https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_47&feed=atom&action=history 1954 AHSME Problems/Problem 47 - Revision history 2022-07-02T01:39:03Z Revision history for this page on the wiki MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=1954_AHSME_Problems/Problem_47&diff=127725&oldid=prev Duck master: created page w/ solution and categorization 2020-07-07T20:43:23Z <p>created page w/ solution and categorization</p> <p><b>New page</b></p><div>At the midpoint of line segment &lt;math&gt;AB&lt;/math&gt; which is &lt;math&gt;p&lt;/math&gt; units long, a perpendicular &lt;math&gt;MR&lt;/math&gt; is erected with length &lt;math&gt;q&lt;/math&gt; units. An arc is described from &lt;math&gt;R&lt;/math&gt; with a radius equal to &lt;math&gt;\frac{1}{2}AB&lt;/math&gt;, meeting &lt;math&gt;AB&lt;/math&gt; at &lt;math&gt;T&lt;/math&gt;. Then &lt;math&gt;AT&lt;/math&gt; and &lt;math&gt;TB&lt;/math&gt; are the roots of:<br /> <br /> &lt;math&gt;\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&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> Note that &lt;math&gt;AT + TB = p&lt;/math&gt;. Note, also, that &lt;math&gt;AT * TB = (AM - TM) * (BM + TM) = (AM - TM) * (AM + TM) = AM^2 - TM^2 = TR^2 - TM^2 = MR^2 = q^2&lt;/math&gt;. Therefore, by Vieta's formulas, we have &lt;math&gt;(x - AT)(x - TB) = x^2 - px + q^2&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(B)}}&lt;/math&gt; and we are done.<br /> <br /> == See also ==<br /> <br /> {{AHSME 50p box|year=1954|num-b=46|num-a=48}}<br /> <br /> {{MAA Notice}}<br /> <br /> [[Category:AHSME]][[Category:AHSME Problems]]</div> Duck master