Difference between revisions of "2011 AMC 10B Problems/Problem 20"

Revision as of 04:22, 5 April 2021

The following problem is from both the 2011 AMC 12B #16 and 2011 AMC 10B #20, so both problems redirect to this page.

Problem

Rhombus $ABCD$ has side length $2$ and $\angle B = 120$ °. Region $R$ consists of all points inside the rhombus that are closer to vertex $B$ than any of the other three vertices. What is the area of $R$ ?

$\textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2$

Solution

Suppose that $P$ is a point in the rhombus $ABCD$ and let $\ell_{BC}$ be the perpendicular bisector of $\overline{BC}$ . Then $PB < PC$ if and only if $P$ is on the same side of $\ell_{BC}$ as $B$ . The line $\ell_{BC}$ divides the plane into two half-planes; let $S_{BC}$ be the half-plane containing $B$ . Let us define similarly $\ell_{BD},S_{BD}$ and $\ell_{BA},S_{BA}$ . Then $R$ is equal to $ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}$ . The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:

$[asy] unitsize(8mm); defaultpen(linewidth(0.8pt)+fontsize(10pt)); dotfactor=4; pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2; fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray); draw(A--B--C--D--cycle); draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G); label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW); label("$E$",E,N);label("$F$",F,SW);label("$G$",G,SW);label("$H$",H,S);label("$I$",I,NE); label("$2$",(D--C),SW); [/asy]$ Since $\triangle BCD$ and $\triangle BAD$ are equilateral, $\ell_{BC}$ contains $D$ , $\ell_{BD}$ contains $A$ and $C$ , and $\ell_{BA}$ contains $D$ . Then $\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH$ with $BE = 1$ and $EF = \frac{1}{\sqrt{3}}$ so $[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}$ . Multiply this by 4 and it turns out that the pentagon has area $\boxed{(C)\frac{2\sqrt{3}}{3}}$ .

Solution 2

We follow the steps shown above until we draw pentagon $BIHFE$ . We know that rhombus $ABCD$ can be divided into equilateral triangles $\triangle ABD$ and $\triangle CBD$ . Using the $30-60-90$ special right triangle rules, we find the height of the equilateral triangles (and the height of the rhombus) to be $\sqrt{3}$ . Therefore, the area of $ABCD$ is $2\sqrt{3}$ . We now have to take off the areas $\triangle CDA$ , $\triangle CEF$ , and $\triangle AIH$ to get the desired shape. $\triangle CDA$ is just half of $ABCD$ $(\sqrt {3})$ and $\triangle AIH$ and $\triangle CEF$ are each $\frac{\sqrt {3}}{6}$ , for a total area of $2\sqrt {3}-\sqrt {3}-\frac{\sqrt{3}}{6}-\frac{\sqrt{3}}{6}=\boxed{(C)\frac{2\sqrt{3}}{3}}$ .

Solution 3

We split rhombus $ABCD$ into two equilateral triangles, $ABD$ and $BCD$ . In triangle $ABD$ , the probability that a randomly selected point is closer to $B$ than either other point is $\frac{1}{3}$ (why?). Similarly, in triangle $BCD$ , the same principle applies. Thus, the area of the region closer to $B$ than $A$ , $C$ , or $D$ is $\frac{1}{3} [ABD] + \frac{1}{3} [BCD]$ . Since $ABD$ and $BCD$ are congruent, we have $\frac{1}{3} [ABD] + \frac{1}{3} [BCD] = \frac{2}{3} [ABD] = \frac{2}{3} \cdot \frac{s^2\sqrt{3}}{4} = \frac{2}{3} \cdot \frac{(2)^2\sqrt{3}}{4} = \boxed{\frac{2\sqrt3}{3} = C}$ , and we are done.

Video Solution by TheBeautyofMath

https://youtu.be/gCmQlaiEG5A

~IceMatrix

@@ Line 1: / Line 1: @@
+{{duplicate|[[2011 AMC 12B Problems|2011 AMC 12B #16]] and [[2011 AMC 10B Problems|2011 AMC 10B #20]]}}
 == Problem==
 Rhombus <math>ABCD</math> has side length <math>2</math> and <math>\angle B = 120</math>°. Region <math>R</math> consists of all points inside the rhombus that are closer to vertex <math>B</math> than any of the other three vertices. What is the area of <math>R</math>?
-<math> \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2</math>
+<math> \textbf{(A)}\ \frac{\sqrt{3}}{3} \qquad\textbf{(B)}\ \frac{\sqrt{3}}{2} \qquad\textbf{(C)}\ \frac{2\sqrt{3}}{3} \qquad\textbf{(D)}\ 1 + \frac{\sqrt{3}}{3} \qquad\textbf{(E)}\ 2</math>
+[[Category: Introductory Geometry Problems]]
 == Solution ==
-Any point on the [[perpendicular bisector]] of two other points is equidistant from both of those points. This line will divide divide the rhombus into two sections. The region containing <math>B</math> is also the region where the points will be closer to <math>B</math> than the other vertex. The region <math>R</math> we are looking for is the intersection of the sections containing R from the perpendicular bisectors of <math>AB, CB,</math> and <math>DB.</math>
+Suppose that <math>P</math> is a point in the rhombus <math>ABCD</math> and let <math>\ell_{BC}</math> be the [[perpendicular bisector]] of <math>\overline{BC}</math>. Then <math>PB < PC</math> if and only if <math>P</math> is on the same side of <math>\ell_{BC}</math> as <math>B</math>. The line <math>\ell_{BC}</math> divides the plane into two half-planes; let <math>S_{BC}</math> be the half-plane containing <math>B</math>. Let us define similarly <math>\ell_{BD},S_{BD}</math> and <math>\ell_{BA},S_{BA}</math>. Then <math>R</math> is equal to <math>ABCD \cap S_{BC} \cap S_{BD} \cap S_{BA}</math>. The region turns out to be an irregular pentagon. We can make it easier to find the area of this region by dividing it into four triangles:
-<center><asy>
+<asy>
 unitsize(8mm);
 defaultpen(linewidth(0.8pt)+fontsize(10pt));
 dotfactor=4;
-pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0);
+pair A=(4,0), B=(2,2sqrt(3)), C=(-2,2sqrt(3)), D=(0,0), E=(B+C)/2, F=(B+C+D)/3, G=(A+C)/2, H=(A+B+D)/3, I=(A+B)/2;
-fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,lightgray);
+fill((0,2sqrt(3))--B--(3,sqrt(3))--(2,(2sqrt(3))/3)--(0,(4sqrt(3))/3)--cycle,gray);
 draw(A--B--C--D--cycle);
-draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C);
+draw(D--(0,2sqrt(3))); draw(D--(3,sqrt(3))); draw(A--C); draw(F--B--H); draw(B--G);
-label("$A$",A,SE);
-label("$B$",B,NE);
-label("$C$",C,NW);
-label("$D$",D,SW);
-label("$2$",(B--C),N);
-</asy></center>
-The region turns to be an irregular pentagon. We can make it easier to find the area of this by dividing it into a triangle and a trapezoid.
-<center><asy>
+label("$A$",A,SE);label("$B$",B,NE);label("$C$",C,NW);label("$D$",D,SW);
-unitsize(8mm);
+label("$E$",E,N);label("$F$",F,SW);label("$G$",G,SW);label("$H$",H,S);label("$I$",I,NE);
-defaultpen(linewidth(0.8pt)+fontsize(10pt));
+label("$2$",(D--C),SW);
-dotfactor=4;
+</asy>
+Since <math>\triangle BCD</math> and <math>\triangle BAD</math> are equilateral, <math>\ell_{BC}</math> contains <math>D</math>, <math>\ell_{BD}</math> contains <math>A</math> and <math>C</math>, and <math>\ell_{BA}</math> contains <math>D</math>. Then <math>\triangle BEF \cong \triangle BGF \cong \triangle BGH \cong \triangle BIH</math> with <math>BE = 1</math> and <math>EF = \frac{1}{\sqrt{3}}</math> so <math>[BEF] = \frac{1}{2}\cdot 1 \cdot \frac{\sqrt{3}}{3}</math>. Multiply this by 4 and it turns out that the pentagon has area <math>\boxed{(C)\frac{2\sqrt{3}}{3}}</math>.
-pair A=(0,2), B=(2sqrt(3),0), C=((4sqrt(3))/3,-2), D=((-4sqrt(3))/3,-2), E=(-2sqrt(3),0);
+== Solution 2==
-pair O=(0,0);
-draw(A--B--C--D--E--cycle); draw(B--E);
-draw(A--O,gray);
-label("$A$",A,N); label("$B$",B); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,W); label("$O$",O,S);
+We follow the steps shown above until we draw pentagon <math>BIHFE</math>. We know that rhombus <math>ABCD</math> can be divided into equilateral triangles <math>\triangle ABD</math> and <math>\triangle CBD</math>. Using the <math>30-60-90</math> special right triangle rules, we find the height of the equilateral triangles (and the height of the rhombus) to be <math>\sqrt{3}</math>. Therefore, the area of <math>ABCD</math> is <math>2\sqrt{3}</math>. We now have to take off the areas <math>\triangle CDA</math>, <math>\triangle CEF</math>, and <math>\triangle AIH</math> to get the desired shape. <math>\triangle CDA</math> is just half of <math>ABCD</math> <math>(\sqrt {3})</math> and <math>\triangle AIH</math> and <math>\triangle CEF</math> are each <math>\frac{\sqrt {3}}{6}</math>, for a total area of <math>2\sqrt {3}-\sqrt {3}-\frac{\sqrt{3}}{6}-\frac{\sqrt{3}}{6}=\boxed{(C)\frac{2\sqrt{3}}{3}}</math>.
-label("$1$",midpoint(E--A),NW); label("$1$",midpoint(B--A),NE); label("$\frac{1}{2}$",midpoint(A--O),E); label("$\frac{\sqrt{3}}{2}$",midpoint(B--O),S); label("$\frac{\sqrt{3}}{2}$",midpoint(E--O),S);
-</asy></center>
-We know <math>AE=AB=1</math> because they are each half of the sides of the rhombus. <math>\triangle EAO</math> and <math>\triangle BAO</math> are <math>30-60-90</math> triangles because <math>AO</math> is an angle bisector of the <math>120^\circ</math> angle <math>EAB.</math> We can solve for those triangles. The area of triangle <math>\triangle AEB</math> is <cmath>\frac{1}{2}bh = \frac{1}{2} \cdot \sqrt{3} \cdot \frac{1}{2} = \frac{\sqrt{3}}{4}</cmath>
+== Solution 3==
-Another thing we can do is find the height of trapezoid <math>BCDE.</math> Diagonal <math>BD</math> back in the first diagram is equal to <math>2</math> because it would form two equilateral triangles. Because the diagonals of a rhombus bisect each other, the distance from <math>A</math> to <math>CD</math> (in the second diagram) is <math>1.</math> Subtract <math>AO</math> from that, and we get the height of trapezoid <math>BCDE</math> is <math>\frac{1}{2}.</math>
+We split rhombus <math>ABCD</math> into two equilateral triangles, <math>ABD</math> and <math>BCD</math>. In triangle <math>ABD</math>, the probability that a randomly selected point is closer to <math>B</math> than either other point is <math>\frac{1}{3}</math> (why?). Similarly, in triangle <math>BCD</math>, the same principle applies. Thus, the area of the region closer to <math>B</math> than <math>A</math>, <math>C</math>, or <math>D</math> is <math>\frac{1}{3} [ABD] + \frac{1}{3} [BCD]</math>. Since <math>ABD</math> and <math>BCD</math> are congruent, we have <math>\frac{1}{3} [ABD] + \frac{1}{3} [BCD] = \frac{2}{3} [ABD] = \frac{2}{3} \cdot \frac{s^2\sqrt{3}}{4} = \frac{2}{3} \cdot \frac{(2)^2\sqrt{3}}{4} = \boxed{\frac{2\sqrt3}{3} = C}</math>, and we are done.
-To find the area of the trapezoid, we still need to find the length of the other base, <math>CD.</math>
+==Video Solution by TheBeautyofMath==
+https://youtu.be/gCmQlaiEG5A
-<center><asy>
+~IceMatrix
-unitsize(8mm);
-defaultpen(linewidth(0.8pt)+fontsize(9pt));
-dotfactor=4;
-pair A=(0,2), B=(2sqrt(3),0), C=((4sqrt(3))/3,-2), D=((-4sqrt(3))/3,-2), E=(-2sqrt(3),0);
-pair O=(0,0), X=((4sqrt(3))/3,0);
-draw(A--B--C--D--E--cycle); draw(B--E);
-draw(X--C,gray);
-label("$A$",A,N); label("$B$",B,NE); label("$C$",C,SE); label("$D$",D,SW); label("$E$",E,NW); label("$X$",X,NW);
-label("$\frac{1}{2}$",midpoint(X--C),W); label("$\sqrt{3}$",midpoint(B--E),N); label("$\frac{\sqrt{3}}{6}$",midpoint(X--B),N);
-</asy></center>
-Since <math>AB \perp BC</math> and <math>\angle ABE = 30^\circ, \triangle XBC</math> is another <math>30-60-90\text{ triangle }</math> and <math>XB = \frac{\sqrt{3}}{6}.</math>
-<math>BCDE</math> is an isosceles trapezoid, so <math>CD = EB - 2XB = \sqrt{3} - \frac{\sqrt{3}}{3} = \frac{2\sqrt{3}}{3}.</math> The area of the trapezoid is <cmath>\frac{h}{2}(b_1+b_2) = \frac{1}{4} (\frac{2\sqrt{3}}{3} + \sqrt{3}) = \frac{1}{4} \cdot \frac{5\sqrt{3}}{3} = \frac{5\sqrt{3}}{12}</cmath>
-Find the total area of the pentagon
-<cmath>\frac{5\sqrt{3}}{12} + \frac{\sqrt{3}}{4} = \frac{5\sqrt{3}}{12} + \frac{3\sqrt{3}}{12} = \frac{8\sqrt{3}}{12} = \boxed{\mathrm{(C) \ } \frac{2\sqrt{3}}{3}}</cmath>
 == See Also==
 {{AMC10 box|year=2011|ab=B|num-b=19|num-a=21}}
+{{AMC12 box|year=2011|ab=B|num-b=15|num-a=17}}
+{{MAA Notice}}

2011 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
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

2011 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 15	Followed by Problem 17
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

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