Difference between revisions of "2022 AMC 12B Problems/Problem 14"
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\textbf{(E)}\ \frac{4}{7} \qquad</math> | \textbf{(E)}\ \frac{4}{7} \qquad</math> | ||
− | == | + | ==Diagram== |
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(250); | ||
− | + | real xMin = -15; | |
+ | real xMax = 15; | ||
+ | real yMin = -17; | ||
+ | real yMax = 17; | ||
− | + | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | |
+ | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | ||
+ | label("$x$",(xMax,0),(2,0)); | ||
+ | label("$y$",(0,yMax),(0,2)); | ||
− | + | real f(real x) { return x^2+2*x-15; } | |
+ | draw(graph(f,-6.75,4.75),red); | ||
− | + | pair A, B, C; | |
+ | A = (-5,0); | ||
+ | B = (0,-15); | ||
+ | C = (3,0); | ||
− | + | draw(A--B--C); | |
+ | dot("$A$",A,1.5NW,linewidth(4.5)); | ||
+ | dot("$B$",B,1.5SE,linewidth(4.5)); | ||
+ | dot("$C$",C,1.5NE,linewidth(4.5)); | ||
− | Alternatively, the | + | label("$y=x^2+2x-15$",(12,9),red); |
+ | </asy> | ||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 1 (Dot Product) == | ||
+ | |||
+ | First, find <math>A=(-5,0)</math>, <math>B=(0,-15)</math>, and <math>C=(3,0)</math>. Create vectors <math>\overrightarrow{BA}</math> and <math>\overrightarrow{BC}.</math> These can be reduced to <math>\langle -1, 3 \rangle</math> and <math>\langle 1, 5 \rangle</math>, respectively. Then, we can use the dot product to calculate the cosine of the angle (where <math>\theta=\angle ABC</math>) between them: | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \langle -1, 3 \rangle \cdot \langle 1, 5 \rangle = 15-1 &= \sqrt{10}\sqrt{26}\cos(\theta),\\ | ||
+ | \implies \cos (\theta) &= \frac{7}{\sqrt{65}}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Thus, <cmath>\tan(\angle ABC) = \sqrt{\frac{65}{49}-1}= \boxed{\textbf{(E)}\ \frac{4}{7}}.</cmath> | ||
+ | |||
+ | ~Indiiiigo | ||
+ | |||
+ | == Solution 2== | ||
+ | |||
+ | Note that <math>y=x^2+2x-15</math> intersects the <math>x</math>-axis at points <math>(-5, 0)</math> and <math>(3, 0)</math>. Without loss of generality, let these points be <math>A</math> and <math>C</math> respectively. Also, the graph intersects the <math>y</math>-axis at point <math>B = (0, -15)</math>. | ||
+ | |||
+ | Let point <math>O=(0, 0)</math>. It follows that <math>\triangle AOB</math> and <math>\triangle BOC</math> are right triangles. | ||
+ | |||
+ | We have <cmath>\tan(\angle ABC) = \tan(\angle ABO + \angle OBC) = \frac{\tan(\angle ABO) + \tan(\angle OBC)}{1 - \tan(\angle ABO) \cdot \tan(\angle OBC)} = \frac{\frac15 + \frac13}{1 - \frac1{15}} = \boxed{\textbf{(E)}\ \frac{4}{7}}.</cmath> | ||
+ | Alternatively, we can use the [[Pythagorean Theorem]] to find that <math>AB = 5 \sqrt{10}</math> and <math>BC = 3 \sqrt{26}</math> and then use the <math>A = \frac12 ab \sin \angle C</math> [[Area of a Triangle|area formula for a triangle]] and the [[Law of Cosines]] to find <math>\tan(\angle ABC)</math>. | ||
+ | |||
+ | ==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}{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 | ||
+ | |||
+ | ==Solution 4== | ||
+ | |||
+ | We can reflect the figure, but still have the same angle. This problem is the same as having points <math>D(0,0)</math>, <math>E(3,15)</math>, and <math>F(-5,15)</math>, where we're solving for angle FED. We can use the formula for <math>\tan(a-b)</math> to solve now where <math>a</math> is the <math>x</math>-axis to angle <math>F</math> and <math>b</math> is the <math>x</math>-axis to angle <math>E</math>. <math>\tan(a) = \textrm{slope of line }DF = -3</math> and <math>\tan(B) = \textrm{slope of line }DE = 5</math>. Plugging these values into the <math>\tan(a-b)</math> formula, we get <math>(-3-5)/(1+(-3\cdot 5))</math> which is <math>\boxed{\textbf{(E)}\ \frac{4}{7}}.</math> | ||
+ | |||
+ | ~mathboy100 (minor LaTeX edits) | ||
+ | |||
+ | ==Solution 5== | ||
+ | |||
+ | We use the formula <math>[ABC]=\frac{1}{2}ab\sin{C}.</math> | ||
+ | |||
+ | Note that <math>\triangle ABC</math> has side-lengths <math>AB=5\sqrt{10}</math> and <math>BC=3\sqrt{26}</math> from Pythagorean theorem, with the area being <math>\frac12\cdot8\cdot15.</math> | ||
+ | |||
+ | We equate the areas together to get: <cmath>\frac12\cdot8\cdot15=\frac12\cdot5\sqrt{10}\cdot3\sqrt{26}\cdot\sin{B},</cmath> | ||
+ | from which <math>\sin{B}=\frac{8}{\sqrt{260}}.</math> | ||
+ | |||
+ | From Pythagorean Identity, <math>\cos{B}=\frac{14}{\sqrt{260}}.</math> | ||
+ | |||
+ | Then we use <math>\tan{B}=\frac{\sin{B}}{\cos{B}}</math>, to obtain <math>\tan{B}=\frac{8}{14}=\boxed{\textbf{(E)}\ \frac{4}{7}}.</math> | ||
+ | |||
+ | - SAHANWIJETUNGA | ||
+ | |||
+ | ==Solution 6 (Complex Numbers)== | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(300); | ||
+ | |||
+ | real xMin = -15; | ||
+ | real xMax = 15; | ||
+ | real yMin = -17; | ||
+ | real yMax = 17; | ||
+ | |||
+ | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | ||
+ | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | ||
+ | label("$x$",(xMax,0),(2,0)); | ||
+ | label("$y$",(0,yMax),(0,2)); | ||
+ | |||
+ | real f(real x) { return x^2+2*x-15; } | ||
+ | draw(graph(f,-6.75,4.75),red); | ||
+ | |||
+ | pair A, B, C, O; | ||
+ | A = (-5,0); | ||
+ | B = (0,-15); | ||
+ | C = (3,0); | ||
+ | O = origin; | ||
+ | |||
+ | markscalefactor=0.1; | ||
+ | draw(rightanglemark(B,O,C)); | ||
+ | draw(A--B--C); | ||
+ | dot("$A$",A,1.5SW,linewidth(4.5)); | ||
+ | dot("$B$",B,1.5SE,linewidth(4.5)); | ||
+ | dot("$C$",C,1.5SE,linewidth(4.5)); | ||
+ | dot("$O$",O,1.5SW,linewidth(4.5)); | ||
+ | |||
+ | label("$y=x^2+2x-15$",(12,9),red); | ||
+ | label("$5$",(-2.5,0),1.5N); | ||
+ | label("$3$",(1.5,0),1.5N); | ||
+ | label("$15$",(0,-6),W); | ||
+ | |||
+ | label("$\theta$",(0,-15),9*dir(100)); | ||
+ | label("$\phi$",(0,-15),9*dir(84)); | ||
+ | </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 = 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> | ||
+ | |||
+ | ~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.