Difference between revisions of "2009 AIME I Problems/Problem 5"
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+ | == Solution 4(Area Ratios) == | ||
+ | Note that we are given that <math>\overline{MK} = \overline{KP}</math>, that <math>\overline{AK} = \overline{CK}.</math> Note then that <math>\angle MKA = \angle CKB</math> by vertical angles. From this, we have <math>\triangle MKA \cong PKC.</math> This means that <math>\overline{CP}</math> is 180. Applying angle bisector theorem on <math>\triangle ACB</math> gives <math>\frac{\overline{AL}}{\overline{LB}} = \frac{450}{300} = \frac{3}{2}.</math> Applying it on <math>\triangle KCB</math> yields | ||
+ | <cmath>\frac{\overline{KP}}{\overline{PB}} = \frac{225}{300} = \frac{3}{4}</cmath> | ||
+ | Now we can proceed with area ratios. Suppose the area of <math>\triangle ACB = A.</math> This means that <cmath>[\triangle AKL] = \frac{225}{225+225}\frac{3}{5}A = \frac{3}{10}A</cmath> | ||
+ | Continuing on <math>\triangle LPB</math> we have <cmath>[\triangle LPB] = \frac{2}{2+3}\frac{1}{2}\frac{4}{4+3} = \frac{4}{35}A</cmath> | ||
+ | Since <math>\overline{AK}=\overline{KC}</math> <math>[\triangle KPL] = [\triangle AKB]-[\triangle AKL] - [\triangle LPB] = \frac{1}{2}A - \frac{3}{10}A - \frac{4}{35}A = \frac{3}{35}A.</math> | ||
+ | Area ratios on <math>\triangle KCP</math> yield <math>[\triangle KCP] = \frac{1}{2}\frac{3}{3+4} = \frac{3}{14}.</math> Now, suppose <math>\overline{LP} = x.</math> We have that the ratio of areas of <math>\triangle LKP</math> and <math>\triangle PKC</math> is <math>\frac{x}{180}</math> | ||
+ | and is also <math>\frac{\frac{3}{35}}{\frac{3}{14}}</math> and equating these gives <cmath>x = \boxed{72}</cmath> | ||
== See also == | == See also == |
Revision as of 18:45, 21 June 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 or .
The only value that satisfies both equations is , and since , we have
Video Solution
~IceMatrix
Video Solution
~Shreyas S
Solution 4(Area Ratios)
Note that we are given that , that Note then that by vertical angles. From this, we have This means that is 180. Applying angle bisector theorem on gives Applying it on yields Now we can proceed with area ratios. Suppose the area of This means that Continuing on we have Since Area ratios on yield Now, suppose We have that the ratio of areas of and is and is also and equating these gives
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.