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:

$x^2+px+q^2=0$

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

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

$x^2-px-q^2=0$

$x^2-px+q=0$

Answer: B

Concepts:Pythagorean Theorem Vieta’s Formulas circle

Difficulty rating: 1740

Small Hint:

The sum $\overline{AT}+\overline{TB}$ is $p$

Big Hint:

Use $RT=\frac{p}{2}=AM$ in right triangle $RMT$ to show $\overline{AT}\cdot\overline{TB}=q^2$

Solution:

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.

1954 AMC 12 Problem 47

Answer: B

Difficulty rating: 1740

Small Hint:

Big Hint:

Solution:

Problem 47 in Other Years