Difference between revisions of "1994 AIME Problems/Problem 6"
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== Problem == | == Problem == | ||
+ | The graphs of the equations | ||
+ | <center><math>y=k, \qquad y=\sqrt{3}x+2k, \qquad y=-\sqrt{3}x+2k,</math></center> | ||
+ | are drawn in the coordinate plane for <math>k=-10,-9,-8,\ldots,9,10.\,</math> These 63 lines cut part of the plane into equilateral triangles of side length <math>\tfrac{2}{\sqrt{3}}.\,</math> How many such triangles are formed? | ||
== 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. | ||
+ | |||
+ | <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 <math>k=\pm 10</math>, we see that the hexagon in question is regular, with side length <math>\frac{20}{\sqrt{3}}</math>. 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 <math>\left(\frac{20/\sqrt{3}}{2/\sqrt{3}}\right)^2 = 100</math>. Thus, the total number of unit triangles is <math>6 \times 100 = 600</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>. | ||
+ | |||
+ | |||
+ | ===Solution 3 Elementary Counting=== | ||
+ | Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is 20 units, and by extending horizontal lines to the sides of the hexagon. This shows that for every side of the hexagon there are 10 spaces. Therefore the side length of the biggest triangle (imagine one of the overlapping triangles in the Star of David) is 30. The total number of triangles in the hexagon can be found by finding the number of triangles in the extended triangle and subtracting the 3 corner triangles. This gives us <math>30^2 - 10^2 - 10^2- 10^2 = 600</math>. That is the number of triangles in the hexagon. The remaining triangles form in groups of 10 on the exterior of each side. <math>600 + 6 * 10 = \boxed{660}</math>. | ||
+ | |||
+ | |||
+ | -jackshi2006 | ||
== See also == | == See also == | ||
− | + | {{AIME box|year=1994|num-b=5|num-a=7}} | |
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] | ||
+ | [[Category:Intermediate Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 15:14, 8 March 2021
Contents
Problem
The graphs of the equations
are drawn in the coordinate plane for These 63 lines cut part of the plane into equilateral triangles of side length 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.
Solving the above equations for , we see that the hexagon in question is regular, with side length . 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 . Thus, the total number of unit triangles is .
There are equilateral triangles formed by lines on the edges of the hexagon. Thus, our answer is .
Solution 2
There are three types of lines: horizontal, upward-slanting diagonal, and downward-slanting diagonal. There are 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. Therefore, if all horizontal lines are drawn, there will be a total of unit equilateral triangles. Of course, we only draw 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.
We see that the lines and would complete several of the unit equilateral triangles. In fact, we can see that the lines complete triangles, or triangles. The positive horizontal lines complete the same number of triangles, hence the answer is .
Solution 3 Elementary Counting
Picturing the diagram in your head should give you an illustration similar to the one above. The distance from parallel sides of the center hexagon is 20 units, and by extending horizontal lines to the sides of the hexagon. This shows that for every side of the hexagon there are 10 spaces. Therefore the side length of the biggest triangle (imagine one of the overlapping triangles in the Star of David) is 30. The total number of triangles in the hexagon can be found by finding the number of triangles in the extended triangle and subtracting the 3 corner triangles. This gives us . That is the number of triangles in the hexagon. The remaining triangles form in groups of 10 on the exterior of each side. .
-jackshi2006
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.