Difference between revisions of "2021 AMC 12A Problems/Problem 21"
MRENTHUSIASM (talk | contribs) m (Moved the remark to the back, right before the video solutions.) |
MRENTHUSIASM (talk | contribs) m (→Solution 2 (Three Variables, Three Equations)) |
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a^2 &= b^2(2h+3). \hspace{26.75mm} (*) | a^2 &= b^2(2h+3). \hspace{26.75mm} (*) | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
− | + | Substituting <math>(1)</math> and <math>(*)</math> into <math>(2),</math> we solve for <math>h:</math> | |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
b^2(1+h)^2 + 3\underbrace{b^2(2h+3)}_{\text{by }(*)} &= \underbrace{(1-h)^2}_{\text{by }(1)}b^2 \\ | b^2(1+h)^2 + 3\underbrace{b^2(2h+3)}_{\text{by }(*)} &= \underbrace{(1-h)^2}_{\text{by }(1)}b^2 \\ |
Revision as of 15:57, 20 June 2021
Contents
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
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 at
and the axes' lengths
and
Plugging the points
and
into
respectively, we have the following system:
Clearing fractions gives
Since
holds for all real numbers
we rewrite the system as
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
~MRENTHUSIASM
Solution 3
After calculating the 5 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
Recalling that
and
, this implies
. From this, we get
so
, giving an answer of
.
~building
Remark
The graph of is shown below.
Graph in Desmos: https://www.desmos.com/calculator/ptdpdzsgyo
~MRENTHUSIASM
Video Solution by OmegaLearn (Using Ellipse Properties & Quadratic)
~ 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.