Difference between revisions of "1953 AHSME Problems/Problem 49"
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| − | + | ==Problem== | |
| − | k will be between 1 and 5 for AC+BC to be | + | |
| + | The coordinates of <math>A,B</math> and <math>C</math> are <math>(5,5),(2,1)</math> and <math>(0,k)</math> respectively. | ||
| + | The value of <math>k</math> that makes <math>\overline{AC}+\overline{BC}</math> as small as possible is: | ||
| + | |||
| + | <math>\textbf{(A)}\ 3 \qquad | ||
| + | \textbf{(B)}\ 4\frac{1}{2} \qquad | ||
| + | \textbf{(C)}\ 3\frac{6}{7} \qquad | ||
| + | \textbf{(D)}\ 4\frac{5}{6}\qquad | ||
| + | \textbf{(E)}\ 2\frac{1}{7} </math> | ||
| + | |||
| + | ==Solution== | ||
| + | <math>k</math> will be between <math>1</math> and <math>5</math> for <math>AC+BC</math> to be the smallest. If we mirror point <math>A</math> across the y-axis to <math>A'</math>, with coordinates <math>(-5,5),</math> the distance <math>A'C+BC</math> will be same as <math>AC+BC</math>. The minimum of <math>A'C+BC</math> will occur when <math>C</math> is on the straight line connecting <math>A'</math> and <math>B</math> (i.e., <math>C</math> lies on the line <math>A'B</math>). Therefore, <math>k</math> is the y-intercept of the line that passes through <math>A'</math> and <math>B</math>. | ||
| + | |||
| + | The slope of the line is <math>\frac{1-5}{2-(-5)}=-\frac 47</math>. Using [[point-slope form]], the equation of the line is <math>y-5=-\frac 47(x+5)</math>. Letting <math>x=0</math> gives <math>y-5=-\frac{20}{7},</math> so <math>y=\frac{15}{7}</math>. Therefore, <math>k = 2\frac17 \Rightarrow \boxed{\textbf{(E) } 2\frac17}.</math> | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME 50p box|year=1953|num-b=48|num-a=50}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Latest revision as of 20:20, 17 February 2020
Problem
The coordinates of 
and 
are 
and 
respectively.
The value of 
that makes 
as small as possible is:
Solution
will be between 
and 
for 
to be the smallest. If we mirror point 
across the y-axis to 
, with coordinates 
the distance 
will be same as . The minimum of will occur when is on the straight line connecting and 
(i.e., lies on the line 
). Therefore, is the y-intercept of the line that passes through and .
The slope of the line is 
. Using point-slope form, the equation of the line is 
. Letting 
gives 
so 
. Therefore, 
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 48 |
Followed by Problem 50 | |
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