Difference between revisions of "2009 AIME I Problems/Problem 5"
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Using the diagram from solution <math>1</math>, we can also utilize the fact that <math>AMCP</math> forms a parallelogram. Because of that, we know that <math>AM = CP = 180</math>. | Using the diagram from solution <math>1</math>, we can also utilize the fact that <math>AMCP</math> forms a parallelogram. Because of that, we know that <math>AM = CP = 180</math>. | ||
− | Applying the angle bisector theorem to <math>\triangle CKB</math>, we get that <math>\frac{KP}{PB} = | + | Applying the angle bisector theorem to <math>\triangle CKB</math>, we get that <math>\frac{KP}{PB} = \frac{225}{300} = \frac{3}{4}.</math> So, we can let <math>MK = KP = 3x</math> and <math>BP = 4x</math>. |
− | Now, apply law of cosines on < | + | Now, apply law of cosines on <math>\triangle CKP</math> and <math>\triangle CPB.</math> |
− | If we let < | + | If we let <math>\angle KCP = \angle PCB = \alpha</math>, then the law of cosines gives the following system of equations: |
<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 a\lpha.</cmath> | <cmath> 16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos a\lpha.</cmath> | ||
− | Bashing those out, we get that < | + | Bashing those out, we get that <math>x = 15 \sqrt{13}</math> and <math>\cos \alpha = \frac{7}{10}.</math> |
− | Since < | + | Since <math>\cos \alpha = \frac{7}{10}</math>, we can use the double angle formula to calculate that <math>\cos 2 \cdot \alpha = -\frac{1}{50}.</math> |
− | Now, apply Law of Cosines on < | + | Now, apply Law of Cosines on <math>\triangle ABC</math> to find <math>AB</math>. |
We get: <cmath>AB^2 = 450^2 + 300^2 - 2 \cdot 450 \cdot 300 \cdot \left(- \frac{1}{50} \right).</cmath> | We get: <cmath>AB^2 = 450^2 + 300^2 - 2 \cdot 450 \cdot 300 \cdot \left(- \frac{1}{50} \right).</cmath> | ||
− | Bashing gives < | + | Bashing gives <math>AB = 30 \sqrt{331}.</math> |
− | From the angle bisector theorem on < | + | 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 < | + | 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>. |
We get the following system: | We get the following system: | ||
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<cmath>(12 \sqrt{331})^2 = LC^2 + 300^2 - 2 \cdot 300 \cdot LC \cdot \frac{7}{10}</cmath> | <cmath>(12 \sqrt{331})^2 = LC^2 + 300^2 - 2 \cdot 300 \cdot LC \cdot \frac{7}{10}</cmath> | ||
− | The first equation gives < | + | The first equation gives <math>LC = 252</math> or <math>378</math> and the second gives <math>LC = 252 or 168</math>. |
− | The only value that satisfies both equations is < | + | The only value that satisfies both equations is <math>LC = 252</math>, and since <math>LP = LC - PC</math>, we have <cmath>LC = 252 - 180 = \boxed{072}.</cmath> |
==Video Solution== | ==Video Solution== |
Revision as of 00:08, 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:
\[16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos a\lpha.\] (Error compiling LaTeX. Unknown error_msg)
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 $\frac{AL}{BL} = \frac{450}{300} = \frac[3}{2}.$ (Error compiling LaTeX. Unknown error_msg) 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.