Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$

Solution: Let us consider the horizontal rows.  Since there are five of them, each with constant sum $x$ , we can add up the 25 numbers in 5 rows for a sum of $5x$ .  Since the sum of the 25 numbers used is $-10-9-8-\cdots{}+12+13+14+15=11+12+13+14+15=50$ $5x=50$ and $x=\boxed{10}$ . 
~cw357

Answer: C 10
Problem: The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$

Solution: The mean of the set of numbers is $(14-10) \div 2 = 2$ . The numbers around it must be equal (i.e. if the mean of $1$ $2$ $3$ $4$ , and $5$ is $3$ , then $2+4=1+5$ .)
One row of the square would be \[\square  \square  2  \square  \square\]
Adding the numbers would be
\[0, 1, 2, 3, 4\]
with a sum of $\boxed{10}$

Answer: C 10
Problem: The $25$ integers from $-10$ to $14,$ inclusive, can be arranged to form a $5$ -by- $5$ square in which the sum of the numbers in each row, the sum of the numbers in each column, and the sum of the numbers along each of the main diagonals are all the same. What is the value of this common sum?
$\textbf{(A) }2 \qquad\textbf{(B) } 5\qquad\textbf{(C) } 10\qquad\textbf{(D) } 25\qquad\textbf{(E) } 50$

Solution: If the sum of each row, column, and diagonal is x, then we have a total of 12x for the sum. The sum of the rows and columns is the sum of all the numbers doubled, which is $50\cdot2=100$ . Therefore $100+2x=12x$ $100=10x$ , and $x=\boxed{10}$ .
~MC413551

Answer:  10
Problem: What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

Solution: Looking at the numbers, you see that every set of $4$ has $3$ positive numbers and 1 negative number. Calculating the sum of the first couple sets gives us $2+10+18...+394$ . Clearly, this pattern is an arithmetic sequence. By using the formula we get $\frac{2+394}{2}\cdot 50=\boxed{9900}$

Answer: B 9900
Problem: What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

Solution: Split the even numbers and the odd numbers apart. If we group every 2 even numbers together and add them, we get a total of $50\cdot (-2)=-100$ . Summing the odd numbers is equivalent to summing the first 100 odd numbers, which is equal to $100^2=10000$ . Adding these two, we obtain the answer of $\boxed{9900}$

Answer: B 9900
Problem: What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

Solution: We can break this entire sum down into $4$ integer bits, in which the sum is $2x$ , where $x$ is the first integer in this bit. We can find that the first sum of every sequence is $4x-3$ , which we plug in for the $50$ bits in the entire sequence is $1+2+3+\cdots+50=1275$ , so then we can plug it into the first term of every sequence equation we got above $4(1275)-3(50)=4950$ , and so the sum of every bit is $2x$ , and we only found the value of $x$ , the sum of the sequence is $4950\cdot2=\boxed{9900}$ .-middletonkids

Answer: B 9900
Problem: What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

Solution: Another solution involves adding everything and subtracting out what is not needed.  The first step involves solving $1+2+3+4+5+6+7+8+\cdots+197+198+199+200$ .  To do this, we can simply multiply $200$ and $201$ and divide by $2$ to get us $20100$ .  The next step involves subtracting out the numbers with minus signs.  We actually have to do this twice, because we need to take out the numbers we weren’t supposed to add and then subtract them from the problem.  Then, we can see that from $4$ to $200$ , incrementing by $4$ , there are $50$ numbers that we have to subtract.  To do this we can do $50$ times $51$ divided by $2$ , and then we can multiply by $4$ , because we are counting by fours, not ones.  Our answer will be $5100$ , but remember, we have to do this twice.  Once we do that, we will get $10200$ .  Finally, we just have to do $20100-10200$ , and our answer is $\boxed{9900}$

Answer: B 9900
Problem: What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

Solution: In this solution, we group every 4 terms. Our groups should be: $1 + 2 + 3 - 4 = 2$ $5 + 6 + 7 - 8 = 10$ $9 + 10 + 11 - 12 = 18$ , ... $197 + 198 + 199 - 200 = 394$ . We add them together to get this expression: $2 + 10 + 18 + ... + 394$ . This can be rewritten as $8 * (0 + 1 + 2 + ... + 49) + 100$ . We add this to get $\boxed{9900}$ .    ~Baolan

Answer: B 9900
Problem: What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

Solution: We can split up this long sum into groups of four integers. Finding the first few sums, we have that $1 + 2 + 3 - 4 = 2$ $5 + 6 + 7 - 8 = 10$ , and $9 + 10 + 11 - 12 = 18$ . Notice that this is an increasing arithmetic sequence, with a common difference of $8$ . We can find the sum of the arithmetic sequence by finding the average of the first and last terms, and then multiplying by the number of terms in the sequence. The first term is $1 + 2 + 3 - 4$ , or $2$ , the last term is $197 + 198 + 199 - 200$ , or $394$ , and there are $200\div 4$ or $50$ terms. So, we have that the sum of the sequence is $\frac{(394+2)\cdot 50}{2}$ , or $\boxed{9900}$ . ~Arctic_Bunny

Answer: B 9900
Problem: What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

Solution: Note that the original expression is equal to \[(1+2+3+\dots+199+200)-2(4(1+2+3+\dots+49+50)).\] Since the sum of the first $n$ positive integers is $\frac{n(n+1)}{2}$ , this is equal to \[\frac{200(201)}{2}-2\left(4\left(\frac{50(51)}{2}\right)\right),\] which can be simplified as \[20100-4(50)(51)=20100-10200.\] Doing the subtraction yields $\boxed{9900}$ . -BorealBear

Answer: B 9900
Problem: What is the value of \[1+2+3-4+5+6+7-8+\cdots+197+198+199-200?\]
$\textbf{(A) } 9,800 \qquad \textbf{(B) } 9,900 \qquad \textbf{(C) } 10,000 \qquad \textbf{(D) } 10,100 \qquad \textbf{(E) } 10,200$

Solution: We can split the sum into 4 groups so that the terms in these groups have the same common difference of 4 $(1+5+9+...+197)+(2+6+10+...+198)+(3+7+11+...+199)-(4+8+12+...+200)$ .
Then, we can use the sum formula. It's equal to $\frac{50\cdot(1+197)}{2} + \frac{50\cdot(2+198)}{2} + \frac{50\cdot(3+199)}{2} - \frac{50\cdot(4+200)}{2}$ .
This can also be expressed as $(25\cdot198) + (25\cdot200) + (25\cdot202) - (25\cdot204)$ .
And it is equal to $25\cdot(198+200+202-204) = 25\cdot396 = 9900$ .
Therefore, the answer is $\boxed{9900}$ . ~kurbanlik_21

Answer: B 9900
Problem: A single bench section at a school event can hold either $7$ adults or $11$ children. When $N$ bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of $N?$
$\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77$

Solution: The least common multiple of $7$ and $11$ is $77$ . Therefore, there must be $77$ adults and $77$ children. The total number of benches is $\frac{77}{7}+\frac{77}{11}=11+7=\boxed{18}$ .~taarunganesh

Answer: B 18
Problem: A single bench section at a school event can hold either $7$ adults or $11$ children. When $N$ bench sections are connected end to end, an equal number of adults and children seated together will occupy all the bench space. What is the least possible positive integer value of $N?$
$\textbf{(A) } 9 \qquad \textbf{(B) } 18 \qquad \textbf{(C) } 27 \qquad \textbf{(D) } 36 \qquad \textbf{(E) } 77$

Solution: Let $x$ denote how many adults there are. Since the number of adults is equal to the number of children we can write $N$ as $\frac{x}{7}+\frac{x}{11}=N$ . Simplifying we get $\frac{18x}{77} = N$ Since both $n$ and $x$ have to be positive integers, $x$ has to equal $77$ . Therefore, $N=\boxed{18}$ is our final answer.

Answer: B 18
Problem: Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$

Solution: The volume of each cube follows the pattern of $n^3$ , for $n$ is between $1$ and $7$
We see that the total surface area can be comprised of three parts: the sides of the cubes, the tops of the cubes, and the bottom of the $7\times 7\times 7$ cube (which is just $7 \times 7 = 49$ ). The sides areas can be measured as the sum $4\sum_{n=1}^{7} n^2$ , giving us $560$ . Structurally, if we examine the tower from the top, we see that it really just forms a $7\times 7$ square of area $49$ . Therefore, we can say that the total surface area is $560 + 49 + 49 = \boxed{658}$ .
Alternatively, for the area of the tops, we could have found the sum $\sum_{n=2}^{7}((n)^{2}-(n-1)^{2})$ , giving us $49$ as well.

Answer: B 658
Problem: Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$

Solution: It can quickly be seen that the side lengths of the cubes are the integers from 1 to 7 inclusive.
First, we will calculate the total surface area of the cubes, ignoring overlap. This value is $6 ( 1^2 + 2^2 + \cdots + 7^2 ) = 6\sum_{n=1}^{7} n^2 = 6 \left( \frac{7(7 + 1)(2 \cdot 7 + 1)}{6} \right) = 7 \cdot 8 \cdot 15 = 840$ . Then, we need to subtract out the overlapped parts of the cubes. Between each consecutive pair of cubes, one of the smaller cube's faces is completely covered, along with an equal area of one of the larger cube's faces. The total area of the overlapped parts of the cubes is thus equal to $2\sum_{n=1}^{6} n^2 = 182$ . Subtracting the overlapped surface area from the total surface area, we get $840 - 182 = \boxed{658}$ . ~ emerald_block

Answer: B 658
Problem: Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$

Solution: It can be seen that the side lengths of the cubes using cube roots are all integers from $1$ to $7$ , inclusive.
Only the cubes with side length $1$ and $7$ have $5$ faces in the surface area and the rest have $4$ . Also, since the cubes are stacked, we have to find the difference between each $n^2$ and $(n-1)^2$ side length as $n$ ranges from $7$ to $2$
We then come up with this: $5(49)+13+4(36)+11+4(25)+9+4(16)+7+4(9)+5+4(4)+3+5(1)$
We then add all of this and get $\boxed{658}$

Answer: B 658
Problem: Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$

Solution: Notice that the surface area of the top cube is $6s^2$ and the others are $4s^2$ . Then we can directly compute. The edge length for the first cube is $7$ and has a surface area of $294$ . The surface area of the next cube is $144$ . The surface area of the next cube $100$ . The surface area of the next cube is $64$ . The surface area of the next cube is $36$ . The surface area of the next cube is $16$ . The surface area of the next cube is $4$ . We then sum up $294+144+100+64+36+16+4$ to get $\boxed{658}$ .
~smartatmath

Answer: B 658
Problem: Seven cubes, whose volumes are $1$ $8$ $27$ $64$ $125$ $216$ , and $343$ cubic units, are stacked vertically to form a tower in which the volumes of the cubes decrease from bottom to top. Except for the bottom cube, the bottom face of each cube lies completely on top of the cube below it. What is the total surface area of the tower (including the bottom) in square units?
$\textbf{(A)}\ 644\qquad\textbf{(B)}\ 658\qquad\textbf{(C)}\ 664\qquad\textbf{(D)}\ 720\qquad\textbf{(E)}\ 749$

Solution: First of all, compute the area of the sides, excluding the top and bottoms, of the cubes. The side lengths (cube root the volumes) are 1, 2, 3, 4, 5, 6, 7. 
Each cube's area of the sides can be calculated with $4($ area of one side $)$ $4(l^2)$ so in total that is $4(1+4+16+...+49)$ so $4(140)=560$ the area of all the sides of the cubes is $560$ . 
Then, calculate the bottom face of the largest cube, $7*7=49$ . 
Now, notice that if you stack the cubes up on top of each other, and look directly down on them, the tops of the cubes showing add up to the area of the bottom cube, the 7x7. Therefore, the sum of the area of the tops of the cubes is $7*7=49$
Now add them all up: $49+49+560=658.$ Therefore, the answer is $\boxed{658}$

Answer: B 658
Problem: What is the median of the following list of $4040$ numbers $?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

Solution: We can see that $44^2=1936$ which is less than 2020. Therefore, there are $2020-44=1976$ of the $4040$ numbers greater than $2020$ . Also, there are $2020+44=2064$ numbers that are less than or equal to $2020$
Since there are $44$ duplicates/extras, it will shift up our median's placement down $44$ . Had the list of numbers been $1,2,3, \dots, 4040$ , the median of the whole set would be $\dfrac{1+4040}{2}=2020.5$
Thus, our answer is $2020.5-44=\boxed{1976.5}$

Answer: C 1976.5
Problem: What is the median of the following list of $4040$ numbers $?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

Solution: As we are trying to find the median of a $4040$ -term set, we must find the average of the $2020$ th and $2021$ st terms.
Since $45^2 = 2025$ is slightly greater than $2020$ , we know that the $44$ perfect squares $1^2$ through $44^2$ are less than $2020$ , and the rest are greater. Thus, from the number $1$ to the number $2020$ , there are $2020 + 44 = 2064$ terms. Since $44^2$ is $44 + 45 = 89$ less than $45^2 = 2025$ and $84$ less than $2020$ , we will only need to consider the perfect square terms going down from the $2064$ th term, $2020$ , after going down $84$ terms. Since the $2020$ th and $2021$ st terms are only $44$ and $43$ terms away from the $2064$ th term, we can simply subtract $44$ from $2020$ and $43$ from $2020$ to get the two terms, which are $1976$ and $1977$ . Averaging the two, we get $\boxed{1976.5}.$

Answer: C 1976.5
Problem: What is the median of the following list of $4040$ numbers $?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

Solution: We want to know the $2020$ th term and the $2021$ st term to get the median.
We know that $44^2=1936$ . So, numbers $1^2, 2^2, \ldots,44^2$ are in between $1$ and $1936$
So, the sum of $44$ and $1936$ will result in $1980$ , which means that $1936$ is the $1980$ th number.
Also, notice that $45^2=2025$ , which is larger than $2021$
Then the $2020$ th term will be $1936+40 = 1976$ , and similarly the $2021$ th term will be $1977$
Solving for the median of the two numbers, we get $\boxed{1976.5}$

Answer: C 1976.5
Problem: What is the median of the following list of $4040$ numbers $?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

Solution: We note that $44^2 = 1936$ , which is the first square less than $2020$ , which means that there are $44$ additional terms before $2020$ . This makes $2020$ the $2064$ th term. To find the median, we need the $2020$ th and $2021$ st term. We note that every term before $2020$ is one less than the previous term (That is, we subtract $1$ to get the previous term.). If $2020$ is the $2064$ th term, than $2020 - 44$ is the $(2064 - 44)$ th term. So, the $2020$ th term is $1976$ , and the $2021$ st term is $1977$ , and the average of these two terms is the median, or $\boxed{1976.5}$

Answer: C 1976.5
Problem: What is the median of the following list of $4040$ numbers $?$ \[1, 2, 3, \ldots, 2020, 1^2, 2^2, 3^2, \ldots, 2020^2\] $\textbf{(A)}\ 1974.5\qquad\textbf{(B)}\ 1975.5\qquad\textbf{(C)}\ 1976.5\qquad\textbf{(D)}\ 1977.5\qquad\textbf{(E)}\ 1978.5$

Solution: To find the median, we sort the $4040$ numbers in decreasing order, then average the $2020$ th and the $2021$ st numbers of the sorted list.
Since $45^2=2025$ and $44^2=1936,$ the first $2021$ numbers of the sorted list are \[\underbrace{2020^2,2019^2,2018^2,\ldots,46^2,45^2}_{1976\mathrm{ \ numbers}}\phantom{ },\phantom{ }\underbrace{2020,2019,2018,\ldots,1977,1976}_{45\mathrm{ \ numbers}}\phantom{ },\] from which the answer is $\frac{1977+1976}{2}=\boxed{1976.5}.$

Answer: C 1976.5
Problem: Triangle $AMC$ is isosceles with $AM = AC$ . Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$ . What is the area of $\triangle AMC?$
[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0, 3.6), S); [/asy]
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$

Solution: Since quadrilateral $UVCM$ has perpendicular diagonals, its area can be found as half of the product of the length of the diagonals. Also note that $\triangle AUV$ has $\frac 14$ the area of triangle $AMC$ by similarity, so $[UVCM]=\frac 34\cdot [AMC].$ Thus, \[\frac 12 \cdot 12\cdot 12=\frac 34 \cdot [AMC]\] \[72=\frac 34\cdot [AMC]\] \[[AMC]=96\rightarrow \boxed{96}.\]

Answer: C 96
Problem: Triangle $AMC$ is isosceles with $AM = AC$ . Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$ . What is the area of $\triangle AMC?$
[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0, 3.6), S); [/asy]
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$

Solution: [asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((-2,6)--(2,6)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0, 3.6), S); [/asy]
We know that $\triangle AUV \sim \triangle AMC$ , and since the ratios of its sides are $\frac{1}{2}$ , the ratio of of their areas is $\left(\frac{1}{2}\right)^2=\frac{1}{4}$
If $\triangle AUV$ is $\frac{1}{4}$ the area of $\triangle AMC$ , then trapezoid $MUVC$ is $\frac{3}{4}$ the area of $\triangle AMC$
Let's call the intersection of $\overline{UC}$ and $\overline{MV}$ $P$ . Let $\overline{UP}=x$ . Then $\overline{PC}=12-x$ . Since $\overline{UC}  \perp \overline{MV}$ $\overline{UP}$ and $\overline{CP}$ are heights of triangles $\triangle MUV$ and $\triangle MCV$ , respectively. Both of these triangles have base $12$
Area of $\triangle MUV = \frac{x\cdot12}{2}=6x$
Area of $\triangle MCV = \frac{(12-x)\cdot12}{2}=72-6x$
Adding these two gives us the area of trapezoid $MUVC$ , which is $6x+(72-6x)=72$
This is $\frac{3}{4}$ of the triangle, so the area of the triangle is $\frac{4}{3}\cdot{72}=\boxed{96}$ ~quacker88, diagram by programjames1

Answer: C 96
Problem: Triangle $AMC$ is isosceles with $AM = AC$ . Medians $\overline{MV}$ and $\overline{CU}$ are perpendicular to each other, and $MV=CU=12$ . What is the area of $\triangle AMC?$
[asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0, 3.6), S); [/asy]
$\textbf{(A) } 48 \qquad \textbf{(B) } 72 \qquad \textbf{(C) } 96 \qquad \textbf{(D) } 144 \qquad \textbf{(E) } 192$

Solution: Draw median $\overline{AB}$ [asy] draw((-4,0)--(4,0)--(0,12)--cycle); draw((-2,6)--(4,0)); draw((2,6)--(-4,0)); draw((0,12)--(0,0)); label("M", (-4,0), W); label("C", (4,0), E); label("A", (0, 12), N); label("V", (2, 6), NE); label("U", (-2, 6), NW); label("P", (0.5, 4), E); label("B", (0, 0), S); [/asy]
Since we know that all medians of a triangle intersect at the centroid, we know that $\overline{AB}$ passes through point $P$ . We also know that medians of a triangle divide each other into segments of ratio $2:1$ . Knowing this, we can see that $\overline{PC}:\overline{UP}=2:1$ , and since the two segments sum to $12$ $\overline{PC}$ and $\overline{UP}$ are $8$ and $4$ , respectively.
Finally knowing that the medians divide the triangle into $6$ sections of equal area, finding the area of $\triangle PUM$ is enough. $\overline{PC} = \overline{MP} = 8$
The area of $\triangle PUM = \frac{4\cdot8}{2}=16$ . Multiplying this by $6$ gives us $6\cdot16=\boxed{96}$

Answer: C 96