Difference between revisions of "2018 AMC 12B Problems/Problem 16"
(→Solution) |
MRENTHUSIASM (talk | contribs) m (→Solution 5 (Trigonometry)) |
||
(20 intermediate revisions by 4 users not shown) | |||
Line 5: | Line 5: | ||
<math>\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1</math> | <math>\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1</math> | ||
− | == Solution == | + | == 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 <math>6</math> units right, so they become the solutions to the equation <math>z^8=81.</math> | |
− | == Solution 2 ( | + | We rewrite <math>z</math> to the polar form <cmath>z=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta,</cmath> where <math>r</math> is the magnitude of <math>z</math> such that <math>r\geq0,</math> and <math>\theta</math> is the argument of <math>z</math> such that <math>0\leq\theta<2\pi.</math> |
+ | |||
+ | By De Moivre's Theorem, we have <cmath>z^8=r^8\operatorname{cis}(8\theta)={\sqrt3}^8(1),</cmath> from which | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li><math>r^8={\sqrt3}^8,</math> so <math>r=\sqrt3.</math></li><p> | ||
+ | <li><math>\begin{cases} | ||
+ | \begin{aligned} | ||
+ | \cos(8\theta) &= 1 \\ | ||
+ | \sin(8\theta) &= 0 | ||
+ | \end{aligned}, | ||
+ | \end{cases}</math> so <math>\theta=0,\frac{\pi}{4},\frac{\pi}{2},\frac{3\pi}{4},\pi,\frac{5\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4}.</math></li><p> | ||
+ | </ol> | ||
+ | In the complex plane, the solutions to the equation <math>z^8=81</math> are the vertices of a regular octagon with center <math>0</math> and radius <math>\sqrt3.</math> | ||
+ | |||
+ | The least possible area of <math>\triangle ABC</math> occurs when <math>A,B,</math> and <math>C</math> are the consecutive vertices of the octagon. For simplicity purposes, let <math>A=\sqrt3\operatorname{cis}\frac{\pi}{4}=\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i, B=\sqrt3\operatorname{cis}\frac{\pi}{2}=\sqrt3i,</math> and <math>C=\sqrt3\operatorname{cis}\frac{3\pi}{4}=-\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i,</math> as shown below. | ||
+ | <asy> | ||
+ | /* Made by MRENTHUSIASM */ | ||
+ | size(200); | ||
+ | |||
+ | int xMin = -2; | ||
+ | int xMax = 2; | ||
+ | int yMin = -2; | ||
+ | int yMax = 2; | ||
+ | int numRays = 24; | ||
+ | |||
+ | //Draws a polar grid that goes out to a number of circles | ||
+ | //equal to big, with numRays specifying the number of rays: | ||
+ | void polarGrid(int big, int numRays) | ||
+ | { | ||
+ | for (int i = 1; i < big+1; ++i) | ||
+ | { | ||
+ | draw(Circle((0,0),i), gray+linewidth(0.4)); | ||
+ | } | ||
+ | for (int i=0;i<numRays;++i) | ||
+ | draw(rotate(i*360/numRays)*((-big,0)--(big,0)), gray+linewidth(0.4)); | ||
+ | } | ||
+ | |||
+ | //Draws the horizontal gridlines | ||
+ | void horizontalLines() | ||
+ | { | ||
+ | for (int i = yMin+1; i < yMax; ++i) | ||
+ | { | ||
+ | draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); | ||
+ | } | ||
+ | } | ||
+ | |||
+ | //Draws the vertical gridlines | ||
+ | void verticalLines() | ||
+ | { | ||
+ | for (int i = xMin+1; i < xMax; ++i) | ||
+ | { | ||
+ | draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); | ||
+ | } | ||
+ | } | ||
+ | |||
+ | horizontalLines(); | ||
+ | verticalLines(); | ||
+ | polarGrid(xMax,numRays); | ||
+ | draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); | ||
+ | draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); | ||
+ | label("Re",(xMax,0),2*E); | ||
+ | label("Im",(0,yMax),2*N); | ||
+ | |||
+ | //The n such that we're taking the nth roots of unity multiplied by 2. | ||
+ | int n = 8; | ||
+ | |||
+ | pair A[]; | ||
+ | for(int i = 0; i <= n-1; i+=1) { | ||
+ | A[i] = rotate(360*i/n)*(sqrt(3),0); | ||
+ | } | ||
+ | |||
+ | label("$A$",A[1],1.5*NE,UnFill); | ||
+ | label("$B$",A[2],1.5*NE,UnFill); | ||
+ | label("$C$",A[3],1.5*NW,UnFill); | ||
+ | fill(A[1]--A[2]--A[3]--cycle,green); | ||
+ | |||
+ | draw(A[1]--A[3]^^A[0]--A[1]--A[2]--A[3]--A[4]--A[5]--A[6]--A[7]--cycle,red); | ||
+ | |||
+ | for(int i = 0; i< n; ++i) dot(A[i],red+linewidth(4.5)); | ||
+ | </asy> | ||
+ | Note that <math>\triangle ABC</math> has base <math>AC=\sqrt6</math> and height <math>\sqrt3-\frac{\sqrt6}{2},</math> so its area is <cmath>\frac12\cdot\sqrt6\cdot\left(\sqrt3-\frac{\sqrt6}{2}\right)=\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.</cmath> | ||
+ | ~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 <math>6</math> units right, so they become the solutions to the equation <math>z^8=81.</math> | ||
+ | |||
+ | We have | ||
+ | <cmath>\begin{align*} | ||
+ | z^8 &= 81 \\ | ||
+ | z^4 &= 9, -9 \\ | ||
+ | z^2 &= 3, -3, 3i, -3i. \\ | ||
+ | \end{align*}</cmath> | ||
+ | Note that: | ||
+ | <ol style="margin-left: 1.5em;"> | ||
+ | <li>For <math>z^2=3,</math> we get <math>\boldsymbol{z=\sqrt3,-\sqrt3}.</math></li><p> | ||
+ | <li>For <math>z^2=-3,</math> we get <math>\boldsymbol{z=\sqrt3i,-\sqrt3i}.</math></li><p> | ||
+ | <li>For <math>z^2=3i,</math> let <math>z=a+bi</math> for some real numbers <math>a</math> and <math>b.</math> We substitute and then expand: | ||
+ | <cmath>\begin{align*} | ||
+ | (a+bi)^2 &= 3i \\ | ||
+ | \left(a^2-b^2\right)+2abi &= 3i. | ||
+ | \end{align*}</cmath> | ||
+ | We equate the real parts and the imaginary parts, respectively, then simplify: | ||
+ | <cmath>\begin{align*} | ||
+ | a^2 &= b^2, &&(1) \\ | ||
+ | ab &= \frac32. &&(2) | ||
+ | \end{align*}</cmath> | ||
+ | Since <math>ab>0</math> in <math>(2),</math> we conclude that <math>a</math> and <math>b</math> must have the same sign. It follows that <math>(1)</math> becomes <math>a=b.</math> By substitution, we get <math>(a,b)=\left(\frac{\sqrt6}{2},\frac{\sqrt6}{2}\right),\left(-\frac{\sqrt6}{2},-\frac{\sqrt6}{2}\right),</math> from which <math>\boldsymbol{z=\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i,-\frac{\sqrt6}{2}-\frac{\sqrt6}{2}i}.</math> | ||
+ | </li><p> | ||
+ | <li>For <math>z^2=-3i,</math> we get <math>\boldsymbol{z=\frac{\sqrt6}{2}-\frac{\sqrt6}{2}i,-\frac{\sqrt6}{2}+\frac{\sqrt6}{2}i}</math> by a similar process.</li><p> | ||
+ | </ol> | ||
+ | We continue with the last two paragraphs of Solution 1 to obtain the answer <math>\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.</math> | ||
+ | |||
+ | ~MRENTHUSIASM | ||
+ | |||
+ | == Solution 3 (Regular Octagon) == | ||
The polygon formed will be a regular octagon since there are <math>8</math> roots of <math>z^8=81</math>. By normal math computation, we can figure out that two roots of <math>z^8=81</math> are <math>\sqrt{3}</math> and <math>-\sqrt{3}</math>. 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 <math>\sqrt{3}i</math> and <math>-\sqrt{3}i</math>. | The polygon formed will be a regular octagon since there are <math>8</math> roots of <math>z^8=81</math>. By normal math computation, we can figure out that two roots of <math>z^8=81</math> are <math>\sqrt{3}</math> and <math>-\sqrt{3}</math>. 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 <math>\sqrt{3}i</math> and <math>-\sqrt{3}i</math>. | ||
Line 17: | Line 132: | ||
The circumcircle of the octagon will have the equation <math>i^2+r^2=3</math>. The coordinates of the point in the first quadrant will be equal in magnitude and both positive, so <math>i=r</math>. Solving gives <math>i=r=\frac{\sqrt{3}}{\sqrt{2}}</math> (meaning that the root represented is <math>\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i</math>). | The circumcircle of the octagon will have the equation <math>i^2+r^2=3</math>. The coordinates of the point in the first quadrant will be equal in magnitude and both positive, so <math>i=r</math>. Solving gives <math>i=r=\frac{\sqrt{3}}{\sqrt{2}}</math> (meaning that the root represented is <math>\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i</math>). | ||
− | This way we can deduce the values of the <math>8</math> roots of the equation to be < | + | This way we can deduce the values of the <math>8</math> roots of the equation to be <cmath>\sqrt{3},-\sqrt{3},-\sqrt{3}i,\sqrt{3}i,\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,-\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i,\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i,-\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i.</cmath> |
− | To get the area, <math>3</math> consecutive points such as <math>\sqrt{3},</math> <math>\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,</math> and <math>\sqrt{3}i</math> can be used. The area can be computed using different methods like using the | + | To get the area, <math>3</math> consecutive points such as <math>\sqrt{3},</math> <math>\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,</math> and <math>\sqrt{3}i</math> 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 <math>\boxed{B}</math>. | + | The answer you get is <math>\boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}</math>. |
(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 <math>x^8=1</math> and multiplying your answer by <math>\sqrt{3}</math>). | (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 <math>x^8=1</math> and multiplying your answer by <math>\sqrt{3}</math>). | ||
+ | ~OlutosinNGA | ||
+ | |||
+ | == Solution 4 (Roots of Unity) == | ||
+ | |||
+ | Now, we need to solve the equation <math>(z+6)^8 = 81</math> where <math>z = a+bi</math>. We can substitute this as <cmath>(a+6+bi)^8 = 81</cmath> Now, let <math>a+6 = q</math> for some <math>q \in \mathbb{Z}</math>. Thus, the equation becomes <math>((q+bi)^2)^4 = 81</math>. Taking it to the other side, we get the equation to be <math>(q+bi)^2 = 3</math>. Rearranging variables, we get <math>(q+bi) = \sqrt{3}</math>. Plotting this in the complex place, this is a circle centered at the origin and of radius <math>\sqrt{3}</math>. | ||
+ | |||
+ | The graph of the original equation <math>(a+6+bi) = \sqrt{3}</math> is merely a transformation which doesn't change affect the area. Thus, we can find the minimum area of the transformed equation <math>(q+bi)^2 = 3</math>. Using Roots of Unity, we know that the roots of the equation lie at <math>0, \frac{\pi}{4}, \frac{\pi}{2}, \frac{3\pi}{4}, \ldots, 2\pi</math> radians from the origin. | ||
+ | |||
+ | 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 (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> |
+ | Further simplifying, we obtain that the desired expression is <math>\frac{3 \sqrt{3 - 2\sqrt{2}}}{2}.</math> Note that <math>(\sqrt{2} - 1)^2 = 3 - 2\sqrt{2},</math> so our answer comes out to be <math>\frac{3(\sqrt{2}-1)}{2} = \boxed{\textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2}}.</math> | ||
+ | |||
+ | ~fidgetboss_4000 | ||
== See Also == | == See Also == |
Latest revision as of 14:12, 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.