \textbf{(D)}\ 2\sqrt {3} \qquad \textbf{(E)}\ \frac {7}{2}$

Solution 1

Let $D$ be the origin. $A$ is the point $(0,4)$ and $M$ is the point $(2,0)$ . We are given the radius of the quarter circle and semicircle as $4$ and $2$ , respectively, so their equations, respectively, are:

$x^2 + (y-4)^2 = 4^2$

$(x-2)^2 + y^2 = 2^2$

Algebraically manipulating the second equation gives:

$y^2 = 2^2 - (x-2)^2$

$y^2 = (2-(x-2)(2+(x-2))$

$y^2 = (4-x)(x)$

$y = \sqrt{4x - x^2}$

Substituting this back into the first equation:

$x^2 + (\sqrt{4x - x^2} - 4)^2 = 4^2$

$x^2 + 4x - x^2 - 8\sqrt{4x - x^2} + 16 = 16$

$4x - 8\sqrt{4x - x^2} = 0$

$4x = 8\sqrt{4x - x^2}$

$16x^2 = 64(4x - x^2)$

$16x^2 = 256x - 64x^2$

$80x^2 - 256x = 0$

$x(80x - 256) = 0$

Solving each factor for 0 yields $x = 0 , \frac{16}{5}$ . The first value of $0$ is obviously referring to the x-coordinate of the point where the circles intersect at the origin, $D$ , so the second value must be referring to the x coordinate of $P$ . Since $\overline{AD}$ is the y-axis, the distance to it from $P$ is the same as the x-value of the coordinate of $P$ , so the distance from $P$ to $\overline{AD}$ is $\frac{16}{5} \Rightarrow B$

Solution 2

Note that $P$ is merely a reflection of $D$ over $AM$ . Call the intersection of $AM$ and $DP$ $X$ . Drop perpendiculars from $X$ and $P$ to $AD$ , and denote their respective points of intersection by $J$ and $K$ . We then have $\triangle DXJ\sim\triangle DPK$ , with a scale factor of 2. Thus, we can find $XJ$ and double it to get our answer. With some analytical geometry, we find that $XJ=\frac{8}{5}$ , implying that $PK=\frac{16}{5}$ .

Solution 3

As in Solution 2, draw in $DP$ and $AM$ and denote their intersection point $X$ . Next, drop a perpendicular from $P$ to $AD$ and denote the foot as $Z$ . $AP \cong AD$ as they are both radii and similarly $DM \cong MP$ so $APMD$ is a kite and $DX \perp XM$ by a well-known theorem.

Pythagorean theorem gives us $AM=2 \sqrt{5}$ . Clearly $\triangle XMD \sim \triangle XDA \sim \triangle DMA \sim \triangle ZDP$ by angle-angle and $\triangle XMD \cong \triangle XMP$ by Hypotenuse Leg. Manipulating similar triangles gives us $PZ=\frac{16}{5}$

@@ Line 50: / Line 50: @@
 Note that <math>P</math> is merely a reflection of <math>D</math> over <math>AM</math>. Call the intersection of <math>AM</math> and <math>DP</math> <math>X</math>. Drop perpendiculars from <math>X</math> and <math>P</math> to <math>AD</math>, and denote their respective points of intersection by <math>J</math> and <math>K</math>. We then have <math>\triangle DXJ\sim\triangle DPK</math>, with a scale factor of 2. Thus, we can find <math>XJ</math> and double it to get our answer. With some analytical geometry, we find that <math>XJ=\frac{8}{5}</math>, implying that <math>PK=\frac{16}{5}</math>.
+==Solution 3==
+As in Solution 2, draw in <math>DP</math> and <math>AM</math> and denote their intersection point <math>X</math>. Next, drop a perpendicular from <math>P</math> to <math>AD</math> and denote the foot as <math>Z</math>. <math>AP \cong AD</math> as they are both radii and similarly <math>DM \cong MP</math> so <math>APMD</math> is a kite and <math>DX \perp XM</math> by a well-known theorem.
+Pythagorean theorem gives us <math>AM=2 \sqrt{5}</math>. Clearly <math>\triangle XMD \sim \triangle XDA \sim \triangle DMA \sim \triangle ZDP</math> by angle-angle and <math>\triangle XMD \cong \triangle XMP</math> by Hypotenuse Leg.
+Manipulating similar triangles gives us <math>PZ=\frac{16}{5}</math>
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