Difference between revisions of "1950 AHSME Problems/Problem 49"
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Thus, D) is valid because <cmath>\sqrt{(2a-b-b)^2+(2y_A-y_B-y_B)^2}=2\sqrt{(a-b)^2+(y_A-y_b)^2}=2AB=4.</cmath> | Thus, D) is valid because <cmath>\sqrt{(2a-b-b)^2+(2y_A-y_B-y_B)^2}=2\sqrt{(a-b)^2+(y_A-y_b)^2}=2AB=4.</cmath> | ||
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+ | ==Video Solution== | ||
+ | https://www.youtube.com/watch?v=l4lAvs2P_YA&t=333s | ||
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+ | ~MathProblemSolvingSkills.com | ||
==See Also== | ==See Also== |
Latest revision as of 15:52, 15 July 2023
Contents
[hide]Problem
A triangle has a fixed base that is inches long. The median from to side is inches long and can have any position emanating from . The locus of the vertex of the triangle is:
Solution
The locus of the median's endpoint on is the circle about and of radius inches. The locus of the vertex is then the circle twice as big and twice as far from , i.e. of radius inches and with center inches from along which means that our answer is: .
Solution 2
Let , and .
Hence, is a midpoint of .
Thus, the equation of needed locus is which is equation of the circle:
Thus, D) is valid because
Video Solution
https://www.youtube.com/watch?v=l4lAvs2P_YA&t=333s
~MathProblemSolvingSkills.com
See Also
1950 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 |
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