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Difference between revisions of "2014 AMC 10A Problems/Problem 13"

(→‎Solution 2)
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==Solution 2==
 
==Solution 2==
−
I have no idea how to draw pictures with LaTex, so someone please help me with this:
+
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, 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>.  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>
 +
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;
  
−
The extra equilateral triangles have sides of <math>\sqrt{3}</math>, so their areas are <math>3(\dfrac{\sqrt{3}}{4}(\sqrt{3})^2)=\dfrac{9\sqrt{3}}{4}</math>. 
+
pair E = rotate(270,A)*B;
 +
pair D = rotate(270,E)*A;
  
−
Now, you subtract the areas of the 3 equilateral triangles:  
+
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 <math>\triangle{DIJ}</math>, <math>\triangle{EFK}</math>, and <math>\triangle{GHL}</math> have sides of <math>\sqrt{3}</math>, so their total area is <math>3(\dfrac{\sqrt{3}}{4}(\sqrt{3})^2)=\dfrac{9\sqrt{3}}{4}</math>. 
 +
 
 +
Now, you subtract their total area from the area of <math>\triangle{JKL}</math>:  
  
 
<math>\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}</math>
 
<math>\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}</math>

Revision as of 11:24, 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}$$ (Error compiling LaTeX. Unknown error_msg)\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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