Difference between revisions of "1968 AHSME Problems/Problem 18"
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<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 == |
Latest revision as of 20:55, 17 July 2024
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
Thus, we choose answer .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.