Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: First, let define a triangle with side lengths $\sqrt{2x}$ $\sqrt{2z}$ , and $l$ , with altitude from $l$ 's equal to $\sqrt{xz}$ $l = \sqrt{2x - xz} + \sqrt{2z - xz}$ , the left side of one equation in the problem.
Let $\theta$ be angle opposite the side with length $\sqrt{2x}$ . Then the altitude has length $\sqrt{2z} \cdot \sin(\theta) = \sqrt{xz}$ and thus $\sin(\theta) = \sqrt{\frac{x}{2}}$ , so $x=2\sin^2(\theta)$ and the side length $\sqrt{2x}$ is equal to $2\sin(\theta)$
We can symmetrically apply this to the two other equations/triangles.
By law of sines, we have $\frac{2\sin(\theta)}{\sin(\theta)} = 2R$ , with $R=1$ as the circumradius, same for all 3 triangles. 
The circumcircle's central angle to a side is $2 \arcsin(l/2)$ , so the 3 triangles' $l=1, \sqrt{2}, \sqrt{3}$ ,  have angles $120^{\circ}, 90^{\circ}, 60^{\circ}$ , respectively.
This means that by half angle arcs, we see that we have in some order, $x=2\sin^2(\alpha)$ $y=2\sin^2(\beta)$ , and $z=2\sin^2(\gamma)$ (not necessarily this order, but here it does not matter due to symmetry), satisfying that $\alpha+\beta=180^{\circ}-\frac{120^{\circ}}{2}$ $\beta+\gamma=180^{\circ}-\frac{90^{\circ}}{2}$ , and $\gamma+\alpha=180^{\circ}-\frac{60^{\circ}}{2}$ . Solving, we get $\alpha=\frac{135^{\circ}}{2}$ $\beta=\frac{105^{\circ}}{2}$ , and $\gamma=\frac{165^{\circ}}{2}$
We notice that \[[(1-x)(1-y)(1-z)]^2=[\sin(2\alpha)\sin(2\beta)\sin(2\gamma)]^2=[\sin(135^{\circ})\sin(105^{\circ})\sin(165^{\circ})]^2\] \[=\left(\frac{\sqrt{2}}{2} \cdot \frac{\sqrt{6}-\sqrt{2}}{4} \cdot \frac{\sqrt{6}+\sqrt{2}}{4}\right)^2 = \left(\frac{\sqrt{2}}{8}\right)^2=\frac{1}{32} \to \boxed{033}. \blacksquare\]

Answer:  033
Problem: Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Let $1-x=a;1-y=b;1-z=c$ , rewrite those equations
$\sqrt{(1-a)(1+b)}+\sqrt{(1+a)(1-b)}=1$
$\sqrt{(1-b)(1+c)}+\sqrt{(1+b)(1-c)}=\sqrt{2}$
$\sqrt{(1-a)(1+c)}+\sqrt{(1-c)(1+a)}=\sqrt{3}$
and solve for $m/n = (abc)^2 = a^2b^2c^2$
Square both sides and simplify, to get three equations:
$2ab-1=2\sqrt{(1-a^2)(1-b^2)}$
$2bc~ ~ ~ ~  ~ ~=2\sqrt{(1-b^2)(1-c^2)}$
$2ac+1=2\sqrt{(1-c^2)(1-a^2)}$
Square both sides again, and simplify to get three equations:
$a^2+b^2-ab=\frac{3}{4}$
$b^2+c^2~ ~ ~ ~ ~ ~=1$
$a^2+c^2+ac=\frac{3}{4}$
Subtract first and third equation, getting $(b+c)(b-c)=a(b+c)$ $a=b-c$
Put it in first equation, getting $b^2-2bc+c^2+b^2-b(b-c)=b^2+c^2-bc=\frac{3}{4}$ $bc=\frac{1}{4}$
Since $a^2=b^2+c^2-2bc=\frac{1}{2}$ $m/n = a^2b^2c^2 = a^2(bc)^2 = \frac{1}{2}\left(\frac{1}{4}\right)^2=\frac{1}{32}$ and so the final answer is $\boxed{033}$

Answer:  033
Problem: Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: Denote $u = 1 - x$ $v = 1 - y$ $w = 1 - z$ .
Hence, the system of equations given in the problem can be written as \begin{align*} \sqrt{(1-u)(1+v)} + \sqrt{(1+u)(1-v)} & = 1 \hspace{1cm} (1) \\ \sqrt{(1-v)(1+w)} + \sqrt{(1+v)(1-w)} & = \sqrt{2} \hspace{1cm} (2) \\ \sqrt{(1-w)(1+u)} + \sqrt{(1+w)(1-u)} & = \sqrt{3} . \hspace{1cm} (3)  \end{align*}
Each equation above takes the following form: \[ \sqrt{(1-a)(1+b)} + \sqrt{(1+a)(1-b)} = k . \]
Now, we simplify this equation by removing radicals.
Denote $p = \sqrt{(1-a)(1+b)}$ and $q = \sqrt{(1+a)(1-b)}$
Hence, the equation above implies \[ \left\{ \begin{array}{l} p + q = k \\ p^2 = (1-a)(1+b) \\ q^2 = (1+a)(1-b) \end{array} \right.. \]
Hence, $q^2 - p^2 = (1+a)(1-b) - (1-a)(1+b) = 2 (a-b)$ .
Hence, $q - p = \frac{q^2 - p^2}{p+q} = \frac{2}{k} (a-b)$
Because $p + q = k$ and $q - p = \frac{2}{k} (a-b)$ , we get $q = \frac{a-b}{k} + \frac{k}{2}$ .
Plugging this into the equation $q^2 = (1+a)(1-b)$ and simplifying it, we get \[ a^2 + \left( k^2 - 2 \right) ab + b^2 = k^2 - \frac{k^4}{4} . \]
Therefore, the system of equations above can be simplified as \begin{align*} u^2 - uv + v^2 & = \frac{3}{4} \\ v^2 + w^2 & = 1 \\ w^2 + wu + u^2 & = \frac{3}{4}  . \end{align*}
Denote $w' = - w$ .
The system of equations above can be equivalently written as \begin{align*} u^2 - uv + v^2 & = \frac{3}{4} \hspace{1cm} (1') \\ v^2 + w'^2 & = 1 \hspace{1cm} (2') \\ w'^2 - w'u + u^2 & = \frac{3}{4} \hspace{1cm} (3') . \end{align*}
Taking $(1') - (3')$ , we get \[ (v - w') (v + w' - u) = 0 . \]
Thus, we have either $v - w' = 0$ or $v + w' - u = 0$
$\textbf{Case 1}$ $v - w' = 0$
Equation (2') implies $v = w' = \pm \frac{1}{\sqrt{2}}$
Plugging $v$ and $w'$ into Equation (2), we get contradiction. Therefore, this case is infeasible.
$\textbf{Case 2}$ $v + w' - u = 0$
Plugging this condition into (1') to substitute $u$ , we get \[ v^2 + v w' + w'^2 = \frac{3}{4} \hspace{1cm} (4) . \]
Taking $(4) - (2')$ , we get \[ v w' = - \frac{1}{4} . \hspace{1cm} (5) . \]
Taking (4) + (5), we get \[ \left( v + w' \right)^2 = \frac{1}{2} . \]
Hence, $u^2 = \left( v + w' \right)^2 = \frac{1}{2}$
Therefore, \begin{align*} \left[ (1-x)(1-y)(1-z) \right]^2 & = u^2 (vw)^2 \\ & = u^2 (vw')^2 \\ & = \frac{1}{2} \left( - \frac{1}{4} \right)^2 \\ & = \frac{1}{32} . \end{align*}
Therefore, the answer is $1 + 32 = \boxed{033}$

Answer:  033
Problem: Let $x,$ $y,$ and $z$ be positive real numbers satisfying the system of equations: \begin{align*} \sqrt{2x-xy} + \sqrt{2y-xy} &= 1 \\ \sqrt{2y-yz} + \sqrt{2z-yz} &= \sqrt2 \\ \sqrt{2z-zx} + \sqrt{2x-zx} &= \sqrt3. \end{align*} Then $\left[ (1-x)(1-y)(1-z) \right]^2$ can be written as $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution: In given equations, $0 \leq x,y,z \leq 2,$ so we define some points: \[\bar {O} = (0, 0), \bar {A} = (1, 0), \bar{M} = \left(\frac {1}{\sqrt{2}},\frac {1}{\sqrt{2}}\right),\] \[\bar {X} = \left(\sqrt {\frac {x}{2}}, \sqrt{1 – \frac{x}{2}}\right), \bar {Y'} = \left(\sqrt {\frac {y}{2}}, \sqrt{1 – \frac{y}{2}}\right),\] \[\bar {Y} = \left(\sqrt {1 – \frac{y}{2}},\sqrt{\frac {y}{2}}\right), \bar {Z} = \left(\sqrt {1 – \frac{z}{2}},\sqrt{\frac {z}{2}}\right).\] Notice, that \[\mid \vec {AO} \mid = \mid \vec {MO} \mid = \mid \vec {XO} \mid =\mid \vec {YO} \mid = \mid \vec {Y'O} \mid =\mid \vec {ZO} \mid = 1\] and each points lies in the first quadrant.
We use given equations and get some scalar products: \[(\vec {XO} \cdot \vec {YO}) = \frac {1}{2} = \cos \angle XOY \implies  \angle XOY = 60 ^\circ,\] \[(\vec {XO} \cdot \vec {ZO}) = \frac {\sqrt{3}}{2} = \cos \angle XOZ \implies  \angle XOZ = 30^\circ,\] \[(\vec {Y'O} \cdot \vec {ZO}) = \frac {1}{\sqrt{2}} = \cos \angle Y'OZ \implies  \angle Y'OZ = 45^\circ.\] So $\angle YOZ =  \angle XOY – \angle XOZ =  60 ^\circ – 30 ^\circ = 30 ^\circ,   \angle Y'OY =  \angle Y'OZ +  \angle YOZ = 45^\circ + 30 ^\circ = 75^\circ.$
Points $Y$ and $Y'$ are symmetric with respect to $OM.$
Case 1 \[\angle YOA = \frac{90^\circ – 75^\circ}{2} = 7.5^\circ, \angle ZOA = 30^\circ + 7.5^\circ = 37.5^\circ, \angle XOA = 60^\circ + 7.5^\circ = 67.5^\circ .\] \[1 – x = \left(\sqrt{1 – \frac{x}{2}} \right)^2– \left(\sqrt{\frac {x}{2}}\right)^2  = \sin^2 \angle XOA – \cos^2 \angle XOA = –\cos 2 \angle XOA = –\cos 135^\circ,\] \[1 – y = \cos 15^\circ, 1 – z = \cos 75^\circ \implies  \left[ (1–x)(1–y)(1–z) \right]^2 = \left[ \sin 45^\circ \cdot \cos 15^\circ \cdot \sin 15^\circ \right]^2 =\] \[=\left[ \frac {\sin 45^\circ \cdot \sin 30^\circ}{2} \right]^2  = \frac {1}{32} \implies \boxed{033}.\] Case 2

Answer:  033
Problem: Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.

Solution: Let $x$ be the number of people at the party before the bus arrives. We know that $x\equiv 0\pmod {12}$ , as $\frac{5}{12}$ of people at the party before the bus arrives are adults. Similarly, we know that $x + 50 \equiv 0 \pmod{25}$ , as $\frac{11}{25}$ of the people at the party are adults after the bus arrives. $x + 50 \equiv 0 \pmod{25}$ can be reduced to $x \equiv 0 \pmod{25}$ , and since we are looking for the minimum amount of people, $x$ is $300$ . That means there are $350$ people at the party after the bus arrives, and thus there are $350 \cdot \frac{11}{25} = \boxed{154}$ adults at the party.

Answer:  154
Problem: Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.

Solution: Since at the beginning, adults make up $\frac{5}{12}$ of the concert, the amount of people must be a multiple of 12.
Call the amount of people in the beginning $x$ .Then $x$ must be divisible by 12, in other words: $x$ must be a multiple of 12.
Since after 50 more people arrived, adults make up $\frac{11}{25}$ of the concert, $x+50$ is a multiple of 25. This means $x+50$ must be a multiple of 5.
Notice that if a number is divisible by 5, it must end with a 0 or 5. Since 5 is impossible (obviously, since multiples of 12 end in 2, 4, 6, 8, 0,...), $x$ must end in 0.
Notice that the multiples of 12 that end in 0 are: 60, 120, 180, etc.. By trying out, you can clearly see that $x=300$ is the minimum number of people at the concert.
So therefore, after 50 more people arrive, there are $300+50=350$ people at the concert, and the number of adults is $350*\frac{11}{25}=154$ . Therefore the answer is $\boxed{154}$

Answer:  154
Problem: Adults made up $\frac5{12}$ of the crowd of people at a concert. After a bus carrying $50$ more people arrived, adults made up $\frac{11}{25}$ of the people at the concert. Find the minimum number of adults who could have been at the concert after the bus arrived.

Solution: Let $a$ be the number of adults before the bus arrived and $x$ be the total number of people at the concert. So, $\frac{a}{x}=\frac{5}{12}$ . Solving for $x$ in terms of $a$ $x = \frac{12}{5}a$ . After the bus arrives, let's say there are an additional $y$ adults out of the 50 more people who enter the concert. From that, we get $\frac{a+y}{x+50}=\frac{11}{25}$ . Replacing $x$ with the value of $a$ , the second equation becomes $\frac{a+y}{\frac{12}{5}a+50}=\frac{11}{25}$
By cross-multiplying and simplifying, we get that $25(y-22)=\frac{7a}{5}$
Observe that we must make sure $y-22$ is positive and divisible by $7$ to have an integer value of $a$ . The smallest possible value of $y$ that satisfies this conditions is $29$ . Plugging this into the equation, $a = 125$ . The question asks for the minimum number of adults that are there after the bus arrives, which is $a+y$ . Thus, the answer is simply $125+29=\boxed{154}$

Answer:  154
Problem: Azar, Carl, Jon, and Sergey are the four players left in a singles tennis tournament. They are randomly assigned opponents in the semifinal matches, and the winners of those matches play each other in the final match to determine the winner of the tournament. When Azar plays Carl, Azar will win the match with probability $\frac23$ . When either Azar or Carl plays either Jon or Sergey, Azar or Carl will win the match with probability $\frac34$ . Assume that outcomes of different matches are independent. The probability that Carl will win the tournament is $\frac{p}{q}$ , where $p$ and $q$ are relatively prime positive integers. Find $p+q$

Solution: Let $A$ be Azar, $C$ be Carl, $J$ be Jon, and $S$ be Sergey. The $4$ circles represent the $4$ players, and the arrow is from the winner to the loser with the winning probability as the label.
2022AIMEIIP2.png
This problem can be solved by using $2$ cases.
$\textbf{Case 1:}$ $C$ 's opponent for the semifinal is $A$
The probability $C$ 's opponent is $A$ is $\frac13$ . Therefore the probability $C$ wins the semifinal in this case is $\frac13 \cdot \frac13$ . The other semifinal game is played between $J$ and $S$ , it doesn't matter who wins because $C$ has the same probability of winning either one. The probability of $C$ winning in the final is $\frac34$ , so the probability of $C$ winning the tournament in case 1 is $\frac13 \cdot \frac13 \cdot \frac34$
$\textbf{Case 2:}$ $C$ 's opponent for the semifinal is $J$ or $S$
It doesn't matter if $C$ 's opponent is $J$ or $S$ because $C$ has the same probability of winning either one. The probability $C$ 's opponent is $J$ or $S$ is $\frac23$ . Therefore the probability $C$ wins the semifinal in this case is $\frac23 \cdot \frac34$ . The other semifinal game is played between $A$ and $J$ or $S$ . In this case it matters who wins in the other semifinal game because the probability of $C$ winning $A$ and $J$ or $S$ is different.
$\textbf{Case 2.1:}$ $C$ 's opponent for the final is $A$
For this to happen, $A$ must have won $J$ or $S$ in the semifinal, the probability is $\frac34$ . Therefore, the probability that $C$ won $A$ in the final is $\frac34 \cdot \frac13$
$\textbf{Case 2.2:}$ $C$ 's opponent for the final is $J$ or $S$
For this to happen, $J$ or $S$ must have won $A$ in the semifinal, the probability is $\frac14$ . Therefore, the probability that $C$ won $J$ or $S$ in the final is $\frac14 \cdot \frac34$
In Case 2 the probability of $C$ winning the tournament is $\frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34)$
Adding case 1 and case 2 together we get $\frac13 \cdot \frac13 \cdot \frac34 + \frac23 \cdot \frac34 \cdot (\frac34 \cdot \frac13 + \frac14 \cdot \frac34) = \frac{29}{96},$ so the answer is $29 + 96 = \boxed{125}$

Answer:  125
Problem: A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Although I can't draw the exact picture of this problem, but it is quite easy to imagine that four vertices of the base of this pyramid is on a circle (Radius $\frac{6}{\sqrt{2}} = 3\sqrt{2}$ ). Since all five vertices are on the sphere, the distances of the spherical center and the vertices are the same: $l$ . Because of the symmetrical property of the pyramid,
we can imagine that the line of the apex and the (sphere's) center will intersect the square at the (base's) center.
Since the volume is $54 = \frac{1}{3} \cdot S \cdot h = \frac{1}{3} \cdot 6^2 \cdot h$ , where $h=\frac{9}{2}$ is the height of this pyramid, we have: $l^2=\left(\frac{9}{2}-l\right)^2+\left(3\sqrt{2}\right)^2$ according to the Pythagorean theorem.
Solve this equation will give us $l = \frac{17}{4},$ therefore $m+n=\boxed{021}.$

Answer:  021
Problem: A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: To start, we find the height of the pyramid. By the volume of a pyramid formula, we have \[\frac13 \cdot 6^2 \cdot h=54 \implies h=\frac92.\] Next, let us find the length of the non-base sides of the pyramid. By the Pythagorean Theorem, noting that the distance from one vertex of the base to the center of the base is $\frac12 \cdot 6\sqrt2=3\sqrt2$ , we have \[x=\sqrt{\left(\frac92\right)^2+(3\sqrt2)^2}=\sqrt{\frac{153}4}=\frac{3\sqrt{17}}2.\] Taking the cross section of the pyramid and transforming the problem into $2$ -d, it suffices to find the radius of the circumcircle of a triangle of side lengths $\frac{3\sqrt{17}}2$ $\frac{3\sqrt{17}}2$ $6\sqrt2$ . This turns out to be easy by the formula $R=\frac{abc}{4A}$ , and through computing this value (the work has been left out) we find that $R=\frac{17}4$ , so our answer is $\boxed{021}$

Answer:  021
Problem: A right square pyramid with volume $54$ has a base with side length $6.$ The five vertices of the pyramid all lie on a sphere with radius $\frac mn$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: By the volume of a pyramid formula, we have that the height of the pyramid is $\frac{9}{2}$ . Since the base is a square with side length 6, the simplest way to place it in the coordinate plane is to put the center of the square at the origin and let the base be on the $xy$ plane. Then, the vertices of the base would be $(3,3,0), (3,-3,0), (-3,3,0), (-3,-3,0)$ in some order. Also, let the vertex be $(0,0,\frac{9}{2})$ . Recall that the formula for a sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ where the center is $(a,b,c)$ and the radius is $r$ . Symmetry gives that $a=b=0$ . Plug in $(3,3,0)$ and $(0,0,\frac{9}{2})$ and you get the system of equation
$18+c^2=r^2$
$(\frac{9}{2}-c)^2=r^2$
Solving gives $c=1/4$ and $r=17/4$ , so our answer is $17+4=\boxed{021}$ .~ Ddk001

Answer:  021
Problem: There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Define $a$ to be $\log_{20x} (22x) = \log_{2x} (202x)$ , what we are looking for. Then, by the definition of the logarithm, \[\begin{cases}  (20x)^{a} &= 22x \\ (2x)^{a} &= 202x.  \end{cases}\] Dividing the first equation by the second equation gives us $10^a = \frac{11}{101}$ , so by the definition of logs, $a = \log_{10} \frac{11}{101}$ . This is what the problem asked for, so the fraction $\frac{11}{101}$ gives us $m+n = \boxed{112}$

Answer:  112
Problem: There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: We could assume a variable $v$ which equals to both $\log_{20x} (22x)$ and $\log_{2x} (202x)$
So that $(20x)^v=22x \textcircled{1}$ and $(2x)^v=202x \textcircled{2}$
Express $\textcircled{1}$ as: $(20x)^v=(2x \cdot 10)^v=(2x)^v \cdot \left(10^v\right)=22x \textcircled{3}$
Substitute $\textcircled{{2}}$ to $\textcircled{3}$ $202x \cdot (10^v)=22x$
Thus, $v=\log_{10} \left(\frac{22x}{202x}\right)= \log_{10} \left(\frac{11}{101}\right)$ , where $m=11$ and $n=101$
Therefore, $m+n = \boxed{112}$

Answer:  112
Problem: There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: We have \begin{align*} \log_{20x} (22x) & = \frac{\log_k 22x}{\log_k 20x} \\ & = \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} . \end{align*}
We have \begin{align*} \log_{2x} (202x) & = \frac{\log_k 202x}{\log_k 2x} \\ & = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . \end{align*}
Because $\log_{20x} (22x)=\log_{2x} (202x)$ , we get \[ \frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} . \]
We denote this common value as $\lambda$
By solving the equality $\frac{\log_k x + \log_k 22}{\log_k x + \log_k 20} = \lambda$ , we get $\log_k x = \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1}$
By solving the equality $\frac{\log_k x + \log_k 202 }{\log_k x + \log_k 2} = \lambda$ , we get $\log_k x = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1}$
By equating these two equations, we get \[ \frac{\log_k 22 - \lambda \log_k 20}{\lambda - 1} = \frac{\log_k 202 - \lambda \log_k 2}{\lambda - 1} . \]
Therefore, \begin{align*} \log_{20x} (22x) & = \lambda \\ & = \frac{\log_k 22 - \log_k 202}{\log_k 20 - \log_k 2} \\ & = \frac{\log_k \frac{11}{101}}{\log_k 10} \\ & = \log_{10} \frac{11}{101} . \end{align*}
Therefore, the answer is $11 + 101 = \boxed{112}$

Answer:  112
Problem: There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Let $a$ be the exponent such that $(20x)^a = 22x$ and $(2x)^a = 202x$ . Dividing, we get \begin{align*} \dfrac{(20x)^a}{(2x)^a} &= \dfrac{22x}{202x}. \\ \left(\dfrac{20x}{2x}\right)^a &= \dfrac{22x}{202x}. \\ 10^a &= \dfrac{11}{101}. \\ \end{align*} Thus, we see that $\log_{10} \left(\dfrac{11}{101}\right) = a = \log_{20x} 22x$ , so the answer is $11 + 101 = \boxed{112}$

Answer:  112
Problem: There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: By the change of base rule, we have $\frac{\log 22x}{\log 20x}=\frac{\log 202x}{\log 2x}$ , or $\frac{\log 22 +\log x}{\log 20 +\log x}=\frac{\log 202 +\log x}{\log 2 +\log x}=k$ . We also know that if $a/b=c/d$ , then this also equals $\frac{a-c}{b-d}$ . We use this identity and find that $k=\frac{\log 202 -\log 22}{\log 2 -\log 20}=-\log\frac{202}{22}=\log\frac{11}{101}$ . The requested sum is $11+101=\boxed{112}.$

Answer:  112
Problem: There is a positive real number $x$ not equal to either $\tfrac{1}{20}$ or $\tfrac{1}{2}$ such that \[\log_{20x} (22x)=\log_{2x} (202x).\] The value $\log_{20x} (22x)$ can be written as $\log_{10} (\tfrac{m}{n})$ , where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: By change of base formula, \[\frac{\log_{2x} 22x}{\log_{2x} 20x} = \frac{{\log_{2x} 11} + 1}{{\log_{2x} 10} + 1} = {\log_{2x} 101} + 1\] \[\log_{2x} 11 + 1 = (\log_{2x} 10)(\log_{2x} 101) + \log{2x} 1010 + 1\] \[\frac{\log_{2x} \frac{11}{1010}}{\log_{2x} 10} = \log_{2x} 101\] \[\log_{10} {\frac{11}{1010}} = \log_{2x} 101\] \[\log_{10} {\frac{11}{1010}} + 1 = \log_{2x} 101 + 1 = \log_{2x} 202x = \log_{20x} {22x}\] Thus, \[\log_{20x} 22x = \log_{10} \left( \frac{11}{1010} \times 10 \right) = \log_{10} \frac{11}{101}\] The requested answer is $11 + 101 = \boxed{112}$

Answer:  112
Problem: Twenty distinct points are marked on a circle and labeled $1$ through $20$ in clockwise order. A line segment is drawn between every pair of points whose labels differ by a prime number. Find the number of triangles formed whose vertices are among the original $20$ points.

Solution: Let $a$ $b$ , and $c$ be the vertex of a triangle that satisfies this problem, where $a > b > c$ \[a - b = p_1\] \[b - c = p_2\] \[a - c = p_3\]
$p_3 = a - c = a - b + b - c = p_1 + p_2$ . Because $p_3$ is the sum of two primes, $p_1$ and $p_2$ $p_1$ or $p_2$ must be $2$ . Let $p_1 = 2$ , then $p_3 = p_2 + 2$ . There are only $8$ primes less than $20$ $2, 3, 5, 7, 11, 13, 17, 19$ . Only $3, 5, 11, 17$ plus $2$ equals another prime. $p_2 \in \{ 3, 5, 11, 17 \}$
Once $a$ is determined, $a = b+2$ and $b = c + p_2$ . There are $18$ values of $a$ where $a+2 \le 20$ , and $4$ values of $p_2$ . Therefore the answer is $18 \cdot 4 = \boxed{072}$

Answer:  072