Difference between revisions of "1983 AHSME Problems/Problem 19"
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==Solution== | ==Solution== | ||
− | Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so | + | Let <math>AD = y</math>. Since <math>AD</math> bisects <math>\angle{BAC}</math>, the Angle Bisector Theorem gives <math>\frac{DB}{CD} = \frac{AB}{AC} = 2</math>, so let <math>CD = x</math> and <math>DB = 2x</math>. Applying the Law of Cosines to <math>\triangle CAD</math> gives <math>x^2 = 3^2 + y^2 - 3y</math>, and to <math>\triangle DAB</math> gives <math>(2x)^2 = 6^2 + y^2 - 6y</math>. Subtracting <math>4</math> times the first equation from the second equation therefore yields <math>0 = 6y - 3y^2 \Rightarrow y(y-2) = 0</math>, so <math>y</math> is <math>0</math> or <math>2</math>. But since <math>y \neq 0</math> (<math>y</math> is the length of a side of a triangle), <math>y</math> must be <math>2</math>, so the answer is <math>\boxed{\textbf{(A)} \ 2}</math>. |
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+ | ==See Also== | ||
+ | {{AHSME box|year=1983|num-b=18|num-a=20}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 00:57, 20 February 2019
Problem
Point is on side of triangle . If , then the length of is
Solution
Let . Since bisects , the Angle Bisector Theorem gives , so let and . Applying the Law of Cosines to gives , and to gives . Subtracting times the first equation from the second equation therefore yields , so is or . But since ( is the length of a side of a triangle), must be , so the answer is .
See Also
1983 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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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