Difference between revisions of "2021 AMC 12A Problems/Problem 21"
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<math>\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15</math> | <math>\textbf{(A) }7 \qquad \textbf{(B) }9 \qquad \textbf{(C) }11 \qquad \textbf{(D) }13\qquad \textbf{(E) }15</math> | ||
− | ==Solution 1== | + | ==Solution 1 (Perpendicular Bisectors)== |
The solutions to this equation are <math>z = 1</math>, <math>z = -1 \pm i\sqrt 3</math>, and <math>z = -2\pm i\sqrt 2</math>. Consider the five points <math>(1,0)</math>, <math>\left(-1,\pm\sqrt 3\right)</math>, and <math>\left(-2,\pm\sqrt 2\right)</math>; these are the five points which lie on <math>\mathcal E</math>. Note that since these five points are symmetric about the <math>x</math>-axis, so must <math>\mathcal E</math>. | The solutions to this equation are <math>z = 1</math>, <math>z = -1 \pm i\sqrt 3</math>, and <math>z = -2\pm i\sqrt 2</math>. Consider the five points <math>(1,0)</math>, <math>\left(-1,\pm\sqrt 3\right)</math>, and <math>\left(-2,\pm\sqrt 2\right)</math>; these are the five points which lie on <math>\mathcal E</math>. Note that since these five points are symmetric about the <math>x</math>-axis, so must <math>\mathcal E</math>. | ||
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
− | ==Solution 3 | + | == Solution 3 (Alternate Version of Solution 2) == |
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Starting from this system of equations from Solution 2: | Starting from this system of equations from Solution 2: | ||
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Therefore, the answer is <math>1+6=\boxed{\textbf{(A) }7}</math>. | Therefore, the answer is <math>1+6=\boxed{\textbf{(A) }7}</math>. | ||
− | ~ wzs26843545602 | + | ~wzs26843545602 |
+ | |||
+ | ==Solution 4 (Four Variables, Three Equations)== | ||
+ | |||
+ | The five roots are <math>1,-1+i\sqrt{3},-1-i\sqrt{3},-2+i\sqrt{2},-2-i\sqrt{2}.</math> | ||
+ | |||
+ | So, we express this conic in the form <math>ax^2+by^2+cx+z=0.</math> Note that this conic cannot have the <math>ky</math> term since the roots are symmetric about the <math>x</math>-axis. | ||
+ | |||
+ | Now we have equations | ||
+ | <cmath>\begin{align*} | ||
+ | a+c+z&=0, \\ | ||
+ | a+3b-c+z&=0, \\ | ||
+ | 4a+2b-2c+z&=0, | ||
+ | \end{align*}</cmath> | ||
+ | from which <math>a:b:c=5:6:9.</math> | ||
+ | |||
+ | So, the conic can be written in the form <math>5x^2+6y^2+9x=14.</math> If it is written in the form of <math>\frac{(x-m)^2}{r^2}+\frac{y^2}{s^2}=1,</math> then <math>r^2:s^2=6:5.</math> | ||
+ | |||
+ | Therefore, the desired eccentricity is <math>\sqrt{\frac{\sqrt{6-5}}{6}}=\sqrt{\frac{1}{6}},</math> and the answer is <math>1+6=\boxed{\textbf{(A) }7}.</math> | ||
+ | |||
+ | ~bluesoul | ||
+ | |||
+ | ==Solution 5 (Transformations)== | ||
+ | |||
+ | After calculating the <math>5</math> points that lie on <math>\mathcal E</math>, we try to find a transformation that sends <math>\mathcal E</math> to the unit circle. Scaling about <math>(1, 0)</math> works, since <math>(1, 0)</math> is already on the unit circle and such a transformation will preserve the ellipse's symmetry about the <math>x</math>-axis. If <math>2a</math> and <math>2b</math> are the lengths of the major and minor axes, respectively, then the ellipse will be scaled by a factor of <math>r := \frac1a</math> in the <math>x</math>-dimension and <math>s := \frac1b</math> in the <math>y</math>-dimension. | ||
+ | |||
+ | The transformation then sends the points <math>\left(-1,\pm\sqrt 3\right)</math> and <math>\left(-2,\pm\sqrt 2\right)</math> to the points <math>\left(1-2r, \pm s\sqrt 3\right)</math> and <math>\left(1-3r, \pm s\sqrt 2\right)</math>, respectively. These points are on the unit circle, so | ||
+ | <cmath>(1-2r)^2 + 3s^2 = 1 \quad \text{and} \quad (1-3r)^2 + 2s^2 = 1.</cmath> | ||
+ | This yields <cmath>4r^2 + 3s^2 = 4r \quad \text{and} \quad 9r^2 + 2s^2 = 6r,</cmath> from which | ||
+ | <cmath>\begin{align*} | ||
+ | 12r^2 + 9s^2 &= 18r^2 + 4s^2 \\ | ||
+ | \frac{r^2}{s^2} &= \frac56. | ||
+ | \end{align*}</cmath> | ||
+ | Recalling that <math>r = \frac1a</math> and <math>s = \frac1b</math>, this implies <math>\frac{b^2}{a^2} = \frac56</math>. From this, we get | ||
+ | <cmath>\frac{c^2}{a^2} = \frac{a^2-b^2}{a^2} = 1 - \frac{b^2}{a^2} = \frac{1}{6},</cmath> | ||
+ | so <math>\frac ca = \sqrt{\frac16}</math>, giving an answer of <math>1 + 6 = \boxed{\textbf{(A) } 7}</math>. | ||
+ | |||
+ | ~building | ||
== Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic) == | == Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic) == | ||
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~ pi_is_3.14 | ~ pi_is_3.14 | ||
+ | |||
+ | == Video Solution by MRENTHUSIASM (English & Chinese) == | ||
+ | https://www.youtube.com/watch?v=PQdz8IBAZig&t=8s | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/zAIcLfye_Mc | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
==See also== | ==See also== |
Latest revision as of 18:33, 1 August 2022
Contents
- 1 Problem
- 2 Solution 1 (Perpendicular Bisectors)
- 3 Solution 2 (Three Variables, Three Equations)
- 4 Solution 3 (Alternate Version of Solution 2)
- 5 Solution 4 (Four Variables, Three Equations)
- 6 Solution 5 (Transformations)
- 7 Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic)
- 8 Video Solution by MRENTHUSIASM (English & Chinese)
- 9 Video Solution
- 10 See also
Problem
The five solutions to the equation may be written in the form
for
where
and
are real. Let
be the unique ellipse that passes through the points
and
. The eccentricity of
can be written in the form
, where
and
are relatively prime positive integers. What is
? (Recall that the eccentricity of an ellipse
is the ratio
, where
is the length of the major axis of
and
is the is the distance between its two foci.)
Solution 1 (Perpendicular Bisectors)
The solutions to this equation are ,
, and
. Consider the five points
,
, and
; these are the five points which lie on
. Note that since these five points are symmetric about the
-axis, so must
.
Now let denote the ratio of the length of the minor axis of
to the length of its major axis. Remark that if we perform a transformation of the plane which scales every
-coordinate by a factor of
,
is sent to a circle
. Thus, the problem is equivalent to finding the value of
such that
,
, and
all lie on a common circle; equivalently, it suffices to determine the value of
such that the circumcenter of the triangle formed by the points
,
, and
lies on the
-axis.
Recall that the circumcenter of a triangle is the intersection point of the perpendicular bisectors of its three sides. The equations of the perpendicular bisectors of the segments
and
are
respectively. These two lines have different slopes for
, so indeed they will intersect at some point
; we want
. Plugging
into the first equation yields
, and so plugging
into the second equation and simplifying yields
Solving yields
.
Finally, recall that the lengths ,
, and
(where
is the distance between the foci of
) satisfy
. Thus the eccentricity of
is
and the requested answer is
.
Solution 2 (Three Variables, Three Equations)
Completing the square in the original equation, we have from which
Now, we will find the equation of an ellipse that passes through
and
in the
-plane. By symmetry, the center of
must be on the
-axis.
The formula of is
with the center
and the axes' lengths
and
Plugging the points and
into
respectively, we have the following system of equations:
Since
holds for all real numbers
we clear fractions and simplify:
Applying the Transitive Property to
and
we isolate
Substituting
and
into
we solve for
Substituting this into
we get
Substituting the current results into we get
Finally, we obtain
from which
The answer is
The graph of is shown below. Note that the foci are at
as shown in the blue points.
~MRENTHUSIASM
Solution 3 (Alternate Version of Solution 2)
Starting from this system of equations from Solution 2:
Let
and
. Therefore, the system can be rewritten as:
Subtracting
from
and
, we get
Plugging the former into the latter and simplifying yields
. Hence
. Since
, we get
, so the eccentricity is
.
Therefore, the answer is .
~wzs26843545602
Solution 4 (Four Variables, Three Equations)
The five roots are
So, we express this conic in the form Note that this conic cannot have the
term since the roots are symmetric about the
-axis.
Now we have equations
from which
So, the conic can be written in the form If it is written in the form of
then
Therefore, the desired eccentricity is and the answer is
~bluesoul
Solution 5 (Transformations)
After calculating the points that lie on
, we try to find a transformation that sends
to the unit circle. Scaling about
works, since
is already on the unit circle and such a transformation will preserve the ellipse's symmetry about the
-axis. If
and
are the lengths of the major and minor axes, respectively, then the ellipse will be scaled by a factor of
in the
-dimension and
in the
-dimension.
The transformation then sends the points and
to the points
and
, respectively. These points are on the unit circle, so
This yields
from which
Recalling that
and
, this implies
. From this, we get
so
, giving an answer of
.
~building
Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic)
~ pi_is_3.14
Video Solution by MRENTHUSIASM (English & Chinese)
https://www.youtube.com/watch?v=PQdz8IBAZig&t=8s
~MRENTHUSIASM
Video Solution
~MathProblemSolvingSkills.com
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.