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Difference between revisions of "1958 AHSME Problems/Problem 42"

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<cmath>(\sqrt{144 - h^2} - \sqrt{64 - h^2})(\sqrt{144 - h^2} + \sqrt{64 - h^2}) = 8(ED)</cmath>
 
<cmath>(\sqrt{144 - h^2} - \sqrt{64 - h^2})(\sqrt{144 - h^2} + \sqrt{64 - h^2}) = 8(ED)</cmath>
 
\\
 
\\
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<math></math>(144 - h^2) - (64 - h^2) = 8(ED)<math>
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<cmath>(144 - h^2) - (64 - h^2) = 8(ED)</cmath>
 
\\
 
\\
 
<cmath>80 = 8(ED)</cmath>
 
<cmath>80 = 8(ED)</cmath>
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<cmath>ED = 10</cmath>
 
<cmath>ED = 10</cmath>
 
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\\
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Adding up </math>AD<math> and </math>ED<math> we get </math>\fbox{E}$.
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Adding up <math>AD</math> and <math>ED</math> we get <math>\fbox{E}</math>.
  
 
== See Also ==
 
== See Also ==

Revision as of 10:01, 29 June 2024

Problem

In a circle with center $O$, chord $\overline{AB}$ equals chord $\overline{AC}$. Chord $\overline{AD}$ cuts $\overline{BC}$ in $E$. If $AC = 12$ and $AE = 8$, then $AD$ equals:

$\textbf{(A)}\ 27\qquad \textbf{(B)}\ 24\qquad \textbf{(C)}\ 21\qquad \textbf{(D)}\ 20\qquad \textbf{(E)}\ 18$

Solution

Let $X$ be a point on $BC$ so $AX \perp BC$. Let $AX = h$, $EX = \sqrt{64 - h^2}$ and $BX = \sqrt{144 - h^2}$. $CE = CX - EX = \sqrt{144 - h^2} - \sqrt{64 - h^2}$. Using Power of a Point on $E$, $(BE)(EC) = (AE)(ED)$ (there isn't much information about the circle so I wanted to use PoP).

\[(\sqrt{144 - h^2} - \sqrt{64 - h^2})(\sqrt{144 - h^2} + \sqrt{64 - h^2}) = 8(ED)\] \\ \[(144 - h^2) - (64 - h^2) = 8(ED)\] \\ \[80 = 8(ED)\] \\ \[ED = 10\] \\ Adding up $AD$ and $ED$ we get $\fbox{E}$.

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 41 		Followed by
Problem 43
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All AHSME Problems and Solutions

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