Difference between revisions of "1980 AHSME Problems/Problem 12"

m (See also)
(Solution)
Line 7: Line 7:
  
 
== Solution ==
 
== Solution ==
<math>\fbox{}</math>
+
<math>4n = m</math>, as stated in the question. In the line <math>L_1</math>, draw a triangle with the coordinates <math>(0,0)</math>, <math>(1,0)</math>
  
 
== See also ==
 
== See also ==

Revision as of 13:30, 17 November 2016

Problem

The equations of $L_1$ and $L_2$ are $y=mx$ and $y=nx$, respectively. Suppose $L_1$ makes twice as large of an angle with the horizontal (measured counterclockwise from the positive x-axis ) as does $L_2$, and that $L_1$ has 4 times the slope of $L_2$. If $L_1$ is not horizontal, then $mn$ is

$\text{(A)} \ \frac{\sqrt{2}}{2} \qquad \text{(B)} \ -\frac{\sqrt{2}}{2} \qquad \text{(C)} \ 2 \qquad \text{(D)} \ -2 \qquad \text{(E)} \ \text{not uniquely determined}$


Solution

$4n = m$, as stated in the question. In the line $L_1$, draw a triangle with the coordinates $(0,0)$, $(1,0)$

See also

1980 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 11
Followed by
Problem 12
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