1954 AMC 12 Problem 47

Question

At the midpoint of line segment

\overline{AB}

which is

p

units long, a perpendicular

\overline{MR}

is erected with length

q

units. An arc is described from

R

with a radius equal to

\dfrac12\overline{AB},

meeting

\overline{AB}

at

T.

Then

\overline{AT}

and

\overline{TB}

are the roots of:

x^2+px+q^2=0

x^2-px+q^2=0

x^2+px-q^2=0

x^2-px-q^2=0

x^2-px+q=0

The sum

\overline{AT}+\overline{TB}

is

p

Use

RT=\frac{p}{2}=AM

in right triangle

RMT

to show

\overline{AT}\cdot\overline{TB}=q^2

Let

MT=t.

Since

AM=BM=RT=\frac{p}{2},

the Pythagorean theorem gives

t^2+q^2=\frac{p^2}{4}.

Also,

AT=\frac p2+t,\qquad TB=\frac p2-t.

Their sum is

p,

and their product is

\frac{p^2}{4}-t^2=q^2.

Therefore they are the roots of

x^2-px+q^2=0.

Thus, the correct answer is B.

Problem 47 in other years