Difference between revisions of "2022 AMC 12B Problems/Problem 14"
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==Solution 3== | ==Solution 3== | ||
− | Like above, we set <math>A</math> to <math>(-5,0)</math>, <math>B</math> to <math>(0, -15)</math>, and <math>C</math> to <math>(3,0)</math>, then finding via the Pythagorean Theorem that <math>AB = 5 \sqrt{10}</math> and <math>BC = 3 \sqrt{26}</math>. Using the Law of Cosines, we see that <cmath>\cos(\angle ABC) = \frac{AB^2 + BC^2 - AC^2}{2 AB BC} = \frac{250 + 234 - 64}{ | + | Like above, we set <math>A</math> to <math>(-5,0)</math>, <math>B</math> to <math>(0, -15)</math>, and <math>C</math> to <math>(3,0)</math>, then finding via the Pythagorean Theorem that <math>AB = 5 \sqrt{10}</math> and <math>BC = 3 \sqrt{26}</math>. Using the Law of Cosines, we see that <cmath>\cos(\angle ABC) = \frac{AB^2 + BC^2 - AC^2}{2 AB BC} = \frac{250 + 234 - 64}{30 \sqrt{260}} = \frac{7}{\sqrt{65}}.</cmath> Then, we use the identity <math>\tan^2(x) = \sec^2(x) - 1</math> to get <cmath>\tan(\angle ABC) = \sqrt{\frac{65}{49} - 1} = \boxed{\textbf{(E)}\ \frac{4}{7}}.</cmath> |
~ jamesl123456 | ~ jamesl123456 | ||
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- SAHANWIJETUNGA | - SAHANWIJETUNGA | ||
− | ==Solution 6 (Complex | + | ==Solution 6 (Complex Numbers)== |
<asy> | <asy> | ||
/* Made by MRENTHUSIASM */ | /* Made by MRENTHUSIASM */ | ||
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label("$\phi$",(0,-15),9*dir(84)); | label("$\phi$",(0,-15),9*dir(84)); | ||
</asy> | </asy> | ||
− | From <math>x^2 + 2x - 15 = (x-3)(x+5)</math>, we may assume, without loss of generality, that <math>x</math>-intercepts of the given parabola are <math>A(-5,0)</math> and <math>C(3,0)</math>. And, point <math>B</math> has coordinates <math>(0,-15)</math>. Consider complex numbers <math>z = | + | From <math>x^2 + 2x - 15 = (x-3)(x+5)</math>, we may assume, without loss of generality, that <math>x</math>-intercepts of the given parabola are <math>A(-5,0)</math> and <math>C(3,0)</math>. And, point <math>B</math> has coordinates <math>(0,-15)</math>. Consider complex numbers <math>z = 3 + i</math> and <math>w = 5 + i</math> whose arguments are <math>\theta \coloneqq \angle OBA</math> and <math>\phi \coloneqq \angle OBC</math>, respectively. Notice that <math>\angle ABC = \theta + \phi</math> is the argument of the product <math>zw</math> which is <cmath> zw = (3+i)(5+i) = 14 + 8i. </cmath> |
Hence <cmath>\tan \angle ABC = \frac{\operatorname{Im}(zw)}{\operatorname{Re}(zw)} = \frac{8}{14} = \boxed{\textbf{(E)}\ \frac{4}{7}}.</cmath> | Hence <cmath>\tan \angle ABC = \frac{\operatorname{Im}(zw)}{\operatorname{Re}(zw)} = \frac{8}{14} = \boxed{\textbf{(E)}\ \frac{4}{7}}.</cmath> | ||
~VensL. | ~VensL. | ||
+ | |||
+ | ==Video Solution by mop 2024== | ||
+ | https://youtu.be/ezGvZgBLe8k&t=458s | ||
+ | |||
+ | ~r00tsOfUnity | ||
+ | |||
+ | ==Video Solution (Under 2 min!)== | ||
+ | https://youtu.be/InJgY_JYBkE | ||
+ | |||
+ | ~<i>Education, the Study of Everything</i> | ||
+ | |||
+ | ==Video Solution(1-16)== | ||
+ | https://youtu.be/SCwQ9jUfr0g | ||
+ | |||
+ | ~~Hayabusa1 | ||
== See Also == | == See Also == | ||
{{AMC12 box|year=2022|ab=B|num-b=13|num-a=15}} | {{AMC12 box|year=2022|ab=B|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:23, 27 October 2023
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 by mop 2024
https://youtu.be/ezGvZgBLe8k&t=458s
~r00tsOfUnity
Video Solution (Under 2 min!)
~Education, the Study of Everything
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.