Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square

Solution: Given $g : x \mapsto \max_{j : 2^j | x} 2^j$ , consider $S_n = g(2) + \cdots + g(2^n)$ . Define $S = \{2, 4, \ldots, 2^n\}$ . There are $2^0$ elements of $S$ that are divisible by $2^n$ $2^1 - 2^0 = 2^0$ elements of $S$ that are divisible by $2^{n-1}$ but not by $2^n, \ldots,$ and $2^{n-1}-2^{n-2} = 2^{n-2}$ elements of $S$ that are divisible by $2^1$ but not by $2^2$
Thus \begin{align*} S_n &= 2^0\cdot2^n + 2^0\cdot2^{n-1} + 2^1\cdot2^{n-2} + \cdots + 2^{n-2}\cdot2^1\\ &= 2^n + (n-1)2^{n-1}\\ &= 2^{n-1}(n+1).\end{align*} Let $2^k$ be the highest power of $2$ that divides $n+1$ . Thus by the above formula, the highest power of $2$ that divides $S_n$ is $2^{k+n-1}$ . For $S_n$ to be a perfect square, $k+n-1$ must be even. If $k$ is odd, then $n+1$ is even, hence $k+n-1$ is odd, and $S_n$ cannot be a perfect square. Hence $k$ must be even. In particular, as $n<1000$ , we have five choices for $k$ , namely $k=0,2,4,6,8$
If $k=0$ , then $n+1$ is odd, so $k+n-1$ is odd, hence the largest power of $2$ dividing $S_n$ has an odd exponent, so $S_n$ is not a perfect square.
In the other cases, note that $k+n-1$ is even, so the highest power of $2$ dividing $S_n$ will be a perfect square. In particular, $S_n$ will be a perfect square if and only if $(n+1)/2^{k}$ is an odd perfect square.
If $k=2$ , then $n<1000$ implies that $\frac{n+1}{4} \le 250$ , so we have $n+1 = 4, 4 \cdot 3^2, \ldots, 4 \cdot 13^2, 4\cdot 3^2 \cdot 5^2$
If $k=4$ , then $n<1000$ implies that $\frac{n+1}{16} \le 62$ , so $n+1 = 16, 16 \cdot 3^2, 16 \cdot 5^2, 16 \cdot 7^2$
If $k=6$ , then $n<1000$ implies that $\frac{n+1}{64}\le 15$ , so $n+1=64,64\cdot 3^2$
If $k=8$ , then $n<1000$ implies that $\frac{n+1}{256}\le 3$ , so $n+1=256$
Comparing the largest term in each case, we find that the maximum possible $n$ such that $S_n$ is a perfect square is $4\cdot 3^2 \cdot 5^2 - 1 = \boxed{899}$

Answer:  899
Problem: For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square

Solution: First note that $g(k)=1$ if $k$ is odd and $2g(k/2)$ if $k$ is even.
so $S_n=\sum_{k=1}^{2^{n-1}}g(2k). = \sum_{k=1}^{2^{n-1}}2g(k) = 2\sum_{k=1}^{2^{n-1}}g(k) = 2\sum_{k=1}^{2^{n-2}} g(2k-1)+g(2k).$ $2k-1$ must be odd so this reduces to $2\sum_{k=1}^{2^{n-2}}1+g(2k) = 2(2^{n-2}+\sum_{k=1}^{2^n-2}g(2k)).$ Thus $S_n=2(2^{n-2}+S_{n-1})=2^{n-1}+2S_{n-1}.$ Further noting that $S_0=1$ we can see that $S_n=2^{n-1}\cdot (n-1)+2^n\cdot S_0=2^{n-1}\cdot (n-1)+2^{n-1}\cdot2=2^{n-1}\cdot (n+1).$ which is the same as above.  To simplify the process of finding the largest square $S_n$ we can note that if $n-1$ is odd then $n+1$ must be exactly divisible by an odd power of $2$ .  However, this means $n+1$ is even but it cannot be.  Thus $n-1$ is even and $n+1$ is a large even square.  The largest even square $< 1000$ is $900$ so $n+1= 900 => n= \boxed{899}$

Answer:  899
Problem: For each even positive integer $x$ , let $g(x)$ denote the greatest power of 2 that divides $x.$ For example, $g(20)=4$ and $g(16)=16.$ For each positive integer $n,$ let $S_n=\sum_{k=1}^{2^{n-1}}g(2k).$ Find the greatest integer $n$ less than 1000 such that $S_n$ is a perfect square

Solution: First, we set intervals. Say that a number $N$ falls strictly within $[2^k, 2^{k+1}]$
$N<2^{k+1}=2^k+2^k$
We can generalize this:
If a number is in the form $N=2^k+2^{R}O$ where $O$ is a positive odd number, $R<k$
$N<2^{k+1}=2^k+2^k\Longrightarrow O<2^{k-R}$ so there are $2^{k-R-1}$ numbers that satisfy this form. 
For all numbers that satisfy this form, $g(N)=2^R$ . The sum of $g(N)$ for all $N$ in this form and interval is $2^{k-1}$ $R$ can vary between $1$ and $k-1$ , so the total sum is $\underbrace{2^{k-1}+2^{k-1}\cdots 2^{k-1}}_{k-1}=(k-1)2^{k-1}$ . The domain of the function we are trying to find is all even integers in the interval $[2^1, 2^n]$ so there are $n-1$ values of $k$ . Now we have the sum $\sum_{k=1}^{n-1}{(k-1)2^{k-1}}=\sum_{i=1}^{n-2}{k\cdot2^{k}}$ . However, we did not consider powers of two yet(since our interval was strictly between powers of 2), so we have to add $\sum_{k=1}^{n}{2^k}$ . 
Note that $\sum_{i=1}^{n-2}{k\cdot2^{k}}=(2^1+2^2\cdots 2^{n-2})+(2^2+2^3\cdots 2^{n-2})\cdots +(2^{n-3}+2^{n-2})+2^{n-2}=2^1(2^{n-2}-1)+2^2(2^{n-3}-1)\cdots +2^{n-3}(2^2-1)+2^{n-2}(2^1-1)=(n-2)(2^{n-1})-\sum_{k=1}^{n-2}{2^k}=(n-2)(2^{n-1})-2(2^{n-2}-1)=(n-3)(2^{n-1})+2$ . Adding $\sum_{k=1}^{n}{2^k}=2(2^n-1)$ , we get $(n+1)(2^{n-1})$ . 
If this sum is a perfect square, $n\not\equiv0\pmod2$ , since that would make $2^{n-1}$ not a perfect square, and $n+1$ odd so it cannot contribute a factor of 2 to make the power of 2 a perfect square.
We want the least odd number less than $1000$ such that $(n+1)$ is an even perfect square. The greatest even square less than $1000$ is $30^2=900$ so $n+1=900 \Longrightarrow n=\boxed{899}$

Answer:  899
Problem: A tripod has three legs each of length $5$ feet. When the tripod is set up, the angle between any pair of legs is equal to the angle between any other pair, and the top of the tripod is $4$ feet from the ground. In setting up the tripod, the lower 1 foot of one leg breaks off. Let $h$ be the height in feet of the top of the tripod from the ground when the broken tripod is set up. Then $h$ can be written in the form $\frac m{\sqrt{n}},$ where $m$ and $n$ are positive integers and $n$ is not divisible by the square of any prime. Find $\lfloor m+\sqrt{n}\rfloor.$ (The notation $\lfloor x\rfloor$ denotes the greatest integer that is less than or equal to $x.$

Solution: Diagram borrowed from Solution 1
Apply Pythagorean Theorem on $\bigtriangleup TOB$ yields \[BO=\sqrt{TB^2-TO^2}=3\] Since $\bigtriangleup ABC$ is equilateral, we have $\angle MOB=60^{\circ}$ and \[BC=2BM=2(OB\sin MOB)=3\sqrt{3}\] Apply Pythagorean Theorem on $\bigtriangleup TMB$ yields \[TM=\sqrt{TB^2-BM^2}=\sqrt{5^2-(\frac{3\sqrt{3}}{2})^2}=\frac{\sqrt{73}}{2}\] Apply Law of Cosines on $\bigtriangleup TBC$ we have \[BC^2=TB^2+TC^2-2(TB)(TC)\cos BTC\] \[(3\sqrt{3})^2=5^2+5^2-2(5)^2\cos BTC\] \[\cos BTC=\frac{23}{50}\] Apply Law of Cosines on $\bigtriangleup STB$ using the fact that $\angle STB=\angle BTC$ we have \[SB^2=ST^2+BT^2-2(ST)(BT)\cos STB\] \[SB=\sqrt{4^2+5^2-2(4)(5)\cos BTC}=\frac{\sqrt{565}}{5}\] Apply Pythagorean Theorem on $\bigtriangleup BSM$ yields \[SM=\sqrt{SB^2-BM^2}=\frac{\sqrt{1585}}{10}\] Let the perpendicular from $T$ hits $SBC$ at $P$ . Let $SP=x$ and $PM=\frac{\sqrt{1585}}{10}-x$ . Apply Pythagorean Theorem on $TSP$ and $TMP$ we have \[TP^2=TS^2-SP^2=TM^2-PM^2\] \[4^2-x^2=(\frac{\sqrt{73}}{2})^2-(\frac{\sqrt{1585}}{10}-x)^2\] Cancelling out the $x^2$ term and solving gets $x=\frac{181}{2\sqrt{1585}}$
Finally, by Pythagorean Theorem, \[TP=\sqrt{TS^2-SP^2}=\frac{144}{\sqrt{1585}}\] so $\lfloor m+\sqrt{n}\rfloor=\boxed{183}$

Answer:  183
Problem: Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$

Solution: Suppose $b_{i} = \frac {x_{i}}3$ .
We have \[\sum_{i = 1}^{2006}b_{i}^{2} = \sum_{i = 0}^{2005}(b_{i} + 1)^{2} = \sum_{i = 0}^{2005}(b_{i}^{2} + 2b_{i} + 1)\] So \[\sum_{i = 0}^{2005}b_{i} = \frac {b_{2006}^{2} - 2006}2\] Now \[\sum_{i = 1}^{2006}b_{i} = \frac {b_{2006}^{2} + 2b_{2006} - 2006}2\] Therefore \[\left|\sum_{i = 1}^{2006}b_{i}\right| = \left|\frac {(b_{2006} + 1)^{2} - 2007}2\right|\geq \left|\frac{2025 - 2007}{2}\right| = 9\] This lower bound can be achieved if we use $b_1 = -1$ $b_2 = 0$ $b_3 = -1$ $b_4 = 0$ , and so on until $b_{1962} = 0$ , after which we let $b_k = b_{k - 1} + 1$ so that $b_{2006} = 44$ . So \[\left|\sum_{i = 1}^{2006}x_{i}\right|\geq \boxed{027}\]

Answer:  027
Problem: Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$

Solution: We know $|x_k| = |x_{k - 1} + 3|$ .  We get rid of the absolute value by squaring both sides: ${x_k}^2 = {x_{k - 1}}^2 + 6{x_{k - 1}} + 9\Rightarrow {x_k}^2 - {x_{k - 1}}^2 = 6{x_{k - 1}} + 9$ . So we set this up:
\begin{align*} {x_1}^2 - {x_0}^2 & = 6{x_0} + 9 \\ {x_2}^2 - {x_1}^2 & = 6{x_1} + 9 \\ & \vdots \\ {x_{2007}}^2 - {x_{2006}}^2 & = 6{x_{2006}} + 9 \end{align*} There are $2007$ equations. Sum them. We get: ${x_{2007}}^2 = 6\left(x_1 + x_2 + \cdots + x_{2006}\right) + 9\cdot{2007}$
So $|x_1 + x_2 + \cdots + x_{2006}| = \tfrac 16 \left|{x_{2007}}^2 - 9\cdot{2007}\right|$
We know $3\ |\ x_{2007}$ and we want to minimize $\left|{x_{2007}}^2 - 9\cdot{2007}\right|$ , so $x_{2007}$ must be $3\cdot{45}$ for it to be minimal ( $45^2 = 2025$ which is closest to $2007$ ). We can achieve this with $x_k=3k$ till $x_{45}=135$ and then alternating $x_{46}=132$ $x_{47}=135$ and so on ... Then $x_{2k}=132$ and $x_{2k+1}=135$ for all $k>22$ . Since $2007$ is odd, we have $x_{2007}=135$
This means that $|x_1 + x_2 + \cdots + x_{2006}| = \tfrac 16 \left|9(2025 - 2007)\right| = \boxed{027}$

Answer:  027
Problem: Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$

Solution: Playing around with a couple numbers, we see that we can generate the sequence $0, 3, -6, 3, -6, \cdots$ , and we can also generate the sequence $3, 6, 9, 12, \cdots$ after each $-6$ value. Thus, we will apply this to try and find some bounds. We can test if the first $1000$ pairs of numbers each sum up to $-3$ , and the rest form an arithmetic sequence, if the first $990$ pairs sum up to $-3$ , and so on. When we get to $980$ , we find that $980(-3) + 3 + 6 + \cdots + 3\cdot 46 = 303$ . If we shift the number of pairs up by $1$ , we get $981(-3) + 3 + 6 + \cdots + 3\cdot 44 = \boxed{027}$

Answer:  027
Problem: In convex hexagon $ABCDEF$ , all six sides are congruent, $\angle A$ and $\angle D$ are right angles , and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent . The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$

Solution: Let the side length be called $x$ , so $x=AB=BC=CD=DE=EF=AF$
2006 II AIME-1.png
The diagonal $BF=\sqrt{AB^2+AF^2}=\sqrt{x^2+x^2}=x\sqrt{2}$ . Then the areas of the triangles AFB and CDE in total are $\frac{x^2}{2}\cdot 2$ ,
and the area of the rectangle BCEF equals $x\cdot x\sqrt{2}=x^2\sqrt{2}$
Then we have to solve the equation
Therefore, $AB$ is $\boxed{046}$

Answer:  046
Problem: In convex hexagon $ABCDEF$ , all six sides are congruent, $\angle A$ and $\angle D$ are right angles , and $\angle B, \angle C, \angle E,$ and $\angle F$ are congruent . The area of the hexagonal region is $2116(\sqrt{2}+1).$ Find $AB$

Solution: Because $\angle B$ $\angle C$ $\angle E$ , and $\angle F$ are congruent, the degree-measure of each of them is ${{720-2\cdot90}\over4}= 135$ . Lines $BF$ and $CE$ divide the hexagonal region into two right triangles and a rectangle. Let $AB=x$ . Then $BF=x\sqrt2$ . Thus \begin{align*} 2116(\sqrt2+1)&=[ABCDEF]\\ &=2\cdot {1\over2}x^2+x\cdot x\sqrt2=x^2(1+\sqrt2), \end{align*} so $x^2=2116$ , and $x=\boxed{046}$

Answer:  046
Problem: The lengths of the sides of a triangle with positive area are $\log_{10} 12$ $\log_{10} 75$ , and $\log_{10} n$ , where $n$ is a positive integer. Find the number of possible values for $n$

Solution: By the Triangle Inequality and applying the well-known logarithmic property $\log_{c} a + \log_{c} b = \log_{c} ab$ , we have that
Also,
Combining these two inequalities:
\[6.25 < n < 900\]
Thus $n$ is in the set $(6.25 , 900)$ ; the number of positive integer $n$ which satisfies this requirement is $\boxed{893}$

Answer:  893
Problem: Let $P$ be the product of the first $100$ positive odd integers . Find the largest integer $k$ such that $P$ is divisible by $3^k .$

Solution: Note that the product of the first $100$ positive odd integers can be written as $1\cdot 3\cdot 5\cdot 7\cdots 195\cdot 197\cdot 199=\frac{1\cdot 2\cdots200}{2\cdot4\cdots200} = \frac{200!}{2^{100}\cdot 100!}$
Hence, we seek the number of threes in $200!$ decreased by the number of threes in $100!.$
There are
$\left\lfloor \frac{200}{3}\right\rfloor+\left\lfloor\frac{200}{9}\right\rfloor+\left\lfloor \frac{200}{27}\right\rfloor+\left\lfloor\frac{200}{81}\right\rfloor =66+22+7+2=97$
threes in $200!$ and
$\left\lfloor \frac{100}{3}\right\rfloor+\left\lfloor\frac{100}{9}\right\rfloor+\left\lfloor \frac{100}{27}\right\rfloor+\left\lfloor\frac{100}{81}\right\rfloor=33+11+3+1=48$
threes in $100!$
Therefore, we have a total of $97-48=\boxed{049}$ threes.

Answer:  049
Problem: Let $P$ be the product of the first $100$ positive odd integers . Find the largest integer $k$ such that $P$ is divisible by $3^k .$

Solution: We are obviously searching for multiples of three set S of odd numbers 1-199. Starting with 3, every number $\equiv 2 \pmod{3}$ in set S will be divisible by 3. In other words, every number $\equiv 3 \pmod{6}$ . This is because the LCM must be divisible by 3, and 2, because the set is comprised of only odd numbers. Using some simple math, there are 33 numbers that fit this description. All you have to do is find the largest odd number that is divisible by 3 and below 200, then add 3 and divide by 6. Next, we find the number of odd digits below 200 that are divisible by 9. The same strategy works, and gives us 11. 27 gives 3, and 81 gives 1. $33 + 11 + 4 + 1 = \boxed{49}$

Answer:  49
Problem: Let $(a_1,a_2,a_3,\ldots,a_{12})$ be a permutation of $(1,2,3,\ldots,12)$ for which
An example of such a permutation is $(6,5,4,3,2,1,7,8,9,10,11,12).$ Find the number of such permutations.

Solution: Clearly, $a_6=1$ .  Now, consider selecting $5$ of the remaining $11$ values.  Sort these values in descending order, and sort the other $6$ values in ascending order.  Now, let the $5$ selected values be $a_1$ through $a_5$ , and let the remaining $6$ be $a_7$ through ${a_{12}}$ .  It is now clear that there is a bijection between the number of ways to select $5$ values from $11$ and ordered 12-tuples $(a_1,\ldots,a_{12})$ .  Thus, there will be ${11 \choose 5}=\boxed{462}$ such ordered 12-tuples.

Answer:  462
Problem: Let $(a_1,a_2,a_3,\ldots,a_{12})$ be a permutation of $(1,2,3,\ldots,12)$ for which
An example of such a permutation is $(6,5,4,3,2,1,7,8,9,10,11,12).$ Find the number of such permutations.

Solution: Clearly, $a_6=1$ , and either $a_1$ or $a_{12}$ is 12.
Case 1: $a_1 = 12$
In this case, there are 4 empty spaces between $a_1$ and $a_6$ , and 6 empty spaces between $a_6$ and $a_{12}$ $\binom{10}{4}$ is 210. This splits the remaining 10 numbers into two distinct sets that are automatically ordered. For this reason, there is no need to multiply by two to count doubles or treat as a permutation.
Case 2: $a_{12} = 12$
In this case, there are 5 empty spaces between $a_1$ and $a_6$ , and 5 empty spaces between $a_6$ and $a_{12}$ $\binom{10}{5}$ is 252. Like last time, this splits the remaining 10 numbers into two distinct sets that are automatically ordered. It is important to realize that the two sets are distinct because one side has 12 and the other does not. There is no need to multiply by two.
$210 + 252 = \boxed{462}$ ordered 12-tuples.

Answer:  462
Problem: When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $47/288$ . Given that the probability of obtaining face $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution: Without loss of generality , assume that face $F$ has a 6, so the opposite face has a 1.  Let $A(n)$ be the probability of rolling a number $n$ on one die and let $B(n)$ be the probability of rolling a number $n$ on the other die. 7 can be obtained by rolling a 2 and 5, 5 and 2, 3 and 4, or 4 and 3. Each has a probability of $\frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36}$ , totaling $4 \cdot \frac{1}{36} = \frac{1}{9}$ . Subtracting all these probabilities from $\frac{47}{288}$ leaves $\frac{15}{288}=\frac{5}{96}$ chance of getting a 1 on die $A$ and a 6 on die $B$ or a 6 on die $A$ and a 1 on die $B$
\[A(6)\cdot B(1)+B(6)\cdot A(1)=\frac{5}{96}\]
Since the two dice are identical, $B(1)=A(1)$ and $B(6)=A(6)$ so
\begin{align*}A(1)\cdot A(6)+A(1)\cdot A(6)&=\frac{5}{96}\\ A(1)\cdot A(6)&=\frac{5}{192}\end{align*}
Also, we know that $A(2)=A(3)=A(4)=A(5)=\frac{1}{6}$ and that the total probability must be $1$ , so:
\[A(1)+4 \cdot \frac{1}{6}+A(6)=\frac{6}{6} \Longrightarrow A(1)+A(6)=\frac{1}{3}\]
Combining the equations:
\begin{align*}A(6)\left(\frac{1}{3}-A(6)\right)&=\frac{5}{192}\\ 0 &= 192 \left(A(6)\right)^2 - 64 \left(A(6)\right) + 5\\ A(6)&=\frac{64\pm\sqrt{64^2 - 4 \cdot 5 \cdot 192}}{2\cdot192} =\frac{5}{24}, \frac{1}{8}\end{align*} We know that $A(6)>\frac{1}{6}$ , so it can't be $\frac{1}{8}$ . Therefore, the probability is $\frac{5}{24}$ and the answer is $5+24=\boxed{29}$

Answer:  29
Problem: When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $47/288$ . Given that the probability of obtaining face $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution: We have that the cube probabilities to land on its faces are $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{6}+x$ $\frac{1}{6}-x$ we also know that the sum could be 7 only when the faces in each of the two tosses are opposite hence the probability to get a 7 is: \[4 \cdot \left(\frac{1}{6} \right)^2+2 \left(\frac{1}{6}+x \right) \left(\frac{1}{6}-x \right)=\frac{47}{288}\] multiplying by 288 we get: \[32+16(1-6x)(6x+1)=47 \Longrightarrow 16(1-36x^2)=15\] dividing by 16 and rearranging we get: \[\frac{1}{16}=36x^2 \longrightarrow x=\frac{1}{24}\] so the probability F  which is greater than $\frac{1}{6}$ is equal $\frac{1}{6}+\frac{1}{24}=\frac{5}{24}\longrightarrow 24+5=\boxed{29}$

Answer:  29
Problem: When rolling a certain unfair six-sided die with faces numbered 1, 2, 3, 4, 5, and 6, the probability of obtaining face $F$ is greater than $1/6$ , the probability of obtaining the face opposite is less than $1/6$ , the probability of obtaining any one of the other four faces is $1/6$ , and the sum of the numbers on opposite faces is 7. When two such dice are rolled, the probability of obtaining a sum of 7 is $47/288$ . Given that the probability of obtaining face $F$ is $m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution: Let $p(a,b)$ denote the probability of obtaining $a$ on the first die and $b$ on the second. Then the probability of obtaining a sum of 7 is \[p(1,6)+p(2,5)+p(3,4)+p(4,3)+p(5,2)+p(6,1).\] Let the probability of obtaining face $F$ be $(1/6)+x$ . Then the probability of obtaining the face opposite face $F$ is $(1/6)-x$ . Therefore
\begin{align*} {{47}\over{288}}&= 4\left({1\over6}\right)^2+2\left({1\over6}+x\right) \left({1\over6}-x\right)\cr&= {4\over36}+2\left({1\over36}-x^2\right)\cr&= {1\over6}-2x^2. \end{align*}
Then $2x^2=1/288$ , and so $x=1/24$ . The probability of obtaining face $F$ is therefore $(1/6)+(1/24)=5/24$ , and $m+n=\boxed{29}$

Answer:  29
Problem: Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$

Solution: [asy] unitsize(32mm); defaultpen(linewidth(.8pt)+fontsize(10pt)); dotfactor=3;  pair B = (0, 0), C = (1, 0), D = (1, 1), A = (0, 1); pair Ep = (2 - sqrt(3), 0), F = (1, sqrt(3) - 1); pair Ap = (0, (3 - sqrt(3))/6); pair Cp = ((3 - sqrt(3))/6, 0); pair Dp = ((3 - sqrt(3))/6, (3 - sqrt(3))/6); pair[] dots = {A, B, C, D, Ep, F, Ap, Cp, Dp};  draw(A--B--C--D--cycle); draw(A--F--Ep--cycle); draw(Ap--B--Cp--Dp--cycle); dot(dots);  label("$A$", A, NW); label("$B$", B, SW); label("$C$", C, SE); label("$D$", D, NE); label("$E$", Ep, SE); label("$F$", F, E); label("$A'$", Ap, W); label("$C'$", Cp, SW); label("$D'$", Dp, E); label("$s$", Ap--B, W); label("$1$", A--D, N); [/asy] Call the vertices of the new square A', B', C', and D', in relation to the vertices of $ABCD$ , and define $s$ to be one of the sides of that square. Since the sides are parallel , by corresponding angles and AA~ we know that triangles $AA'D'$ and $D'C'E$ are similar. Thus, the sides are proportional: $\frac{AA'}{A'D'} = \frac{D'C'}{C'E} \Longrightarrow \frac{1 - s}{s} = \frac{s}{1 - s - CE}$ . Simplifying, we get that $s^2 = (1 - s)(1 - s - CE)$
$\angle EAF$ is $60$ degrees, so $\angle BAE = \frac{90 - 60}{2} = 15$ . Thus, $\cos 15 = \cos (45 - 30) = \frac{\sqrt{6} + \sqrt{2}}{4} = \frac{1}{AE}$ , so $AE = \frac{4}{\sqrt{6} + \sqrt{2}} \cdot \frac{\sqrt{6} - \sqrt{2}}{\sqrt{6} - \sqrt{2}} = \sqrt{6} - \sqrt{2}$ . Since $\triangle AEF$ is equilateral $EF = AE = \sqrt{6} - \sqrt{2}$ $\triangle CEF$ is a $45-45-90 \triangle$ , so $CE = \frac{AE}{\sqrt{2}} = \sqrt{3} - 1$ . Substituting back into the equation from the beginning, we get $s^2 = (1 - s)(2 - \sqrt{3} - s)$ , so $(3 - \sqrt{3})s = 2 - \sqrt{3}$ . Therefore, $s = \frac{2 - \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{3 - \sqrt{3}}{6}$ , and $a + b + c = 3 + 3 + 6 = \boxed{12}$

Answer:  12
Problem: Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$

Solution: Suppose $\overline{AB} = \overline{AD} = x.$ Note that $\angle EAF = 60$ since the triangle is equilateral, and by symmetry, $\angle BAE = \angle DAF = 15.$ Note that if $\overline{AD} = x$ and $\angle BAE = 15$ , then $\overline{AA'}=\frac{x}{\tan(15)}.$ Also note that \[AB = 1 = \overline{AA'} + \overline{A'B} = \frac{x}{\tan(15)} + x\] Using the fact $\tan(15) = 2-\sqrt{3}$ , this yields \[x = \frac{1}{3+\sqrt{3}} = \frac{3-\sqrt{3}}{6} \rightarrow 3 + 3 + 6 = \boxed{12}\]

Answer:  12
Problem: Square $ABCD$ has sides of length 1. Points $E$ and $F$ are on $\overline{BC}$ and $\overline{CD},$ respectively, so that $\triangle AEF$ is equilateral . A square with vertex $B$ has sides that are parallel to those of $ABCD$ and a vertex on $\overline{AE}.$ The length of a side of this smaller square is $\frac{a-\sqrt{b}}{c},$ where $a, b,$ and $c$ are positive integers and $b$ is not divisible by the square of any prime. Find $a+b+c.$

Solution: Why not solve in terms of the side $x$ only (single-variable beauty)? By similar triangles we obtain that $BE=\frac{x}{1-x}$ , therefore $CE=\frac{1-2x}{1-x}$ . Then $AE=\sqrt{2}*\frac{1-2x}{1-x}$ . Using Pythagorean Theorem on $\triangle{ABE}$ yields $\frac{x^2}{(1-x)^2} + 1 = 2 * \frac{(1-2x)^2}{(1-x)^2}$ . This means $6x^2-6x+1=0$ , and it's clear we take the smaller root: $x=\frac{3-\sqrt{3}}{6}$ . Answer: $\boxed{12}$

Answer:  12