1968 AHSME Problems/Problem 18: Difference between revisions

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 $\fbox{D}$ .

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

@@ Line 13: / Line 13: @@
 <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>.
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1968 AHSME Problems/Problem 18: Difference between revisions

Latest revision as of 19:55, 17 July 2024

Problem

Solution

See also