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Difference between revisions of "1994 AIME Problems/Problem 6"

m (→‎Solution)
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== Solution ==
 
== Solution ==
 +
 +
=== Solution 1 ===
 
We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.  
 
We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.  
  
Line 23: Line 25:
  
 
There are <math>6 \cdot 10</math> equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is <math>600+60 = \boxed{660}</math>.
 
There are <math>6 \cdot 10</math> equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is <math>600+60 = \boxed{660}</math>.
 +
 +
=== Solution 2 ===
 +
There are three types of lines: horizontal, upward-slanting diagonal, and downward-slanting diagonal. There are <math>21</math> of each type of line, and a unit equilateral triangle is determined by exactly one of each type of line. Given an upward-slanting diagonal and a downward-slanting diagonal, they determine exactly two unit equilateral triangles as shown below.
 +
<asy>
 +
size(60);
 +
pair u=rotate(60)*(1,0),d=rotate(-60)*(1,0),h=(1,0);
 +
draw((0,0)--4*u^^-2*h+4*u--(-2*h+4*u+4*d));
 +
draw(u--2*u+d,dotted);
 +
draw(3*u--3*u-h,dotted);
 +
</asy>
 +
Therefore, if all horizontal lines are drawn, there will be a total of <math>2\cdot 21^2=882</math> unit equilateral triangles. Of course, we only draw <math>21</math> horizontal lines, so we are overcounting the triangles that are caused by the undrawn horizontal lines. In the below diagram, we draw the diagonal lines and the highest and lowest horizontal lines.
 +
<asy>
 +
size(200);
 +
pair u=rotate(60)*(2/sqrt(3),0),d=rotate(-60)*(2/sqrt(3),0),h=(2/sqrt(3),0);
 +
for (int i=0;i<21;++i) {path v=(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d;draw(shift(0,-2*i)*v);}
 +
for (int i=0;i<21;++i) {path v=rotate(180)*((-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d);draw(shift(0,2*i)*v);}
 +
draw((-15,-10)--(15,-10));
 +
draw((-15,10)--(15,10));
 +
</asy>
 +
 +
We see that the lines <math>y=-21,-20,\dots, -11</math> and <math>y=11,12,\dots,21</math> would complete several of the <math>882</math> unit equilateral triangles. In fact, we can see that the lines <math>y=-21,-20,\dots,-11</math> complete <math>1,2,(1+3),(2+4),(3+5),(4+6),\dots,(9+11)</math> triangles, or <math>111</math> triangles. The positive horizontal lines complete the same number of triangles, hence the answer is <math>882-2\cdot 111=\boxed{660}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 15:13, 28 February 2016

Problem

The graphs of the equations

$y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,$

are drawn in the coordinate plane for $k=-10,-9,-8,\ldots,9,10.\,$ These 63 lines cut part of the plane into equilateral triangles of side $2/\sqrt{3}.\,$ How many such triangles are formed?

Solution

Solution 1

We note that the lines partition the hexagon of the six extremal lines into disjoint unit regular triangles, and forms a series of unit regular triangles along the edge of the hexagon.

[asy] size(200); picture pica, picb, picc; int i; for(i=-10;i<=10;++i){ if((i%10) == 0){draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i),black+0.7);} else{draw(pica,(-20/sqrt(3)-abs((0,i))/sqrt(3),i)--(20/sqrt(3)+abs((0,i))/sqrt(3),i));} } picb = rotate(120,origin)*pica; picc = rotate(240,origin)*pica; add(pica);add(picb);add(picc); [/asy]

Solving the above equations for $k=\pm 10$, we see that the hexagon in question is regular, with side length $\frac{20}{\sqrt{3}}$. Then, the number of triangles within the hexagon is simply the ratio of the area of the hexagon to the area of a regular triangle. Since the ratio of the area of two similar figures is the square of the ratio of their side lengths, we see that the ratio of the area of one of the six equilateral triangles composing the regular hexagon to the area of a unit regular triangle is just $\left(\frac{20/\sqrt{3}}{2/\sqrt{3}}\right)^2 = 100$. Thus, the total number of unit triangles is $6 \times 100 = 600$.

There are $6 \cdot 10$ equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is $600+60 = \boxed{660}$.

Solution 2

There are three types of lines: horizontal, upward-slanting diagonal, and downward-slanting diagonal. There are $21$ of each type of line, and a unit equilateral triangle is determined by exactly one of each type of line. Given an upward-slanting diagonal and a downward-slanting diagonal, they determine exactly two unit equilateral triangles as shown below. [asy] size(60); pair u=rotate(60)*(1,0),d=rotate(-60)*(1,0),h=(1,0); draw((0,0)--4*u^^-2*h+4*u--(-2*h+4*u+4*d)); draw(u--2*u+d,dotted); draw(3*u--3*u-h,dotted); [/asy] Therefore, if all horizontal lines are drawn, there will be a total of $2\cdot 21^2=882$ unit equilateral triangles. Of course, we only draw $21$ horizontal lines, so we are overcounting the triangles that are caused by the undrawn horizontal lines. In the below diagram, we draw the diagonal lines and the highest and lowest horizontal lines. [asy] size(200); pair u=rotate(60)*(2/sqrt(3),0),d=rotate(-60)*(2/sqrt(3),0),h=(2/sqrt(3),0); for (int i=0;i<21;++i) {path v=(-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d;draw(shift(0,-2*i)*v);} for (int i=0;i<21;++i) {path v=rotate(180)*((-20/sqrt(3),0)-2*u+i*u--(0,20)--(20/sqrt(3),0)+2*d-i*d);draw(shift(0,2*i)*v);} draw((-15,-10)--(15,-10)); draw((-15,10)--(15,10)); [/asy]

We see that the lines $y=-21,-20,\dots, -11$ and $y=11,12,\dots,21$ would complete several of the $882$ unit equilateral triangles. In fact, we can see that the lines $y=-21,-20,\dots,-11$ complete $1,2,(1+3),(2+4),(3+5),(4+6),\dots,(9+11)$ triangles, or $111$ triangles. The positive horizontal lines complete the same number of triangles, hence the answer is $882-2\cdot 111=\boxed{660}$.

See also

1994 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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