2013 AIME II Problems/Problem 13
Contents
Problem 13
In , , and point is on so that . Let be the midpoint of . Given that and , the area of can be expressed in the form , where and are positive integers and is not divisible by the square of any prime. Find .
Video Solution by Punxsutawney Phil
https://www.youtube.com/watch?v=IXPT0vHgt_c
Solution 1
We can set . Set , therefore . Thereafter, by Stewart's Theorem on and cevian , we get . Also apply Stewart's Theorem on with cevian . After simplification, . Therefore, . Finally, note that (using [] for area) , because of base-ratios. Using Heron's Formula on , as it is simplest, we see that , so your answer is .
Solution 2
After drawing the figure, we suppose , so that , , and .
Using Law of Cosines for and ,we get
So, , we get
Using Law of Cosines in , we get
So,
Using Law of Cosines in and , we get
, and according to , we can get
Using and , we can solve and .
Finally, we use Law of Cosines for ,
then , so the height of this is .
Then the area of is , so the answer is .
Solution 3
Let be the foot of the altitude from with other points labelled as shown below. Now we proceed using mass points. To balance along the segment , we assign a mass of and a mass of . Therefore, has a mass of . As is the midpoint of , we must assign a mass of as well. This gives a mass of and a mass of .
Now let be the base of the triangle, and let be the height. Then as , and as , we know that Also, as , we know that . Therefore, by the Pythagorean Theorem on , we know that
Also, as , we know that . Furthermore, as , and as , we know that and , so . Therefore, by the Pythagorean Theorem on , we get Solving this system of equations yields and . Therefore, the area of the triangle is , giving us an answer of .
Solution 4
Let the coordinates of , and be , and respectively. Then and implies ; implies Solve this system of equations simultaneously, and . Area of the triangle is , giving us an answer of .
Solution 5
Let . Then and . Also, let . Using Stewart's Theorem on gives us the equation or, after simplifying, . We use Stewart's again on : , which becomes . Substituting , we see that , or . Then .
We now use Law of Cosines on . . Plugging in for and , , so . Using the Pythagorean trig identity , , so .
, and our answer is .
Note to writter: Couldn't we just use Heron's formula for after is solved then noticing that ?
Solution 6 (Barycentric Coordinates)
Let ABC be the reference triangle, with , , and . We can easily calculate and subsequently . Using distance formula on and gives
But we know that , so we can substitute and now we have two equations and two variables. So we can clear the denominators and prepare to cancel a variable:
Then we add the equations to get
Then plugging gives and . Then the height from is , and the area is and our answer is .
Solution 7
Let and . It is trivial to show that and . Thus, since and , we get that
Multiplying both equations by , we get that
Solving these equations, we get that and .
Thus, the area of is , so our answer is .
Solution 8
The main in solution is to prove that .
Let be midpoint Let be cross point of and
We use the formula for crossing segments in and get:
By Stewart's Theorem on and cevian , we get after simplification trapezium is cyclic vladimir.shelomovskii@gmail.com, vvsss
Solution 9
Let and let Then and
By the Law of Cosines on triangle Then by the Law of Cosines on triangle \begin{align*} AD^2 &= 16y^2 + 9y^2 - 2 \cdot 4y \cdot 3y \cdot \cos C \\ &= 25y^2 - 24y^2 \cdot \frac{8y^2 - x^2}{8y^2} \\ &= 3x^2 + y^2. \end{align*}Applying Stewart's Theorem to median in triangle we get Thus, This simplifies to
Applying Stewart's Theorem to median in triangle we get Thus, This simplifies to
Solving the system and we find and so and
Plugging this back in for our equation for gives us , so We can apply the alternative area of a triangle formula, where Therefore, our answer is .
Video Solution
https://youtu.be/jVV4pYDGxGE?si=fDGGUOvCZRfdwUEz
~MathProblemSolvingSkills.com
See Also
2013 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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