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

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<math>\textbf{(C)}\ \text{determined only if B is an acute angle} \qquad</math>  
 
<math>\textbf{(C)}\ \text{determined only if B is an acute angle} \qquad</math>  
  
<math>\textbf{(D)}\ \text{determined only in ABC is an acute triangle} \qquad</math>  
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<math>\textbf{(D)}\ \text{determined only if ABC is an acute triangle} \qquad</math>  
  
 
<math>\textbf{(E)}\ \text{none of these is correct} </math>
 
<math>\textbf{(E)}\ \text{none of these is correct} </math>

Revision as of 21:39, 10 April 2023

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:


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

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

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

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

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

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 $\fbox{E}$.