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 "1985 AHSME Problems/Problem 13"
(Improved solutions, formatting, and LaTeX)
 
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
Pegs are put in a board <math> 1 </math> unit apart both horizontally and vertically. A rubber band is stretched over <math> 4 </math> pegs as shown in the figure, forming a [[quadrilateral]]. Its [[area]] in square units is
+
Pegs are put in a board <math>1</math> unit apart both horizontally and vertically. A rubber band is stretched over <math>4</math> pegs as shown in the figure, forming a quadrilateral. Its area in square units is
  
 
<asy>
 
<asy>
Line 12: Line 12:
 
<math> \mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \  } 5 \qquad \mathrm{(D) \  } 5.5 \qquad \mathrm{(E) \  }6 </math>
 
<math> \mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \  } 5 \qquad \mathrm{(D) \  } 5.5 \qquad \mathrm{(E) \  }6 </math>
  
−
==Solution==
+
==Solution 1==
−
===Solution 1===
 
−
We see that the number of interior points is <math> 5 </math> and the number of boundary points is <math> 4 </math>. Therefore, by [[Pick's Theorem]], the area is <math> 5+\frac{4}{2}-1=6, \boxed{\text{E}} </math>.
 
−
 
 
−
===Solution 2===
 
 
<asy>
 
<asy>
 
int i,j;
 
int i,j;
Line 34: Line 30:
 
label("$H$",(0,1),W);
 
label("$H$",(0,1),W);
 
</asy>
 
</asy>
−
Draw in the perimeter of the rectangle and label the points as shown. We have <math> [ABCD]=(3)(4)=12 </math>, <math> [AEH]=\frac{1}{2}(1)(2)=1 </math>, <math> [EBF]=\frac{1}{2}(3)(2)=3 </math>, <math> [FCG]=\frac{1}{2}(1)(1)=\frac{1}{2} </math>, and <math> [GDH]=\frac{1}{2}(3)(1)=\frac{3}{2} </math>. Therefore, <math> [EFGH]=[ABCD]-([AEH]+[EBF]+[FCG]+[GDH]) </math> <math> =12-(1+3+\frac{1}{2}+\frac{3}{2})=6, \boxed{\text{E}} </math>.
+
We draw in the rectangle bounding the given quadrilateral and label the points as shown. The area of rectangle <math>ABCD</math> is <math>(3)(4) = 12</math>, while the areas of the triangles <math>AEH</math>, <math>EBF</math>, <math>FCG</math>, and <math>GDH</math> are, respectively, <cmath>\begin{align*}&\text{area of } \triangle AEH = \frac{1}{2}(1)(2) = 1, \\ &\text{area of } \triangle EBF = \frac{1}{2}(3)(2) = 3, \\ &\text{area of } \triangle FCG = \frac{1}{2}(1)(1) = \frac{1}{2}, \text{ and} \\ &\text{area of } \triangle GDH = \frac{1}{2}(3)(1) = \frac{3}{2}.\end{align*}</cmath>
 +
Hence the area of the given quadrilateral <math>EFGH</math> is <math>12-\left(1+3+\frac{1}{2}+\frac{3}{2}\right) = \boxed{\text{(E)} \ 6}</math>.
 +
 
 +
==Solution 2==
 +
The number of lattice points (i.e. pegs on the board) strictly inside the quadrilateral is <math>5</math> and the number of lattice points on its boundary is <math>4</math>. Therefore, by [[Pick's theorem]], its area is <math>5+\frac{4}{2}-1 = \boxed{\text{(E)} \ 6}</math>.
  
 
==See Also==
 
==See Also==
 
{{AHSME box|year=1985|num-b=12|num-a=14}}
 
{{AHSME box|year=1985|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 20:05, 19 March 2024

Problem

Pegs are put in a board $1$ unit apart both horizontally and vertically. A rubber band is stretched over $4$ pegs as shown in the figure, forming a quadrilateral. Its area in square units is

[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7));[/asy]

$\mathrm{(A)\ } 4 \qquad \mathrm{(B) \ }4.5 \qquad \mathrm{(C) \ } 5 \qquad \mathrm{(D) \ } 5.5 \qquad \mathrm{(E) \ }6$

Solution 1

[asy] int i,j; for(i=0; i<5; i=i+1) { for(j=0; j<4; j=j+1) { dot((i,j)); }} draw((0,1)--(1,3)--(4,1)--(3,0)--cycle, linewidth(0.7)); draw((0,0)--(4,0)--(4,3)--(0,3)--cycle); label("$A$",(0,3),NW); label("$B$",(4,3),NE); label("$C$",(4,0),SE); label("$D$",(0,0),SW); label("$E$",(1,3),N); label("$F$",(4,1),E); label("$G$",(3,0),S); label("$H$",(0,1),W); [/asy] We draw in the rectangle bounding the given quadrilateral and label the points as shown. The area of rectangle $ABCD$ is $(3)(4) = 12$, while the areas of the triangles $AEH$, $EBF$, $FCG$, and $GDH$ are, respectively, \begin{align*}&\text{area of } \triangle AEH = \frac{1}{2}(1)(2) = 1, \\ &\text{area of } \triangle EBF = \frac{1}{2}(3)(2) = 3, \\ &\text{area of } \triangle FCG = \frac{1}{2}(1)(1) = \frac{1}{2}, \text{ and} \\ &\text{area of } \triangle GDH = \frac{1}{2}(3)(1) = \frac{3}{2}.\end{align*} Hence the area of the given quadrilateral $EFGH$ is $12-\left(1+3+\frac{1}{2}+\frac{3}{2}\right) = \boxed{\text{(E)} \ 6}$.

Solution 2

The number of lattice points (i.e. pegs on the board) strictly inside the quadrilateral is $5$ and the number of lattice points on its boundary is $4$. Therefore, by Pick's theorem, its area is $5+\frac{4}{2}-1 = \boxed{\text{(E)} \ 6}$.

See Also

1985 AHSME (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 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1985_AHSME_Problems/Problem_13&oldid=217117"