Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2008 AMC 12B Problems/Problem 15"

(Created page with '==Problem== On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of…')
 
 
(5 intermediate revisions by 4 users not shown)
Line 4: Line 4:
 
<math> \textbf{(A)} \; \frac{1}{4} \qquad \textbf{(B)} \; \frac{\sqrt{2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt{3} \qquad \textbf{(E)} \; 2 \sqrt{3}</math>
 
<math> \textbf{(A)} \; \frac{1}{4} \qquad \textbf{(B)} \; \frac{\sqrt{2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt{3} \qquad \textbf{(E)} \; 2 \sqrt{3}</math>
  
−
==Solution==
+
==Solution 1==
 
<asy>
 
<asy>
 
real a = 1/2, b = sqrt(3)/2;
 
real a = 1/2, b = sqrt(3)/2;
Line 15: Line 15:
 
filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9));
 
filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9));
 
</asy>
 
</asy>
−
The equilateral triangles form trapezoids with side lengths <math> 1, 1, 1, 2</math> (half a unit hexagon) on each face of the unit square.  The four triangles "in between" these trapezoids are isosceles triangles with two side lengths <math> 1</math> and an angle of <math> 30^{\circ}</math> in between them, so the total area of these triangles (which is the area of <math> S \minus{} R</math>) is, by the [[Law of Sines]], <math> 4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1</math> which makes the answer <math> \boxed{C}</math>.
 
  
 +
The equilateral triangles form trapezoids with side lengths <math> 1, 1, 1, 2</math> (half a unit hexagon) on each face of the unit square.  The four triangles "in between" these trapezoids are isosceles triangles with two side lengths <math> 1</math> and an angle of <math> 30^{\circ}</math> in between them, so the total area of these triangles (which is the area of <math> S - R</math>) is, <math> 4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1</math> which makes the answer <math> \boxed{C}</math>.
 +
==Solution 2==
 +
<asy>
 +
real a = 1/2, b = sqrt(3)/2;
 +
draw((0,0)--(1,0)--(1,1)--(0,1)--cycle);
 +
draw((a+b+a,a+b+a)--(a+b+a,-b)--(-b,-b)--(-b,a+a+b)--cycle,dashed);
 +
draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0));
 +
draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,1)--(-b,1+a)--(-b,-1+a)--(0,0));
 +
filldraw((1+a,-b)--(1,0)--(1+b,-1+a)--cycle,gray(0.9));
 +
filldraw((1+b,1+a)--(1,1)--(1+a,1+b)--cycle,gray(0.9));
 +
filldraw((-1+a,1+b)--(0,1)--(-b,1+a)--cycle,gray(0.9));
 +
filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9));
 +
</asy>
 +
 +
The area of the largest square is <math>(1+\sqrt{3})^2=4+2\sqrt{3}</math>. The area of region <math>R</math> is <math>1+12\frac{1}{2}\frac{\sqrt{3}}{2}=1+3\sqrt{3}</math>. The total area of four small 45-45-90 triangles at corner is <math>4*\frac{1}{2}(\frac{\sqrt{3}+1-2}{2})^2=2-\sqrt{3}</math>. <math>S=4+2\sqrt{3}-(2-\sqrt{3})=2+3\sqrt{3}</math>, <math>S-R=1</math>.
 +
 +
 +
~Bran Qin
 
==See Also==  
 
==See Also==  
 
[http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1051031#p1051031 Region with Squares and Triangles]
 
[http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1051031#p1051031 Region with Squares and Triangles]
  
 
{{AMC12 box|year=2008|ab=B|num-b=14|num-a=16}}
 
{{AMC12 box|year=2008|ab=B|num-b=14|num-a=16}}
 +
{{MAA Notice}}

Latest revision as of 21:58, 7 October 2021

Problem

On each side of a unit square, an equilateral triangle of side length 1 is constructed. On each new side of each equilateral triangle, another equilateral triangle of side length 1 is constructed. The interiors of the square and the 12 triangles have no points in common. Let $R$ be the region formed by the union of the square and all the triangles, and $S$ be the smallest convex polygon that contains $R$. What is the area of the region that is inside $S$ but outside $R$?

$\textbf{(A)} \; \frac{1}{4} \qquad \textbf{(B)} \; \frac{\sqrt{2}}{4} \qquad \textbf{(C)} \; 1 \qquad \textbf{(D)} \; \sqrt{3} \qquad \textbf{(E)} \; 2 \sqrt{3}$

Solution 1

[asy] real a = 1/2, b = sqrt(3)/2; draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0)); draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,1)--(-b,1+a)--(-b,-1+a)--(0,0)); filldraw((1+a,-b)--(1,0)--(1+b,-1+a)--cycle,gray(0.9)); filldraw((1+b,1+a)--(1,1)--(1+a,1+b)--cycle,gray(0.9)); filldraw((-1+a,1+b)--(0,1)--(-b,1+a)--cycle,gray(0.9)); filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9)); [/asy]

The equilateral triangles form trapezoids with side lengths $1, 1, 1, 2$ (half a unit hexagon) on each face of the unit square. The four triangles "in between" these trapezoids are isosceles triangles with two side lengths $1$ and an angle of $30^{\circ}$ in between them, so the total area of these triangles (which is the area of $S - R$) is, $4 \left( \frac {1}{2} \sin 30^{\circ} \right) = 1$ which makes the answer $\boxed{C}$.

Solution 2

[asy] real a = 1/2, b = sqrt(3)/2; draw((0,0)--(1,0)--(1,1)--(0,1)--cycle); draw((a+b+a,a+b+a)--(a+b+a,-b)--(-b,-b)--(-b,a+a+b)--cycle,dashed); draw((0,0)--(a,-b)--(1,0)--(1+b,a)--(1,1)--(a,1+b)--(0,1)--(-b,a)--(0,0)); draw((0,0)--(-1+a,-b)--(1+a,-b)--(1,0)--(1+b,-1+a)--(1+b,1+a)--(1,1)--(1+a,1+b)--(-1+a,1+b)--(0,1)--(-b,1+a)--(-b,-1+a)--(0,0)); filldraw((1+a,-b)--(1,0)--(1+b,-1+a)--cycle,gray(0.9)); filldraw((1+b,1+a)--(1,1)--(1+a,1+b)--cycle,gray(0.9)); filldraw((-1+a,1+b)--(0,1)--(-b,1+a)--cycle,gray(0.9)); filldraw((-b,-1+a)--(0,0)--(-1+a,-b)--cycle,gray(0.9)); [/asy]

The area of the largest square is $(1+\sqrt{3})^2=4+2\sqrt{3}$. The area of region $R$ is $1+12\frac{1}{2}\frac{\sqrt{3}}{2}=1+3\sqrt{3}$. The total area of four small 45-45-90 triangles at corner is $4*\frac{1}{2}(\frac{\sqrt{3}+1-2}{2})^2=2-\sqrt{3}$. $S=4+2\sqrt{3}-(2-\sqrt{3})=2+3\sqrt{3}$, $S-R=1$.


~Bran Qin

See Also

Region with Squares and Triangles

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Problem 16
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_15&oldid=163279"