Creative Thinking (Fall 2021 AMC 12A #8, 9, 13, 14)

Notable Early Problems from Fall 2021 AMC 12A

Fall 2021 AMC 12A #8

Problem: Let $M$ be the least common multiple of all the integers $10$ through $30,$ inclusive. Let $N$ be the least common multiple of $M,32,33,34,35,36,37,38,39,$ and $40$. What is the value of $\frac{N}{M}$?

Solution: We can see that $N$ contains all the factors of the $M$, which in turn contains all the factors of the numbers from 10 to 30. So, we must consider the additional terms $32,33,34,35,36,37,38,39,$ and $40$ to determine what additional factors $N$ has other than $M$.

Beginning with $32$, we see that $32 = 2^5$, but $M$ only contains $16 = 2^4$, so $N$ has one extra factor of $2$ than $M$. $33$ is already included as $3*11$ is a factor of two relatively prime numbers, $18*11$. $34=2*17$, which is included in $10*17$. $35=5*7$ is included in $10*21$ and $36=4*9$ is included in $16*27$. $37$ is prime, so $N$ has an additional factor of $37$ as well. Finally, $38=2*19$ is included in $10*19$ and $39=3*13$ is included in $12*13$.

So, we see that $N$ contains the extra factors of $2$ and $37$, so $\frac{N}{M} = 2*37=74$.

Fall 2021 AMC 12A #9

Problem: A right rectangular prism whose surface area and volume are numerically equal has edge lengths $\log_{2}x, \log_{3}x,$ and $\log_{4}x$. What is $x?$

Solution: This is a typical use of logarithms and their properties in early AMC 12 problems. Equating surface area and volume, we get $$2[(\log_{2}x)(\log_{3}x)+(\log_{2}x)(\log_{4}x)+(\log_{3}x)(\log_{4}x)] = (\log_{2}x)(\log_{3}x)(\log_{4}x)$$ Dividing both sides by $(\log_{2}x)(\log_{3}x)(\log_{4}x)$, we obtain $$\frac{1}{\log_{4}x}+\frac{1}{\log_{3}x}+\frac{1}{\log_{2}x} = \frac{1}{2}$$ We now use the property that $\frac{1}{\log_{y}x} = \log_{x}y$. To see why, we can let $\log_{y}x=n$, which means $y^n=x$. So, $y=x^{1/n}$ and $\log_{x}y=\frac{1}{n}$, which gives us $\frac{1}{\log_{y}x} = \frac{1}{n} = \log_{x}y$. (Or we could also use the change of base formula with $\frac{1}{\log_{y}x} = \frac{\log_{y}y}{\log_{y}x} = \log_{x}{y}$.)

So, we have $\log_{x}4+\log_{x}3+\log_{x}2 = \log_{x}(4*3*2) = \frac{1}{2}$ by logarithmic addition. This gives us $x^{\frac{1}{2}} = 24$, which means $x = 24^2 = 576$.

Fall 2021 AMC 12A #13

Problem: The angle bisector of the acute angle formed at the origin by the graphs of the lines $y = x$ and $y=3x$ has equation $y=kx$. What is $k$?

Solution: We can draw the $3 \times 1$ rectangle as shown below with the points $(1,1)$ and $(1,3)$ on lines $y = x$ and $y=3x$, respectively. We place the point $(1,k)$ on $y=kx$ between the two points, so that we can apply the angle bisector theorem to get $$\frac{3-k}{\sqrt{10}} = \frac{k-1}{\sqrt{2}}$$ This can be simplified to $3-k = \sqrt{5}(k-1)$ and solving for $k$, we get $$k = \frac{3+\sqrt{5}}{1+\sqrt{5}} = \frac{(3+\sqrt{5})(\sqrt{5}-1)}{(1+\sqrt{5})(\sqrt{5}-1)} = \frac{5-3+2\sqrt{5}}{4} = \frac{1+\sqrt{5}}{2}$$

Fall 2021 AMC 12A #14

Problem: In the figure, equilateral hexagon $ABCDEF$ has three nonadjacent acute interior angles that each measure $30^\circ$. The enclosed area of the hexagon is $6\sqrt{3}$. What is the perimeter of the hexagon?

Solution: Let the side lengths of the equilateral hexagon be $y$, and the sides of the inner equilateral triangle be $x$ as shown in the diagram below. We can express the area as $$3[\frac{1}{2}y^2\sin(30^\circ)] + \frac{\sqrt{3}}{4}x^2 = 6\sqrt{3}$$ since the area of a triangle is given by $\frac{1}{2}ab\sin{C}$ and the area of an equilateral triangle is $\frac{\sqrt{3}}{4}s^2$. By the law of cosines, we have that $$x^2 = y^2 + y^2 - 2y^2\cos(30^\circ) = y^2(2-\sqrt{3})$$

Substituting, $x^2$, we get $3[\frac{1}{2}y^2\sin(30^\circ)] + \frac{\sqrt{3}}{4}(y^2)(2-\sqrt{3}) = 6\sqrt{3}$, which simplifies to $$y^2(\frac{3}{4} + \frac{\sqrt{3}}{2} - \frac{3}{4}) = y^2 (\frac{\sqrt{3}}{2}) = 6\sqrt{3}$$

So, we get that $y = 2\sqrt{3}$, which means the perimeter equals $6y = 12\sqrt{3}$.

Extra - Diagram for #14

How did I create the diagram? Well, besides finding $y$, I used polar coordinates to graph the points of the equilateral hexagon, with its center at the origin. Then, I needed to find the distance the inner vertices and outer vertices were from the origin for the $r$ values, and graph them $60^\circ$ apart. After some calculations (use $a+b\sqrt{3} = \sqrt{2-\sqrt{3}}$ and solve to help find $x$), we find that points $B, F, D$ are all a distance $r_1 = \sqrt{6}-\sqrt{2}$ from the center and the points $A, C, E$ are all a distance $r_2 = \sqrt{6}+\sqrt{2}$ from the center. And of course, polar coordinates $(r, \theta)$ can be expressed as $(x,y) = (r\cos{\theta}, r\sin{\theta})$ in rectangular coordinates.