AMC10 2021 B

We see that $1, 2, 3,$ and $4$ each appear in the ones, tens, hundreds, and thousands digit exactly once. Since $1+2+3+4=10$, we find that the sum is equal to \[10\cdot(1+10+100+1000)=\boxed{\textbf{(E)} \: 11{,}110}.\] Note that it is equally valid to manually add all four numbers together to get the answer.

The line of symmetry divides the shaded figure into two congruent triangles, each with base $3$ and height $2.$ Therefore, the area of the shaded figure is \[2\cdot\left(\frac12\cdot3\cdot2\right)=2\cdot3=\boxed{\textbf{(B)} \: 6}.\]

We write the given expression as a single fraction: \[\frac{2021}{2020} - \frac{2020}{2021} = \frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}\] by cross multiplication. Then by factoring the numerator, we get \[\frac{2021\cdot2021-2020\cdot2020}{2020\cdot2021}=\frac{(2021-2020)(2021+2020)}{2020\cdot2021}.\] The question is asking for the numerator, so our answer is $2021+2020=4041,$ giving $\boxed{\textbf{(E) }4041}$.

At noon on a certain day, let $M$ and $L$ be the temperatures (in degrees) in Minneapolis and St. Louis, respectively. It follows that $M=L+N.$ At $4{:}00,$ we get \begin{align*} |(M-5)-(L+3)| &= 2 \\ |M-L-8| &= 2 \\ |N-8| &= 2. \end{align*} We have two cases: 1. If $N-8=2,$ then $N=10.$ 2. If $N-8=-2,$ then $N=6.$ Together, the product of all possible values of $N$ is $10\cdot6=\boxed{\textbf{(C)} \: 60}.$

We have \[n=8^{2022}= \left(8^\frac{2}{3}\right)^{3033}=4^{3033}.\] Therefore, \[\frac{n}4=\boxed{\textbf{(E)} \: 4^{3032}}.\]

Let this positive integer be written as $p_1^{e_1}\cdot p_2^{e_2}$. The number of factors of this number is therefore $(e_1+1) \cdot (e_2+1)$, and this must equal 2021. The prime factorization of 2021 is $43 \cdot 47$, so $e_1+1 = 43 \implies e_1=42$ and $e_2+1=47\implies e_2=46$. To minimize this integer, we set $p_1 = 3$ and $p_2 = 2$. Then this integer is $3^{42} \cdot 2^{46} = 2^4 \cdot 2^{42} \cdot 3^{42} = 16 \cdot 6^{42}$. Now $m=16$ and $k=42$ so $m+k = 16 + 42 = \boxed{\textbf{(B) }58}$

The special fractions are \[\frac{1}{14},\frac{2}{13},\frac{3}{12},\frac{4}{11},\frac{5}{10},\frac{6}{9},\frac{7}{8},\frac{8}{7},\frac{9}{6},\frac{10}{5},\frac{11}{4},\frac{12}{3},\frac{13}{2},\frac{14}{1}.\] We rewrite them in the simplest form: \[\frac{1}{14},\frac{2}{13},\frac{1}{4},\frac{4}{11},\frac{1}{2},\frac{2}{3},\frac{7}{8},1\frac{1}{7},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] Note that two unlike fractions in the simplest form cannot sum to an integer. So, we only consider the fractions whose denominators appear more than once: \[\frac{1}{4},\frac{1}{2},1\frac{1}{2},2,2\frac{3}{4},4,6\frac{1}{2},14.\] For the set $\{2,4,14\},$ two elements (not necessarily different) can sum to $4,6,8,16,18,28.$ For the set $\left\{\frac{1}{2},1\frac{1}{2},6\frac{1}{2}\right\},$ two elements (not necessarily different) can sum to $1,2,3,7,8,13.$ For the set $\left\{\frac{1}{4},2\frac{3}{4}\right\},$ two elements (not necessarily different) can sum to $3.$ Together, there are $\boxed{\textbf{(C)}\ 11}$ distinct integers that can be written as the sum of two, not necessarily different, special fractions: \[1,2,3,4,6,7,8,13,16,18,28.\]

We have \begin{align*} 16383 & = 2^{14} - 1 \\ & = \left( 2^7 + 1 \right) \left( 2^7 - 1 \right) \\ & = 129 \cdot 127 \\ \end{align*} Since $129$ is composite, $127$ is the largest prime which can divide $16383$. The sum of $127$'s digits is $1+2+7=\boxed{\textbf{(C) }10}$. Note that you can quickly tell that $2^7 -1$ is prime because it is a Mersenne prime.

Let $k$ be the number of knights: then the number of red knights is $\frac{2}{7}k$ and the number of blue knights is $\frac{5}{7}k$. Let $b$ be the fraction of blue knights that are magical - then $2b$ is the fraction of red knights that are magical. Thus we can write the equation $b \cdot \frac{5}{7}k + 2b \cdot \frac{2}{7}k = \frac{k}{6}\implies \frac{5}{7}b + \frac{4}{7}b = \frac{1}{6}$ $\implies \frac{9}{7}b = \frac{1}{6} \implies b=\frac{7}{54}$ We want to find the fraction of red knights that are magical, which is $2b = \frac{7}{27} = \boxed{\textbf{(C) }\frac{7}{27}}$

Let Alice have the number A, Bob B. When Alice says that she can't tell who has the larger number, it means that $A$ cannot equal $1$. Therefore, it makes sense that Bob has $2$ because he now knows that Alice has the larger number. $2$ is also prime. The last statement means that $200+A$ is a perfect square. The three squares in the range $200-300$ are $225$, $256$, and $289$. So, $A$ could equal $25$, $56$, or $89$, so $A+B$ is $27$, $58$, or $91$, of only $\boxed{\textbf{(A) }27}$ is below $40$, therefore it is the correct answer since the maximum number one can have is 40.

This is the graph of the original Hexagon. After reflecting each minor arc over the sides of the hexagon it will look like this; This bounded region is the same as the area of the hexagon minus the area of each of the reflect arcs. From the first diagram, the area of each arc is the area of the $60^{\circ}$ sector minus the area of the equilateral triangle, so each arc has an area of $\frac{\pi r^2}{6} - \frac{s^2\sqrt{3}}{4} \implies \frac{\pi}{6} - \frac{\sqrt{3}}{4}$. There are 6 total arcs, so the total area of the arcs is $6\cdot (\frac{\pi}{6} - \frac{\sqrt{3}}{4}) = \pi - \frac{3\sqrt{3}}{2}$. The area of the hexagon is $6\cdot \frac{\sqrt{3}}{4} = \frac{3\sqrt{3}}{2}$, so the area of the bounded region is: $\frac{3\sqrt{3}}{2} - (\pi - \frac{3\sqrt{3}}{2}) = 3\sqrt{3} - \pi = \boxed{B}$

It is obvious $x$, $y$, and $z$ are symmetrical. We are going to solve the problem by Completing the Square. $x ^ 2 + y ^ 2 + z ^ 2 - xy - yz - zx = 1$ $2x ^ 2 + 2y ^ 2 + 2z ^ 2 - 2xy - 2yz - 2zx = 2$ $(x-y)^2 + (y-z)^2 + (z-x)^2 = 2$ Because $x, y, z$ are integers, $(x-y)^2$, $(y-z)^2$, and $(z-x)^2$ can only equal $0, 1, 1$. So one variable must equal another, and the third variable is $1$ different from those $2$ equal variables. So the answer is $\boxed{D}$.

Let's split the triangle down the middle and label it: We see that $\bigtriangleup ADG \sim \bigtriangleup BEG \sim \bigtriangleup CFG$ by AA similarity. $BE = \frac{3}{2}$ because $AG$ cuts the side length of the square in half; similarly, $CF = 1$. Let $CG = h$: then by side ratios, \[\frac{h+2}{h} = \frac{\frac{3}{2}}{1} \implies 2(h+2) = 3h \implies h = 4\]. Now the height of the triangle is $AG = 4+2+3 = 9$. By side ratios, \[\frac{9}{4} = \frac{AD}{1} \implies AD = \frac{9}{4}\]. The area of the triangle is $AG\cdot AD = 9 \cdot \frac{9}{4} = \frac{81}{4} = \boxed{\textbf{(B) }20 \frac{1}{4}}$

We will use complementary counting to find the probability that the product is not divisible by $4$. Then, we can find the probability that we want by subtracting this from 1. We split this into $2$ cases. Case 1: The product is not divisible by $2$. We need every number to be odd, and since the chance we roll an odd number is $\frac12,$ our probability is $\left(\frac12\right)^6=\frac1{64}.$ Case 2: The product is divisible by $2$, but not by $4$. We need $5$ numbers to be odd, and one to be divisible by $2$, but not by $4$. There is a $\frac12$ chance that an odd number is rolled, a $\frac13$ chance that we roll a number satisfying the second condition (only $2$ and $6$ work), and $6$ ways to choose the order in which the even number appears. Our probability is $\left(\frac12\right)^5\left(\frac13\right)\cdot6=\frac1{16}.$ Therefore, the probability the product is not divisible by $4$ is $\frac1{64}+\frac1{16}=\frac{5}{64}$. Our answer is $1-\frac{5}{64}=\boxed{\textbf{(C)}\ \frac{59}{64}}$.

Note that $\triangle APB \cong \triangle BQC$ by ASA. ($\angle PAB = \angle QBC = 90^\circ, AB=CB,$ and $\angle PBA = \angle QCB.$) Then, it follows that $\overline{PB} \cong \overline{QC}.$ Thus, $QC = PB = PR + RB = 7 + 6 = 13.$ Define $x$ to be the length of side $CR,$ then $RQ = 13-x.$ Because $\overline{BR}$ is the altitude of the triangle, we can use the property that $QR \cdot RC = BR^2.$ Substituting the given lengths, we have \[(13-x) \cdot x = 36.\] Solving, gives $RQ = 4$ and $RC = 9.$ We eliminate the possibility of $x=4$ because $RC > QR.$ Thus, the side length of the square, by Pythagorean Theorem, is \[\sqrt{9^2 +6^2} = \sqrt{81+36} = \sqrt{117}.\] Thus, the area of the square is $(\sqrt{117})^2 = 117,$ so the answer is $\boxed{\textbf{(D) }117}.$ Please note that there is another way to prove that $CR = 4$ is impossible. If $CR = 4,$ then the side length would be $\sqrt{4^2 + 6^2} = \sqrt{52},$ and the area would be $52,$ but that isn't in the answer choices. Thus, $CR$ must be $9.$ Extra Note: Another way to prove $4$ is impossible. The side length of the square, $S$, is equal to $\sqrt{4^2 + 6^2} = \sqrt{52}$. Because $x = 4$, $RQ = 9$. Because $QB = \sqrt{RB^2 + RQ^2} = \sqrt{6^2 + 9^2} = \sqrt{117}$ and $QB < S$ but $\sqrt{117} > \sqrt{52}$, we have proof by contradiction. And so $x = 9$.

After the first swap, we do casework on the next swap. Case 1: Silva swaps the two balls that were just swapped There is only one way for Silva to do this, and it leaves 5 balls occupying their original position. Case 2: Silva swaps one ball that has just been swapped with one that hasn't been swapped There are two ways for Silva to do this, and it leaves 2 balls occupying their original positions. Case 3: Silva swaps two balls that have not been swapped There are two ways for Silva to do this, and it leaves 1 ball occupying their original position. Our answer is the average of all 5 possible swaps, so we get \[\frac{5+2\cdot2+2\cdot1}{5}=\frac{11}5=\boxed{(\textbf{D}) \: 2.2}.\]

Denote $O$ as the origin. Even though the problem is phrased as a coordinate bash, that looks disgusting. Instead, let's try to phrase this problem in terms of Euclidean geometry, using the observation that $\angle POP'' = 90^{\circ}$, and that both $\ell$ and $m$ must pass through $O$ in order to preserve the distance from $P$ to the origin. ($A$ and $B$ are just defined as points on lines $\ell$ and $m$.) Because of how reflections work, we have that $\angle AOP' = \angle POA$ and $\angle P'OB = \angle BOP''$; adding these two equations together and using angle addition, we have that $\angle AOB = \angle POA + \angle BOP''$. Since the sum of both sides combined must be $90^{\circ}$ by angle addition, \[\angle AOB = 45^{\circ}.\] This is helpful! We can now return to using coordinates, with this piece of information in mind: The $45^{\circ}$ angle is a little bit unwieldy in the coordinate plane, so we should try to make a $45-45-90$ triangle. Let $A$ be a point on $\ell$; to make $A$ fit nicely in the diagram, let it be $(0.5,2.5)$. Now, let's draw a perpendicular to $\ell$ through point $A$, intersecting $m$ at point $B$. $OAB$ is a $45-45-90$ triangle, so $B$ is a $90$ degree counterclockwise rotation from $O$ about $A$. Therefore, the coordinates of $B$ are \[(0.5+2.5,2.5-0.5) = (3,2).\] So, $(3,2)$ is a point on line $m$, which we already know passes through the origin; therefore, $m$'s equation is $y=\frac{2x}{3} \implies \boxed{\textbf{(D) } 2x-3y = 0}.$ (We never actually had to use the information of the exact coordinates of $P$; as long as $\angle POP'' = 90^{\circ}$, when we move $P$ around, this will not affect $m$'s equation.)

The $24$-sided polygon is made out of $24$ shapes like $\triangle ABC$. Then $\angle BAC=360^\circ/24=15^\circ$, and $\angle EAC = 45^\circ$, so $\angle{EAB} = 30^{\circ}$. Then $EB=AE\tan 30^\circ = \sqrt{3}$; therefore $BC=EC-EB=3-\sqrt{3}$. Thus \[[ABC] = \frac{BC}{EC}\cdot [ACE] = \frac{3-\sqrt{3}}{3}\cdot \frac 92\]and the required area is $24\cdot[ABC] =108-36\sqrt{3}$. Finally $108+36+3=\boxed{(\textbf{E})\ 147}$.

We can rewrite $N$ as $\frac{7}{9}\cdot 9999\ldots999 = \frac{7}{9}\cdot(10^{313}-1)$. When approximating values, as we will shortly do, the minus one will become negligible so we can ignore it. When we take the power of ten out of the square root, we’ll be multiplying by another power of ten, so the leading digit will not change. Thus the leading digit of $f(r)$ will be equal to the leading digit of $\sqrt[r]{\frac{7}{9} \cdot 10^{313(\mod r)}}$. Then $f(2)$ is the first digit of $\sqrt{\frac{7}{9}\cdot(10)} = \sqrt{\frac{70}{9}} = \sqrt{7. \ldots} \approx 2$ $f(3) = \sqrt[3]{\frac{7}{9} \cdot 10} = \sqrt[3]{\frac{70}{9}} = \sqrt[3]{7.\ldots} \approx 1$. $f(4) = \sqrt[4]{\frac{7}{9} \cdot 10} = \sqrt[4]{\frac{70}{9}} = \sqrt[4]{7.\ldots} \approx 1$. $f(5) = \sqrt[5]{\frac{7}{9} \cdot 1000} = \sqrt[5]{\frac{7000}{9}} = \sqrt[5]{777.\ldots} \approx 3$. $f(6) = \sqrt[6]{\frac{7}{9} \cdot 10} = \sqrt[6]{\frac{70}{9}} = \sqrt[6]{7.\ldots} \approx 1$. The final answer is therefore $2+1+1+3+1 = \boxed{\textbf{(A) }8}$. Note: in all of the divisions, we omitted the decimal places, because they are irrelevant to finding the leading digit.

The conditional probability formula states that $P(A|B) = \frac{P(A\cap B)}{P(B)}$, where $A|B$ means A given B and $A\cap B$ means A and B. Therefore the probability that Hugo rolls a five given he won is $\tfrac{P(A \cap B)}{P(B)}$, where A is the probability that he rolls a five and B is the probability that he wins. In written form, \[\text{P(Hugo rolled a 5 given he won)}=\frac{\text{P(Hugo rolls a 5 and wins)}}{\text{P(Hugo wins)}}.\] The probability that Hugo wins is $\frac{1}{4}$ by symmetry since there are four people playing and there is no bias for any one player. The probability that he gets a 5 and wins is more difficult; we will have to consider cases on how many players tie with Hugo. We consider the cases in which Hugo wins first, find the probabilities of those, and then account for the probability that Hugo rolled a 5, at the end. $\textbf{Case 1:}$ No Players Tie With Hugo In this case, all other players must have numbers from 1 through four, and Hugo had a probability of 1 to win after the first round. There is a $\left(\frac{4}{6}\right)^{3} \cdot 1 = \frac{8}{27}$ chance of this happening. $\textbf{Case 2:}$ One Player Ties With Hugo In this case, there are $3 \choose 1$ $= 3$ ways to choose which other player ties with Hugo, and the probability that this happens is $\tfrac{1}{6} \cdot \left(\tfrac{4}{6}\right)^2$. The probability that Hugo wins the game is then $\tfrac{1}{2}$ because there are now two players rolling die. Therefore the total probability in this case is $3 \cdot \frac{1}{2} \cdot \frac{1}{6} \cdot \left(\frac{4}{6}\right)^2= \frac{1}{9}$. $\textbf{Case 3:}$ Two Players Tie With Hugo In this case, there are $3 \choose 2$ $= 3$ ways to choose which other players tie with Hugo, and the probability that this happens is $\left(\tfrac{1}{6}\right)^2 \cdot \tfrac{4}{6}$. The probability that Hugo wins the game is then $\tfrac{1}{3}$ because there are now three players rolling the die. Therefore the total probability in this case is $3 \cdot \frac{1}{3} \cdot \left(\frac{1}{6}\right)^2 \cdot \frac{4}{6} = \frac{1}{54}$. $\textbf{Case 4:}$ All Three Players Tie With Hugo In this case, the probability that all three players tie with Hugo is $\left(\tfrac{1}{6}\right)^3$. The probability that Hugo wins the game is $\tfrac{1}{4}$, so the total probability is $\frac{1}{4} \cdot \left(\frac{1}{6}\right)^3 = \frac{1}{864}$. Hugo has a $\frac{1}{6}$ probability of rolling a five himself, so the total probability that Hugo rolls a 5 and wins is \[\frac{1}{6}\left(\frac{8}{27} + \frac{1}{9} + \frac{1}{54} + \frac{1}{864}\right) = \frac{1}{6}\left(\frac{369}{864}\right) = \frac{1}{6}\left(\frac{41}{96}\right).\] Finally, the total probability is this probability divided by $\frac{1}{4}$ which is this probability times four; the final answer is \[4 \cdot \frac{1}{6}\left(\frac{41}{96}\right) = \frac{2}{3} \cdot \frac{41}{96} = \frac{41}{48 \cdot 3} = \frac{41}{144} = \boxed{C}.\]

Imagine we have $2$ regular polygons with $m$ and $n$ sides and $m>n$ inscribed in a circle without sharing a vertex. We see that each side of the polygon with $n$ sides (the polygon with fewer sides) will be intersected twice. (We can see this because to have a vertex of the $m$-gon on an arc subtended by a side of the $n$-gon, there will be one intersection to “enter” the arc and one to “exit” the arc. ~KingRavi) This means that we will end up with $2$ times the number of sides in the polygon with fewer sides. If we have polygons with $5,$ $6,$ $7,$ and $8$ sides, we need to consider each possible pair of polygons and count their intersections. Throughout $6$ of these pairs, the $5$-sided polygon has the least number of sides $3$ times, the $6$-sided polygon has the least number of sides $2$ times, and the $7$-sided polygon has the least number of sides $1$ time. Therefore the number of intersections is $2\cdot(3\cdot5+2\cdot6+1\cdot7)=\boxed{(\textbf{E}) \:68}$.

To get from $S_n$ to $S_{n+1}$, we add $1(n+1)+2(n+1)+\cdots +n(n+1)=(1+2+\cdots +n)(n+1)=\frac{n(n+1)^2}{2}$. Now, we can look at the different values of $n$ mod $3$. For $n\equiv 0\pmod{3}$ and $n\equiv 2\pmod{3}$, then we have $\frac{n(n+1)^2}{2}\equiv 0\pmod{3}$. However, for $n\equiv 1\pmod{3}$, we have \[\frac{1\cdot {2}^2}{2}\equiv 2\pmod{3}.\] Clearly, $S_2\equiv 2\pmod{3}.$ Using the above result, we have $S_5\equiv 1\pmod{3}$, and $S_8$, $S_9$, and $S_{10}$ are all divisible by $3$. After $3\cdot 3=9$, we have $S_{17}$, $S_{18}$, and $S_{19}$ all divisible by $3$, as well as $S_{26}, S_{27}, S_{28}$, and $S_{35}$. Thus, our answer is $8+9+10+17+18+19+26+27+28+35=27+54+81+35=162+35=\boxed{\mathrm{(B)}\ 197}$ .

Instead of finding the probability of a same-colored triangle appearing, let us find the probability that one does not appear. After drawing the regular pentagon out, note the topmost vertex; it has 4 sides/diagonals emanating outward from it. We do casework on the color distribution of these sides/diagonals. $\textbf{Case 1}$: all 4 are colored one color. In that case, all of the remaining sides must be of the other color to not have a triangle where all three sides are of the same color. We can correspondingly fill out each color based on this constraint, but in this case you will always end up with a triangle where all three sides have the same color by inspection. $\textbf{Case 2}$: 3 are one color and one is the other. Following the steps from the previous case, you can try filling out the colors, but will always arrive at a contradiction so this case does not work either. $\textbf{Case 3}$: 2 are one color and 2 are of the other color. Using the same logic as previously, we can color the pentagon 2 different ways by inspection to satisfy the requirements. There are ${4\choose2}$ ways to color the original sides/diagonals and 2 ways after that to color the remaining ones for a total of $6\cdot 2 = 12$ ways to color the pentagon so that no such triangle has the same color for all of its sides. These are all the cases, and there are a total of $2^{10}$ ways to color the pentagon. Therefore the answer is $1-\frac{12}{1024} = 1-\frac{3}{256} = \frac{253}{256} = \boxed{D}$

This problem is about the relationships between the white unit cubes and the blue unit cubes, which can be solved by Graph Theory. We use a Planar Graph to represent the larger cube. Each vertex of the planar graph represents a unit cube. Each edge of the planar graph represents a shared face between $2$ neighboring unit cubes. Each face of the planar graph represents a face of the larger cube. Now the problem becomes a Graph Coloring problem of how many ways to assign $4$ vertices blue and $4$ vertices white with Topological Equivalence. For example, in Figure $(1)$, as long as the $4$ blue vertices belong to the same planar graph face, the different planar graphs are considered to be topological equivalent by rotating the larger cube. Here is how the $4$ blue unit cubes are arranged: In Figure $(1)$: $4$ blue unit cubes are on the same layer (horizontal or vertical). In Figure $(2)$: $4$ blue unit cubes are in $T$ shape. In Figure $(3)$ and $(4)$: $4$ blue unit cubes are in $S$ shape. In Figure $(5)$: $3$ blue unit cubes are in $L$ shape, and the other is isolated without a shared face. In Figure $(6)$: $2$ pairs of neighboring blue unit cubes are isolated from each other without a shared face. In Figure $(7)$: $4$ blue unit cubes are isolated from each other without a shared face. So the answer is $\boxed{\textbf{(A)}\ 7}$

We see that the polygon bounded by the small square, large square, and rectangle of known lengths is an isosceles triangle. Let’s draw a perpendicular from the vertex of this triangle to its opposing side; We see that this creates two congruent triangles. Let the smaller side of the triangle have length $a$ and let the larger side of the triangle have length $b$. Now we see by AAS congruency that if we draw perpendiculars that surround the smaller square, each outer triangle will be congruent to these two triangles. Now notice that these small triangles are also similar to the large triangle bounded by the bigger square and the rectangle by AA, and the ratio of the sides are 1:3, so we can fill in the lengths of that triangle. Similarly, the small triangle on the right bounded by the rectangle and the square is also congruent to the other small triangles by AAS, so we can fill in those sides; Since the larger square by definition has all equal sides, we can set the sum of the lengths of the sides equal to each other. $3a+b+b+a = 3b+a \implies 3a = b$. Now let's draw some more perpendiculars and rename the side lengths. By AA similarity, when we draw a perpendicular from the intersection of the two rectangles to the large square, we create a triangle below that is similar to the small congruent triangles with length $a,3a$. Since we don't know its scale, we'll label its sides $c,3c$. The triangle that is created above the perpendicular is congruent to the triangle on the opposite of the rectangle with unknown dimensions because they share the same hypotenuse and have two angles in common. Thus we can label these two triangles accordingly. The side length of the big square is $10a$, so we can find the remaining dimensions of the triangle bounded by the rectangle with unknown dimensions and the large square in terms of $a$ and $c$: This triangle with side lengths $4a-c$ and $6a-3c$ is similar to the triangle directly below it with side lengths $3a$ and $3c$ by AA similarity, so we can set up a ratio equation: $\frac{3a}{3c} = \frac{6a-3c}{4a-c} \implies 4a^2-ac = -3c^2 + 6ac \implies 4a^2 - 7ac + 3c^2 = 0 \implies (4a-3c)(a-c) = 0$. There are two solutions to this equation; $c = \frac{4}{3}a$ and $c = a$. For the first solution, the triangle in the corner has sides $2a$ and $\frac{8}{3}a$. Using Pythagorean theorem on that triangle, the hypotenuse has length $\frac{10}{3}a$. The triangle directly below has side lengths $3a$ and $4a$ in this case, so special right triangle yields the hypotenuse to be $5a$. The area of the rectangle is thus $5a\cdot\frac{10}{3}a = \frac{50}{3}a^2$. For the second solution, the side lengths of the corner triangle are $3a$ and $3a$, so the hypotenuse of the triangle is $3\sqrt{2}a$. The triangle below that also has side lengths $3a$ and $3a$, so its hypotenuse is the same. Then the area of the rectangle is $(3\sqrt{2}a)^2 = 18a^2$. The sum of the possible areas of the rectangle is therefore $18a^2+\frac{50}{3}a^2 = \frac{104}{3}a^2$. Using Pythagorean theorem on the original small congruent triangles, $a^2+9a^2 = 1$ or $a^2 = \frac{1}{10}$. Therefore the sum of the possible areas of the rectangle is $\frac{104}{3}\cdot\frac{1}{10} = \frac{52}{15}$. Therefore $m = 52$, $n = 15$, and $m + n = 67 = \boxed{E}$