Difference between revisions of "1953 AHSME Problems/Problem 49"
| Line 1: | Line 1: | ||
| − | + | ==Problem== | |
| + | |||
| + | 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== | ||
k will be between 1 and 5 for AC+BC to be minimum. If we mirror A across the Y axis as A' (-5,5), the distance A'C+BC will be same as AC+BC. The minimum of A'C+BC will occur when C is on the straight line connecting A' and B, i.e., C lies on the line A'B. So, we can find the Y-intercept of the line connecting A'(-5,5) and B(2,1), which is 15/7 = 2 1/7. so, the answer is <math>\boxed{(E) 2 1/7}.</math> | k will be between 1 and 5 for AC+BC to be minimum. If we mirror A across the Y axis as A' (-5,5), the distance A'C+BC will be same as AC+BC. The minimum of A'C+BC will occur when C is on the straight line connecting A' and B, i.e., C lies on the line A'B. So, we can find the Y-intercept of the line connecting A'(-5,5) and B(2,1), which is 15/7 = 2 1/7. so, the answer is <math>\boxed{(E) 2 1/7}.</math> | ||
| + | |||
| + | ==See Also== | ||
| + | |||
| + | {{AHSME 50p box|year=1953|num-b=48|num-a=50}} | ||
| + | |||
| + | {{MAA Notice}} | ||
Revision as of 20:08, 17 February 2020
Problem
The coordinates of 
and 
are 
and 
respectively.
The value of 
that makes 
as small as possible is:
Solution
k will be between 1 and 5 for AC+BC to be minimum. If we mirror A across the Y axis as A' (-5,5), the distance A'C+BC will be same as AC+BC. The minimum of A'C+BC will occur when C is on the straight line connecting A' and B, i.e., C lies on the line A'B. So, we can find the Y-intercept of the line connecting A'(-5,5) and B(2,1), which is 15/7 = 2 1/7. so, the answer is 
See Also
| 1953 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 48 |
Followed by Problem 50 | |
| 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. 
