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 10B Problems/Problem 7"

m (→‎Solution)
m (→‎Solution 2)
 
(20 intermediate revisions by 6 users not shown)
Line 4: Line 4:
 
<math>\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000</math>
 
<math>\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000</math>
  
−
==Solution==
+
==Solution 1==
−
The area of the large triangle is <math>\frac{10^2\sqrt3}{4}</math>, while the area each small triangle is <math>\frac{1^2\sqrt3}{4}</math>. Dividing these two quantities, we get 100, therefore <math>\boxed{100}</math> small triangles can fit in the large one.
+
The area of the large triangle is <math>\frac{10^2\sqrt3}{4}</math>, while the area of each small triangle is <math>\frac{1^2\sqrt3}{4}</math>. Dividing these two quantities results in <math>100</math>, therefore <math>\boxed{100} \mathrm{(C)}</math> small triangles can fill the large one without overlap.
  
−
 
+
==Solution 2-4==
−
Another Solution:
 
 
<asy>
 
<asy>
 
unitsize(0.5cm);
 
unitsize(0.5cm);
Line 18: Line 17:
  
 
The number of triangles is <math>1+3+\dots+19 = \boxed{100}</math>.
 
The number of triangles is <math>1+3+\dots+19 = \boxed{100}</math>.
 +
 +
Also, another way to do it is to notice that as you go row by row (from the bottom), the number of triangles decrease by 2 from 19, so we have:
 +
<math>19+17+15...+3+1 = \frac{19+1}{2}\cdot 10 = \boxed{100}</math>
 +
 +
A fourth solution is to notice that the small triangles are similar to the large triangle as they are both equilateral. Therefore, the ratio of their areas is the square of the ratio of their side lengths.  Hence the ratio of their areas is <math>(1/10)^2=1/100</math>, so the answer is <math>\boxed{100}</math>.
 +
 +
==Solution 5==
 +
 +
The side length of the large equilateral triangle is <math>10</math>. The number of equilateral triangles with side length <math>1</math> that can fill it is <math>10^2</math>. This is equal to the sum of the first <math>10</math> odd integers. The sum of the first <math>n</math> odd integers is <math>n^2</math>. In general, an equilateral triangle with side length <math>n</math> can be filled by <math>n^2</math> triangles. The answer is <math>100\ \mathrm{(C)}</math>.
 +
 +
~mobius247
  
 
==See also==
 
==See also==
 
{{AMC10 box|year=2008|ab=B|num-b=6|num-a=8}}
 
{{AMC10 box|year=2008|ab=B|num-b=6|num-a=8}}
 +
{{MAA Notice}}

Latest revision as of 19:23, 14 July 2021

Problem

An equilateral triangle of side length $10$ is completely filled in by non-overlapping equilateral triangles of side length $1$. How many small triangles are required?

$\mathrm{(A)}\ 10\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 100\qquad\mathrm{(D)}\ 250\qquad\mathrm{(E)}\ 1000$

Solution 1

The area of the large triangle is $\frac{10^2\sqrt3}{4}$, while the area of each small triangle is $\frac{1^2\sqrt3}{4}$. Dividing these two quantities results in $100$, therefore $\boxed{100} \mathrm{(C)}$ small triangles can fill the large one without overlap.

Solution 2-4

[asy] unitsize(0.5cm); defaultpen(0.8); for (int i=0; i<10; ++i) { draw( (i*dir(60)) -- ( (10,0) + (i*dir(120)) ) ); } for (int i=0; i<10; ++i) { draw( (i*dir(0)) -- ( 10*dir(60) + (i*dir(-60)) ) ); } for (int i=0; i<10; ++i) { draw( ((10-i)*dir(60)) -- ((10-i)*dir(0)) ); } [/asy]

The number of triangles is $1+3+\dots+19 = \boxed{100}$.

Also, another way to do it is to notice that as you go row by row (from the bottom), the number of triangles decrease by 2 from 19, so we have: $19+17+15...+3+1 = \frac{19+1}{2}\cdot 10 = \boxed{100}$

A fourth solution is to notice that the small triangles are similar to the large triangle as they are both equilateral. Therefore, the ratio of their areas is the square of the ratio of their side lengths. Hence the ratio of their areas is $(1/10)^2=1/100$, so the answer is $\boxed{100}$.

Solution 5

The side length of the large equilateral triangle is $10$. The number of equilateral triangles with side length $1$ that can fill it is $10^2$. This is equal to the sum of the first $10$ odd integers. The sum of the first $n$ odd integers is $n^2$. In general, an equilateral triangle with side length $n$ can be filled by $n^2$ triangles. The answer is $100\ \mathrm{(C)}$.

~mobius247

See also

2008 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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

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_10B_Problems/Problem_7&oldid=158524"