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Difference between revisions of "2006 AMC 10B Problems/Problem 24"

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== Problem ==
 
== Problem ==
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Circles with centers <math>O</math> and <math>P</math> have radii <math>2</math> and <math>4</math>, respectively, and are externally tangent. Points <math>A</math> and <math>B</math> on the circle with center <math>O</math> and points <math>C</math> and <math>D</math> on the circle with center <math>P</math> are such that <math>AD</math> and <math>BC</math> are common external tangents to the circles. What is the area of the concave hexagon <math>AOBCPD</math>?
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[[Circle]]s with centers <math>O</math> and <math>P</math> have radii <math>2</math> and <math>4</math>, respectively, and are externally tangent. Points <math>A</math> and <math>B</math> on the circle with center <math>O</math> and points <math>C</math> and <math>D</math> on the circle with center <math>P</math> are such that <math>AD</math> and <math>BC</math> are common external tangents to the circles. What is the area of the [[concave]] [[hexagon]] <math>AOBCPD</math>?
  
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[[Image:2006amc10b24.gif]]
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<asy>size(200);defaultpen(linewidth(0.8));
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pair X=(-6,0), O=origin, P=(6,0), B=tangent(X, O, 2, 1), A=tangent(X, O, 2, 2), C=tangent(X, P, 4, 1), D=tangent(X, P, 4, 2);
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pair top=X+15*dir(X--A), bottom=X+15*dir(X--B);
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draw(Circle(O, 2)^^Circle(P, 4));
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draw(bottom--X--top);
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draw(A--O--B^^O--P^^D--P--C);
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pair point=X;
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label("$2$", midpoint(O--A), dir(point--midpoint(O--A)));
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label("$4$", midpoint(P--D), dir(point--midpoint(P--D)));
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label("$O$", O, SE);
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label("$P$", P, dir(point--P));
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pair point=O;
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label("$A$", A, dir(point--A));
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label("$B$", B, dir(point--B));
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pair point=P;
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label("$C$", C, dir(point--C));
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label("$D$", D, dir(point--D));
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fill((-3,7)--(-3,-7)--(-7,-7)--(-7,7)--cycle, white);</asy>
  
 
<math> \mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2} </math>
 
<math> \mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2} </math>
  
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== Solution ==
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==Video Solution 1==
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Since a tangent line is perpendicular to the radius containing the tangent point, <math>\angle OAD = \angle PDA = 90^\circ</math>.
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https://youtu.be/cdjZ9Xd3Yt8
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~ Education, the Study of Everything
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== Solution 1 (Similar Triangles) ==
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When we see this problem, it practically screams similar triangles at us. Extend <math>OP</math> to the left until it intersects lines <math>AD</math> and <math>BC</math> at point <math>E</math>. Triangles <math>EBO</math> and <math>ECP</math> are similar, and by symmetry, so are triangles <math>EAO</math> and <math>EDP</math>. Then, the area of kite <math>EAOB</math> is just <math>4 \sqrt{2} \times 2</math> and the area of kite <math>EDPC</math> is <math>8 \sqrt{2} \times 4</math> (using our similarity ratios). The difference of these yields the area of hexagon <math>AOBCPD</math> or <math>24\sqrt{2} \Longrightarrow \boxed{\mathrm{(B)}}</math>.
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~icecreamrolls8
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== Solution 2 (Perpendiculars) ==
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 +
File is too big, so go to https://www.imgur.com/a/7aphGaa
 +
 
 +
Sorry for the wrong point names, I didn't know how to change them.
 +
 
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Since a [[tangent line]] is [[perpendicular]] to the [[radius]] containing the point of tangency, <math>\angle OAD = \angle PDA = 90^\circ</math>.
  
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Construct a perpendicular to <math>DP</math> that goes through point <math>O</math>. Label the point of intersection <math>X</math>.
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Construct a perpendicular to <math>DP</math> that goes through point <math>O</math>. Label the point of [[intersection]] <math>X</math>.
  
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Clearly <math>OADX</math> is a rectangle.  
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Clearly <math>OADX</math> is a [[rectangle]], so <math>DX=2</math> and <math>PX=2</math>. By the [[Pythagorean Theorem]], <math>OX = \sqrt{6^2 - 2^2} = 4\sqrt{2}</math>.
  
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Therefore <math>DX=2</math> and <math>PX=2</math>.
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The area of <math>OADX</math> is <math>2\cdot4\sqrt{2}=8\sqrt{2}</math>. The area of <math>OXP</math> is <math>\frac{1}{2}\cdot2\cdot4\sqrt{2}=4\sqrt{2}</math>, so the area of quadrilateral <math>OADP</math> is <math>8\sqrt{2}+4\sqrt{2}=12\sqrt{2}</math>. Using similar steps, the area of quadrilateral <math>OBCP</math> is also <math>12\sqrt{2}</math>. Therefore, the area of hexagon <math>AOBCPD</math> is <math>2\cdot12\sqrt{2}= 24\sqrt{2} \Longrightarrow \boxed{\mathrm{(B)}}</math>.
  
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By the [[Pythagorean Theorem]]:
 
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<math>OX = \sqrt{6^2 - 2^2} = 4\sqrt{2}</math>.
 
  
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The area of <math>OADX</math> is <math>2\cdot4\sqrt{2}=8\sqrt{2}</math>.
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Note: Quadrilaterals <math>OADP</math> and <math>OBCP</math> are congruent, so they have equal areas.
  
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The area of <math>OXP</math> is <math>\frac{1}{2}\cdot2\cdot4\sqrt{2}=4\sqrt{2}</math>.
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== Video Solution ==
  
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So the area of quadrilateral <math>OADP</math> is <math>8\sqrt{2}+4\sqrt{2}=12\sqrt{2}</math>.
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https://www.youtube.com/watch?v=EmUfNAxLZ7A    ~David
  
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Using similar steps, the area of quadrilateral <math>OBCP</math> is also <math>12\sqrt{2}</math>
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== See also ==
 +
{{AMC10 box|year=2006|ab=B|num-b=23|num-a=25}}
  
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Therefore, the area of hexagon <math>AOBCPD</math> is <math>2\cdot12\sqrt{2}= 24\sqrt{2} \Rightarrow B </math>
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[[Category:Introductory Geometry Problems]]
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== See Also ==
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[[Category:Area Problems]]
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*[[2006 AMC 10B Problems]]
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[[Category:Circle Problems]]
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{{MAA Notice}}

Latest revision as of 13:38, 6 April 2024

Problem

Circles with centers $O$ and $P$ have radii $2$ and $4$, respectively, and are externally tangent. Points $A$ and $B$ on the circle with center $O$ and points $C$ and $D$ on the circle with center $P$ are such that $AD$ and $BC$ are common external tangents to the circles. What is the area of the concave hexagon $AOBCPD$?

[asy]size(200);defaultpen(linewidth(0.8)); pair X=(-6,0), O=origin, P=(6,0), B=tangent(X, O, 2, 1), A=tangent(X, O, 2, 2), C=tangent(X, P, 4, 1), D=tangent(X, P, 4, 2); pair top=X+15*dir(X--A), bottom=X+15*dir(X--B); draw(Circle(O, 2)^^Circle(P, 4)); draw(bottom--X--top); draw(A--O--B^^O--P^^D--P--C); pair point=X; label("$2$", midpoint(O--A), dir(point--midpoint(O--A))); label("$4$", midpoint(P--D), dir(point--midpoint(P--D))); label("$O$", O, SE); label("$P$", P, dir(point--P)); pair point=O; label("$A$", A, dir(point--A)); label("$B$", B, dir(point--B)); pair point=P; label("$C$", C, dir(point--C)); label("$D$", D, dir(point--D)); fill((-3,7)--(-3,-7)--(-7,-7)--(-7,7)--cycle, white);[/asy]

$\mathrm{(A) \ } 18\sqrt{3}\qquad \mathrm{(B) \ } 24\sqrt{2}\qquad \mathrm{(C) \ } 36\qquad \mathrm{(D) \ } 24\sqrt{3}\qquad \mathrm{(E) \ } 32\sqrt{2}$

Video Solution 1

https://youtu.be/cdjZ9Xd3Yt8

~ Education, the Study of Everything

Solution 1 (Similar Triangles)

When we see this problem, it practically screams similar triangles at us. Extend $OP$ to the left until it intersects lines $AD$ and $BC$ at point $E$. Triangles $EBO$ and $ECP$ are similar, and by symmetry, so are triangles $EAO$ and $EDP$. Then, the area of kite $EAOB$ is just $4 \sqrt{2} \times 2$ and the area of kite $EDPC$ is $8 \sqrt{2} \times 4$ (using our similarity ratios). The difference of these yields the area of hexagon $AOBCPD$ or $24\sqrt{2} \Longrightarrow \boxed{\mathrm{(B)}}$.

~icecreamrolls8

Solution 2 (Perpendiculars)

File is too big, so go to https://www.imgur.com/a/7aphGaa

Sorry for the wrong point names, I didn't know how to change them.

Since a tangent line is perpendicular to the radius containing the point of tangency, $\angle OAD = \angle PDA = 90^\circ$.

Construct a perpendicular to $DP$ that goes through point $O$. Label the point of intersection $X$.

Clearly $OADX$ is a rectangle, so $DX=2$ and $PX=2$. By the Pythagorean Theorem, $OX = \sqrt{6^2 - 2^2} = 4\sqrt{2}$.

The area of $OADX$ is $2\cdot4\sqrt{2}=8\sqrt{2}$. The area of $OXP$ is $\frac{1}{2}\cdot2\cdot4\sqrt{2}=4\sqrt{2}$, so the area of quadrilateral $OADP$ is $8\sqrt{2}+4\sqrt{2}=12\sqrt{2}$. Using similar steps, the area of quadrilateral $OBCP$ is also $12\sqrt{2}$. Therefore, the area of hexagon $AOBCPD$ is $2\cdot12\sqrt{2}= 24\sqrt{2} \Longrightarrow \boxed{\mathrm{(B)}}$.


Note: Quadrilaterals $OADP$ and $OBCP$ are congruent, so they have equal areas.

Video Solution

https://www.youtube.com/watch?v=EmUfNAxLZ7A ~David

See also

2006 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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 10 Problems and Solutions

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