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Difference between revisions of "1950 AHSME Problems/Problem 49"

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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>.
  
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==Solution 2==
 
Let <math>A(a,y_A)</math>, <math>B(b,y_B)</math> and <math>C(x,y)</math>.
 
Let <math>A(a,y_A)</math>, <math>B(b,y_B)</math> and <math>C(x,y)</math>.
  

Revision as of 22:09, 9 August 2017

Problem

A triangle has a fixed base $AB$ that is $2$ inches long. The median from $A$ to side $BC$ is $1\frac{1}{2}$ inches long and can have any position emanating from $A$. The locus of the vertex $C$ of the triangle is:

$\textbf{(A)}\ \text{A straight line }AB,1\dfrac{1}{2}\text{ inches from }A \qquad\\ \textbf{(B)}\ \text{A circle with }A\text{ as center and radius }2\text{ inches} \qquad\\ \textbf{(C)}\ \text{A circle with }A\text{ as center and radius }3\text{ inches} \qquad\\ \textbf{(D)}\ \text{A circle with radius }3\text{ inches and center }4\text{ inches from }B\text{ along } BA \qquad\\ \textbf{(E)}\ \text{An ellipse with }A\text{ as focus}$

Solution

The locus of the median's endpoint on $BC$ is the circle about $A$ and of radius $1\frac{1}{2}$ inches. The locus of the vertex $C$ is then the circle twice as big and twice as far from $B$, i.e. of radius $3$ inches and with center $4$ inches from $B$ along $BA$ which means that our answer is: $\textbf{(D)}$.


Solution 2

Let $A(a,y_A)$, $B(b,y_B)$ and $C(x,y)$.

Hence, $D\left(\frac{x+b}{2},\frac{y+y_B}{2}\right)$ is a midpoint of $BC$.

Thus, the equation of needed locus is \[\left(\frac{x+b}{2}-a\right)^2+\left(\frac{y+y_B}{2}-y_A\right)^2=\left(\frac{3}{2}\right)^2,\] which is equation of the circle: \[(x-(2a-b))^2+(y-(2y_A-y_B))^2=3^2.\]

Thus, D) is valid because \[\sqrt{(2a-b-b)^2+(2y_A-y_B-y_B)^2}=2\sqrt{(a-b)^2+(y_A-y_b)^2}=2AB=4.\]

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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