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Difference between revisions of "2021 AMC 12A Problems/Problem 21"

m (→‎Solution 2 (Three Variables, Three Equations))
(→‎Solution 2 (Three Variables, Three Equations))
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Now, we will find the equation of an ellipse <math>\mathcal E</math> that passes through <math>(1,0),\left(-1,\pm\sqrt3\right),</math> and <math>\left(-2,\pm\sqrt2\right)</math> in the <math>xy</math>-plane. By symmetry, the center of <math>\mathcal E</math> must be on the <math>x</math>-axis.
 
Now, we will find the equation of an ellipse <math>\mathcal E</math> that passes through <math>(1,0),\left(-1,\pm\sqrt3\right),</math> and <math>\left(-2,\pm\sqrt2\right)</math> in the <math>xy</math>-plane. By symmetry, the center of <math>\mathcal E</math> must be on the <math>x</math>-axis.
  
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The formula of <math>\mathcal E</math> is <cmath>\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1, \hspace{44.5mm} (\bigstar)</cmath> with the center <math>(h,0)</math> and the axes' lengths <math>2a</math> and <math>2b.</math>
+
The formula of <math>\mathcal E</math> is <cmath>\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1, </cmath> with the center <math>(h,0)</math> and the axes' lengths <math>2a</math> and <math>2b.</math>
  
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Plugging the points <math>(1,0),\left(-1,\sqrt3\right),</math> and <math>\left(-2,\sqrt2\right)</math> into <math>(\bigstar),</math> respectively, we have the following system of equations:
+
Plugging in the points <math>(1,0),\left(-1,\sqrt3\right),</math> and <math>\left(-2,\sqrt2\right)</math> we have the following system of equations:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
\frac{(1-h)^2}{a^2}&=1, \\
 
\frac{(1-h)^2}{a^2}&=1, \\
Line 27: Line 27:
 
\frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1.
 
\frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Clearing fractions gives
+
Simplifying, we get
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<cmath>\begin{align*}
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<cmath>\begin{align}
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(1-h)^2&=a^2, \\
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(1-h)^2&=a^2 \\
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b^2(-1-h)^2 + 3a^2 &= a^2b^2, \\
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b^2(1+h)^2 + 3a^2 &= a^2b^2 \\
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b^2(-2-h)^2 + 2a^2 &= a^2b^2.
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b^2(2+h)^2 + 2a^2 &= a^2b^2.  
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\end{align*}</cmath>
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\end{align}</cmath>
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Since <math>t^2=(-t)^2</math> holds for all real numbers <math>t,</math> we rewrite the system as
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Setting <math>(2) = (3)</math>, we isolate <math>a^2:</math>
−
<cmath>\begin{align*}
 
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(1-h)^2&=a^2, \hspace{30.25mm} &(1)\\
 
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b^2(1+h)^2 + 3a^2 &= a^2b^2, &(2)\\
 
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b^2(2+h)^2 + 2a^2 &= a^2b^2. &(3)
 
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\end{align*}</cmath>
 
−
Applying the Transitive Property to <math>(2)</math> and <math>(3),</math> we isolate <math>a^2:</math>
 
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
b^2(1+h)^2 + 3a^2 &= b^2(2+h)^2 + 2a^2 \\
 
b^2(1+h)^2 + 3a^2 &= b^2(2+h)^2 + 2a^2 \\
 
a^2 &= b^2\left((2+h)^2-(1+h)^2\right) \\
 
a^2 &= b^2\left((2+h)^2-(1+h)^2\right) \\
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a^2 &= b^2(2h+3). \hspace{26.75mm} (*)
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a^2 &= b^2(2h+3).  
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
Substituting <math>(1)</math> and <math>(*)</math> into <math>(2),</math> we solve for <math>h:</math>  
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Now we substitute these into <math>(2)</math> to solve for <math>h</math>. 
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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b^2(1+h)^2 + 3\underbrace{b^2(2h+3)}_{\text{by }(*)} &= \underbrace{(1-h)^2}_{\text{by }(1)}b^2 \\
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b^2(1+h)^2 + 3b^2(2h+3) &= (1-h)^2b^2 \\
 
(1+h)^2+3(2h+3)&=(1-h)^2 \\
 
(1+h)^2+3(2h+3)&=(1-h)^2 \\
 
1+2h+h^2+6h+9&=1-2h+h^2 \\
 
1+2h+h^2+6h+9&=1-2h+h^2 \\
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h&=-\frac{9}{10}.
 
h&=-\frac{9}{10}.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
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Substituting this into <math>(1),</math> we get <math>a^2=\frac{361}{100}.</math>
 
  
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Substituting the current results into <math>(*),</math> we get <math>b^2=\frac{361}{120}.</math>
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Substituting this into <math>(1),</math> we get <math>a^2=\frac{361}{100}</math>, and thus <math>b^2 = \frac{361}{120}</math>.
  
−
Finally, we obtain
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Finally, note that <math>c^2 = a^2 - b^2</math>, hence
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<cmath>c^2 = a^2-b^2 = 361\left(\frac{1}{100}-\frac{1}{120}\right) = \frac{361}{600},</cmath>
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<cmath>c^2 =361\left(\frac{1}{100}-\frac{1}{120}\right) = \frac{361}{600},</cmath>
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from which <cmath>\frac{c}{a}=\sqrt{\frac{c^2}{a^2}}=\sqrt{\frac{361/600}{361/100}}=\sqrt{\frac 16}.</cmath>
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thus <cmath>\frac{c}{a}=\sqrt{\frac{c^2}{a^2}}=\sqrt{\frac{361/600}{361/100}}=\sqrt{\frac 16}.</cmath>
 
The answer is <math>1+6=\boxed{\textbf{(A) } 7}.</math>
 
The answer is <math>1+6=\boxed{\textbf{(A) } 7}.</math>
  
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~MRENTHUSIASM
+
~MRENTHUSIASM (edits by eagleye)
  
 
==Solution 3==
 
==Solution 3==

Revision as of 14:35, 16 July 2021

Problem

The five solutions to the equation\[(z-1)(z^2+2z+4)(z^2+4z+6)=0\] may be written in the form $x_k+y_ki$ for $1\le k\le 5,$ where $x_k$ and $y_k$ are real. Let $\mathcal E$ be the unique ellipse that passes through the points $(x_1,y_1),(x_2,y_2),(x_3,y_3),(x_4,y_4),$ and $(x_5,y_5)$. The eccentricity of $\mathcal E$ can be written in the form $\sqrt{\frac mn}$ where $m$ and $n$ are relatively prime positive integers. What is $m+n$? (Recall that the eccentricity of an ellipse $\mathcal E$ is the ratio $\frac ca$, where $2a$ is the length of the major axis of $\mathcal E$ and $2c$ is the is the distance between its two foci.)

$\textbf{(A) } 7\qquad\textbf{(B) } 9\qquad\textbf{(C) } 11\qquad\textbf{(D) } 13\qquad\textbf{(E) } 15\qquad$

Solution 1

The solutions to this equation are $z = 1$, $z = -1 \pm i\sqrt 3$, and $z = -2\pm i\sqrt 2$. Consider the five points $(1,0)$, $(-1,\pm\sqrt 3)$, and $(-2,\pm\sqrt 2)$; these are the five points which lie on $\mathcal E$. Note that since these five points are symmetric about the $x$-axis, so must $\mathcal E$.

Now let $r=b/a$ denote the ratio of the length of the minor axis of $\mathcal E$ to the length of its major axis. Remark that if we perform a transformation of the plane which scales every $x$-coordinate by a factor of $r$, $\mathcal E$ is sent to a circle $\mathcal E'$. Thus, the problem is equivalent to finding the value of $r$ such that $(r,0)$, $(-r,\pm\sqrt 3)$, and $(-2r,\pm\sqrt 2)$ all lie on a common circle; equivalently, it suffices to determine the value of $r$ such that the circumcenter of the triangle formed by the points $P_1 = (r,0)$, $P_2 = (-r,\sqrt 3)$, and $P_3 = (-2r,\sqrt 2)$ lies on the $x$-axis.

Recall that the circumcenter of a triangle $ABC$ is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments $\overline{P_1P_2}$ and $\overline{P_1P_3}$ are\[y = \tfrac{\sqrt 3}2 + \tfrac{2r}{\sqrt 3}x\qquad\text{and}\qquad y = \tfrac{\sqrt 2}2 + \tfrac{3r}{\sqrt 2}(x + \tfrac r2)\]respectively. These two lines have different slopes for $r\neq 0$, so indeed they will intersect at some point $(x_0,y_0)$; we want $y_0 = 0$. Plugging $y = 0$ into the first equation yields $x = -\tfrac{3}{4r}$, and so plugging $y=0$ into the second equation and simplifying yields\[-\tfrac{1}{3r} = x + \tfrac r2 = -\tfrac{3}{4r} + \tfrac{r}{2}.\]Solving yields $r=\sqrt{\tfrac 56}$.

Finally, recall that the lengths $a$, $b$, and $c$ (where $c$ is the distance between the foci of $\mathcal E$) satisfy $c = \sqrt{a^2 - b^2}$. Thus the eccentricity of $\mathcal E$ is $\tfrac ca = \sqrt{1 - (\tfrac ba)^2} = \sqrt{\tfrac 16}$ and the requested answer is $\boxed{\textbf{(A) } 7}$.

Solution 2 (Three Variables, Three Equations)

Completing the square in the original equation, we have \[(z-1)\left((z+1)^2+3\right)\left((z+2)^2+2\right)=0,\] from which $z=1,-1\pm\sqrt{3}i,-2\pm\sqrt{2}i.$

Now, we will find the equation of an ellipse $\mathcal E$ that passes through $(1,0),\left(-1,\pm\sqrt3\right),$ and $\left(-2,\pm\sqrt2\right)$ in the $xy$-plane. By symmetry, the center of $\mathcal E$ must be on the $x$-axis.

The formula of $\mathcal E$ is \[\frac{(x-h)^2}{a^2}+\frac{y^2}{b^2}=1,\] with the center $(h,0)$ and the axes' lengths $2a$ and $2b.$

Plugging in the points $(1,0),\left(-1,\sqrt3\right),$ and $\left(-2,\sqrt2\right)$ we have the following system of equations: \begin{align*} \frac{(1-h)^2}{a^2}&=1, \\ \frac{(-1-h)^2}{a^2}+\frac{{\sqrt3}^2}{b^2}&=1, \\ \frac{(-2-h)^2}{a^2}+\frac{{\sqrt2}^2}{b^2}&=1. \end{align*} Simplifying, we get \begin{align} (1-h)^2&=a^2 \\ b^2(1+h)^2 + 3a^2 &= a^2b^2 \\ b^2(2+h)^2 + 2a^2 &= a^2b^2. \end{align} Setting $(2) = (3)$, we isolate $a^2:$ \begin{align*} b^2(1+h)^2 + 3a^2 &= b^2(2+h)^2 + 2a^2 \\ a^2 &= b^2\left((2+h)^2-(1+h)^2\right) \\ a^2 &= b^2(2h+3). \end{align*} Now we substitute these into $(2)$ to solve for $h$. \begin{align*} b^2(1+h)^2 + 3b^2(2h+3) &= (1-h)^2b^2 \\ (1+h)^2+3(2h+3)&=(1-h)^2 \\ 1+2h+h^2+6h+9&=1-2h+h^2 \\ 10h&=-9 \\ h&=-\frac{9}{10}. \end{align*}

Substituting this into $(1),$ we get $a^2=\frac{361}{100}$, and thus $b^2 = \frac{361}{120}$.

Finally, note that $c^2 = a^2 - b^2$, hence \[c^2 =361\left(\frac{1}{100}-\frac{1}{120}\right) = \frac{361}{600},\] thus \[\frac{c}{a}=\sqrt{\frac{c^2}{a^2}}=\sqrt{\frac{361/600}{361/100}}=\sqrt{\frac 16}.\] The answer is $1+6=\boxed{\textbf{(A) } 7}.$

~MRENTHUSIASM (edits by eagleye)

Solution 3

After calculating the 5 points that lie on $\mathcal E$, we try to find a transformation that sends $\mathcal E$ to the unit circle. Scaling about $(1, 0)$ works, since $(1, 0)$ is already on the unit circle and such a transformation will preserve the ellipse's symmetry about the $x$-axis. If $2a$ and $2b$ are the lengths of the major and minor axes, respectively, then the ellipse will be scaled by a factor of $r := 1/a$ in the $x$-dimension and $s := 1/b$ in the $y$-dimension.

The transformation then sends the points $(-1,\pm\sqrt 3)$ and $(-2,\pm\sqrt 2)$ to the points $(1-2r, \pm s\sqrt 3)$ and $(1-3r, \pm s\sqrt 2)$, respectively. These points are on the unit circle, so \[(1-2r)^2 + 3s^2 = 1 \quad \text{and} \quad (1-3r)^2 + 2s^2 = 1.\] This yields \begin{align*} 4r^2 + 3s^2 = 4r \quad \text{and} \quad 9r^2 + 2s^2 = 6r& \\ \implies \enskip 12r^2 + 9s^2 = 18r^2 + 4s^2& \\ \implies \enskip r^2/s^2 = 5/6.& \end{align*} Recalling that $r = 1/a$ and $s = 1/b$, this implies $b^2/a^2 = 5/6$. From this, we get \[\frac{c^2}{a^2} = \frac{a^2-b^2}{a^2} = 1 - \frac{b^2}{a^2} = \frac{1}{6},\] so $c/a = \sqrt{1/6}$, giving an answer of $1 + 6 = \boxed{\textbf{(A) } 7}$.

~building

Remark

The graph of $\mathcal E$ is shown below.

File:2021 AMC 12A Problem 21.png

Graph in Desmos: https://www.desmos.com/calculator/ptdpdzsgyo

~MRENTHUSIASM

Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic)

https://youtu.be/eIYFQSeIRzM

~ pi_is_3.14

See also

2021 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 20 		Followed by
Problem 22
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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