2022 AMC 12B Problems/Problem 14
Contents
Problem
The graph of intersects the -axis at points and and the -axis at point . What is ?
Diagram
~MRENTHUSIASM
Solution 1 (Dot Product)
First, find , , and . Create vectors and These can be reduced to and , respectively. Then, we can use the dot product to calculate the cosine of the angle (where ) between them:
Thus,
~Indiiiigo
Solution 2
Note that intersects the -axis at points and . Without loss of generality, let these points be and respectively. Also, the graph intersects the -axis at point .
Let point . It follows that and are right triangles.
We have Alternatively, we can use the Pythagorean Theorem to find that and and then use the area formula for a triangle and the Law of Cosines to find .
Solution 3
Like above, we set to , to , and to , then finding via the Pythagorean Theorem that and . Using the Law of Cosines, we see that Then, we use the identity to get
~ jamesl123456
Solution 4
We can reflect the figure, but still have the same angle. This problem is the same as having points , , and , where we're solving for angle FED. We can use the formula for to solve now where is the -axis to angle and is the -axis to angle . and . Plugging these values into the formula, we get which is
~mathboy100 (minor LaTeX edits)
Solution 5
We use the formula
Note that has side-lengths and from Pythagorean theorem, with the area being
We equate the areas together to get: from which
From Pythagorean Identity,
Then we use , to obtain
- SAHANWIJETUNGA
Solution 6 (Complex Numbers)
From , we may assume, without loss of generality, that -intercepts of the given parabola are and . And, point has coordinates . Consider complex numbers and whose arguments are and , respectively. Notice that is the argument of the product which is Hence
~VensL.
Video Solution(1-16)
~~Hayabusa1
See Also
2022 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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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