Difference between revisions of "2013 AMC 12B Problems/Problem 24"
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==Solution== | ==Solution== | ||
+ | Let <math>BN=x</math> and <math>NA=y</math>. From the conditions, let's deduct some convenient conditions that seems sufficient to solve the problem. | ||
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+ | '''<math>M</math> is the midpoint of side <math>AC</math>.''' | ||
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+ | This implies that <math>[ABX]=[CBX]</math>. Given that angle <math>ABX</math> is <math>60</math> degrees and angle <math>BXC</math> is <math>120</math> degrees, we can use the area formula to get | ||
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+ | <cmath>\frac{1}{2}(x+y)x \frac{\sqrt{3}}{2} = \frac{1}{2} x \cdot CX \frac{\sqrt{3}}{2}</cmath> | ||
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+ | So, <math>x+y=CX</math> .....(1) | ||
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+ | '''<math>CN</math> is angle bisector.''' | ||
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+ | In the triangle <math>ABC</math>, one has <math>BC/AC=x/y</math>, therefore <math>BC=2x/y</math>.....(2) | ||
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+ | Furthermore, triangle <math>BCN</math> is similar to triangle <math>MCX</math>, so <math>BC/CM=CN/CX</math>, therefore <math>BC = (CX+x)/CX = (2x+y)/(x+y)</math>....(3) | ||
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+ | By (2) and (3) and the subtraction law of ratios, we get | ||
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+ | <cmath>BC=2x/y = (2x+y)/(y+x) = y/x</cmath> | ||
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+ | Therefore <math>2x^2=y^2</math>, or <math>y=\sqrt{2}x</math>. So <math>BC = 2x/(\sqrt{2}x) = \sqrt{2}</math>. | ||
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+ | Finally, using the law of cosine for triangle <math>BCN</math>, we get | ||
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+ | <cmath>2 = BC^2 = x^2 + (2x+y)^2 - x(2x+y) = 3x^2 + 3xy + y^2 = (5+3\sqrt{2})x^2</cmath> | ||
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+ | <cmath>x^2 = \frac{2}{5+3\sqrt{2}} = \frac{10-6\sqrt{2}}{7}.</cmath> | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}} | {{AMC12 box|year=2013|ab=B|num-b=23|num-a=25}} |
Revision as of 13:28, 25 February 2013
Problem
Let be a triangle where is the midpoint of , and is the angle bisector of with on . Let be the intersection of the median and the bisector . In addition is equilateral with . What is ?
Solution
Let and . From the conditions, let's deduct some convenient conditions that seems sufficient to solve the problem.
is the midpoint of side .
This implies that . Given that angle is degrees and angle is degrees, we can use the area formula to get
So, .....(1)
is angle bisector.
In the triangle , one has , therefore .....(2)
Furthermore, triangle is similar to triangle , so , therefore ....(3)
By (2) and (3) and the subtraction law of ratios, we get
Therefore , or . So .
Finally, using the law of cosine for triangle , we get
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 | |
All AMC 12 Problems and Solutions |