2023 AMC 10A Problems/Problem 24

Problem

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. $\dfrac{3\sqrt{3}}{2}(3^2-6\cdot1^2) = \dfrac{3\sqrt{3}}{2}(3) = \boxed{\textbf{(C)}~\frac{9 \sqrt{3}}{2}}$

~Technodoggo

Solution 2 (Not rigorous)

Note that one can "slide' the small hexagons along their respective edges, and either by sliding them to the center or to the corners, and thus getting that the side length of the larger hexagon is 3. The rest proceeds the same as solution 1.

Solution 2.1 (Clarification)

Notice that when sliding the smaller hexagon along the edge, we see that the contact edge withe the smaller hexagon "in front" of it is $60^{\circ}$ , thus meaning the hexagon "in front" is pushed at a speed $\sin{60^{\circ}}$ times the actual speed of the hexagon. We can preform a similar analysis on the hexagon that is being pushed and get that the speed at which that hexagon is moving is $\frac{1}{\sin{60^{\circ}}}$ times the speed it pushed by. As we can see, the two factors cancel out and by the same argument, every small hexagon can move at the same speed while mantaining an edge of contact with the two adjacent hexagons.

$[asy] unitsize(1cm); draw(scale(3)*polygon(6)); filldraw(shift(dir(0)*2+dir(120)*0.4)*polygon(6), pink); filldraw(shift(dir(60)*2+dir(180)*0.4)*polygon(6), palered); filldraw(shift(dir(120)*2+dir(240)*0.4)*polygon(6), paleyellow); filldraw(shift(dir(180)*2+dir(300)*0.4)*polygon(6), palegreen); filldraw(shift(dir(240)*2+dir(360)*0.4)*polygon(6), palecyan); filldraw(shift(dir(300)*2+dir(420)*0.4)*polygon(6), paleblue); draw(dir(120)*3+dir(240)*0.4--dir(120)*3,red+linewidth(1.5),EndArrow(size=12)); draw(dir(180)*3+dir(300)*0.4--dir(180)*3,red+linewidth(1.5),EndArrow(size=12)); draw(dir(240)*3+dir(0)*0.4--dir(240)*3,red+linewidth(1.5),EndArrow(size=12)); draw(dir(300)*3+dir(60)*0.4--dir(300)*3,red+linewidth(1.5),EndArrow(size=12)); draw(dir(0)*3+dir(120)*0.4--dir(360)*3,red+linewidth(1.5),EndArrow(size=12)); draw(dir(60)*3+dir(180)*0.4--dir(60)*3,red+linewidth(1.5),EndArrow(size=12)); [/asy]$ $[asy] unitsize(1cm); draw(scale(3)*polygon(6)); filldraw(shift(dir(0)*2+dir(120)*0)*polygon(6), pink); filldraw(shift(dir(60)*2+dir(180)*0)*polygon(6), palered); filldraw(shift(dir(120)*2+dir(240)*0)*polygon(6), paleyellow); filldraw(shift(dir(180)*2+dir(300)*0)*polygon(6), palegreen); filldraw(shift(dir(240)*2+dir(360)*0)*polygon(6), palecyan); filldraw(shift(dir(300)*2+dir(420)*0)*polygon(6), paleblue); [/asy]$ Diagram by ihatemath123

Note

The number $\frac{3}{7}$ is irrelevant to solve the problem. In fact, if the smaller hexagons have side length $x$ , the side length of the large hexagon will always be $3x$ .

Solution

We put the diagram to a complex plane, with the center of the outside hexagon at the origin. We denote by $s$ the length of each side of the outside hexagon.

The complex number of the upper left vertex of the upper right small hexagon is $\begin{align*} s e^{i 30^\circ} + \frac{3}{7} e^{i \left( 90^\circ + 60^\circ \right)} - \sqrt{3} . \end{align*}$

The complex number of the upper right vertex of the top small hexagon is $\begin{align*} s i + \frac{3}{7} e^{i \left( \pi + \frac{\pi}{6} \right)} + e^{- i \frac{\pi}{6}} . \end{align*}$

The above two vertices are on the same vertical line. So their real part values are the same. By solving this equation, we get $s = 3$ .

Therefore, the area of the region not occupied by the blocks is $\begin{align*} 6 \cdot \frac{\sqrt{3}}{4} s^2 - 6 \cdot 6 \cdot \frac{\sqrt{3}}{4} 1^2 = \boxed{\textbf{(C) } \frac{9 \sqrt{3}}{2}} . \end{align*}$

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by epicbird08

https://youtu.be/7XoekZvtQuU

~EpicBird08

Video Solution by Sohil Rathi

https://youtu.be/pObolGKEKc0

Video Solution

https://youtu.be/tLTTuQ7tOb4

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by TheBeautyofMath

https://youtu.be/uaNbxfuas5o

~IceMatrix

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
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