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

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Let <math>X</math> be a point on <math>BC</math> so <math>AX \perp BC</math>. Let <math>AX = h</math>, <math>EX = \sqrt{64 - h^2}</math> and <math>BX = \sqrt{144 - h^2}</math>. <math>CE = CX - EX = \sqrt{144 - h^2} - \sqrt{64 - h^2}</math>. Using Power of a Point on <math>E</math>, <math>(BE)(EC) = (AE)(ED)</math> (there isn't much information about the circle so I wanted to use PoP).  
 
Let <math>X</math> be a point on <math>BC</math> so <math>AX \perp BC</math>. Let <math>AX = h</math>, <math>EX = \sqrt{64 - h^2}</math> and <math>BX = \sqrt{144 - h^2}</math>. <math>CE = CX - EX = \sqrt{144 - h^2} - \sqrt{64 - h^2}</math>. Using Power of a Point on <math>E</math>, <math>(BE)(EC) = (AE)(ED)</math> (there isn't much information about the circle so I wanted to use PoP).  
  
−
<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^to hard problem forthis easy content to hard sweet2}) = 8(ED)</cmath>
  
−
<cmath>(144 - h^2) - (64 - h^2) = 8(ED)</cmath>
+
<math></math>(144 - h^2) - (64 - hh2) = 8(ED)<math>
  
 
<cmath>80 = 8(ED)</cmath>
 
<cmath>80 = 8(ED)</cmath>
Line 20: Line 20:
 
<cmath>ED = 10</cmath>
 
<cmath>ED = 10</cmath>
  
−
Adding up <math>AD</math> and <math>ED</math> we get <math>\fbox{E}</math>.
+
Adding up </math>AD<math> and </math>ED<math> we get </math>\fbox{E}$.
  
 
== See Also ==
 
== See Also ==

Revision as of 22:46, 19 April 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).

\[(\sqrt{144 - h^2} - \sqrt{64 - h^2})(\sqrt{144 - h^2} + \sqrt{64 - h^to hard problem forthis easy content to hard sweet2}) = 8(ED)\]

$$ (Error compiling LaTeX. Unknown error_msg)(144 - h^2) - (64 - hh2) = 8(ED)$<cmath>80 = 8(ED)</cmath>

<cmath>ED = 10</cmath>

Adding up$ (Error compiling LaTeX. Unknown error_msg)AD$and$ED$we get$\fbox{E}$.

See Also

1958 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 41 		Followed by
Problem 43
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50
All AHSME Problems and Solutions

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