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 $\overline{AD}$ ?

$[asy] pair A,B,C,D,M,P; D=(0,0); C=(10,0); B=(10,10); A=(0,10); M=(5,0); P=(8,4); dot(M); dot(P); draw(A--B--C--D--cycle,linewidth(0.7)); draw((5,5)..D--C..cycle,linewidth(0.7)); draw((7.07,2.93)..B--A--D..cycle,linewidth(0.7)); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$M$",M,S); label("$P$",P,N); [/asy]$

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

Solution 2

$APMD$ obviously forms a kite. Let the intersection of the diagonals be $E$ . $AE+EM=AM=2\sqrt{5}$ Let $AE=x$ . Then, $EM=2\sqrt{5}-x$ .

By Pythagorean Theorem, $DE^2=4^2-AE^2=2^2-EM^2$ . Thus, $16-x^2=4-(2\sqrt{5}-x)^2$ . Simplifying, $x=\frac{8}{\sqrt{5}}$ . By Pythagoras again, $DE=\frac{4}{\sqrt{5}}$ . Then, the area of $ADP$ is $DE\cdot AE=\frac{32}{5}$ .

Using $4$ instead as the base, we can drop a altitude from P. $\frac{32}{5}=\frac{bh}{2}$ . $\frac{32}{5}=\frac{4h}{2}$ . Thus, the horizontal distance is $\frac{16}{5} \implies \boxed{\textbf{(B)}\frac{16}{5}}$

~BJHHar

Solution 3

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 4

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

Solution 5

Using the double-angle formula for sine, what we need to find is $AP\cdot \sin(DAP) = AP\cdot 2\sin( DAM) \cos(DAM) = 4\cdot 2\cdot \frac{2}{\sqrt{20}}\cdot\frac{4}{\sqrt{20}} = \frac{16}{5}$ .

Solution 6(LoC)

We use the Law of Cosines:

$32-32 \cos \theta = 8 + 8 \cos \theta$

$\frac{3}{5} = \cos \theta$

$2 + 2*\frac{3}{5} = \frac{16}{5}$

Solution 7

Let $H$ be the foot of the perpendicular from $P$ to $CD$ , and let $HD = x$ . Then we have $HC = 4-x$ , and $PH = 4 - \sqrt{16 - x^2}$ . Since $\triangle DHP \sim \triangle PHC$ , we have $HP^2 = DH \cdot HC$ , or $-x^2 + 4x = 16 - 8\sqrt{16-x^2}$ . Solving gives $x = \frac{16}{5}$ .

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$",R,W); draw(rightanglemark(M, Q, P), linewidth(.5)); [/asy]$

Draw $AM$ , $DP$ , and $PR$ . $PR$ is parallel with $CD$

$[AMD] = \frac12 \cdot AD \cdot DM = 4$ , $AM = \sqrt{AD^2 + DM^2} = 2 \sqrt{5}$

$\triangle ADQ \sim \triangle AMD$ by $AA$ , $[ADQ] = [AMD] \cdot \left( \frac{AD}{AM} \right) ^2 = 4 \cdot \left( \frac{2 \sqrt{5}}{5} \right)^2 = \frac{16}{5}$

$\triangle ADQ \cong \triangle APQ$ , $[APD] = 2 \cdot [ADQ] = 2 \cdot \frac{16}{5} = \frac{32}{5}$

$PR = \frac{2 \cdot [APD]}{AD} = \frac{2 \cdot \frac{32}{5}}{4} = \boxed{\textbf{(B) } \frac{16}{5} }$

~isabelchen

2003 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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