Difference between revisions of "2018 AMC 12B Problems/Problem 16"
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We can quickly notice that the area of the roots will be smallest with points at <math>\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}</math>. Using trigonometry, we get the respective roots to be <math>(\operatorname{Re}(z), \operatorname{Im}(z)) \in \left\{\left(\sqrt{3},0\right), \left(\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right), \left(0,\sqrt{3}\right)\right\}</math>. Using the Shoelace Theorem, the area quickly comes out to be <math>\frac{3\sqrt{2}-3}{2} = \boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.</math> | We can quickly notice that the area of the roots will be smallest with points at <math>\frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}</math>. Using trigonometry, we get the respective roots to be <math>(\operatorname{Re}(z), \operatorname{Im}(z)) \in \left\{\left(\sqrt{3},0\right), \left(\frac{\sqrt{6}}{2}, \frac{\sqrt{6}}{2}\right), \left(0,\sqrt{3}\right)\right\}</math>. Using the Shoelace Theorem, the area quickly comes out to be <math>\frac{3\sqrt{2}-3}{2} = \boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.</math> | ||
− | ==Solution 5 ( | + | ==Solution 5 (Trigonometry)== |
It is easy to see that the figure the solutions form is that of an regular octagon <math>ABCDEFGH</math> With center <math>O</math> such that <math>AO = \sqrt{3}.</math> Also, the minimum area of a triangle formed by the solutions would be that of Triangle <math>ABC</math>, since the three points are consecutive vertices of the octagon. By dropping an altitude from <math>O</math> to <math>AB</math>, note that <math>\frac{1}{2} AB = \sqrt{3} \sin(22.5).</math> Thus, <math>AB = 2\sqrt{3} \sin(22.5).</math> Now given the length of <math>AB</math> we need to find the area of <math>ABC,</math> which, by simple trigonometry, is given as <cmath>AB \cos(22.5) \cdot AB \sin(22.5) = 12 \sin^3(22.5) \cdot \cos(22.5).</cmath> By the half angle formulas, <math>\sin(22.5) = \sqrt{\frac{1 - \sin(45)}{2}} = \frac{\sqrt{2-\sqrt{2}}}{2},</math> and <math>\cos(22.5)</math> is its conjugate <math>\frac{\sqrt{2+\sqrt{2}}}{2}.</math> Doing some expanding and simplifying, <cmath>12 \cdot \left ( \frac{\sqrt{2-\sqrt{2}}}{2} \right)^3 \cdot \left (\frac{\sqrt{2+\sqrt{2}}}{2} \right) = \frac{3\sqrt{12-8\sqrt{2}}}{4}.</cmath> | It is easy to see that the figure the solutions form is that of an regular octagon <math>ABCDEFGH</math> With center <math>O</math> such that <math>AO = \sqrt{3}.</math> Also, the minimum area of a triangle formed by the solutions would be that of Triangle <math>ABC</math>, since the three points are consecutive vertices of the octagon. By dropping an altitude from <math>O</math> to <math>AB</math>, note that <math>\frac{1}{2} AB = \sqrt{3} \sin(22.5).</math> Thus, <math>AB = 2\sqrt{3} \sin(22.5).</math> Now given the length of <math>AB</math> we need to find the area of <math>ABC,</math> which, by simple trigonometry, is given as <cmath>AB \cos(22.5) \cdot AB \sin(22.5) = 12 \sin^3(22.5) \cdot \cos(22.5).</cmath> By the half angle formulas, <math>\sin(22.5) = \sqrt{\frac{1 - \sin(45)}{2}} = \frac{\sqrt{2-\sqrt{2}}}{2},</math> and <math>\cos(22.5)</math> is its conjugate <math>\frac{\sqrt{2+\sqrt{2}}}{2}.</math> Doing some expanding and simplifying, <cmath>12 \cdot \left ( \frac{\sqrt{2-\sqrt{2}}}{2} \right)^3 \cdot \left (\frac{\sqrt{2+\sqrt{2}}}{2} \right) = \frac{3\sqrt{12-8\sqrt{2}}}{4}.</cmath> | ||
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== See Also == | == See Also == | ||
Revision as of 15:11, 22 November 2021
Contents
Problem
The solutions to the equation are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled
and
. What is the least possible area of
Solution 1 (Complex Numbers in Polar Form)
Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation units right, so they become the solutions to the equation
We rewrite to the polar form
where
is the magnitude of
such that
and
is the argument of
such that
By De Moivre's Theorem, we have from which
so
so
In the complex plane, the solutions to the equation are the vertices of a regular octagon with center
and radius
The least possible area of occurs when
and
are the consecutive vertices of the octagon. For simplicity purposes, let
and
as shown below.
Note that
has base
and height
so its area is
~MRENTHUSIASM
Solution 2 (Complex Numbers in Rectangular Form)
Recall that translations preserve the shapes and the sizes for all objects. We translate the solutions to the given equation units right, so they become the solutions to the equation
We have
Note that:
- For
we get
- For
we get
- For
let
for some real numbers
and
We substitute and then expand:
We equate the real parts and the imaginary parts, respectively, then simplify:
Since
in
we conclude that
and
must have the same sign. It follows that
becomes
By substitution, we get
from which
- For
we get
by a similar process.
We continue with the last two paragraphs of Solution 1 to obtain the answer
~MRENTHUSIASM
Solution 3 (Regular Octagon)
The polygon formed will be a regular octagon since there are roots of
. By normal math computation, we can figure out that two roots of
are
and
. These will lie on the real axis of the plane. Since it's a regular polygon, there has to be points on the vertical plane also which will be
and
.
Clearly, the rest of the points will lie in each quadrant. The next thing is to get their coordinates (note that to answer this question, we do not need all the coordinates, only 3 consecutive ones are needed).
The circumcircle of the octagon will have the equation . The coordinates of the point in the first quadrant will be equal in magnitude and both positive, so
. Solving gives
(meaning that the root represented is
).
This way we can deduce the values of the roots of the equation to be
To get the area, consecutive points such as
and
can be used. The area can be computed using different methods like using the Shoelace Theorem, or subtracting areas to find the area.
The answer you get is
.
(This method is not actually as long as it seems if you understand what you're doing while doing it. Also calculations can be made a little easier by solving using and multiplying your answer by
).
~OlutosinNGA
Solution 4 (Roots of Unity)
Now, we need to solve the equation where
. We can substitute this as
Now, let
for some
. Thus, the equation becomes
. Taking it to the other side, we get the equation to be
. Rearranging variables, we get
. Plotting this in the complex place, this is a circle centered at the origin and of radius
.
The graph of the original equation is merely a transformation which doesn't change affect the area. Thus, we can find the minimum area of the transformed equation
. Using Roots of Unity, we know that the roots of the equation lie at
radians from the origin.
We can quickly notice that the area of the roots will be smallest with points at . Using trigonometry, we get the respective roots to be
. Using the Shoelace Theorem, the area quickly comes out to be
Solution 5 (Trigonometry)
It is easy to see that the figure the solutions form is that of an regular octagon With center
such that
Also, the minimum area of a triangle formed by the solutions would be that of Triangle
, since the three points are consecutive vertices of the octagon. By dropping an altitude from
to
, note that
Thus,
Now given the length of
we need to find the area of
which, by simple trigonometry, is given as
By the half angle formulas,
and
is its conjugate
Doing some expanding and simplifying,
Further simplifying, we obtain that the desired expression is
Note that
so our answer comes out to be
-fidgetboss_4000
See Also
2018 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 15 |
Followed by Problem 17 |
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.