1954 AHSME Problems/Problem 47

At the midpoint of line segment $AB$ which is $p$ units long, a perpendicular $MR$ is erected with length $q$ units. An arc is described from $R$ with a radius equal to $\frac{1}{2}AB$ , meeting $AB$ at $T$ . Then $AT$ and $TB$ are the roots of:

$\textbf{(A)}\ x^2+px+q^2=0\\ \textbf{(B)}\ x^2-px+q^2=0\\ \textbf{(C)}\ x^2+px-q^2=0\\ \textbf{(D)}\ x^2-px-q^2=0\\ \textbf{(E)}\ x^2-px+q=0$

Solution

Note that $AT + TB = p$ . Note, also, that $AT * TB = (AM - TM) * (BM + TM) = (AM - TM) * (AM + TM) = AM^2 - TM^2 = TR^2 - TM^2 = MR^2 = q^2$ . Therefore, by Vieta's formulas, we have $(x - AT)(x - TB) = x^2 - px + q^2$ , so our answer is $\boxed{\textbf{(B)}}$ and we are done.

1954 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 46	Followed by Problem 48
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1954 AHSME Problems/Problem 47

Solution

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