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Difference between revisions of "2014 AMC 10A Problems/Problem 13"
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==Solution 2==
 
==Solution 2==
−
As seen in the previous solution, segment <math>GH</math> is <math>\sqrt{3}</math>.  Think of the picture as one large equilateral triangle, <math>\triangle{JKL} with the sides of </math>2\sqrt{3}+1<math>, by extending </math>EF<math>, </math>GH<math>, and </math>DI<math> to points </math>J<math>, </math>K<math>, and </math>L<math>, respectively.  This makes the area of </math>\triangle{JKL}<math>  </math>\dfrac{\sqrt{3}}{4}(2\sqrt{3}+1)^2=\dfrac{12+13\sqrt{3}}{4}$.   
+
As seen in the previous solution, segment <math>GH</math> is <math>\sqrt{3}</math>.  Think of the picture as one large equilateral triangle, <math>\triangle{JKL}</math> with the sides of <math>2\sqrt{3}+1</math>, by extending <math>EF</math>, <math>GH</math>, and <math>DI</math> to points <math>J</math>, <math>K</math>, and <math>L</math>, respectively.  This makes the area of <math>\triangle{JKL}</math>  <math>\dfrac{\sqrt{3}}{4}(2\sqrt{3}+1)^2=\dfrac{12+13\sqrt{3}}{4}</math>.   
  
 
<asy>
 
<asy>

Revision as of 12:04, 13 February 2014

Problem

Equilateral $\triangle ABC$ has side length $1$, and squares $ABDE$, $BCHI$, $CAFG$ lie outside the triangle. What is the area of hexagon $DEFGHI$?

[asy] import graph; size(6cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,W); label("$F$",F,E); label("$G$",G,E); label("$H$",H,SE); label("$I$",I,SW); [/asy]

$\textbf{(A)}\ \dfrac{12+3\sqrt3}4\qquad\textbf{(B)}\ \dfrac92\qquad\textbf{(C)}\ 3+\sqrt3\qquad\textbf{(D)}\ \dfrac{6+3\sqrt3}2\qquad\textbf{(E)}\ 6$

Solution 1

The area of the equilateral triangle is $\dfrac{\sqrt{3}}{4}$. The area of the three squares is $3\times 1=3$.

Since $\angle C=360$, $\angle GCH=360-90-90-60=120$.

Dropping an altitude from $C$ to $GH$ allows to create a $30-60-90$ triangle since $\triangle GCH$ is isosceles. This means that the height of $\triangle GCH$ is $\dfrac{1}{2}$ and half the length of $GH$ is $\dfrac{\sqrt{3}}{2}$. Therefore, the area of each isosceles triangle is $\dfrac{1}{2}\times\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{4}$. Multiplying by $3$ yields $\dfrac{3\sqrt{3}}{4}$ for all three isosceles triangles.

Therefore, the total area is $3+\dfrac{\sqrt{3}}{4}+\dfrac{3\sqrt{3}}{4}=3+\dfrac{4\sqrt{3}}{4}=3+\sqrt{3}\implies\boxed{\textbf{(C)}\ 3+\sqrt3}$.

Solution 2

As seen in the previous solution, segment $GH$ is $\sqrt{3}$. Think of the picture as one large equilateral triangle, $\triangle{JKL}$ with the sides of $2\sqrt{3}+1$, by extending $EF$, $GH$, and $DI$ to points $J$, $K$, and $L$, respectively. This makes the area of $\triangle{JKL}$ $\dfrac{\sqrt{3}}{4}(2\sqrt{3}+1)^2=\dfrac{12+13\sqrt{3}}{4}$.

[asy] import graph; size(10cm); pen dps = linewidth(0.7) + fontsize(8); defaultpen(dps); pair B = (0,0); pair C = (1,0); pair A = rotate(60,B)*C; pair E = rotate(270,A)*B; pair D = rotate(270,E)*A; pair F = rotate(90,A)*C; pair G = rotate(90,F)*A; pair I = rotate(270,B)*C; pair H = rotate(270,I)*B; pair J = rotate(60,I)*D; pair K = rotate(60,E)*F; pair L = rotate(60,G)*H; draw(A--B--C--cycle); draw(A--E--D--B); draw(A--F--G--C); draw(B--I--H--C); draw(E--F); draw(D--I); draw(I--H); draw(H--G); draw(I--J--D); draw(E--K--F); draw(G--L--H); label("$A$",A,N); label("$B$",B,SW); label("$C$",C,SE); label("$D$",D,W); label("$E$",E,W); label("$F$",F,E); label("$G$",G,E); label("$H$",H,SE); label("$I$",I,SW); label("$J$",J,SW); label("$K$",K,N); label("$L$",L,SE); [/asy]

Triangles $\triangle{DIJ}$, $\triangle{EFK}$, and $\triangle{GHL}$ have sides of $\sqrt{3}$, so their total area is $3(\dfrac{\sqrt{3}}{4}(\sqrt{3})^2)=\dfrac{9\sqrt{3}}{4}$.

Now, you subtract their total area from the area of $\triangle{JKL}$:

$\dfrac{12+13\sqrt{3}}{4}-\dfrac{9\sqrt{3}}{4}=\dfrac{12+13\sqrt{3}-9\sqrt{3}}{4}=\dfrac{12+4\sqrt{3}}{4}=3+\sqrt{3}\implies\boxed{\textbf{(C)}\ 3+\sqrt3}$


(Solution by Pyson)

See Also

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
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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