2013 AMC 12B Problems/Problem 24
Contents
Problem
Let be a triangle where is the midpoint of , and is the angle bisector of with on . Let be the intersection of the median and the bisector . In addition is equilateral with . What is ?
Solution
Let and . From the conditions, let's deduct some convenient conditions that seem sufficient to solve the problem.
is the midpoint of side .
This implies that . Given that angle is degrees and angle is degrees, we can use the area formula to get
So, .....(1)
is angle bisector.
In the triangle , one has , therefore .....(2)
Furthermore, triangle is similar to triangle , so , therefore ....(3)
By (2) and (3) and the subtraction law of ratios, we get
Therefore , or . So .
Finally, using the law of cosine for triangle , we get
Solution 2 (Analytic)
Let us dilate triangle so that the sides of equilateral triangle are all equal to The purpose of this is to ease the calculations we make in the problem. Given this, we aim to find the length of segment so that we can un-dilate triangle by dividing each of its sides by . Doing so will make it so that , as desired, and doing so will allow us to get the length of , whose square is our final answer.
Let the foot of the altitude from to On the coordinate plane, position at , and make lie on the x-axis. Since points , , and , are collinear, must also lie on the x-axis. Additionally, since , , meaning that we can position point at . Now, notice that line has the equation and that line has the equation because angles and are both . We can then position at point and at point . Quickly note that, because is an angle bisector, must pass through the point .
We proceed to construct a system of equations. First observe that the midpoint of must lie on , with the equation . The coordinates of are , and we can plug in these coordinates into the equation of line , yielding that For our second equation, notice that line has equation . Midpoint must also lie on this line, and we can substitute coordinates again to get
Setting both equations equal to each other and multiplying both sides by , we have that , which in turn simplifies into when dividing the entire equation by Using the quadratic formula, we have that Here, we discard the positive root since must lie to the left of the y-axis. Then, the coordinates of are , and the coordinates of are Seeing that segment has half the length of side , we have that the length of is
Now, we divide each side length of by , and from this, will equal
Solution 3
By some angle-chasing, we find that . From here, construct a point on such that . Now, let , which means that and , and let . Note that we want to compute . Because , we have:
However, we have more similar triangles. In fact, going back to our original pair of similar triangles - and - gives us more similarity ratios:
Since we constructed point such that is parallel to , we now examine trapezoid . From the variables that we already set up, we know that , and . Let denote the foot of the perpendicular from to base and define similarly.
Because is a triangle, and . Thus, and . By the Pythagorean Theorem on ,
.
Solution 4
Since is equilateral, let's assume the sides of them are all , and denote the length of is . Since bisects , applying the angle bisector theorem and we can get ;. Now applying LOC, we can get . We get . Now applying the Stewart theorem in , we can find that , after simplifying, we get . After observation, the main key for this problem is , so we can solve in term of . Let's see the equation , we can find that so . Now back solving the first equation we can get that cuz the negative one can't work. After solving, we can get that so and we get ~ bluesoul
Solution 5 (Similar Triangles)
Denote the length of as and the length of as
Let be the midpoint of Denote the intersection of and as Note that and As we have that or is equilateral and Thus, and
Observe that
By the angle bisector theorem, we have that
We apply the Law of Cosines on as follows: or
~ASAB
Solution 6
, , ,
, , , ,
Let , , , ,
Video Solution by MOP 2024
~r00tsOfUnity
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
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