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Difference between revisions of "1968 AHSME Problems/Problem 18"
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== Solution ==
 
== Solution ==
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== Solution ==
 
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<math>\fbox{D}</math>
 
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Draw a line passing through <math>A</math> and parallel to <math>BC</math>. Let <math>\angle FEC = 2n</math>. By alternate-interior-angles or whatever, <math>\angle BAE = n</math>, so <math>BAE</math> is an isosceles triangle, and it follows that <math>BE = 8</math>. <math>\triangle ABC \sim \triangle DEC</math>. Let <math>CE = x</math>. We have
 
Draw a line passing through <math>A</math> and parallel to <math>BC</math>. Let <math>\angle FEC = 2n</math>. By alternate-interior-angles or whatever, <math>\angle BAE = n</math>, so <math>BAE</math> is an isosceles triangle, and it follows that <math>BE = 8</math>. <math>\triangle ABC \sim \triangle DEC</math>. Let <math>CE = x</math>. We have
 
<cmath>\frac{8}{8+x} = \frac{5}{x} \Rightarrow 40+5x = 8x \Rightarrow x = CE =\boxed{\frac{40}{3}}.</cmath>
 
<cmath>\frac{8}{8+x} = \frac{5}{x} \Rightarrow 40+5x = 8x \Rightarrow x = CE =\boxed{\frac{40}{3}}.</cmath>
  
 +
Thus, we choose answer <math>\fbox{D}</math>.
  
 
== See also ==
 
== See also ==
−
{{AHSME box|year=1968|num-b=17|num-a=19}}   
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{{AHSME 35p box|year=1968|num-b=17|num-a=19}}   
  
 
[[Category: Introductory Geometry Problems]]
 
[[Category: Introductory Geometry Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 19:55, 17 July 2024

Problem

Side $AB$ of triangle $ABC$ has length 8 inches. Line $DEF$ is drawn parallel to $AB$ so that $D$ is on segment $AC$, and $E$ is on segment $BC$. Line $AE$ extended bisects angle $FEC$. If $DE$ has length $5$ inches, then the length of $CE$, in inches, is:

$\text{(A) } \frac{51}{4}\quad \text{(B) } 13\quad \text{(C) } \frac{53}{4}\quad \text{(D) } \frac{40}{3}\quad \text{(E) } \frac{27}{2}$

Solution

Draw a line passing through $A$ and parallel to $BC$. Let $\angle FEC = 2n$. By alternate-interior-angles or whatever, $\angle BAE = n$, so $BAE$ is an isosceles triangle, and it follows that $BE = 8$. $\triangle ABC \sim \triangle DEC$. Let $CE = x$. We have \[\frac{8}{8+x} = \frac{5}{x} \Rightarrow 40+5x = 8x \Rightarrow x = CE =\boxed{\frac{40}{3}}.\]

Thus, we choose answer $\fbox{D}$.

See also

1968 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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
All AHSME Problems and Solutions

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