Difference between revisions of "2020 AMC 10B Problems/Problem 14"

Revision as of 21:27, 27 January 2021

The following problem is from both the 2020 AMC 10B #14 and 2020 AMC 12B #11, so both problems redirect to this page.

Problem 14

size(140); fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4)); fill(arc((2,0),1,180,0)--(2,0)--cycle,white); fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white); fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white); fill(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); label(" $2$ ",(3.5,3sqrt(3)/2),NE); </asy>

Solution 1

$[asy] real x=sqrt(3); real y=2sqrt(3); real z=3.5; real a=x/2; real b=0.5; real c=3a; pair A, B, C, D, E, F; A = (1,0); B = (3,0); C = (4,x); D = (3,y); E = (1,y); F = (0,x); fill(A--B--C--D--E--F--A--cycle,grey); fill(arc((2,0),1,0,180)--cycle,white); fill(arc((2,y),1,180,360)--cycle,white); fill(arc((z,a),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); label("2",(z,c),NE); pair X,Y,Z; X = (1,0); Y = (2,0); Z = (1.5,a); pair d = 1.9*dir(7); dot(X); dot(Y); dot(Z); label("A",X,SW); label("B",Y,SE); label("C",Z,N); draw(X--Y--Z--A); label("1",(1.5,0),S); label("1",(1.75,a/2),dir(30)); draw(anglemark(Y,X,Z,8),blue); label("$60^\circ$",anglemark(Y,X,Z),d); [/asy]$

Let point A be a vertex of the regular hexagon, let point B be the midpoint of the line connecting point A and a neighboring vertex, and let point C be the second intersection of the two semicircles that pass through point A. Then, $BC = 1$ , since B is the center of the semicircle with radius 1 that C lies on, $AB = 1$ , since B is the center of the semicircle with radius 1 that A lies on, and $\angle BAC = 60^\circ$ , as a regular hexagon has angles of 120 $^\circ$ , and $\angle BAC$ is half of any angle in this hexagon. Now, using the sine law, $\frac{1}{\sin \angle ACB} = \frac{1}{\sin 60^\circ}$ , so $\angle ACB = 60^\circ$ . Since the angles in a triangle sum to 180 $^\circ$ , $\angle ABC$ is also 60 $^\circ$ . Therefore, $\triangle ABC$ is an equilateral triangle with side lengths of 1.

$[asy] real x=sqrt(3); real y=2sqrt(3); real z=3.5; real a=x/2; real b=0.5; real c=3a; pair A, B, C, D, E, F; A = (1,0); B = (3,0); C = (4,x); D = (3,y); E = (1,y); F = (0,x); fill(A--B--C--D--E--F--A--cycle,grey); fill(arc((2,0),1,0,180)--cycle,white); fill(arc((2,y),1,180,360)--cycle,white); fill(arc((z,a),1,60,240)--cycle,white); fill(arc((b,a),1,300,480)--cycle,white); fill(arc((b,c),1,240,420)--cycle,white); fill(arc((z,c),1,120,300)--cycle,white); draw(A--B--C--D--E--F--A); draw(arc((z,c),1,120,300)); draw(arc((b,c),1,240,420)); draw(arc((b,a),1,300,480)); draw(arc((z,a),1,60,240)); draw(arc((2,y),1,180,360)); draw(arc((2,0),1,0,180)); pair G,H,I,J,K; G = (2,0); H = (2.5,a); I = (1.5,a); J = (1,0); K = (3,0); pair d = 2.7*dir(78); dot(G); dot(H); dot(I); dot(J); dot(K); label("2",(z,c),NE); label("1",(1.5,0),S); label("1",(2.5,0),S); add(pathticks(G--J,1,0.5,0,3,red)); add(pathticks(I--J,1,0.5,0,3,red)); add(pathticks(G--I,1,0.5,0,3,red)); add(pathticks(G--H,1,0.5,0,3,red)); add(pathticks(G--K,1,0.5,0,3,red)); add(pathticks(K--H,1,0.5,0,3,red)); label("$60^\circ$",anglemark(H,G,I),d); draw(anglemark(H,G,I,8),blue); label("$60^\circ$",G,2*dir(146)); draw(anglemark(I,G,J,8),blue); label("$60^\circ$",G,2.8*dir(28)); draw(anglemark(K,G,H,8),blue); draw(G--J--I--G); draw(G--H--K--G); [/asy]$

Since the area of a regular hexagon can be found with the formula $\frac{3\sqrt{3}s^2}{2}$ , where $s$ is the side length of the hexagon, the area of this hexagon is $\frac{3\sqrt{3}(2^2)}{2} = 6\sqrt{3}$ . Since the area of an equilateral triangle can be found with the formula $\frac{\sqrt{3}}{4}s^2$ , where $s$ is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is $\frac{\sqrt{3}}{4}(1^2) = \frac{\sqrt{3}}{4}$ . Since the area of a circle can be found with the formula $\pi r^2$ , the area of a sixth of a circle with radius 1 is $\frac{\pi(1^2)}{6} = \frac{\pi}{6}$ . In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of $\frac{\sqrt{3}}{4}$ , and one sixth of a circle with radius 1 colored white, with an area of $\frac{\pi}{6}$ . The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is $2(\frac{\sqrt{3}}{4}) + \frac{\pi}{6}$ , which equals $\frac{\sqrt{3}}{2} + \frac{\pi}{6}$ , and the total area colored white is $6(\frac{\sqrt{3}}{2} + \frac{\pi}{6})$ , which equals $3\sqrt{3} + \pi$ . Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is $6\sqrt{3} - (3\sqrt{3} + \pi)$ , which equals $\boxed{\textbf{(D) }3\sqrt{3} - \pi}$ .

Solution 2

First, subdivide the hexagon into 24 equilateral triangles with side length 1: $[asy] size(140); fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4)); fill(arc((2,0),1,180,0)--(2,0)--cycle,white); fill(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle,white); fill(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle,white); fill(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle,white); fill(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle,white); fill(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle,white); draw((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--(1,0)); draw(arc((2,0),1,180,0)--(2,0)--cycle); draw(arc((3.5,sqrt(3)/2),1,60,240)--(3.5,sqrt(3)/2)--cycle); draw(arc((3.5,3sqrt(3)/2),1,120,300)--(3.5,3sqrt(3)/2)--cycle); draw(arc((2,2sqrt(3)),1,180,360)--(2,2sqrt(3))--cycle); draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle); draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle); label("$2$",(3.5,3sqrt(3)/2),NE); draw((1,0)--(3,2sqrt(3))); draw((3,0)--(1,2sqrt(3))); draw((4,sqrt(3))--(0,sqrt(3))); draw((2,0)--(3.5,3sqrt(3)/2)); draw((3.5,sqrt(3)/2)--(2,2sqrt(3))); draw((3.5,3sqrt(3)/2)--(0.5,3sqrt(3)/2)); draw((2,2sqrt(3))--(0.5,sqrt(3)/2)); draw((2,0)--(0.5,3sqrt(3)/2)); draw((3.5,sqrt(3)/2)--(0.5,sqrt(3)/2)); [/asy]$ Now note that the entire shaded region is just 6 times this part: $[asy] size(100); fill((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle,gray(0.4)); fill(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle,white); draw(arc((2,2sqrt(3)),1,240,300)--(2,2sqrt(3))--cycle); label("$1$",(2.25,7sqrt(3)/4),NE); draw((2,sqrt(3))--(2.5,3sqrt(3)/2)--(2,2sqrt(3))--(1.5,3sqrt(3)/2)--cycle); draw((2.5,3sqrt(3)/2)--(1.5,3sqrt(3)/2)); [/asy]$ The entire rhombus is just 2 equilatrial triangles with side lengths of 1, so it has an area of: $2\cdot\frac{\sqrt{3}}{4}=\frac{\sqrt{3}}{2}$ The arc that is not included has an area of: $\frac16 \cdot\pi \cdot1^2 = \frac{\pi}{6}$ Hence, the area of the shaded region in that section is $\frac{\sqrt{3}}{2}-\frac{\pi}{6}$ For a final area of: $6\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)=3\sqrt{3}-\pi\Rightarrow \boxed{\mathrm{(D)}}$ ~N828335

Video Solution

https://youtu.be/t6yjfKXpwDs?t=786 (for AMC 10)

https://youtu.be/0xgTR3UEqbQ (for AMC 12)

~IceMatrix

https://youtu.be/oTqx8OqSMQI ~DSA_Catachu

@@ Line 1: / Line 1: @@
 {{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #14]] and [[2020 AMC 12B Problems|2020 AMC 12B #11]]}}
-== Solution 6 ==
+== Problem 14 ==
-We can see there are six congruent "leaves" on the original figure and let's say the total area of them is <math>S</math>. If we make a circle with radius one in the circle(the dashed one), we can get another six congruent "leaves", also they are same with the old six ones. So the area of the six new leaves can also be represented by <math>S</math>.
-Then the area of the shaded region can be expressed in the following two ways:
-. <math>Area</math> of hexagon - (<math>Area</math> of six semicircles <math>-</math> <math>Area</math> of six <math>old</math> "leaves") <math>=</math> <math>\frac{\sqrt{3}}{4}*4*6 - (3\pi - S);</math>
-. <math>Area</math>  of  dashed  circle <math>-</math> <math>Area</math> of six <math>new</math> "leaves" = <math>\pi - S;</math>
-Then we have <math>\frac{\sqrt{3}}{4}*4*6 - (3\pi - S) = \pi - S</math>
-<math>S = 2\pi - 3\sqrt{3}</math>
-Plug in <math>S</math> back to either <math>1</math> or <math>2</math>, we can slove shaded area <math>= 3\sqrt{3} - \pi</math>  <math>\rightarrow\boxed{\mathrm{(D)}}</math>
---RyanZ@BRS
-<asy>
 size(140);
 fill((1,0)--(3,0)--(4,sqrt(3))--(3,2sqrt(3))--(1,2sqrt(3))--(0,sqrt(3))--cycle,gray(0.4));
@@ Line 33: / Line 18: @@
 draw(arc((0.5,3sqrt(3)/2),1,240,420)--(0.5,3sqrt(3)/2)--cycle);
 draw(arc((0.5,sqrt(3)/2),1,300,480)--(0.5,sqrt(3)/2)--cycle);
-label("$2$",(3.5,3sqrt(3)/2),NE);
+label("<math>2</math>",(3.5,3sqrt(3)/2),NE);
-draw(Arc((2,sqrt(3)), 1, 0, 360),dashed+linewidth(.5));
 </asy>

2020 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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

2020 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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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