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> : <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== | ||
{{AHSME 50p box|year=1950|num-b=48|num-a=50}} | {{AHSME 50p box|year=1950|num-b=48|num-a=50}} | ||
[[Category:Introductory Geometry Problems]] | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 16:52, 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
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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