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

Latest revision as of 20:43, 19 September 2024

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

@@ Line 2: / Line 2: @@
 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>?
-[[Image:5d50417537c6cddfb70810403c62787b889cdcb1.png]]
+<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>
-== Solution 1==
+== 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>.
-Let <math>D</math> be the origin. <math>A</math> is the point <math>(0,4)</math> and <math>M</math> is the point <math>(2,0)</math>. We are given the radius of the quarter circle and semicircle as <math>4</math> and <math>2</math>, respectively, so their equations, respectively, are:
-<math>x^2 + (y-4)^2 = 4^2</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>.
-<math>(x-2)^2 + y^2 = 2^2</math>
-Subtract the second equation from the first:
+Using <math>4</math> instead as the base, we can drop a altitude from P. <math>\frac{32}{5}=\frac{bh}{2}</math>. <math>\frac{32}{5}=\frac{4h}{2}</math>. Thus, the horizontal distance is  <math>\frac{16}{5} \implies \boxed{\textbf{(B)}\frac{16}{5}}</math>
-<math>x^2 + (y - 4)^2 - (x - 2)^2 - y^2 = 12</math>
-<math>4x - 8y + 12 = 12</math>
-<math>x = 2y.</math>
+~BJHHar
-Then substitute:
+== Solution 3 ==
-<math>(2y)^2 + (y - 4)^2 = 16</math>
+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>.
-<math>4y^2 + y^2 - 8y + 16 = 16</math>
+== Solution 4 ==
+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.
-<math>5y^2 - 8y = 0</math>
+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>
-<math>y(5y - 8) = 0.</math>
+== Solution 5 ==
+Using the double-angle formula for sine, what we need to find is <math>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}</math>.
-Thus <math>y = 0</math> and <math>y = \frac{8}{5}</math> making <math>x = 0</math> and <math>x = \frac{16}{5}</math>.
+==Solution 6(LoC)==
+We use the Law of Cosines:
-The first value of <math>0</math> is obviously referring to the x-coordinate of the point where the circles intersect at the origin, <math>D</math>, so the second value must be referring to the x coordinate of <math>P</math>. Since <math>\overline{AD}</math> is the y-axis, the distance to it from <math>P</math> is the same as the x-value of the coordinate of <math>P</math>, so the distance from <math>P</math> to <math>\overline{AD}</math> is <math>\frac{16}{5} \Rightarrow B</math>
+<math>32-32 \cos \theta = 8 + 8 \cos \theta </math>
-==Solution 2==
+<math>\frac{3}{5} = \cos \theta </math>
-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>.
+<math>2 + 2*\frac{3}{5} = \frac{16}{5}</math>
+== Solution 7 ==
+Let <math>H</math> be the foot of the perpendicular from <math>P</math> to <math>CD</math>, and let <math>HD = x</math>. Then we have <math>HC = 4-x</math>, and <math>PH = 4 - \sqrt{16 - x^2}</math>. Since <math>\triangle DHP \sim \triangle PHC</math>, we have <math>HP^2 = DH \cdot HC</math>, or <math>-x^2 + 4x = 16 - 8\sqrt{16-x^2}</math>. Solving gives <math>x = \frac{16}{5}</math>.
+==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 <math>AM</math>, <math>DP</math>, and <math>PR</math>. <math>PR</math> is parallel with <math>CD</math>
+<math>[AMD] = \frac12 \cdot AD \cdot DM = 4</math>, <math>AM = \sqrt{AD^2 + DM^2} = 2 \sqrt{5}</math>
+<math>\triangle ADQ \sim \triangle AMD</math> by <math>AA</math>, <math>[ADQ] = [AMD] \cdot \left( \frac{AD}{AM} \right) ^2 = 4 \cdot \left( \frac{2 \sqrt{5}}{5} \right)^2 = \frac{16}{5}</math>
+<math>\triangle ADQ \cong \triangle APQ</math>, <math>[APD] = 2 \cdot [ADQ] = 2 \cdot \frac{16}{5} = \frac{32}{5}</math>
-==Solution 3==
+<math>PR = \frac{2 \cdot [APD]}{AD} = \frac{2 \cdot \frac{32}{5}}{4} = \boxed{\textbf{(B) } \frac{16}{5} }</math>
-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.
+~[https://artofproblemsolving.com/wiki/index.php/User:Isabelchen isabelchen]
-Manipulating similar triangles gives us <math>PZ=\frac{16}{5}</math>
 == See Also ==
-*[[2003 AMC 12A Problems]]
+{{AMC12 box|year=2003|ab=A|num-b=16|num-a=18}}
-*[[2003 AMC 12A Problems/Problem 16|Previous Problem]]
-*[[2003 AMC 12A Problems/Problem 18|Next Problem]]
 {{MAA Notice}}

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

Page

Toolbox

Search