Difference between revisions of "1950 AHSME Problems/Problem 49"
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==Solution== | ==Solution== | ||
The locus of the median's endpoint on <math>BC</math> is the circle about <math>A</math> and of radius <math>1\frac{1}{2}</math> inches. The locus of the vertex <math>C</math> is then the circle twice as big and twice as far from <math>B</math>, i.e. of radius <math>3</math> inches and with center <math>4</math> inches from <math>B</math> along <math>BA</math> which means that our answer is: <math>\textbf{(D)}</math>. | The locus of the median's endpoint on <math>BC</math> is the circle about <math>A</math> and of radius <math>1\frac{1}{2}</math> inches. The locus of the vertex <math>C</math> is then the circle twice as big and twice as far from <math>B</math>, i.e. of radius <math>3</math> inches and with center <math>4</math> inches from <math>B</math> along <math>BA</math> which means that our answer is: <math>\textbf{(D)}</math>. | ||
| + | |||
| + | |||
| + | ==Solution 2== | ||
| + | Let <math>A(a,y_A)</math>, <math>B(b,y_B)</math> and <math>C(x,y)</math>. | ||
| + | |||
| + | Hence, <math>D\left(\frac{x+b}{2},\frac{y+y_B}{2}\right)</math> is a midpoint of <math>BC</math>. | ||
| + | |||
| + | Thus, the equation of needed locus is | ||
| + | <cmath>\left(\frac{x+b}{2}-a\right)^2+\left(\frac{y+y_B}{2}-y_A\right)^2=\left(\frac{3}{2}\right)^2,</cmath> | ||
| + | which is equation of the circle: | ||
| + | <cmath>(x-(2a-b))^2+(y-(2y_A-y_B))^2=3^2.</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> | ||
| + | |||
| + | |||
| + | ==Video Solution== | ||
| + | https://www.youtube.com/watch?v=l4lAvs2P_YA&t=333s | ||
| + | |||
| + | ~MathProblemSolvingSkills.com | ||
==See Also== | ==See Also== | ||
Latest revision as of 15:52, 15 July 2023
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
![\[\left(\frac{x+b}{2}-a\right)^2+\left(\frac{y+y_B}{2}-y_A\right)^2=\left(\frac{3}{2}\right)^2,\]](http://latex.artofproblemsolving.com/8/b/1/8b107cf25ef1a86da1ab76ededbfbb0935c23a49.png)
which is equation of the circle:
![\[(x-(2a-b))^2+(y-(2y_A-y_B))^2=3^2.\]](http://latex.artofproblemsolving.com/4/1/a/41a4c4a789bf630a348a4fecb8ef3ebbfb417eba.png)
Thus, D) is valid because ![\[\sqrt{(2a-b-b)^2+(2y_A-y_B-y_B)^2}=2\sqrt{(a-b)^2+(y_A-y_b)^2}=2AB=4.\]](http://latex.artofproblemsolving.com/8/d/1/8d1cf2f95352488d9fe02195dada98d3927eb496.png)
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
