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 "1951 AHSME Problems/Problem 30"
(→‎Solution 1)
 
(5 intermediate revisions by the same user not shown)
Line 6: Line 6:
  
 
==Solution 1==
 
==Solution 1==
−
The two pole formula says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is <math>\frac1{\frac1{20}+\frac1{80}}</math>, or <math>\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}</math>.
+
The [[two poles formula]] says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is <math>\frac1{\frac1{20}+\frac1{80}}</math>, or <math>\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}</math>.
  
 
==Solution 2==
 
==Solution 2==
Line 14: Line 14:
 
<math>\frac{-x}{5}+20=\frac{4x}{5}</math>
 
<math>\frac{-x}{5}+20=\frac{4x}{5}</math>
  
−
<math>20=x</math>
+
<math>20=x.</math>
  
−
So the lines meet at an <math>x-coordinate</math> of 20.
+
So the lines meet at an <math>x</math>-coordinate of 20.
  
 
Solving for the height they meet,
 
Solving for the height they meet,
  
−
<math>y=\frac{4\cdot 20}{5}</math>
+
<cmath>y=\frac{4\cdot 20}{5}</cmath>
−
<math>y=16</math> <math>\boxed{16 \textbf{ (C)}}</math>
+
<cmath>y=\boxed{16 \textbf{ (C)}}.</cmath>
  
 
== See Also ==
 
== See Also ==

Latest revision as of 18:41, 17 August 2021

Problem

If two poles $20''$ and $80''$ high are $100''$ apart, then the height of the intersection of the lines joining the top of each pole to the foot of the opposite pole is:

$\textbf{(A)}\ 50''\qquad\textbf{(B)}\ 40''\qquad\textbf{(C)}\ 16''\qquad\textbf{(D)}\ 60''\qquad\textbf{(E)}\ \text{none of these}$

Solution 1

The two poles formula says this height is half the harmonic mean of the heights of the two poles. (The distance between the poles is irrelevant.) So the answer is $\frac1{\frac1{20}+\frac1{80}}$, or $\frac1{\frac1{16}}=\boxed{16 \textbf{ (C)}}$.

Solution 2

The two lines can be represented as $y=\frac{-x}{5}+20$ and $y=\frac{4x}{5}$. Solving the system,

$\frac{-x}{5}+20=\frac{4x}{5}$

$20=x.$

So the lines meet at an $x$-coordinate of 20.

Solving for the height they meet,

\[y=\frac{4\cdot 20}{5}\] \[y=\boxed{16 \textbf{ (C)}}.\]

See Also

1951 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Problem 31
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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
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=1951_AHSME_Problems/Problem_30&oldid=160506"