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2003 AMC 12A Problems/Problem 17

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 $\overline{AD}$?

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

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 $PZ=\frac{16}{5}$

See Also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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