AMC12 2024 A

The likely fastest method will be direct computation. $9901\cdot101$ evaluates to $1000001$ and $99\cdot10101$ evaluates to $999999$. The difference is $\boxed{\textbf{(A) }2}.$ Solution by juwushu.

Plug in the values into the equation to give you the following two equations: 69=1.5a+800b,69=1.2a+1100b. Solving for the values $a$ and $b$ gives you that $a=30$ and $b=\frac{3}{100}$. These values can be plugged back in showing that these values are correct. Now, using the given length of the trail, $4.2$, and the given vertical increase, $4000$ , we get a final answer of $\boxed{\textbf{(B) }246}.$ Solution by juwushu. Minor edits by ParticlePhysics and TigerSenju

Since we want the least number of two-digit numbers, we maximize the two-digit numbers by choosing as many $99$s as possible. Since $2024=99\cdot20+44\cdot1,$ we choose twenty $99$s and one $44,$ for a total of $\boxed{\textbf{(B) }21}$ two-digit numbers.

Note that $2024=2^3\cdot11\cdot23$ in the prime factorization. Since $23!$ is a multiple of $2^3, 11,$ and $23,$ we conclude that $23!$ is a multiple of $2024.$ Therefore, we have $n=\boxed{\textbf{(D) } 23}.$ Remark Memorizing the prime factorization of the current year is useful for the AMC 8/10/12 Exams. Remark

Because the set has $20$ numbers and mean $45$, the sum of the terms in the set is $45\cdot 20=900$. Let there be $s$ sixes in the set. Then, the mean of the set without the sixes is $\frac{900-6s}{20-s}$. Equating this expression to $66$ and solving yields $s=7$, so we choose answer choice $\boxed{\textbf{(D) }7}$.

We notice that the optimal solution involves two negative numbers and a positive number. Thus we may split $60$ into three factors and choose negativity. We notice that $10\cdot6\cdot1=10\cdot(-6)\cdot(-1)=60$, and trying other combinations does not yield lesser results so the answer is $10-6-1=\boxed{\textbf{(B) }3}$.

Let us find an expression for the $x$- and $y$-components of $\overrightarrow{BP_i}$. Note that $AP_1+P_1P_2+\dots+P_{2023}P_{2024}+P_{2024}C=AC=2$, so $AP_1=P_1P_2=\dots=P_{2023}P_{2024}=P_{2024}C=\dfrac2{2025}$. All of the vectors $\overrightarrow{AP_1},\overrightarrow{P_1P_2},$ and so on up to $\overrightarrow{P_{2024}C}$ are equal; moreover, they equal $\textbf v=\left\langle\dfrac1{\sqrt2}\cdot\dfrac2{2025},-\dfrac1{\sqrt2}\cdot\dfrac2{2025}\right\rangle=\left\langle\dfrac{\sqrt2}{2025},-\dfrac{\sqrt2}{2025}\right\rangle$. We now note that $\overrightarrow{AP_i}=i\textbf v=\left\langle\dfrac{i\sqrt2}{2025},-\dfrac{i\sqrt2}{2025}\right\rangle$ ($i$ copies of $\textbf v$ added together). Furthermore, note that $\overrightarrow{BP_i}=\overrightarrow{BA}+\overrightarrow{AP_i}=\left\langle0,\sqrt2\right\rangle+\left\langle\dfrac{i\sqrt2}{2025},-\dfrac{i\sqrt2}{2025}\right\rangle=\left\langle\dfrac{i\sqrt2}{2025},\sqrt2-\dfrac{i\sqrt2}{2025}\right\rangle.$ We want $\sum_{i=1}^{2024}\overrightarrow{BP_i}$'s length, which can be determined from the $x$- and $y$-components. Note that the two values should actually be the same - in this problem, everything is symmetric with respect to the line $x=y$, so the magnitudes of the $x$- and $y$-components should be identical. The $x$-component is easier to calculate. \[\sum_{i=1}^{2024}\left(\overrightarrow{BP_i}\right)_x=\sum_{i=1}^{2024}\dfrac{i\sqrt2}{2025}=\dfrac{\sqrt2}{2025}\sum_{i=1}^{2024}i=\dfrac{\sqrt2}{2025}\cdot\dfrac{2024\cdot2025}2=1012\sqrt2.\] One can similarly evaulate the $y$-component and obtain an identical answer; thus, our desired length is $\sqrt{\left(1012\sqrt2\right)^2+\left(1012\sqrt2\right)^2}=\sqrt{4\cdot1012^2}=\boxed{\textbf{(D) }2024}$.

Note that this is equivalent to $\sin(3\theta)\cos(2\theta)=1$, which is clearly only possible when $\sin(3\theta)=\cos(2\theta)=\pm1$. (If either one is between $1$ and $-1$, the other one must be greater than $1$ or less than $-1$ to offset the product, which is impossible for sine and cosine.) They cannot be both $-1$ since we cannot take logarithms of negative numbers, so they are both $+1$. Then $3\theta$ is $\dfrac\pi2$ more than a multiple of $2\pi$ and $2\theta$ is a multiple of $2\pi$, so $\theta$ is $\dfrac\pi6$ more than a multiple of $\dfrac23\pi$ and also a multiple of $\pi$. However, a multiple of $\dfrac23\pi$ will always have a denominator of $1$ or $3$, and never $6$; it can thus never add with $\dfrac\pi6$ to form an integral multiple of $\pi$. Thus, there are $\boxed{\textbf{(A) }0}$ solutions.

Let $M+1213=P^2$ and $M+3773=Q^2$ for some positive integers $P$ and $Q.$ We subtract the first equation from the second, then apply the difference of squares: \[(Q+P)(Q-P)=2560.\] Note that $Q+P$ and $Q-P$ have the same parity, and $Q+P>Q-P.$ We wish to maximize both $P$ and $Q,$ so we maximize $Q+P$ and minimize $Q-P.$ It follows that \begin{align*} Q+P&=1280, \\ Q-P&=2, \end{align*} from which $(P,Q)=(639,641).$ Finally, we get $M=P^2-1213=Q^2-3773\equiv1-3\equiv8\pmod{10},$ so the units digit of $M$ is $\boxed{\textbf{(E) }8}.$

We are given that \[\tan\alpha=\frac{3}{4} \text{ and } \tan\beta=\frac{7}{24}.\] We have \begin{align*} \tan(\alpha+\beta) &= \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} \\ &= \frac{\frac{3}{4}+\frac{7}{24}}{1-\frac{3}{4} \cdot \frac{7}{24}} \\ &= \frac{4}{3}. \end{align*} It follows that \begin{align*} \alpha+\beta&=\tan^{-1}\left(\frac{4}{3}\right) \\ &=\frac{\pi}{2}-\tan^{-1}\left(\frac{3}{4}\right) \\ &=\frac{\pi}{2}-\alpha. \end{align*} Therefore, the answer is \[\beta=\boxed{\textbf{(C) }\frac{\pi}{2}-2\alpha}.\]

$2024_b = 2b^3+2b+4\equiv 0\pmod{16}\implies b^3+b+2\equiv 0\pmod 8$. If $b$ is even, then $b+2\equiv 0\pmod 8\implies b\equiv 6\pmod 8$. If $b$ is odd, then $b^2\equiv 1\pmod 8\implies b^3+b+2\equiv 2b+2\pmod 8$ so $2b+2\equiv 0\pmod 8\implies b+1\equiv 0\pmod 4\implies b\equiv 3,7\pmod 8$. Now $8\mid 2024$ so $\frac38\cdot 2024=759$ but one of the answers we got from that, $3$, is too small, so $759 - 1 = 758\implies\boxed{\textbf{(D) }20}$.

For a geometric sequence, we have $ab=720^2=2^8 3^4 5^2$, and we can test values for $b$. We find that $b=768$ and $a=675$ works, and we can test multiples of $5$ in between the two values. Finding that none of the multiples of 5 divide $720^2$ besides $720$ itself, we know that the answer is $7+6+8=\boxed{\textbf{(E) } 21}$. (Note: To find the value of $b$ without bashing, we can observe that $2^8=256$, and that multiplying it by $3$ gives us $768$, which is really close to $720$. ~ YTH) Note: The reason why $ab=720^2$ is because $b/720 = 720/a$. Rearranging this gives $ab = 720^2$ Note: Another reason that $ab=720^2$ is because the $\sqrt{ab}=720$ (as the middle term in a geometric series is always the geometric mean [the geometric mean is the square root of the product of the first and last terms of the series]) and squaring on both sides results in $ab=720^2$.

The line of symmetry is probably of the form $x=a$ for some constant $a$. A vertical line of symmetry at $x=a$ for a function $f$ exists if and only if $f(a-b)=f(a+b)$; we substitute $a-b$ and $a+b$ into our given function and see that we must have \[e^{a-b+1}+e^{-(a-b)}-2=e^{a+b+1}+e^{-(a+b)}-2\] for all real $b$. Simplifying: ea−b+1+e−(a−b)−2=ea+b+1+e−(a+b)−2ea−b+1+eb−a=ea+b+1+e−a−bea−b+1−e−a−b=ea+b+1−eb−ae−b(ea+1−e−a)=eb(ea+1−e−a). If $e^{a+1}-e^{-a}\neq0$, then $e^{-b}=e^b$ for all real $b$; this is clearly impossible, so let $e^{a+1}-e^{-a}=0\implies a+1=-a\implies a=-\dfrac12$. Thus, our line of symmetry is $x=-\dfrac12$, and reflecting $\left(-1,\dfrac12\right)$ over this line gives $\boxed{\textbf{(D) }\left(0,\dfrac12\right)}.$

Start from the $0$. Going up, let the common difference be $a$, and going left, let the common difference be $b$. Therefore, we have \[\begin{bmatrix} . & ? &.&.&4a \\ .&.&.&48&3a\\ 12&.&.&.&2a\\ .&.&16&.&a\\ 4b&3b&2b&b&0\end{bmatrix}\] Looking at the third column, we can see that the common difference going up is $16-2b$. We fill this in: \[\begin{bmatrix} . & ? &64-6b&.&4a \\ .&.&48-4b&48&3a\\ 12&.&32-2b&.&2a\\ .&.&16&.&a\\ 4b&3b&2b&b&0\end{bmatrix}\] Looking at the second row, $48$ has two values beside it, so we can write \[48=\dfrac{48-4b+3a}{2}\rightarrow96=48-4b+3a\rightarrow48=-4b+3a,\] and we can do the same with the third row, which gives \[32-2b=\dfrac{12+2a}{2}\rightarrow32-2b=6+a\rightarrow26=a+2b.\] Now we have the system of equations \[48=-4b+3a\]\[26=a+2b,\] and solving it gives $a=20,b=3$, therefore we can now fill in the grid with actual numbers. But before doing that, note that we're only looking for a value in the first row, and because we already have two known values in that row, we can find the common difference for that row and not focus on anything else. Focusing only on the first row yields \[\begin{bmatrix} . & ? &46&.&80\end{bmatrix}\] This means that the common difference from right to left is $\dfrac{80-46}{2}=17$. Therefore, the desired value is $46-17=\boxed{\text{(C) }29}$

You can factor $(p^2 + 4)(q^2 + 4)(r^2 + 4)$ as $(p+2i)(p-2i)(q+2i)(q-2i)(r+2i)(r-2i)$. For any polynomial $f(x)$, you can create a new polynomial $f(x+a)$, which will have roots that instead have the value $a$ subtracted. Substituting $x-2i$ and $x+2i$ into $x$ for the first polynomial, gives you $10i-5$ and $-10i-5$ as $c$ for both equations. Multiplying $10i-5$ and $-10i-5$ together gives you $\boxed{\textbf{(D) }125}$.

We have $\binom{12}{4,4,4}$ ways to handle the red/white/blue tokens distribution on the denominator. Now we simply $\binom{6}{1}$ $\binom{5}{2}$ $3!$ for the numerator in order to handle the black tokens and distinguishable persons. The solution is therefore $\frac {6 \cdot 6 \cdot 10}{70 \cdot 45 \cdot 11} = \frac {4}{385}$ or $4+385=\boxed{\textbf{(C) }389}.$

Subtracting the first two equations yields $(a-c)(b-1)=13$. Notice that both factors are integers, so $b-1$ could equal one of $13,1,-1,-13$ and $b=14,2,0,-12$. We consider each case separately: For $b=0$, from the second equation, we see that $a=87$. Then $87c=60$, which is not possible as $c$ is an integer, so this case is invalid. For $b=2$, we have $2c+a=87$ and $ca=58$, which by experimentation on the factors of $58$ has no solution, so this is also invalid. For $b=14$, we have $14c+a=87$ and $ca=46$, which by experimentation on the factors of $46$ has no solution, so this is also invalid. Thus, we must have $b=-12$, so $a=12c+87$ and $ca=72$. Thus $c(12c+87)=72$, so $c(4c+29)=24$. We can simply trial and error this to find that $c=-8$ so then $a=-9$. The answer is then $(-9)(-12)+(-12)(-8)+(-8)(-9)=108+96+72=\boxed{\textbf{(D) }276}$.

Let the midpoint of $AC$ be $P$. We see that no matter how many moves we do, $P$ stays where it is. Now we can find the angle of rotation ($\angle APB$) per move with the following steps: \[AP^2=(\frac{1}{2})^2+(1+\frac{\sqrt{3}}{2})^2=2+\sqrt{3}\] \[1^2=AP^2+AP^2-2(AP)(AP)\cos\angle APB\] \[1=2(2+\sqrt{3})(1-\cos\angle APB)\] \[\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\] \[\cos\angle APB=\frac{3+2\sqrt{3}}{4+2\sqrt{3}}\cdot\frac{4-2\sqrt{3}}{4-2\sqrt{3}}\] \[\cos\angle APB=\frac{2\sqrt{3}}{4}=\frac{\sqrt{3}}{2}\] \[\angle APB=30^\circ\] Since Vertex $C$ is the closest one and \[\angle BPC=360-180-30=150\] Vertex C will land on Vertex B when $\frac{150}{30}+1=\fbox{(A) 6}$ cards are placed.

Since $ABCD$ is a cyclic quadrilateral, opposite angles sum to $180^\circ$, so \[ \angle CBA = 60^\circ. \] Let $AC = u$. Apply the Law of Cosines in $\triangle ACD$ (where $\angle ACD = 120^\circ$): \begin{align*} u^2 &= 3^2 + 5^2 - 2 \cdot 3 \cdot 5 \cdot \cos 120^\circ \\ u^2 &= 9 + 25 - 30 \cdot \left(-\dfrac{1}{2}\right) \\ u^2 &= 34 + 15 = 49 \\ u &= 7. \end{align*} Let $AB = v$. Apply the Law of Cosines in $\triangle ABC$ (where $\angle BAC = 60^\circ$): \begin{align*} 7^2 &= 3^2 + v^2 - 2 \cdot 3 \cdot v \cdot \cos 60^\circ \\ 49 &= 9 + v^2 - 6v \cdot \dfrac{1}{2} \\ 49 &= 9 + v^2 - 3v \\ v^2 - 3v - 40 &= 0. \end{align*} Solve the quadratic equation: \[ v = \dfrac{3 \pm \sqrt{9 + 160}}{2} = \dfrac{3 \pm \sqrt{169}}{2} = \dfrac{3 \pm 13}{2}. \] The positive root is \[ v = \dfrac{3 + 13}{2} = 8. \] (We discard $v = -5$ since length is positive.) Now apply **Ptolemy's theorem** for cyclic quadrilateral $ABCD$: \[ AB \cdot CD + AD \cdot BC = AC \cdot BD \] \[ 8 \cdot 3 + 5 \cdot 3 = 7 \cdot BD \] \[ 24 + 15 = 7 \cdot BD \] \[ 39 = 7 \cdot BD \implies BD = \dfrac{39}{7}. \] Since \[ \dfrac{39}{7} \approx 5.571 < 7, \] the answer is $\boxed{\dfrac{39}{7}}$.

Let $\overline{AP}=x$ and $\overline{AQ}=y$. Applying the sine formula for a triangle's area, we see that \[[\Delta APQ]=\dfrac12\cdot x\cdot y\sin(\angle PAQ)=\dfrac{xy}2\sin(60^\circ)=\dfrac{\sqrt3}4xy.\] Without loss of generality, we let $AB=BC=CA=1$, and thus $[\Delta ABC]=\dfrac{\sqrt3}4$; we therefore require $\dfrac{\sqrt3}4xy\le\dfrac12\cdot\dfrac{\sqrt3}4\implies xy\le\dfrac12$ for $0\le x,y\le1$. (Note: You can skip most of this by using triangle area ratios ~A_MatheMagician.) A quick rough sketch of $y=\dfrac1{2x}$ on the square given by $x,y\in[0,1]$ reveals that the curve intersects the boundaries at $(0.5,1)$ and $(1,0.5)$, and it is actually quite (very) obvious that the area bounded by the inequality $xy\le0.5$ and the aforementioned unit square is more than $\dfrac34$ but less than $\dfrac78$ (cf. the diagram below). Thus, our answer is $\boxed{\textbf{(D) }\left(\dfrac34,\dfrac78\right]}$.

Multiply both sides of the recurrence to find that $n(a_n-1)=(n-1)(a_{n-1}+1)=(n-1)(a_{n-1}-1)+(n-1)(2)$. Let $b_n=n(a_n-1)$. Then the previous relation becomes \[b_n=b_{n-1}+2(n-1)\] We can rewrite this relation for values of $n$ until $1$ and use telescoping to derive an explicit formula: \[b_n=b_{n-1}+2(n-1)\] \[b_{n-1}=b_{n-2}+2(n-2)\] \[b_{n-2}=b_{n-3}+2(n-3)\] \[\cdot\] \[\cdot\] \[\cdot\] \[b_2=b_1+2(1)\] Summing the equations yields: \[b_n+b_{n-1}+\cdots+b_2=b_{n-1}+b_{n-2}+\cdots+b_1+2((n-1)+(n-2)+\cdots+1)\] \[b_n-b_1=2\cdot\frac{n(n-1)}{2}\] \[b_n-1=n(n-1)\] \[b_n=n(n-1)+1\] Now we can substitute $a_n$ back into our equation: \[n(a_n-1)=n(n-1)+1\] \[a_n-1=n-1+\frac{1}{n}\] \[a_n=n+\frac{1}{n}\] \[a_n^2=n^2+\frac{1}{n^2}+2\] Thus the sum becomes \[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} n^2+ \sum^{100}_{n=1} \frac{1}{n^2}+ \sum^{100}_{n=1} 2\] We know that $\sum^{100}_{n=1} n^2=\frac{100\cdot101\cdot201}{6}=338350$, and we also know that $\sum^{100}_{n=1} 2=200$, so the requested sum is equivalent to $\sum^{100}_{n=1} \frac{1}{n^2}+338550$. All that remains is to calculate $\sum^{100}_{n=1} \frac{1}{n^2}$, and we know that this value lies between $1$ and $2$ (see the note below for a proof). Thus, \[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550<2+338550=338552\] \[\sum^{100}_{n=1} a_n^2= \sum^{100}_{n=1} \frac{1}{n^2}+338550>1+338550=338551\] so \[\sum^{100}_{n=1} a_n^2\in(338551,338552)\] and thus the answer is $\boxed{\textbf{(B) }338551}$. Note: $\sum^{100}_{n=1} \frac{1}{n^2}< \sum^{\infty}_{n=1} \frac{1}{n^2}=\frac{\pi^2}{6}<2$. It is obvious that the sum is greater than 1 (since it contains $\frac{1}{1^2}$ as one of its terms).

Observations: 1. You can not have a vertical line in any place other than the first two columns and the last two columns. If we did, we would have at least one of the middle cells with toothpicks along more than one side, which would violate the conditions of the problem. 2. There are two cases that look completely different. We can have a long horizontal box that spans all eight cells either on top of or below the middle cells, or we have to have a shape that looks like a rectangle, except with a few places "pushed" in. Thus, using casework, we can split the task of finding those rectangles with squiggly edges into 3 cases. For case 1, we assume that the green lines shown below are given (always have toothpicks on them). In effect, we will use all eight columns. The only toothpicks we can place that will connect to the red lines are to go horizontally inward: Now, concentrate on the first row of squares. A toothpick can be placed on either the bottom or top and connected to a continuous squiggle by adding vertical toothpicks: How many squiggles are possible? We can summarize this by giving a high squiggle position a 1 and a low position a 0, thus we have a 6-digit binary sequence. Thus, we can have $2^6=64$ ways to make this squiggle. (The binary is not absolutely necessary, but it works.) Case 2: We can also pull in one of the sides, thus we can have a squiggle with 5 binary digits, which only uses the first or last 7 columns: Here, we only have 5 binary digits to work with, so there are $2^5=32$ ways to make this squiggle for each individual subcase. There are two subcases, one with the first 7 columns, and the other with the last 7, so we have a total of $32\cdot 2 = 64$ arrangements in this case. Case 3: We can use an even smaller section. Using only the middle 6 columns gives us a 4-wide squiggle: Thus, there are $2^4=16$ ways to make this squiggle. These three cases together cover all loops of this form. If we try to bring the square bracket like shapes on each side any closer, there will be some middle cells that do not touch any toothpicks. Adding up all our cases for these types of shapes: $64+32+32+16=144$. However, there are two more ways to draw a qualifying shape: We can draw a rectangle like that in the first row or third row. Thus, we have a total of $144+2=\boxed{\textbf{(C) }146}$ ways. A note to (potential) editors: This answer was not made to be concise or especially professional. It was made to explicitly explain this problem in a way so that it is easy to understand and follow.

First, notice that \[\tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {3\pi}{16} + \tan^2 \frac {\pi}{16} \cdot \tan^2 \frac {5\pi}{16}+\tan^2 \frac {3\pi}{16} \cdot \tan^2 \frac {7\pi}{16}+\tan^2 \frac {5\pi}{16} \cdot \tan^2 \frac {7\pi}{16}\] \[=(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})\] Here, we make use of the fact that \[\tan^2 x+\tan^2 (\frac{\pi}{2}-x)\] \[=(\tan x+\tan (\frac{\pi}{2}-x))^2-2\] \[=\left(\frac{\sin x}{\cos x}+\frac{\sin (\frac{\pi}{2}-x)}{\cos (\frac{\pi}{2}-x)}\right)^2-2\] \[=\left(\frac{\sin x \cos (\frac{\pi}{2}-x)+\sin (\frac{\pi}{2}-x) \cos x}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\] \[=\left(\frac{\sin \frac{\pi}{2}}{\cos x \cos (\frac{\pi}{2}-x)}\right)^2-2\] \[=\left(\frac{1}{\cos x \sin x}\right)^2-2\] \[=\left(\frac{2}{\sin 2x}\right)^2-2\] \[=\frac{4}{\sin^2 2x}-2\] Hence, \[(\tan^2\frac{\pi}{16}+\tan^2 \frac{7\pi}{16})(\tan^2\frac{3\pi}{16}+\tan^2 \frac{5\pi}{16})\] \[=\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\] Note that \[\sin^2 \frac{\pi}{8}=\frac{1-\cos \frac{\pi}{4}}{2}=\frac{2-\sqrt{2}}{4}\] \[\sin^2 \frac{3\pi}{8}=\frac{1-\cos \frac{3\pi}{4}}{2}=\frac{2+\sqrt{2}}{4}\] Hence, \[\left(\frac{4}{\sin^2 \frac{\pi}{8}}-2\right)\left(\frac{4}{\sin^2 \frac{3\pi}{8}}-2\right)\] \[=\left(\frac{16}{2-\sqrt{2}}-2\right)\left(\frac{16}{2+\sqrt{2}}-2\right)\] \[=(14+8\sqrt{2})(14-8\sqrt{2})\] \[=68\] Therefore, the answer is $\fbox{\textbf{(B) } 68}$.

Notice that any scalene $\textit{acute}$ triangle can be the faces of a $\textit{disphenoid}$. (See proof in Solution 2.) As a result, we simply have to find the smallest area a scalene acute triangle with integer side lengths can take on. This occurs with a $4,5,6$ triangle (notice that if you decrease the value of any of the sides the resulting triangle will either be isosceles, degenerate, or non-acute). For this triangle, the semiperimeter is $\frac{15}{2}$, so by Heron’s Formula: A=152⋅72⋅52⋅32=152⋅716=1547 The surface area is simply four times the area of one of the triangles, or $\boxed{\textbf{(D) }15\sqrt{7}}$.

Symmetric about the line $y=x$ implies that the inverse function $y^{-1}=y$. Then we split the question into several cases to find the final answer. Case 1: $c=0$ Then $y=\frac{a}{d}x+\frac{b}{d}$ and $y^{-1}=\frac{d}{a}x-\frac{b}{a}$. Giving us $\frac{a}{d}=\frac{d}{a}$ and $\frac{b}{d}=-\frac{b}{a}$ Therefore, we obtain 2 subcases: $b\neq 0, a+d=0$ and $b=0, a^2=d^2$ Case 2: $c\neq 0$ Then $y^{-1}=\frac{b-dx}{cx-a}=\frac{(cx-a)(-\frac{d}{c})+b-\frac{ad}{c}}{cx-a}=-\frac{d}{c}+\frac{b-\frac{ad}{c}}{cx-a}$ And $y=\frac{(cx+d)(\frac{a}{c})+b-\frac{ad}{c}}{cx+d}=\frac{a}{c}+\frac{b-\frac{ad}{c}}{cx+d}$ So $\frac{a}{c}=-\frac{d}{c}$, or $a=-d$ ($c\neq 0$), and substitute that into $\frac{b-\frac{ad}{c}}{cx-a}=\frac{b-\frac{ad}{c}}{cx+d}$ gives us: $bc-ad\neq 0$ (Otherwise $y=\frac{a}{c}$, $y^{-1}=-\frac{d}{c}=\frac{a}{c}$, and is not symmetric about $y=x$) Therefore we get three cases: Case 1.1: $c= 0, b\neq 0, d\neq 0, a+d=0$ We have 10 choice of $b$, 10 choice of $d$ and each choice of $d$ has one corresponding choice of $a$. In total $10\times 10=100$ ways. Case 1.2: $c= 0, b = 0, d\neq 0, a^2=d^2$ We have 10 choice for $d$ ($d\neq 0$), each choice of $d$ has 2 corresponding choice of $a$, thus $10\times 2=20$ ways. Case 2: $c\neq 0, bc-ad\neq 0, a=-d$ $a=0$: $10\times 10=100$ ways. $a=\pm 1$: $(11\times 10-2)\times 2=216$ ways. $a=\pm 2$: $(11\times 10-6)\times 2=208$ ways. $a=\pm 3$: $(11\times 10-2)\times 2=216$ ways. $a=\pm 4$: $(11\times 10-2)\times 2=216$ ways. $a=\pm 5$: $(11\times 10-2)\times 2=216$ ways. In total $100+208+216\times 4= 1172$ ways. So the answer is $100+20+1172= \boxed{\textbf{(B) }1292}$