Difference between revisions of "2021 AMC 12A Problems/Problem 19"

Latest revision as of 19:10, 5 November 2022

Problem

How many solutions does the equation $\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)$ have in the closed interval $[0,\pi]$ ?

$\textbf{(A) }0 \qquad \textbf{(B) }1 \qquad \textbf{(C) }2 \qquad \textbf{(D) }3\qquad \textbf{(E) }4$

Solution 1 (Inverse Trigonometric Functions)

The ranges of $\frac{\pi}2 \sin x$ and $\frac{\pi}2 \cos x$ are both $\left[-\frac{\pi}2, \frac{\pi}2 \right],$ which is included in the range of $\arcsin,$ so we can use it with no issues. $\begin{align*} \frac{\pi}2 \cos x &= \arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right) \\ \frac{\pi}2 \cos x &= \arcsin \left( \sin \left( \frac{\pi}2 - \frac{\pi}2 \sin x\right)\right) \\ \frac{\pi}2 \cos x &= \frac{\pi}2 - \frac{\pi}2 \sin x \\ \cos x &= 1 - \sin x \\ \cos x + \sin x &= 1. \end{align*}$ This only happens at $x = 0, \frac{\pi}2$ on the interval $[0,\pi],$ because one of $\sin$ and $\cos$ must be $1$ and the other $0.$ Therefore, the answer is $\boxed{\textbf{(C) }2}.$

~Tucker

Solution 2 (Cofunction Identity)

By the Cofunction Identity $\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),$ we rewrite the given equation: $\sin \left(\frac{\pi}2 \cos x\right) = \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right).$ Recall that if $\sin\theta=\sin\phi,$ then $\theta=\phi+2n\pi$ or $\theta=\pi-\phi+2n\pi$ for some integer $n.$ Therefore, we have two cases:

$\boldsymbol{\frac{\pi}2 \cos x = \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}$ for some integer $\boldsymbol{n}$
We rearrange and simplify: $\sin x + \cos x = 1 + 4n.$ By rough constraints, we know that $-2 < \sin x + \cos x < 2,$ from which $-2 < 1 - 4n < 2.$ The only possibility is $n=0,$ so $\begin{align*} \sin x + \cos x &= 1 && (*) \\ \sin^2 x + \cos^2 x + 2\sin x \cos x &= 1 \\ 2\sin x \cos x &= 0 \\ \sin(2x) &= 0 \\ 2x &= k\pi \\ x &= \frac{k\pi}{2} \end{align*}$ for some integer $k.$
We get $x=0,\frac{\pi}{2}$ for this case. Note that $x=\pi$ is an extraneous solution by squaring $(*).$

$\boldsymbol{\frac{\pi}2 \cos x = \pi - \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}$ for some integer $\boldsymbol{n}$
Similar to Case 1, we conclude that $n=0,$ so $\cos x - \sin x = 1.$ We get $x=0$ for this case.

Together, we obtain $\boxed{\textbf{(C) }2}$ solutions: $x=0,\frac{\pi}{2}.$

~MRENTHUSIASM

Solution 3 (Graphs and Analyses)

This problem is equivalent to counting the intersections of the graphs of $y=\sin\left(\frac{\pi}{2}\cos x\right)$ and $y=\cos\left(\frac{\pi}{2}\sin x\right)$ in the closed interval $[0,\pi].$ We construct a table of values, as shown below: $\begin{array}{c|ccc} & & & \\ [-2ex] & \boldsymbol{x=0} & \boldsymbol{x=\frac{\pi}{2}} & \boldsymbol{x=\pi} \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex] \boldsymbol{\sin\left(\frac{\pi}{2}\cos x\right)} & 1 & 0 & -1 \\ [1.5ex] \hline & & & \\ [-1ex] \boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex] \boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex] \boldsymbol{\cos\left(\frac{\pi}{2}\sin x\right)} & 1 & 0 & 1 \\ [1ex] \end{array}$ For $x\in[0,\pi],$ note that:

$\frac{\pi}{2}\cos x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],$ so $\sin\left(\frac{\pi}{2}\cos x\right)\in[-1,1].$

$\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],$ so $\cos\left(\frac{\pi}{2}\sin x\right)\in[0,1].$

For the graphs to intersect, we need $\sin\left(\frac{\pi}{2}\cos x\right)\in[0,1].$ This occurs when $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right].$

By the Cofunction Identity $\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),$ we rewrite the given equation: $\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).$ Since $\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right]$ and $\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],$ it follows that $x\in\left[0,\frac{\pi}{2}\right]$ and $\frac{\pi}{2}-\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right].$

We can apply the arcsine function to both sides, then rearrange and simplify: $\begin{align*} \frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\ \sin x + \cos x &= 1. \end{align*}$ From Case 1 in Solution 2, we conclude that $(0,1)$ and $\left(\frac{\pi}{2},0\right)$ are the only points of intersection, as shown below: $[asy] /* Made by MRENTHUSIASM */ size(600,200); real f(real x) { return sin(pi/2*cos(x)); } real g(real x) { return cos(pi/2*sin(x)); } draw(graph(f,0,pi),red,"$y=\sin\left(\frac{\pi}{2}\cos x\right)$"); draw(graph(g,0,pi),blue,"$y=\cos\left(\frac{\pi}{2}\sin x\right)$"); real xMin = 0; real xMax = 5/4*pi; real yMin = -2; real yMax = 2; //Draws the horizontal gridlines void horizontalLines() { for (real i = yMin+1; i < yMax; ++i) { draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4)); } } //Draws the vertical gridlines void verticalLines() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4)); } } //Draws the horizontal ticks void horizontalTicks() { for (real i = yMin+1; i < yMax; ++i) { draw((-1/8,i)--(1/8,i), black+linewidth(1)); } } //Draws the vertical ticks void verticalTicks() { for (real i = xMin+pi/2; i < xMax; i+=pi/2) { draw((i,-1/8)--(i,1/8), black+linewidth(1)); } } horizontalLines(); verticalLines(); horizontalTicks(); verticalTicks(); draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5)); draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5)); label("$x$",(xMax,0),(2,0)); label("$y$",(0,yMax),(0,2)); pair A[]; A[0] = (pi/2,0); A[1] = (pi,0); A[2] = (0,1); A[3] = (0,0); A[4] = (0,-1); label("$\frac{\pi}{2}$",A[0],(0,-2.5)); label("$\pi$",A[1],(0,-2.5)); label("$1$",A[2],(-2.5,0)); label("$0$",A[3],(-2.5,0)); label("$-1$",A[4],(-2.5,0)); dot((0,1),linewidth(5)); dot((pi/2,0),linewidth(5)); add(legend(),point(E),40E,UnFill); [/asy]$ Therefore, the answer is $\boxed{\textbf{(C) }2}.$

~MRENTHUSIASM (credit given to TheAMCHub)

Video Solution by OmegaLearn (Using Triangle Inequality & Trigonometry)

https://youtu.be/wJxN1YPuyCg

~ pi_is_3.14

Video Solution (Quick and Easy)

https://youtu.be/6AWb9cqFblU

~Education, the Study of Everything

@@ Line 5: / Line 5: @@
 ==Solution 1 (Inverse Trigonometric Functions)==
-<math>\sin \left( \frac{\pi}2 \cos x\right)=\cos \left( \frac{\pi}2 \sin x\right)</math>
+The ranges of <math>\frac{\pi}2 \sin x</math> and <math>\frac{\pi}2 \cos x</math> are both <math>\left[-\frac{\pi}2, \frac{\pi}2 \right],</math> which is included in the range of <math>\arcsin,</math> so we can use it with no issues.
+<cmath>\begin{align*}
-The ranges of <math>\frac{\pi}2 \sin x</math> and <math>\frac{\pi}2 \cos x</math> are both <math>\left[-\frac{\pi}2, \frac{\pi}2 \right]</math>, which is included in the range of <math>\arcsin</math>, so we can use it with no issues.
+\frac{\pi}2 \cos x &= \arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right) \\
+\frac{\pi}2 \cos x &= \arcsin \left( \sin \left( \frac{\pi}2 - \frac{\pi}2 \sin x\right)\right) \\
-<math>\frac{\pi}2 \cos x=\arcsin \left( \cos \left( \frac{\pi}2 \sin x\right)\right)</math>
+\frac{\pi}2 \cos x &= \frac{\pi}2 - \frac{\pi}2 \sin x \\
+\cos x &= 1 - \sin x \\
-<math>\frac{\pi}2 \cos x=\arcsin \left( \sin \left( \frac{\pi}2 - \frac{\pi}2 \sin x\right)\right)</math>
+\cos x + \sin x &= 1.
+\end{align*}</cmath>
-<math>\frac{\pi}2 \cos x=\frac{\pi}2 - \frac{\pi}2 \sin x</math>
+This only happens at <math>x = 0, \frac{\pi}2</math> on the interval <math>[0,\pi],</math> because one of <math>\sin</math> and <math>\cos</math> must be <math>1</math> and the other <math>0.</math> Therefore, the answer is <math>\boxed{\textbf{(C) }2}.</math>
-<math>\cos x = 1 - \sin x</math>
-<math>\cos x + \sin x = 1</math>
-This only happens at <math>x = 0, \frac{\pi}2</math> on the interval <math>[0,\pi]</math>, because one of <math>\sin</math> and <math>\cos</math> must be <math>1</math> and the other <math>0</math>. Therefore, the answer is <math>\boxed{C) 2}</math>
 ~Tucker
 ==Solution 2 (Cofunction Identity)==
-By the cofunction identity <math>\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right)</math> for all <math>\theta,</math> we simplify the given equation:
+By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we rewrite the given equation: <cmath>\sin \left(\frac{\pi}2 \cos x\right) = \sin \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right).</cmath>
-<cmath>\begin{align*}
+Recall that if <math>\sin\theta=\sin\phi,</math> then <math>\theta=\phi+2n\pi</math> or <math>\theta=\pi-\phi+2n\pi</math> for some integer <math>n.</math> Therefore, we have two cases:
-\sin \left( \frac{\pi}2 \cos x\right) &= \cos \left( \frac{\pi}2 \sin x\right) \\
+<ol style="margin-left: 1.5em;">
-\cos \left(\frac{\pi}2-\frac{\pi}2 \cos x\right) &= \cos \left( \frac{\pi}2 \sin x\right) \\
+  <li><b><math>\boldsymbol{\frac{\pi}2 \cos x = \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}</math> for some integer <math>\boldsymbol{n}</math></b> <p>
-\cos \left(\frac{\pi}2 \left(1 - \cos x \right)\right) &= \cos \left( \frac{\pi}2 \sin x\right) \\
+We rearrange and simplify: <cmath>\sin x + \cos x = 1 + 4n.</cmath>
-\frac{\pi}2 \left(1 - \cos x \right) &= \frac{\pi}2 \sin x + 2n\pi,
+By rough constraints, we know that <math>-2 < \sin x + \cos x < 2,</math> from which <math>-2 < 1 - 4n < 2.</math> The only possibility is <math>n=0,</math> so
-\end{align*}</cmath>
-for some integer <math>n.</math> We keep simplifying:
 <cmath>\begin{align*}
-- \cos x &= \sin x + 4n \\
+\sin x + \cos x &= 1 && (*) \\
-- 4n &= \sin x + \cos x.
-\end{align*}</cmath>
-By rough constraints, we know that <math>-2 < \sin x + \cos x < 2,</math> from which <math>-2 < 1 - 4n < 2.</math> The only possibility is <math>n=0.</math> From here, we get
-<cmath>\begin{align*}
-\sin x + \cos x &= 1 \ \ \ \ \ (*) \\
 \sin^2 x + \cos^2 x + 2\sin x \cos x &= 1 \\
 \sin x \cos x &= 0 \\
 \sin(2x) &= 0 \\
 x &= k\pi \\
-x &= \frac{k\pi}{2},
+x &= \frac{k\pi}{2}
 \end{align*}</cmath>
-for some integer <math>k.</math>
+for some integer <math>k.</math> <p>
+We get <math>x=0,\frac{\pi}{2}</math> for this case. Note that <math>x=\pi</math> is an extraneous solution by squaring <math>(*).</math></li><p>
-The <i><b>possible</b></i> solutions in <math>[0,\pi]</math> are <math>x=0,\frac{\pi}{2},\pi,</math> but only <math>x=0,\frac{\pi}{2}</math> check the original equation (Note that <math>x=\pi</math> is an extraneous solution formed by squaring <math>(*)</math> above.). Therefore, the answer is <math>\boxed{\textbf{(C) }2}.</math>
+  <li><b><math>\boldsymbol{\frac{\pi}2 \cos x = \pi - \left(\frac{\pi}2 - \frac{\pi}2 \sin x\right) + 2n\pi}</math> for some integer <math>\boldsymbol{n}</math></b> <p>
+Similar to Case 1, we conclude that <math>n=0,</math> so <cmath>\cos x - \sin x = 1.</cmath>
+We get <math>x=0</math> for this case.</li><p>
+</ol>
+Together, we obtain <math>\boxed{\textbf{(C) }2}</math> solutions: <math>x=0,\frac{\pi}{2}.</math>
 ~MRENTHUSIASM
-==Solution 3 (Graphs and Analysis)==
+==Solution 3 (Graphs and Analyses)==
-Let <math>f(x)=\sin\left(\frac{\pi}{2}\cos x\right)</math> and <math>g(x)=\cos \left( \frac{\pi}2 \sin x\right).</math> This problem is equivalent to counting the intersections of the graphs of <math>f(x)</math> and <math>g(x)</math> in the closed interval <math>[0,\pi].</math> We make a table of values, as shown below:
+This problem is equivalent to counting the intersections of the graphs of <math>y=\sin\left(\frac{\pi}{2}\cos x\right)</math> and <math>y=\cos\left(\frac{\pi}{2}\sin x\right)</math> in the closed interval <math>[0,\pi].</math> We construct a table of values, as shown below:
 <cmath>\begin{array}{c|ccc}
 & & & \\ [-2ex]
@@ Line 60: / Line 51: @@
 \boldsymbol{\cos x} & 1 & 0 & -1 \\ [1.5ex]
 \boldsymbol{\frac{\pi}{2}\cos x} & \frac{\pi}{2} & 0 & -\frac{\pi}{2} \\ [1.5ex]
-\boldsymbol{f(x)} & 1 & 0 & -1 \\ [1.5ex]
+\boldsymbol{\sin\left(\frac{\pi}{2}\cos x\right)} & 1 & 0 & -1 \\ [1.5ex]
 \hline
 & & & \\ [-1ex]
 \boldsymbol{\sin x} & 0 & 1 & 0 \\ [1.5ex]
 \boldsymbol{\frac{\pi}{2}\sin x} & 0 & \frac{\pi}{2} & 0 \\ [1.5ex]
-\boldsymbol{g(x)} & 1 & 0 & 1 \\ [1ex]
+\boldsymbol{\cos\left(\frac{\pi}{2}\sin x\right)} & 1 & 0 & 1 \\ [1ex]
 \end{array}</cmath>
-For the graphs of <math>f(x)</math> and <math>g(x),</math> we will analyze their increasing/decreasing behaviors in <math>[0,\pi]:</math>
+For <math>x\in[0,\pi],</math> note that:
-*The graph of <math>f(x)</math> in <math>[0,\pi]</math> (from left to right) has the same behavior as the graph of <math>y=\sin x</math> in <math>\left[-\frac{\pi}{2},\frac{\pi}{2}\right]</math> (from right to left): the output is from <math>1</math> to <math>-1</math> (from left to right), inclusive, and strictly decreasing.
+* <math>\frac{\pi}{2}\cos x\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right],</math> so <math>\sin\left(\frac{\pi}{2}\cos x\right)\in[-1,1].</math>
-*The graph of <math>g(x)</math> in <math>[0,\pi]</math> (from left to right) has two parts:
+* <math>\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],</math> so <math>\cos\left(\frac{\pi}{2}\sin x\right)\in[0,1].</math>
-<ol style="margin-left: 3em;">
-  <li>The graph of <math>g(x)</math> in <math>\left[0,\frac{\pi}{2}\right]</math> has the same behavior as the graph of <math>y=\cos x</math> in <math>\left[0,\frac{\pi}{2}\right]</math> (from left to right): the output is from <math>1</math> to <math>0</math> (from left to right), inclusive, and strictly decreasing.</li><p>
-  <li>The graph of <math>g(x)</math> in  <math>\left[\frac{\pi}{2},\pi\right]</math> has the same behavior as the graph of <math>y=\cos x</math> in <math>\left[0,\frac{\pi}{2}\right]</math> (from right to left): the output is from <math>0</math> to <math>1</math> (from left to right), inclusive, and strictly increasing.</li><p>
-</ol>
-If <math>x\in\left(\frac{\pi}{2},\pi\right],</math> then <math>f(x)<0</math> and <math>g(x)>0.</math> So, their graphs do not intersect.
+For the graphs to intersect, we need <math>\sin\left(\frac{\pi}{2}\cos x\right)\in[0,1].</math> This occurs when <math>\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right].</math>
-If <math>x\in\left[0,\frac{\pi}{2}\right],</math> then <math>0\leq f(x),g(x)\leq1.</math> Clearly, the graphs intersect at <math>x=0</math> and <math>x=\frac{\pi}{2}</math> (at points <math>(0,1)</math> and <math>\left(\frac{\pi}{2},0\right),</math> respectively), but we will prove or disprove that they are the <i><b>only</b></i> points of intersection:
+By the Cofunction Identity <math>\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),</math> we rewrite the given equation:
+<cmath>\sin\left(\frac{\pi}{2}\cos x\right) = \sin\left(\frac{\pi}{2}-\frac{\pi}{2}\sin x\right).</cmath>
+Since <math>\frac{\pi}{2}\cos x\in\left[0,\frac{\pi}{2}\right]</math> and <math>\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right],</math> it follows that <math>x\in\left[0,\frac{\pi}{2}\right]</math> and <math>\frac{\pi}{2}-\frac{\pi}{2}\sin x\in\left[0,\frac{\pi}{2}\right].</math>
-Let <math>A=\frac{\pi}{2}\cos x</math> and <math>B=\frac{\pi}{2}\sin x.</math> It follows that <math>A,B\in\left[0,\frac{\pi}{2}\right].</math> Since <math>\sin A = \cos B,</math> we know that <math>A+B=\frac{\pi}{2}</math> by the cofunction identity:
+We can apply the arcsine function to both sides, then rearrange and simplify:
 <cmath>\begin{align*}
-\frac{\pi}{2}\cos x + \frac{\pi}{2}\sin x &= \frac{\pi}{2} \\
+\frac{\pi}{2}\cos x &= \frac{\pi}{2}-\frac{\pi}{2}\sin x \\
-\cos x + \sin x &=1.
+\sin x + \cos x &= 1.
 \end{align*}</cmath>
+From Case 1 in Solution 2, we conclude that <math>(0,1)</math> and <math>\left(\frac{\pi}{2},0\right)</math> are the only points of intersection, as shown below:
+<asy>
+/* Made by MRENTHUSIASM */
+size(600,200);
+real f(real x) { return sin(pi/2*cos(x)); }
+real g(real x) { return cos(pi/2*sin(x)); }
+draw(graph(f,0,pi),red,"$y=\sin\left(\frac{\pi}{2}\cos x\right)$");
+draw(graph(g,0,pi),blue,"$y=\cos\left(\frac{\pi}{2}\sin x\right)$");
+real xMin = 0;
+real xMax = 5/4*pi;
+real yMin = -2;
+real yMax = 2;
+//Draws the horizontal gridlines
+void horizontalLines()
+{
+  for (real i = yMin+1; i < yMax; ++i)
+  {
+    draw((xMin,i)--(xMax,i), mediumgray+linewidth(0.4));
+  }
+}
+//Draws the vertical gridlines
+void verticalLines()
+{
+  for (real i = xMin+pi/2; i < xMax; i+=pi/2)
+  {
+    draw((i,yMin)--(i,yMax), mediumgray+linewidth(0.4));
+  }
+}
+//Draws the horizontal ticks
+void horizontalTicks()
+{
+  for (real i = yMin+1; i < yMax; ++i)
+  {
+    draw((-1/8,i)--(1/8,i), black+linewidth(1));
+  }
+}
+//Draws the vertical ticks
+void verticalTicks()
+{
+  for (real i = xMin+pi/2; i < xMax; i+=pi/2)
+  {
+    draw((i,-1/8)--(i,1/8), black+linewidth(1));
+  }
+}
+horizontalLines();
+verticalLines();
+horizontalTicks();
+verticalTicks();
+draw((xMin,0)--(xMax,0),black+linewidth(1.5),EndArrow(5));
+draw((0,yMin)--(0,yMax),black+linewidth(1.5),EndArrow(5));
+label("$x$",(xMax,0),(2,0));
+label("$y$",(0,yMax),(0,2));
+pair A[];
+A[0] = (pi/2,0);
+A[1] = (pi,0);
+A[2] = (0,1);
+A[3] = (0,0);
+A[4] = (0,-1);
+label("$\frac{\pi}{2}$",A[0],(0,-2.5));
+label("$\pi$",A[1],(0,-2.5));
+label("$1$",A[2],(-2.5,0));
+label("$0$",A[3],(-2.5,0));
+label("$-1$",A[4],(-2.5,0));
+dot((0,1),linewidth(5));
+dot((pi/2,0),linewidth(5));
+add(legend(),point(E),40E,UnFill);
+</asy>
+Therefore, the answer is <math>\boxed{\textbf{(C) }2}.</math>
-We apply Solution 2's argument (starts from its last block of equations) to deduce that <math>(0,1)</math> and <math>\left(\frac{\pi}{2},0\right)</math> are the only points of intersection. So, the answer is <math>\boxed{\textbf{(C) }2}.</math>
 ~MRENTHUSIASM (credit given to TheAMCHub)
@@ Line 95: / Line 162: @@
 ~ pi_is_3.14
+==Video Solution (Quick and Easy)==
+https://youtu.be/6AWb9cqFblU
+~Education, the Study of Everything
 ==See also==
 {{AMC12 box|year=2021|ab=A|num-b=18|num-a=20}}
 {{MAA Notice}}

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