Difference between revisions of "1968 AHSME Problems/Problem 18"
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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> |
Revision as of 11:48, 30 June 2021
Problem
Side of triangle has length 8 inches. Line is drawn parallel to so that is on segment , and is on segment . Line extended bisects angle . If has length inches, then the length of , in inches, is:
Solution
Draw a line passing through and parallel to . Let . By alternate-interior-angles or whatever, , so is an isosceles triangle, and it follows that . . Let . We have
See also
1968 AHSME (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 | ||
All AHSME Problems and Solutions |
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