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
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "1953 AHSME Problems/Problem 41"

(Created page with "==Problem== A girls' camp is located <math>300</math> rods from a straight road. On this road, a boys' camp is located <math>500</math> rods from the girls' camp. It is desi...")
 
 
Line 11: Line 11:
  
 
==Solution==
 
==Solution==
 +
 +
<asy>
 +
draw((0,0)--(6,0));
 +
draw((1,0)--(1,3)--(5,0));
 +
draw((1,3)--(1.875,0));
 +
label("$A$",(1,0),S);
 +
label("$B$",(5,0),S);
 +
label("$C$",(1.875,0),S);
 +
label("$G$",(1,3),NW);
 +
label("$r$",(0,0),SW);
 +
</asy>
 +
Let <math>r</math> be the straight road, <math>G</math> be the girls' camp, <math>B</math> be the boys' camp, and <math>C</math> be the water canteen. <math>\overline{AG}</math> is the perpendicular from <math>G</math> to <math>r</math>. Suppose each rod is one unit long. <math>AG=300</math> and <math>BG=500</math>. Since <math>\angle GAB</math> is a right angle, <math>\triangle GAB</math> is a <math>3-4-5</math> right triangle, so <math>AB=400</math>.
 +
 +
Let <math>x</math> be the distance from the canteen to the girls' and boys' camps. We have <math>BC=CG=x</math> and <math>AC=400-x</math>. Using the [[Pythagorean Theorem]] on <math>\triangle GAC</math>, we have
 +
<cmath>300^2+(400-x)^2=x^2</cmath>
 +
Simplifying the left side gives
 +
<cmath>90000+160000-800x+x^2=x^2</cmath>
 +
Subtracting <math>x^2</math> from both sides and rearranging gives
 +
<cmath>800x=250000</cmath>
 +
Therefore, <math>x=\frac{250000}{800}=312.5</math>. The answer is <math>\boxed{\textbf{(E)}\ \text{none of these}}</math>.
 +
==See Also==
 +
 +
{{AHSME 50p box|year=1953|num-b=40|num-a=42}}
 +
 +
{{MAA Notice}}

Latest revision as of 23:43, 25 January 2020

Problem

A girls' camp is located $300$ rods from a straight road. On this road, a boys' camp is located $500$ rods from the girls' camp. It is desired to build a canteen on the road which shall be exactly the same distance from each camp. The distance of the canteen from each of the camps is:

$\textbf{(A)}\ 400\text{ rods} \qquad \textbf{(B)}\ 250\text{ rods} \qquad \textbf{(C)}\ 87.5\text{ rods} \qquad \textbf{(D)}\ 200\text{ rods}\\ \textbf{(E)}\ \text{none of these}$

Solution

[asy] draw((0,0)--(6,0)); draw((1,0)--(1,3)--(5,0)); draw((1,3)--(1.875,0)); label("$A$",(1,0),S); label("$B$",(5,0),S); label("$C$",(1.875,0),S); label("$G$",(1,3),NW); label("$r$",(0,0),SW); [/asy] Let $r$ be the straight road, $G$ be the girls' camp, $B$ be the boys' camp, and $C$ be the water canteen. $\overline{AG}$ is the perpendicular from $G$ to $r$. Suppose each rod is one unit long. $AG=300$ and $BG=500$. Since $\angle GAB$ is a right angle, $\triangle GAB$ is a $3-4-5$ right triangle, so $AB=400$.

Let $x$ be the distance from the canteen to the girls' and boys' camps. We have $BC=CG=x$ and $AC=400-x$. Using the Pythagorean Theorem on $\triangle GAC$, we have \[300^2+(400-x)^2=x^2\] Simplifying the left side gives \[90000+160000-800x+x^2=x^2\] Subtracting $x^2$ from both sides and rearranging gives \[800x=250000\] Therefore, $x=\frac{250000}{800}=312.5$. The answer is $\boxed{\textbf{(E)}\ \text{none of these}}$.

See Also

1953 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 40 		Followed by
Problem 42
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


These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1953_AHSME_Problems/Problem_41&oldid=115576"