Difference between revisions of "2022 AMC 12B Problems/Problem 14"

Revision as of 21:23, 27 October 2023

Problem

The graph of $y=x^2+2x-15$ intersects the $x$ -axis at points $A$ and $C$ and the $y$ -axis at point $B$ . What is $\tan(\angle ABC)$ ?

$\textbf{(A)}\ \frac{1}{7} \qquad \textbf{(B)}\ \frac{1}{4} \qquad \textbf{(C)}\ \frac{3}{7} \qquad \textbf{(D)}\ \frac{1}{2} \qquad \textbf{(E)}\ \frac{4}{7} \qquad$

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$",C,1.5NE,linewidth(4.5)); label("$y=x^2+2x-15$",(12,9),red); [/asy]$ ~MRENTHUSIASM

Solution 1 (Dot Product)

First, find $A=(-5,0)$ , $B=(0,-15)$ , and $C=(3,0)$ . Create vectors $\overrightarrow{BA}$ and $\overrightarrow{BC}.$ These can be reduced to $\langle -1, 3 \rangle$ and $\langle 1, 5 \rangle$ , respectively. Then, we can use the dot product to calculate the cosine of the angle (where $\theta=\angle ABC$ ) between them:

$\begin{align*} \langle -1, 3 \rangle \cdot \langle 1, 5 \rangle = 15-1 &= \sqrt{10}\sqrt{26}\cos(\theta),\\ \implies \cos (\theta) &= \frac{7}{\sqrt{65}}. \end{align*}$

Thus, $\tan(\angle ABC) = \sqrt{\frac{65}{49}-1}= \boxed{\textbf{(E)}\ \frac{4}{7}}.$

~Indiiiigo

Solution 2

Note that $y=x^2+2x-15$ intersects the $x$ -axis at points $(-5, 0)$ and $(3, 0)$ . Without loss of generality, let these points be $A$ and $C$ respectively. Also, the graph intersects the $y$ -axis at point $B = (0, -15)$ .

Let point $O=(0, 0)$ . It follows that $\triangle AOB$ and $\triangle BOC$ are right triangles.

We have $\tan(\angle ABC) = \tan(\angle ABO + \angle OBC) = \frac{\tan(\angle ABO) + \tan(\angle OBC)}{1 - \tan(\angle ABO) \cdot \tan(\angle OBC)} = \frac{\frac15 + \frac13}{1 - \frac1{15}} = \boxed{\textbf{(E)}\ \frac{4}{7}}.$ Alternatively, we can use the Pythagorean Theorem to find that $AB = 5 \sqrt{10}$ and $BC = 3 \sqrt{26}$ and then use the $A = \frac12 ab \sin \angle C$ area formula for a triangle and the Law of Cosines to find $\tan(\angle ABC)$ .

Solution 3

Like above, we set $A$ to $(-5,0)$ , $B$ to $(0, -15)$ , and $C$ to $(3,0)$ , then finding via the Pythagorean Theorem that $AB = 5 \sqrt{10}$ and $BC = 3 \sqrt{26}$ . Using the Law of Cosines, we see that $\cos(\angle ABC) = \frac{AB^2 + BC^2 - AC^2}{2 AB BC} = \frac{250 + 234 - 64}{30 \sqrt{260}} = \frac{7}{\sqrt{65}}.$ Then, we use the identity $\tan^2(x) = \sec^2(x) - 1$ to get $\tan(\angle ABC) = \sqrt{\frac{65}{49} - 1} = \boxed{\textbf{(E)}\ \frac{4}{7}}.$

~ jamesl123456

Solution 4

We can reflect the figure, but still have the same angle. This problem is the same as having points $D(0,0)$ , $E(3,15)$ , and $F(-5,15)$ , where we're solving for angle FED. We can use the formula for $\tan(a-b)$ to solve now where $a$ is the $x$ -axis to angle $F$ and $b$ is the $x$ -axis to angle $E$ . $\tan(a) = \textrm{slope of line }DF = -3$ and $\tan(B) = \textrm{slope of line }DE = 5$ . Plugging these values into the $\tan(a-b)$ formula, we get $(-3-5)/(1+(-3\cdot 5))$ which is $\boxed{\textbf{(E)}\ \frac{4}{7}}.$

~mathboy100 (minor LaTeX edits)

Solution 5

We use the formula $[ABC]=\frac{1}{2}ab\sin{C}.$

Note that $\triangle ABC$ has side-lengths $AB=5\sqrt{10}$ and $BC=3\sqrt{26}$ from Pythagorean theorem, with the area being $\frac12\cdot8\cdot15.$

We equate the areas together to get: $\frac12\cdot8\cdot15=\frac12\cdot5\sqrt{10}\cdot3\sqrt{26}\cdot\sin{B},$ from which $\sin{B}=\frac{8}{\sqrt{260}}.$

From Pythagorean Identity, $\cos{B}=\frac{14}{\sqrt{260}}.$

Then we use $\tan{B}=\frac{\sin{B}}{\cos{B}}$ , to obtain $\tan{B}=\frac{8}{14}=\boxed{\textbf{(E)}\ \frac{4}{7}}.$

- 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 $x^2 + 2x - 15 = (x-3)(x+5)$ , we may assume, without loss of generality, that $x$ -intercepts of the given parabola are $A(-5,0)$ and $C(3,0)$ . And, point $B$ has coordinates $(0,-15)$ . Consider complex numbers $z = 3 + i$ and $w = 5 + i$ whose arguments are $\theta \coloneqq \angle OBA$ and $\phi \coloneqq \angle OBC$ , respectively. Notice that $\angle ABC = \theta + \phi$ is the argument of the product $zw$ which is $zw = (3+i)(5+i) = 14 + 8i.$ Hence $\tan \angle ABC = \frac{\operatorname{Im}(zw)}{\operatorname{Re}(zw)} = \frac{8}{14} = \boxed{\textbf{(E)}\ \frac{4}{7}}.$

~VensL.

Video Solution by mop 2024

https://youtu.be/ezGvZgBLe8k&t=458s

~r00tsOfUnity

Video Solution (Under 2 min!)

https://youtu.be/InJgY_JYBkE

~Education, the Study of Everything

Video Solution(1-16)

https://youtu.be/SCwQ9jUfr0g

~~Hayabusa1

@@ Line 58: / Line 58: @@
 == Solution 2==
-<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 y-axis at point <math>B = (0, -15)</math>.
+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</math> denote the origin <math>(0, 0)</math>. Note that triangles <math>AOB</math> and <math>BOC</math> are right.
+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>
+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}{15 \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>
+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
@@ Line 82: / Line 79: @@
 ==Solution 5==
-We use the identity <math>[ABC]=\frac{1}{2}ab\sin{C}.</math>
+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>
@@ Line 89: / Line 86: @@
 from which <math>\sin{B}=\frac{8}{\sqrt{260}}.</math>
-From Pythagorean identity, <math>\cos{B}=\frac{14}{\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 ==
 {{AMC12 box|year=2022|ab=B|num-b=13|num-a=15}}
 {{MAA Notice}}

2022 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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