Difference between revisions of "2009 AIME I Problems/Problem 5"
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By the SAS congruence, <math>\triangle MKA</math> = <math>\triangle PKC</math>. So, <math>MA</math> = <math>CP</math> = <math>180</math>. | By the SAS congruence, <math>\triangle MKA</math> = <math>\triangle PKC</math>. So, <math>MA</math> = <math>CP</math> = <math>180</math>. | ||
Since <math>LP=\frac{2}{5}CP</math>, <math>LP = \frac{2}{5}(180) = \boxed{072}</math> | Since <math>LP=\frac{2}{5}CP</math>, <math>LP = \frac{2}{5}(180) = \boxed{072}</math> | ||
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+ | |||
+ | ==Solution 3 (Law of Cosines Bash)== | ||
+ | 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} = </math>\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 </math>\triangle CKP<math> and </math>\triangle CPB.<math> | ||
+ | |||
+ | 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> 16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos a\lpha.</cmath> | ||
+ | |||
+ | Bashing those out, we get that </math>x = 15 \sqrt{13}<math> and </math>\cos \alpha = \frac{7}{10}.<math> | ||
+ | |||
+ | 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 </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> | ||
+ | |||
+ | 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[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>. | ||
+ | |||
+ | We get the following system: | ||
+ | |||
+ | <cmath>(18 \sqrt{331})^2 = 450^2 + LC^2 - 2 \cdot 450 \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 </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 </math>LC = 252<math>, and since </math>LP = LC - PC$, we have <cmath>LC = 252 - 180 = \boxed{072}.</cmath> | ||
==Video Solution== | ==Video Solution== |
Revision as of 00:07, 6 March 2021
Contents
Problem
Triangle has and . Points and are located on and respectively so that , and is the angle bisector of angle . Let be the point of intersection of and , and let be the point on line for which is the midpoint of . If , find .
Diagram
Solution 1
Since is the midpoint of and , quadrilateral is a parallelogram, which implies and is similar to
Thus,
Now let's apply the angle bisector theorem.
Solution 2
Using the diagram above, we can solve this problem by using mass points. By angle bisector theorem: So, we can weight as and as and as . Since is the midpoint of and , the weight of is equal to the weight of , which equals . Also, since the weight of is and is , we can weight as .
By the definition of mass points, By vertical angles, angle angle . Also, it is given that and .
By the SAS congruence, = . So, = = . Since ,
Solution 3 (Law of Cosines Bash)
Using the diagram from solution , we can also utilize the fact that forms a parallelogram. Because of that, we know that .
Applying the angle bisector theorem to , we get that \frac{225}{300} = \frac{3}{4}.MK = KP = 3xBP = 4x$.
Now, apply law of cosines on$ (Error compiling LaTeX. ! Missing $ inserted.)\triangle CKP\triangle CPB.\angle KCP = \angle PCB = \alpha$, then the law of cosines gives the following system of equations:
<cmath>9x^2 = 225^2 + 180^2 - 2\cdot 225 \cdot 180 \cdot \cos \alpha</cmath> <cmath> 16x^2 = 180^2 + 300^2 - 2 \cdot 180 \cdot 300 \cdot \cos a\lpha.</cmath>
Bashing those out, we get that$ (Error compiling LaTeX. ! Missing $ inserted.)x = 15 \sqrt{13}\cos \alpha = \frac{7}{10}.\cos \alpha = \frac{7}{10}\cos 2 \cdot \alpha = -\frac{1}{50}.\triangle ABCAB$.
We get: <cmath>AB^2 = 450^2 + 300^2 - 2 \cdot 450 \cdot 300 \cdot \left(- \frac{1}{50} \right).</cmath>
Bashing gives$ (Error compiling LaTeX. ! Missing $ inserted.)AB = 30 \sqrt{331}.\triangle ABC\frac{AL}{BL} = \frac{450}{300} = \frac[3}{2}.AL = 18 \sqrt{331}BL = 12 \sqrt{331}.\triangle ALC\triangle BLCLC$.
We get the following system:
<cmath>(18 \sqrt{331})^2 = 450^2 + LC^2 - 2 \cdot 450 \cdot LC \cdot \frac{7}{10}</cmath> <cmath>(12 \sqrt{331})^2 = LC^2 + 300^2 - 2 \cdot 300 \cdot LC \cdot \frac{7}{10}</cmath>
The first equation gives$ (Error compiling LaTeX. ! Missing $ inserted.)LC = 252378LC = 252 or 168$.
The only value that satisfies both equations is$ (Error compiling LaTeX. ! Missing $ inserted.)LC = 252LP = LC - PC$, 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.