Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Triangle $AB_0C_0$ has side lengths $AB_0 = 12$ $B_0C_0 = 17$ , and $C_0A = 25$ . For each positive integer $n$ , points $B_n$ and $C_n$ are located on $\overline{AB_{n-1}}$ and $\overline{AC_{n-1}}$ , respectively, creating three similar triangles $\triangle AB_nC_n \sim \triangle B_{n-1}C_nC_{n-1} \sim \triangle AB_{n-1}C_{n-1}$ . The area of the union of all triangles $B_{n-1}C_nB_n$ for $n\geq1$ can be expressed as $\tfrac pq$ , where $p$ and $q$ are relatively prime positive integers. Find $q$

Solution: Let $k$ be the coefficient of the similarity of triangles \[\triangle B_0 C_1 C_0 \sim \triangle AB_0 C_0 \implies k = \frac {B_0 C_0}{AC_0} = \frac {17}{25}.\] Then area $\frac {[B_0 C_1 C_0]}{[AB_0 C_0 ]} = k^2 \implies \frac {[AB_0 C_1]}{[AB_0 C_0]} = 1 – k^2.$
The height of triangles $\triangle B_0C_1A$ and $\triangle AB_0C_0$ from $B_0$ is the same $\implies \frac {AC_1}{AC_0} = 1 – k^2.$
The coefficient of the similarity of triangles $\triangle AB_1C_1 \sim \triangle AB_0C_0$ is $\frac {AC_1}{AC_0} = 1 – k^2 \implies \frac {[B_1C_1C_2 ]}{[AB_0C_0 ]} = k^2 (1 – k^2)^2.$
Analogically the coefficient of the similarity of triangles $\triangle AB_2C_2 \sim \triangle AB_0C_0$ is $(1 – k^2)^2 \implies \frac {[B_2C_2C_3]}{[AB_0C_0 ]} = k^2 (1 – k^2)^4$ and so on.
The yellow area $[Y]$ is $\frac {[Y]}{[AB_0C_0 ]} = k^2 + k^2 (1 – k^2)^2 + k^2 (1 – k^2)^4 +.. = \frac {k^2}{1 – (1 – k^2)^2} = \frac{1}{2 – k^2}.$
The required area is $[AB_0C_0 ] – [Y] = [AB_0C_0 ] \cdot (1 –  \frac{1}{2 – k^2}) = [AB_0C_0 ] \cdot \frac  {1 –  k^2}{2 – k^2} = [AB_0C_0 ] \cdot \frac {25^2 – 17^2} {2 \cdot 25^2 – 17^2} =  [AB_0C_0 ] \cdot  \frac {336}{961}.$
The number $961$ is prime, $[AB_0C_0]$ is integer but not $961,$ therefore the answer is $\boxed{961}$

Answer:  961
Problem: For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Noticing the $\sin$ and $\cos$ in both $P$ and $Q,$ we think of the angle addition identities:
\[\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b - \sin a \sin b\]
With this in mind, we multiply $P$ by $\sin \theta$ and $Q$ by $\cos \theta$ to try and use some angle addition identities. Indeed, we get \begin{align*} P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\ &= \cos \theta - \dfrac{1}{4} \cos \theta + \dfrac{1}{8} \sin{2 \theta} + \dfrac{1}{16} \cos{3 \theta} + \cdots \\ &= \cos \theta - \dfrac{1}{2}P \end{align*} after adding term-by-term. Similar term-by-term adding yields \[P \cos \theta + Q \sin \theta = -2(Q - 1).\] This is a system of equations; rearrange and rewrite to get \[P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta\] and \[P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.\] Subtract the two and rearrange to get \[\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.\] Then, square both sides and use Pythagorean Identity to get a quadratic in $\sin \theta.$ Factor that quadratic and solve for $\sin \theta = -17/19, 1/3.$ Since we're given $\pi\leq\theta<2\pi,$ $\sin\theta$ is nonpositive. We therefore use the negative solution, and our desired answer is $17 + 19 = \boxed{036}.$

Answer:  036
Problem: For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Use sum to product formulas to rewrite $P$ and $Q$
\[P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ...\]
Therefore, \[P \sin \theta - Q \cos \theta = -2P\]
Using \[\frac{P}{Q} =  \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P\]
Plug in to the previous equation and cancel out the "P" terms to get: \[\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2\]
Then use the pythagorean identity to solve for $\sin\theta$ \[\sin\theta = -\frac{17}{19} \implies \boxed{036}\]

Answer:  036
Problem: For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Note that \[e^{i\theta}=\cos(\theta)+i\sin(\theta)\]
Thus, the following identities follow immediately: \[ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)\] \[i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)\] \[i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)\]
Consider, now, the sum $Q+iP$ . It follows fairly immediately that:
\[Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}\] \[Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\]
This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:
\[Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)\] \[Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}\]
Comparing real and imaginary parts, we find: \[\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}\]
Squaring this equation and letting $\sin^2(\theta)=x$
$\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}$
Clearing denominators and solving for $x$ gives sine as $x=-\frac{17}{19}$
$017+019=\boxed{036}$

Answer:  036
Problem: For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: A bit similar to Solution 3. We use $\phi = \theta+90^\circ$ because the progression cycles in $P\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)$ . So we could rewrite that as $P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)$
Similarly, $Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, \cos 3\phi\dots)$
Setting complex $z=q_1+p_1i$ , we get $z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)$
$(Q,P)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}=\frac{1}{1-\frac{1}{2}\cos\phi-\frac{i}{2}\sin\phi}=\frac{1-0.5\cos\phi+0.5i\sin\phi}{\dots}$
The important part is the ratio of the imaginary part $i$ to the real part. To cancel out the imaginary part from the denominator, we must add $0.5i\sin\phi$ to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find $\frac{P}{Q}=\tan\text{arg}(\Sigma)$ a PROPORTION of values. So denominators would cancel out.
$\frac{P}{Q}=\frac{\text{Re}\Sigma}{\text{Im}\Sigma}=\frac{0.5\sin\phi}{1-0.5\cos\phi}=\frac{\sin\phi}{2-\cos\phi}=\frac{2\sqrt{2}}{7}$
Setting $\sin\theta=y$ , we obtain \[\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}\] \[7\sqrt{1-y^2}=2\sqrt{2}(2+y)\] \[49-49y^2=8y^2+32y+32\] \[57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}\] \[y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}\]
Since $y<0$ because $\pi<\theta<2\pi$ $y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}$ . Adding up, $17+19=\boxed{036}$

Answer:  036
Problem: For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: We notice $\sin\theta=-\frac{i}{2}(e^{i\theta}-e^{-i\theta})$ and $\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$
We observe that both $P$ and $Q$ can be split into $2$ parts, namely the terms which contain the $\cos$ and the terms which contain the $\sin .$
The $\cos$ part of $P$ can be expressed as: \begin{align*}\frac12\cos\theta-\frac18\cos3\theta+\cdots&=\frac14\left(e^{i\theta}\left(1-\frac{e^{i2\theta}}{4}+\cdots\right)+e^{-i\theta}\left(1-\frac{e^{-i2\theta}}{4}+\cdots\right)\right) \\ &= \frac{1}{4}\left(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}}\right)\\  &= \frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}.\end{align*}
Repeating the above process, we find that the $\sin$ part of $P$ is \[\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}},\] the $\cos$ part of $Q$ is \[\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}},\] and finally, the $\sin$ part of $Q$ is \[\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}.\]
Converting back to trigonometric form, we have \begin{align*}\frac{2\sqrt{2}}{7}&=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}\\ &=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}.\end{align*} Using the $\sin$ double identity and simplifying, we have \[\frac{2\sqrt2}{7}=\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}.\] Factoring the denominator, we have \[10-4\sin^2{\theta}-3\sin{\theta}=(5-4\sin\theta)(2+\sin\theta).\] Simplifying \begin{align*}\frac{2\sqrt2}{7}&= \frac{\cos{\theta}(5-4\sin{\theta})}{(5-4\sin\theta)(2+\sin\theta)}\\ &=\frac{\cos\theta}{2+\sin\theta}.\end{align*} We set $\sin\theta$ as $x$ , and by the Pythagorean Identity, we have $57x^2+32x-17=0$ . This factors into $(19x+17)(3x-1)=0$ , which yields the 2 solutions $x=-\frac{17}{19}, x=\frac{1}{3}$ . As $\pi\leq\theta<2\pi$ , the latter root is erroneous, and we are left with $\sin\theta=-\frac{17}{19}$ . Thus, our final answer is $17+19=\boxed{036}$

Answer:  036
Problem: For $\pi \le \theta < 2\pi$ , let
\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}
and
\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}
so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$ . Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$

Solution: Follow solution 3, up to the point of using the geometric series formula \[Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}\]
Moving everything to the other side, and considering only the imaginary part, we get \[Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0\]
We can then write $P = 2 \sqrt{2} k$ , and $Q = 7k$ , ( $k \neq 0$ ). Thus, we can substitute and divide out by k. \[2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0\] \[2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\sqrt{1-\sin^{2}\theta}=\ 0\] \[2\sqrt{2}+\sqrt{2}\sin\theta\ =\frac{7}{2}\left(\sqrt{1-\sin^{2}\theta}\right)\] \[8+8\sin\theta+2\sin^{2}\theta=\frac{49}{4}-\frac{49}{7}\sin^{2}\theta\] \[\frac{57}{4}\sin^{2}\theta+8\sin\theta-\frac{17}{4} = 0\] \[57\sin^{2}\theta+32\sin\theta-17 = 0\] \[\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0\]
Since $\pi \le \theta < 2\pi$ , we get $\sin \theta < 0$ , and thus, $\sin\theta = \frac{-17}{19} \implies \boxed{036}$

Answer:  036
Problem: Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions 
(a) $0\le A<B<C\le99$ , 
(b) there exist integers $a$ $b$ , and $c$ , and prime $p$ where $0\le b<a<c<p$ , 
(c) $p$ divides $A-a$ $B-b$ , and $C-c$ , and 
(d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$

Solution: From condition (d), we have $(A,B,C)=(B-D,B,B+D)$ and $(b,a,c)=(a-d,a,a+d)$ . Condition $\text{(c)}$ states that $p\mid B-D-a$ $p | B-a+d$ , and $p\mid B+D-a-d$ . We subtract the first two to get $p\mid-d-D$ , and we do the same for the last two to get $p\mid 2d-D$ . We subtract these two to get $p\mid 3d$ . So $p\mid 3$ or $p\mid d$ . The second case is clearly impossible, because that would make $c=a+d>p$ , violating condition $\text{(b)}$ . So we have $p\mid 3$ , meaning $p=3$ . Condition $\text{(b)}$ implies that $(b,a,c)=(0,1,2)$ or $(a,b,c)\in (1,0,2)\rightarrow (-2,0,2)\text{ }(D\equiv 2\text{ mod 3})$ . Now we return to condition $\text{(c)}$ , which now implies that $(A,B,C)\equiv(-2,0,2)\pmod{3}$ . Now, we set $B=3k$ for increasing positive integer values of $k$ $B=0$ yields no solutions. $B=3$ gives $(A,B,C)=(1,3,5)$ , giving us $1$ solution. If $B=6$ , we get $2$ solutions, $(4,6,8)$ and $(1,6,11)$ . Proceeding in the manner, we see that if $B=48$ , we get 16 solutions. However, $B=51$ still gives $16$ solutions because $C_\text{max}=2B-1=101>100$ . Likewise, $B=54$ gives $15$ solutions. This continues until $B=96$ gives one solution. $B=99$ gives no solution. Thus, $N=1+2+\cdots+16+16+15+\cdots+1=2\cdot\frac{16(17)}{2}=16\cdot 17=\boxed{272}$

Answer:  272
Problem: Let $N$ be the number of ordered triples $(A,B,C)$ of integers satisfying the conditions 
(a) $0\le A<B<C\le99$ , 
(b) there exist integers $a$ $b$ , and $c$ , and prime $p$ where $0\le b<a<c<p$ , 
(c) $p$ divides $A-a$ $B-b$ , and $C-c$ , and 
(d) each ordered triple $(A,B,C)$ and each ordered triple $(b,a,c)$ form arithmetic sequences. Find $N$

Solution: Let $(A, B, C)$ $(B-x, B, B+x)$ and $(b, a, c) = (a-y, a, a+y)$ . Now the 3 differences would be \begin{align} \label{1} &A-a = B-x-a \\ \label{2} &B - b = B-a+y \\ \label{3} &C - c = B+x-a-y  \end{align}
Adding equations $(1)$ and $(3)$ would give $2B - 2a - y$ . Then doubling equation $(2)$ would give $2B - 2a + 2y$ . The difference between them would be $3y$ . Since $p|\{(1), (2), (3)\}$ , then $p|3y$ . Since $p$ is prime, $p|3$ or $p|y$ . However, since $p > y$ , we must have $p|3$ , which means $p=3$
If $p=3$ , the only possible values of $(b, a, c)$ are $(0, 1, 2)$ . Plugging this into our differences, we get \begin{align*}  &A-a = B-x-1 \hspace{4cm}(4)\\  &B - b = B \hspace{5.35cm}(5)\\  &C - c = B+x-2 \hspace{4cm}(6) \end{align*} The difference between $(4)$ and $(5)$ is $x+1$ , which should be divisible by 3. So $x \equiv 2 \mod 3$ . Also note that since $3|(5)$ $3|B$ . Now we can try different values of $x$ and $B$
When $x=2$ $B=3, 6, ..., 96 \Rightarrow 17$ triples.
When $x=5$ $B=6, 9, ..., 93\Rightarrow 15$ triples..
... and so on until
When $x=44$ $B=45\Rightarrow 1$ triple.
So the answer is $17 + 15 + \cdots + 1 = \boxed{272}$

Answer:  272
Problem: Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$ , where $\text{A}$ $\text{B}$ , and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$

Solution: There are $24 \cdot 60=1440$ normal minutes in a day , and $10 \cdot 100=1000$ metric minutes in a day. The ratio of normal to metric minutes in a day is $\frac{1440}{1000}$ , which simplifies to $\frac{36}{25}$ . This means that every time 36 normal minutes pass, 25 metric minutes pass. From midnight to $\text{6:36}$ AM, $6 \cdot 60+36=396$ normal minutes pass. This can be viewed as $\frac{396}{36}=11$ cycles of 36 normal minutes, so 11 cycles of 25 metric minutes pass. Adding $25 \cdot 11=275$ to $\text{0:00}$ gives $\text{2:75}$ , so the answer is $\boxed{275}$

Answer:  275
Problem: Suppose that the measurement of time during the day is converted to the metric system so that each day has $10$ metric hours, and each metric hour has $100$ metric minutes. Digital clocks would then be produced that would read $\text{9:99}$ just before midnight, $\text{0:00}$ at midnight, $\text{1:25}$ at the former $\text{3:00}$ AM, and $\text{7:50}$ at the former $\text{6:00}$ PM. After the conversion, a person who wanted to wake up at the equivalent of the former $\text{6:36}$ AM would set his new digital alarm clock for $\text{A:BC}$ , where $\text{A}$ $\text{B}$ , and $\text{C}$ are digits. Find $100\text{A}+10\text{B}+\text{C}$

Solution: First we want to find out what fraction of a day has passed at 6:36 AM. One hour is $\frac{1}{24}$ of a day, and 36 minutes is $\frac{36}{60}=\frac{3}{5}$ of an hour, so at 6:36 AM, $6 \cdot \frac{1}{24} + \frac{1}{24} \cdot \frac{3}{5}=\frac{1}{4}+\frac{1}{40}=\frac{11}{40}$ of a day has passed. Now the metric timing equivalent of $\frac{11}{40}$ of a day is $\frac{11}{40}\cdot 1000=275$ metric minutes, which is equivalent to 2 metric hours and 75 metric minutes, so our answer is $\boxed{275}$ - mathleticguyyy

Answer:  275
Problem: Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$

Solution: To simplify, we write this logarithmic expression as an exponential one. Just looking at  the first log, it has a base of 2 and an argument of the expression in parenthesis. Therefore, we can make 2 the base, 0 the exponent, and the argument the result. That means $\log_{2^a}(\log_{2^b}(2^{1000}))=1$ (because $2^0=1$ ). Doing this again, we get $\log_{2^b}(2^{1000})=2^a$ . Doing the process one more time, we finally eliminate all of the logs, getting ${(2^{b})}^{(2^a)}=2^{1000}$ . Using the property that ${(a^x)^{y}}=a^{xy}$ , we simplify to $2^{b\cdot2^{a}}=2^{1000}$ . Eliminating equal bases leaves $b\cdot2^a=1000$ . The largest $a$ such that $2^a$ divides $1000$ is $3$ , so we only need to check $1$ $2$ , and $3$ . When $a=1$ $b=500$ ; when $a=2$ $b=250$ ; when $a=3$ $b=125$ . Summing all the $a$ 's and $b$ 's gives the answer of $\boxed{881}$

Answer:  881
Problem: Positive integers $a$ and $b$ satisfy the condition \[\log_2(\log_{2^a}(\log_{2^b}(2^{1000}))) = 0.\] Find the sum of all possible values of $a+b$

Solution: We proceed as in Solution 1, raising $2$ to both sides to achieve $\log_{2^a}(\log_{2^b}(2^{1000})) = 1.$ We raise $2^a$ to both sides to get $\log_{2^b}(2^{1000})=2^a$ , then simplify to get $\dfrac{1000}b=2^a$
At this point, we want both $a$ and $b$ to be integers. Thus, $2^a$ can only be a power of $2$ . To help us see the next step, we factorize $1000$ $\dfrac{2^35^3}b=2^a.$ It should be clear that $a$ must be from $1$ to $3$ ; when $a=1$ $b=500$ ; when $a=2$ $b=250$ ; and finally, when $a=3$ $b=125.$ We sum all the pairs to get $\boxed{881}.$

Answer:  881
Problem: A large candle is $119$ centimeters tall.  It is designed to burn down more quickly when it is first lit and more slowly as it approaches its bottom.  Specifically, the candle takes $10$ seconds to burn down the first centimeter from the top, $20$ seconds to burn down the second centimeter, and $10k$ seconds to burn down the $k$ -th centimeter.  Suppose it takes $T$ seconds for the candle to burn down completely.  Then $\tfrac{T}{2}$ seconds after it is lit, the candle's height in centimeters will be $h$ .  Find $10h$

Solution: We find that $T=10(1+2+\cdots +119)$ . From Gauss's formula, we find that the value of $T$ is $10(7140)=71400$ . The value of $\frac{T}{2}$ is therefore $35700$ . We find that $35700$ is $10(3570)=10\cdot \frac{k(k+1)}{2}$ , so $3570=\frac{k(k+1)}{2}$ . As a result, $7140=k(k+1)$ , which leads to $0=k^2+k-7140$ . We notice that $k=84$ , so the answer is $10(119-84)=\boxed{350}$

Answer:  350
Problem: In the Cartesian plane let $A = (1,0)$ and $B = \left( 2, 2\sqrt{3} \right)$ .  Equilateral triangle $ABC$ is constructed so that $C$ lies in the first quadrant.  Let $P=(x,y)$ be the center of $\triangle ABC$ .  Then $x \cdot y$ can be written as $\tfrac{p\sqrt{q}}{r}$ , where $p$ and $r$ are relatively prime positive integers and $q$ is an integer that is not divisible by the square of any prime.  Find $p+q+r$

Solution: The distance from point $A$ to point $B$ is $\sqrt{13}$ . The vector that starts at point A and ends at point B is given by $B - A = (1, 2\sqrt{3})$ . Since the center of an equilateral triangle, $P$ , is also the intersection of the perpendicular bisectors of the sides of the triangle, we need first find the equation for the perpendicular bisector to $\overline{AB}$ . The line perpendicular to $\overline{AB}$ through the midpoint, $M =  \left(\dfrac{3}{2},\sqrt{3}\right)$ $\overline{AB}$ can be parameterized by $\left(\dfrac{2\sqrt{3}}{\sqrt{13}}, \dfrac{-1}{\sqrt{13}}\right) t + \left(\dfrac{3}{2},\sqrt{3}\right)$ . At this point, it is useful to note that $\Delta BMP$ is a 30-60-90 triangle with $\overline{MB}$ measuring $\dfrac{\sqrt{13}}{2}$ . This yields the length of $\overline{MP}$ to be $\dfrac{\sqrt{13}}{2\sqrt{3}}$ . Therefore, $P =\left(\dfrac{2\sqrt{3}}{\sqrt{13}},\dfrac{-1}{\sqrt{13}}\right)\left(\dfrac{\sqrt{13}}{2\sqrt{3}}\right) + \left(\dfrac{3}{2},\sqrt{3}\right) = \left(\dfrac{5}{2}, \dfrac{5}{2\sqrt{3}}\right)$ . Therefore $xy = \dfrac{25\sqrt{3}}{12}$ yielding an answer of $p + q + r  = 25 + 2 + 12 = \boxed{040}$

Answer:  040