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Difference between revisions of "2023 AMC 10A Problems/Problem 24"
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<math>\textbf{(A)}~\frac{13 \sqrt{3}}{3}\qquad\textbf{(B)}~\frac{216 \sqrt{3}}{49}\qquad\textbf{(C)}~\frac{9 \sqrt{3}}{2} \qquad\textbf{(D)}~ \frac{14 \sqrt{3}}{3}\qquad\textbf{(E)}~\frac{243 \sqrt{3}}{49}</math>
 
<math>\textbf{(A)}~\frac{13 \sqrt{3}}{3}\qquad\textbf{(B)}~\frac{216 \sqrt{3}}{49}\qquad\textbf{(C)}~\frac{9 \sqrt{3}}{2} \qquad\textbf{(D)}~ \frac{14 \sqrt{3}}{3}\qquad\textbf{(E)}~\frac{243 \sqrt{3}}{49}</math>
  
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==Solution 1==
  
 
<asy>
 
<asy>
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~Technodoggo
 
~Technodoggo
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==See Also==
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{{AMC10 box|year=2023|ab=A|num-b=23|num-a=25}}
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{{MAA Notice}}

Revision as of 20:42, 9 November 2023

Six regular hexagonal blocks of side length 1 unit are arranged inside a regular hexagonal frame. Each block lies along an inside edge of the frame and is aligned with two other blocks, as shown in the figure below. The distance from any corner of the frame to the nearest vertex of a block is $\frac{3}{7}$ unit. What is the area of the region inside the frame not occupied by the blocks? [asy] unitsize(1cm); draw(scale(3)*polygon(6)); filldraw(shift(dir(0)*2+dir(120)*0.4)*polygon(6), lightgray); filldraw(shift(dir(60)*2+dir(180)*0.4)*polygon(6), lightgray); filldraw(shift(dir(120)*2+dir(240)*0.4)*polygon(6), lightgray); filldraw(shift(dir(180)*2+dir(300)*0.4)*polygon(6), lightgray); filldraw(shift(dir(240)*2+dir(360)*0.4)*polygon(6), lightgray); filldraw(shift(dir(300)*2+dir(420)*0.4)*polygon(6), lightgray); [/asy] $\textbf{(A)}~\frac{13 \sqrt{3}}{3}\qquad\textbf{(B)}~\frac{216 \sqrt{3}}{49}\qquad\textbf{(C)}~\frac{9 \sqrt{3}}{2} \qquad\textbf{(D)}~ \frac{14 \sqrt{3}}{3}\qquad\textbf{(E)}~\frac{243 \sqrt{3}}{49}$

Solution 1

[asy] unitsize(1cm); pair A, B, C, D, E, F, W,X,Y,Z; real bigSide = 3; real smallSide = 1; real angle = 60; // Each external angle for the hexagon real offset = 3/7; // Offset for the smaller hexagons // Function to draw a hexagon given a starting point and side length void drawHexagon(pair start, real side) { pair current = start; for (int i = 0; i < 6; ++i) { pair next = current + side * dir(angle * i); draw(current--next); current = next; } draw(current--start); // Close the hexagon } // Define the first vertex of the big hexagon A = (0,0); // Calculate the other vertices of the big hexagon B = A + bigSide * dir(0); C = B + bigSide * dir(angle); D = C + bigSide * dir(2*angle); E = D + bigSide * dir(3*angle); F = E + bigSide * dir(4*angle); // Draw the big hexagon drawHexagon(A, bigSide); // Function to calculate the center of a side given two vertices pair sideCenter(pair start, pair end) { return (start + end)/2; } // Draw the smaller hexagons drawHexagon(A + offset * dir(0), smallSide); drawHexagon(B - smallSide * dir(0)+offset*dir(60), smallSide); drawHexagon(C - smallSide * dir(0)-dir(60)+dir(120)*3/7, smallSide); drawHexagon(D - 2*smallSide*dir(120)-(2+3/7)*smallSide, smallSide); drawHexagon(E - 2*smallSide*dir(60)+smallSide-3/7*dir(60), smallSide); drawHexagon(F + smallSide*dir(-60)+(3/7)*dir(-60), smallSide); // Optionally, label the vertices of the big hexagon label("$A$", A, SW); label("$B$", B, SE); label("$C$", C, E); label("$D$", D, NE); label("$E$", E, NW); label("$F$", F, W); void drawTrap(pair W, real side, pen p) { X = W+(3/7)*side*dir(0); Y = X+(4/7)*side*dir(60); Z = Y - side*dir(0); draw(W--X, p); draw(X--Y,p); draw(Y--Z,p); draw(Z--W,p); } W = A+smallSide*dir(120); drawTrap(W,1, red+2); pair W2,W3,W4,W5; W2 = A+3*dir(-90); W3 = W2+dir(90)*4*sqrt(3)/7; W4 = W3+dir(0)*6/7; W5 = W2+dir(0)*6/7; drawTrap(W2,2,blue+1); draw(W2--W3,blue+0.5); draw(W4--W5,blue+0.5); label("2/7",W3,NW); label("3/7",W3,NE); W4 = W3+6/7*dir(0); label("2/7",W4,NE); label("4/7",W2+dir(160)*0.5,W); draw(A -1.5*dir(45)-- F -1.5*dir(45), green+0.5); pair J,K,L,M,N; J = ((A/10+9*F/10))-0.25*dir(45); L = ((A+F)/2)-0.25*dir(45); K = ((J+L)/2)-0.25*dir(45); M = ((L+A)/2)-0.25*dir(45); N = ((A+F)/2)-1.6*dir(45); label("3/7",J,SW); label("4/7",L,SW); label("1",K,SW); label("1",M,SW); label("3",N,SW); [/asy]


Examining the red isosceles trapezoid with $1$ and $\dfrac{3}{7}$ as two bases, we know that the side lengths are $\dfrac{4}{7}$ from $30-60-90$ triangle.

We can conclude that the big hexagon has side length 3.

Thus the target area is: area of big hexagon - 6 * area of small hexagon. $\dfrac{3\sqrt{3}}{2}(3^2-6\cdot1^2) = \dfrac{3\sqrt{3}}{2}(3) = \dfrac{9\sqrt{3}}{2}$

~Technodoggo

See Also

2023 AMC 10A (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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