Difference between revisions of "2011 AMC 10B Problems/Problem 20"
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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. | 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. | ||
+ | |||
+ | == Solution 4 == | ||
+ | <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+B+D)/3, H=(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,S);label("$H$",H,E); | ||
+ | label("$2$",(D--C),SW); | ||
+ | </asy> | ||
+ | Since <math>H</math> and <math>E</math> are halfway between <math>AB</math> and <math>CB</math>, respectively, we know that <math>\overline{BH}=\overline{BE}=1</math>. By symmetry, <math>\Delta BFG</math> is equilateral, so <math>\angle FBG=60^\circ\implies\angle EBF=\angle HBG=30^\circ</math> and therefore <math>\Delta EBF</math> and <math>\Delta HBG</math> are 30-60-90 right triangles. | ||
+ | Thus, <math>[\Delta EBF]=[\Delta BFG]=\dfrac1{2\sqrt3}</math>. | ||
+ | We know that <math>\overline{FB}=\overline{GB}=\dfrac2{\sqrt3}</math>, so therefore <math>[\Delta BFG]=\dfrac{\sqrt3}4\left(\dfrac2{\sqrt3}\right)^2=\dfrac1{\sqrt3}</math>. | ||
+ | Summing these three regions, we get <math>\dfrac1{2\sqrt3}+\dfrac1{2\sqrt3}+\dfrac1{\sqrt3}=\boxed{\textbf{(C)}~\dfrac{2\sqrt3}3}</math>. | ||
+ | ~ Technodoggo, Asymptote diagram modified from Solution 1 | ||
+ | |||
+ | == Solution 5 == | ||
+ | <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+B+D)/3, H=(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--F); draw(D--G); draw(F--C); draw(A--G); draw(F--B); draw(B--G); | ||
+ | draw(B--D); | ||
+ | |||
+ | 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,S);label("$H$",H,E); | ||
+ | label("$2$",(D--C),SW); | ||
+ | </asy> | ||
+ | To keep it simple, break rhombus <math>ABCD</math> into two triangles, <math>ABD</math> and <math>BCD</math>. To see the area closest to the point <math>B</math>, notice that a third of each triangle, which contains all the points nearest to <math>B</math> in each triangle, is easily visualizable. Thus, a third of rhombus <math>ABCD</math> must be found. | ||
+ | |||
+ | We find the total area of rhombus <math>ABCD</math>, which we can again split into two congruent equilateral triangles with side length <math>2</math>. Using the formula of equilateral triangles and then multiplying by <math>\dfrac{1}{3}</math>: | ||
+ | <cmath>\dfrac{\sqrt{3}}{4}\cdot 2^2 \cdot 2 \cdot \dfrac{1}{3} = 2\sqrt{3} \cdot \dfrac{1}{3} = \boxed{\textbf{(C)}~\dfrac{2\sqrt{3}}{3}}</cmath> | ||
+ | -NSAoPS, diagram modified from Solution 1. | ||
==Video Solution by TheBeautyofMath== | ==Video Solution by TheBeautyofMath== |
Latest revision as of 20:05, 19 June 2024
- The following problem is from both the 2011 AMC 12B #16 and 2011 AMC 10B #20, so both problems redirect to this page.
Contents
Problem
Rhombus has side length and °. Region consists of all points inside the rhombus that are closer to vertex than any of the other three vertices. What is the area of ?
Solution
Suppose that is a point in the rhombus and let be the perpendicular bisector of . Then if and only if is on the same side of as . The line divides the plane into two half-planes; let be the half-plane containing . Let us define similarly and . Then is equal to . 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:
Since and are equilateral, contains , contains and , and contains . Then with and so . Multiply this by 4 and it turns out that the pentagon has area .
Solution 2
We follow the steps shown above until we draw pentagon . We know that rhombus can be divided into equilateral triangles and . Using the special right triangle rules, we find the height of the equilateral triangles (and the height of the rhombus) to be . Therefore, the area of is . We now have to take off the areas , , and to get the desired shape. is just half of and and are each , for a total area of .
Solution 3
We split rhombus into two equilateral triangles, and . In triangle , the probability that a randomly selected point is closer to than either other point is (why?). Similarly, in triangle , the same principle applies. Thus, the area of the region closer to than , , or is . Since and are congruent, we have , and we are done.
Solution 4
Since and are halfway between and , respectively, we know that . By symmetry, is equilateral, so and therefore and are 30-60-90 right triangles. Thus, . We know that , so therefore . Summing these three regions, we get . ~ Technodoggo, Asymptote diagram modified from Solution 1
Solution 5
To keep it simple, break rhombus into two triangles, and . To see the area closest to the point , notice that a third of each triangle, which contains all the points nearest to in each triangle, is easily visualizable. Thus, a third of rhombus must be found.
We find the total area of rhombus , which we can again split into two congruent equilateral triangles with side length . Using the formula of equilateral triangles and then multiplying by : -NSAoPS, diagram modified from Solution 1.
Video Solution by TheBeautyofMath
~IceMatrix
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.