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 "1992 AJHSME Problems/Problem 6"

(Created page with '==Solution== The first triangle represents <math>1+3-4</math> The 2nd triangle represents <math>2+5-6</math> Solving the first triangle, we get <math>0</math> Solving the 2nd tr…')
 
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 +
==Problem==
 +
 +
Suppose that
 +
<asy>
 +
unitsize(18);
 +
draw((0,0)--(2,0)--(1,sqrt(3))--cycle);
 +
label("$a$",(1,sqrt(3)-0.2),S);
 +
label("$b$",(sqrt(3)/10,0.1),ENE);
 +
label("$c$",(2-sqrt(3)/10,0.1),WNW);
 +
</asy>
 +
means <math>a+b-c</math>.
 +
For example,
 +
<asy>
 +
unitsize(18);
 +
draw((0,0)--(2,0)--(1,sqrt(3))--cycle);
 +
label("$5$",(1,sqrt(3)-0.2),S);
 +
label("$4$",(sqrt(3)/10,0.1),ENE);
 +
label("$6$",(2-sqrt(3)/10,0.1),WNW);
 +
</asy>
 +
is <math>5+4-6 = 3</math>.
 +
Then the sum
 +
<asy>
 +
unitsize(18);
 +
draw((0,0)--(2,0)--(1,sqrt(3))--cycle);
 +
label("$1$",(1,sqrt(3)-0.2),S);
 +
label("$3$",(sqrt(3)/10,0.1),ENE);
 +
label("$4$",(2-sqrt(3)/10,0.1),WNW);
 +
draw((3,0)--(5,0)--(4,sqrt(3))--cycle);
 +
label("$2$",(4,sqrt(3)-0.2),S);
 +
label("$5$",(3+sqrt(3)/10,0.1),ENE);
 +
label("$6$",(5-sqrt(3)/10,0.1),WNW);
 +
label("$+$",(2.5,-0.1),N);
 +
</asy>
 +
is
 +
 +
<math>\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2</math>
 +
 
==Solution==
 
==Solution==
 
The first triangle represents <math>1+3-4</math>
 
The first triangle represents <math>1+3-4</math>
Line 7: Line 44:
  
 
Since we have to add the 2 triangles the final answer is <math>1</math>, which is <math>\boxed{D}</math>.
 
Since we have to add the 2 triangles the final answer is <math>1</math>, which is <math>\boxed{D}</math>.
 +
 +
==See Also==
 +
{{AJHSME box|year=1992|num-b=5|num-a=7}}
 +
 +
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:09, 4 July 2013

Problem

Suppose that [asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("$a$",(1,sqrt(3)-0.2),S); label("$b$",(sqrt(3)/10,0.1),ENE); label("$c$",(2-sqrt(3)/10,0.1),WNW); [/asy] means $a+b-c$. For example, [asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("$5$",(1,sqrt(3)-0.2),S); label("$4$",(sqrt(3)/10,0.1),ENE); label("$6$",(2-sqrt(3)/10,0.1),WNW); [/asy] is $5+4-6 = 3$. Then the sum [asy] unitsize(18); draw((0,0)--(2,0)--(1,sqrt(3))--cycle); label("$1$",(1,sqrt(3)-0.2),S); label("$3$",(sqrt(3)/10,0.1),ENE); label("$4$",(2-sqrt(3)/10,0.1),WNW); draw((3,0)--(5,0)--(4,sqrt(3))--cycle); label("$2$",(4,sqrt(3)-0.2),S); label("$5$",(3+sqrt(3)/10,0.1),ENE); label("$6$",(5-sqrt(3)/10,0.1),WNW); label("$+$",(2.5,-0.1),N); [/asy] is

$\text{(A)}\ -2 \qquad \text{(B)}\ -1 \qquad \text{(C)}\ 0 \qquad \text{(D)}\ 1 \qquad \text{(E)}\ 2$

Solution

The first triangle represents $1+3-4$ The 2nd triangle represents $2+5-6$

Solving the first triangle, we get $0$ Solving the 2nd triangle, we get $1$

Since we have to add the 2 triangles the final answer is $1$, which is $\boxed{D}$.

See Also

1992 AJHSME (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 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 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=1992_AJHSME_Problems/Problem_6&oldid=55626"