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Difference between revisions of "1962 AHSME Problems/Problem 23"

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==Solution==
 
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We can actually determine the length of <math>DB</math> no matter what type of angles <math>A</math> and <math>B</math> are. This can be easily proved through considering all possible cases, though for the purposes of this solution, we'll show that we can determine <math>DB</math> if <math>A</math> is an obtuse angle.
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Let's see what happens when <math>A</math> is an obtuse angle. <math>\triangle AEB\sim \triangle CDB</math> by SSS, so <math>\frac{AE}{CD}=\frac{AB}{DB}</math>. Hence <math>DB=\frac{AE}{CD\times AB}</math>. Since we've determined the length of <math>DB</math> even though we have an obtuse angle, <math>DB</math> is not <math>\bf{only}</math> determined by what type of angle <math>A</math> may be. Hence our answer is <math>\fbox{E}</math>.

Revision as of 07:27, 5 July 2018

Problem

In triangle $ABC$, $CD$ is the altitude to $AB$ and $AE$ is the altitude to $BC$. If the lengths of $AB$, $CD$, and $AE$ are known, the length of $DB$ is:


$\textbf{(A)}\ \text{not determined by the information given} \qquad$

$\textbf{(B)}\ \text{determined only if A is an acute angle} \qquad$

$\textbf{(C)}\ \text{determined only if B is an acute angle} \qquad$

$\textbf{(D)}\ \text{determined only in ABC is an acute triangle} \qquad$

$\textbf{(E)}\ \text{none of these is correct}$

Solution

We can actually determine the length of $DB$ no matter what type of angles $A$ and $B$ are. This can be easily proved through considering all possible cases, though for the purposes of this solution, we'll show that we can determine $DB$ if $A$ is an obtuse angle.

Let's see what happens when $A$ is an obtuse angle. $\triangle AEB\sim \triangle CDB$ by SSS, so $\frac{AE}{CD}=\frac{AB}{DB}$. Hence $DB=\frac{AE}{CD\times AB}$. Since we've determined the length of $DB$ even though we have an obtuse angle, $DB$ is not $\bf{only}$ determined by what type of angle $A$ may be. Hence our answer is $\fbox{E}$.

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