Difference between revisions of "1953 AHSME Problems/Problem 41"
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==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 rods from a straight road. On this road, a boys' camp is located 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:
Solution
Let be the straight road, be the girls' camp, be the boys' camp, and be the water canteen. is the perpendicular from to . Suppose each rod is one unit long. and . Since is a right angle, is a right triangle, so .
Let be the distance from the canteen to the girls' and boys' camps. We have and . Using the Pythagorean Theorem on , we have Simplifying the left side gives Subtracting from both sides and rearranging gives Therefore, . The answer is .
See Also
1953 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 40 |
Followed by Problem 42 | |
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All AHSME Problems and Solutions |
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