Difference between revisions of "2009 AIME I Problems/Problem 5"
(→Solution 3 (Law of Cosines Bash)) |
(→Solution 3 (Law of Cosines Bash)) |
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<cmath>9x^2 = 225^2 + 180^2 - 2\cdot 225 \cdot 180 \cdot \cos \alpha</cmath> | <cmath>9x^2 = 225^2 + 180^2 - 2\cdot 225 \cdot 180 \cdot \cos \alpha</cmath> | ||
− | <cmath> 16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos | + | <cmath> 16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos \alpha.</cmath> |
Bashing those out, we get that <math>x = 15 \sqrt{13}</math> and <math>\cos \alpha = \frac{7}{10}.</math> | Bashing those out, we get that <math>x = 15 \sqrt{13}</math> and <math>\cos \alpha = \frac{7}{10}.</math> | ||
Line 103: | Line 103: | ||
Bashing gives <math>AB = 30 \sqrt{331}.</math> | Bashing gives <math>AB = 30 \sqrt{331}.</math> | ||
− | From the angle bisector theorem on <math>\triangle ABC</math>, we know that <math>\frac{AL}{BL} = \frac{450}{300} = \frac | + | From the angle bisector theorem on <math>\triangle ABC</math>, we know that <math>\frac{AL}{BL} = \frac{450}{300} = \frac{3}{2}.</math> So, <math>AL = 18 \sqrt{331}</math> and <math>BL = 12 \sqrt{331}.</math> |
Now, we apply Law of Cosines on <math>\triangle ALC</math> and <math>\triangle BLC</math> in order to solve for the length of <math>LC</math>. | Now, we apply Law of Cosines on <math>\triangle ALC</math> and <math>\triangle BLC</math> in order to solve for the length of <math>LC</math>. |
Revision as of 00:09, 6 March 2021
Contents
Problem
Triangle has and . Points and are located on and respectively so that , and is the angle bisector of angle . Let be the point of intersection of and , and let be the point on line for which is the midpoint of . If , find .
Diagram
Solution 1
Since is the midpoint of and , quadrilateral is a parallelogram, which implies and is similar to
Thus,
Now let's apply the angle bisector theorem.
Solution 2
Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: So, we can weight as and as and as . Since is the midpoint of and , the weight of is equal to the weight of , which equals . Also, since the weight of is and is , we can weight as .
By the definition of mass points, By vertical angles, angle angle . Also, it is given that and .
By the SAS congruence, = . So, = = . Since ,
Solution 3 (Law of Cosines Bash)
Using the diagram from solution , we can also utilize the fact that forms a parallelogram. Because of that, we know that .
Applying the angle bisector theorem to , we get that So, we can let and .
Now, apply law of cosines on and
If we let , then the law of cosines gives the following system of equations:
Bashing those out, we get that and
Since , we can use the double angle formula to calculate that
Now, apply Law of Cosines on to find .
We get:
Bashing gives
From the angle bisector theorem on , we know that So, and
Now, we apply Law of Cosines on and in order to solve for the length of .
We get the following system:
The first equation gives or and the second gives .
The only value that satisfies both equations is , and since , we have
Video Solution
~IceMatrix
Video Solution
~Shreyas S
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.