Difference between revisions of "2020 AMC 10B Problems/Problem 14"
Somebody62 (talk | contribs) (→Solution) |
Somebody62 (talk | contribs) (→Solution: Explain why two equilateral triangles exist in each sixth of the hexagon) |
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==Solution== | ==Solution== | ||
+ | <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, <math>BC = 1</math>, since B is the center of the semicircle with radius 1 that C lies on, <math>AB = 1</math>, since B is the center of the semicircle with radius 1 that A lies on, and <math>\angle BAC = 60^\circ</math>, as a regular hexagon has angles of 120<math>^\circ</math>, and <math>\angle BAC</math> is half of any angle in this hexagon. Now, using the sine law, <math>\frac{1}{\sin \angle ACB} = \frac{1}{\sin 60^\circ}</math>, so <math>\angle ACB = 60^\circ</math>. Since the angles in a triangle sum to 180<math>^\circ</math>, <math>\angle ABC</math> is also 60<math>^\circ</math>. Therefore, <math>\triangle ABC</math> is an equilateral triangle with side lengths of 1. | ||
+ | |||
<asy> | <asy> | ||
real x=sqrt(3); | real x=sqrt(3); | ||
Line 89: | Line 140: | ||
label("$60^\circ$",anglemark(H,G,I),d); | label("$60^\circ$",anglemark(H,G,I),d); | ||
draw(anglemark(H,G,I,8),blue); | 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--J--I--G); | ||
draw(G--H--K--G); | draw(G--H--K--G); |
Revision as of 08:51, 8 February 2020
Contents
[hide]Problem
As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?
Solution
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, , since B is the center of the semicircle with radius 1 that C lies on, , since B is the center of the semicircle with radius 1 that A lies on, and , as a regular hexagon has angles of 120, and is half of any angle in this hexagon. Now, using the sine law, , so . Since the angles in a triangle sum to 180, is also 60. Therefore, is an equilateral triangle with side lengths of 1.
Since the area of a regular hexagon can be found with the formula , where is the side length of the hexagon, the area of this hexagon is . Since the area of an equilateral triangle can be found with the formula , where is the side length of the equilateral triangle, the area of an equilateral triangle with side lengths of 1 is . Since the area of a circle can be found with the formula , the area of a sixth of a circle with radius 1 is . In each sixth of the hexagon, there are two equilateral triangles colored white, each with an area of , and one sixth of a circle with radius 1 colored white, with an area of . The rest of the sixth is colored gray. Therefore, the total area that is colored white in each sixth of the hexagon is , which equals , and the total area colored white is , which equals . Since the area colored gray equals the total area of the hexagon minus the area colored white, the area colored gray is , which equals .
Video Solution
~IceMatrix
See Also
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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