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The circle centered at \((2,-1)\) and with radius \(4\) intersects the circle centered at \((2,5)\) and with radius \(\sqrt{10}\) at two points \(A\) and \(B\).
Find \(\left(\overline{AB}\right)^2\).

\(\begin{array}{|lrcll|} \hline (1) & (x-2)^2 +(y-5)^2 &=& 10 \\ & (x-2)^2 +(y-(-1))^2 &=& 4^2 \\ (2) & (x-2)^2 +(y+1)^2 &=& 16 \\ \hline (2)-(1): & (x-2)^2 +(y+1)^2 -((x-2)^2 +(y-5)^2) &=& 16-10 \\ & (x-2)^2 +(y+1)^2 -(x-2)^2 -(y-5)^2 &=& 6 \\ & (y+1)^2 -(y-5)^2 &=& 6 \\ & y^2+2y+1-(y^2-10y+25) &=& 6 \\ & y^2+2y+1-y^2+10y-25 &=& 6 \\ & 12y+1-25 &=& 6 \\ & 12y-24 &=& 6 \quad &| \quad : 6 \\ & 2y-4 &=& 1 \\ & 2y &=& 1+4 \\ & 2y &=& 5 \\ & \mathbf{ y } &=& \mathbf{\dfrac{5}{2}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1) & (x-2)^2 +(y-5)^2 &=& 10 \quad | \quad y=\dfrac{5}{2} \\ & (x-2)^2 +\left(\dfrac{5}{2}-5\right)^2 &=& 10 \\ & (x-2)^2 +\left(-\dfrac{5}{2}\right)^2 &=& 10 \\ & (x-2)^2 + \dfrac{25}{4} &=& 10 \\ & (x-2)^2 &=& 10 - \dfrac{25}{4} \\ & (x-2)^2 &=& \dfrac{15}{4} \\ & x-2 &=& \pm\dfrac{\sqrt{15}}{2} \\ & x &=& 2 \pm \dfrac{\sqrt{15}}{2} \\ \\ & \mathbf{x_1} &=& 2 + \dfrac{\sqrt{15}}{2} \\ & \mathbf{x_2} &=& 2 - \dfrac{\sqrt{15}}{2} \\\\ & \mathbf{ \left(\overline{AB}\right)^2} &=& (x_1-x_2)^2 \\ & &=& \left(2 + \dfrac{\sqrt{15}}{2}- \left(2 - \dfrac{\sqrt{15}}{2}\right) \right)^2 \\ & &=& \left(2 + \dfrac{\sqrt{15}}{2}-2 + \dfrac{\sqrt{15}}{2} \right)^2 \\ & &=& \left(\dfrac{\sqrt{15}}{2}+ \dfrac{\sqrt{15}}{2} \right)^2 \\ & &=& \left(2\dfrac{\sqrt{15}}{2} \right)^2 \\ & &=& \left(\sqrt{15}\right)^2 \\ & &=& \mathbf{15} \\ \hline \end{array}\)

Points\( A\) and \(B\) are on parabola \(y=3x^2-5x-3\), and the origin is the midpoint of \(\overline{AB}\).
Find the square of the length of \(\overline{AB}\).

\(\text{ Let $A(x_A,\ y_A)$ and $B(x_B,\ y_B)$ }\)

\(\begin{array}{|lrcll|} \hline (1) & y_A&=&3x_A^2-5x_A-3 \\ (2) & y_B&=&3x_B^2-5x_B-3 \\ (3) & x_B&=&-x_A\\ (4) & y_B&=&-y_A\\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (1) & y_A&=&3x_A^2-5x_A-3 \\ \hline (2) & y_B&=&3x_B^2-5x_B-3 \quad &| \quad y_B=-y_A,\ x_B=-x_A \\ & -y_A&=&3(-x_A)^2-5(-x_A)-3 \\ (2)^* & -y_A&=&3x_A^2+5x_A-3 \\ \hline (1)+(2)^*: & y_A-y_A&=&3x_A^2-5x_A-3+3x_A^2+5x_A-3\\ & 0&=&6x_A^2-6 \quad &| \quad : 6 \\ & 0&=&x_A^2-1 \\ & x_A^2 &=& 1 \\ & \mathbf{x_A} &=& \mathbf{1} \\ & \mathbf{x_B} &=& -x_A = \mathbf{-1} \\ \hline & y_A &=& 3x_A^2-5x_A-3 \\ & &=& 3(1)^2-5(1)-3 \\ & &=& 3-5-3 \\ & \mathbf{y_A} &=& \mathbf{-5} \\ & \mathbf{y_B} &=& -y_A = -(-5) = \mathbf{5} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline A(1,\ -5),\ B(-1,\ 5) \\\\ \text{The square of the length of } \overline{AB} : \\ \left(1-(-1)\right)^2 + \left(-5-(5)\right)^2\\ =2^2+(-10)^2 \\ =4+100 \\ \mathbf{= 104} \\ \hline \end{array} \)

If a, b, and c are positive integers less than 13 such that
\(\large{\begin{align*} 2ab+bc+ca&\equiv 0\pmod{13}\\ ab+2bc+ca&\equiv 6abc\pmod{13}\\ ab+bc+2ca&\equiv 8abc\pmod {13} \end{align*}}\)
then determine the remainder when a+b+c is divided by 13.

\(\begin{array}{|lrcll|} \hline & 2ab+bc+ca &\equiv& 0\pmod{13} \quad |\quad : (abc)\\ (1) & \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b} &\equiv& 0 \pmod{13} \\ \hline & ab +2bc+ ca &\equiv& 6abc\pmod{13} \quad |\quad : (abc)\\ (2) & \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b} &\equiv& 6 \pmod{13} \\ \hline & ab + bc+2ca &\equiv& 8abc\pmod{13} \quad |\quad : (abc)\\ (3) & \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b} &\equiv& 8 \pmod{13} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3*(1)-(2)-(3): & 3*\left( \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b}\right) \\ & &\equiv& 3*0 - 6 - 8 \pmod{13} \\ & \dfrac{6}{c} + \dfrac{3}{a} + \dfrac{3}{b} \\ & -\dfrac{1}{c} - \dfrac{2}{a} - \dfrac{1}{b} \\ & -\dfrac{1}{c} - \dfrac{1}{a} - \dfrac{2}{b} \\ & &\equiv& -14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& -14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& 13-14 \pmod{13} \\ & \dfrac{4}{c} &\equiv& -1 \pmod{13} \quad | \quad +1 \\ & \dfrac{4}{c}+1 &\equiv& 0 \pmod{13} \quad | \quad *c \\ & 4+c &\equiv& 0 \pmod{13} \quad | \quad -4 \\ & c &\equiv& -4 \pmod{13} \\ & c &\equiv& 13-4 \pmod{13} \\ & \mathbf{ c } &\equiv& \mathbf{ 9 \pmod{13} } \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline 3*(3)-(2)-(1): & 3*\left( \dfrac{1}{c} + \dfrac{1}{a} + \dfrac{2}{b}\right) \\ & -\left( \dfrac{1}{c} + \dfrac{2}{a} + \dfrac{1}{b}\right) \\ & -\left( \dfrac{2}{c} + \dfrac{1}{a} + \dfrac{1}{b}\right) \\ & &\equiv& 3*8 - 6 - 0 \pmod{13} \\ & \dfrac{3}{c} + \dfrac{3}{a} + \dfrac{6}{b} \\ & -\dfrac{1}{c} - \dfrac{2}{a} - \dfrac{1}{b} \\ & -\dfrac{2}{c} - \dfrac{1}{a} - \dfrac{1}{b} \\ & &\equiv& 18 \pmod{13} \\ & \dfrac{4}{b} &\equiv& 18-13 \pmod{13} \\ & \dfrac{4}{b} &\equiv& 5 \pmod{13} \quad | \quad -5 \\ & \dfrac{4}{b}-5 &\equiv& 0 \pmod{13} \quad | \quad *b \\ & 4-5b &\equiv& 0 \pmod{13} \quad | \quad *(-1) \\ & 5b &\equiv& 0 \pmod{13} \quad | \quad +4 \\ & 5b &\equiv& 4 \pmod{13} \quad | \quad : 5 \\ & b &\equiv& 4*\dfrac{1}{5} \pmod{13} \\ & && \boxed{\dfrac{1}{5} \pmod{13} \\ \equiv 5^{\varphi(13)-1}\pmod{13} \\ \equiv 5^{12-1}\pmod{13} \\ \equiv 5^{11}\pmod{13} \\ \equiv 48828125\pmod{13} \\ \equiv 8\pmod{13} } \\ & b &\equiv& 4*8 \pmod{13} \\ & b &\equiv& 32 \pmod{13} \\ & b &\equiv& 32-2*13 \pmod{13} \\ & b &\equiv& 6 \pmod{13} \\ & \mathbf{ b } &\equiv& \mathbf{ 6 \pmod{13} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2ab+bc+ca &\equiv& 0\pmod{13} \quad & | \quad b=6, \ c=9 \\ 2a*6+6*9+9*a &\equiv& 0\pmod{13} \\ 21a+54 &\equiv& 0\pmod{13} \quad & | \quad 21\equiv 8\pmod{13},\ \quad 54\equiv 2\pmod{13} \\ 8a +2 &\equiv& 0\pmod{13} \quad & | \quad :2 \\ 4a+1 &\equiv& 0 \pmod{13} \quad & | \quad -1 \\ 4a &\equiv& -1 \pmod{13} \quad | \quad : 4 \\ a &\equiv& (-1)*\dfrac{1}{4} \pmod{13} \\ && \boxed{\dfrac{1}{4} \pmod{13} \\ \equiv 4^{\varphi(13)-1}\pmod{13} \\ \equiv 4^{12-1}\pmod{13} \\ \equiv 4^{11}\pmod{13} \\ \equiv 4194304\pmod{13} \\ \equiv 10\pmod{13} } \\ a &\equiv& (-1)*10 \pmod{13} \\ a &\equiv& -10 \pmod{13} \\ a &\equiv& 13-10 \pmod{13} \\ \mathbf{ a } &\equiv& \mathbf{ 3 \pmod{13} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{a+b+c \pmod{13}} \\ &\equiv& 3+6+9 \pmod{13} \\ &\equiv& 18 \pmod{13} \\ &\equiv& 18-13 \pmod{13} \\ &\equiv& \mathbf{ 5 \pmod{13} } \\ \hline \end{array}\)

The sum of three consecutive positive integers is a perfect cube. Find the least possible value of the smallest number,

assuming that the middle number ends in at least 5 zeros.

Yesterday, Teddy played a game repeatedly on his computer and won exactly 85% of the games he played. Today, Teddy played the same game repeatedly, winning every game he played, until his overall winning percentage for yesterday and today was exactly 94%. What is the minimum number of games that he played today?

John noticed that the angle formed by the minute hand and hour hand on a standard 12-hour clock

was 110 degrees when he left home after 6 p.m.;

it was also 110 degrees when he returned before 7 p.m. that same night.

If he left home for more than five minutes, for how many minutes was he away?

\(\text{Let the angle formed by the minute hand and hour hand in degrees $\mathbf{ \Delta \alpha }$ } \\ \text{Let the time in hours $\mathbf{ t }$ }\)

The formula between the two values \(\Delta \alpha \) and \(t\) is:

\(\boxed{~ \Delta \alpha +360^\circ \cdot n = 330 \cdot t ,\ n\in \mathbb{Z} \\or\\ (360^\circ-\Delta \alpha) +360^\circ \cdot m = 330 \cdot t ,\ m\in \mathbb{Z}~ } \)

\(\begin{array}{|rl|rl|} \hline n & t = \dfrac{110^\circ +360^\circ \cdot n} {330} & m & t = \dfrac{(360^\circ-110^\circ) +360^\circ \cdot m} {330} \\ \hline \ldots&&\ldots \\ 5 & 5^h\ 47^m\ 16^s.\overline{36} & 4 & 5^h\ 7^m\ 16^s.\overline{36} \\ 6 & \mathbf{6^h\ 52^m\ 43^s.\overline{63}} & 5 & \mathbf{6^h\ 12^m\ 43^s.\overline{63}} \\ 7 & 7^h\ 58^m\ 10^s.\overline{90} & 6 & 7^h\ 18^m\ 10^s.\overline{90} \\ \ldots&&\ldots \\ \hline \end{array} \)

He was between \(6^h\ 12^m\ 43^s.\overline{63}\) and \(6^h\ 52^m\ 43^s.\overline{63}\) away.
So he left home for exact \(\mathbf{40}\) minutes

Find the sum of all the integer values of  \(m\) that make the following equation true: \(\left(2^m3^5\right)^m 9^7=\dfrac{\left(256\cdot3^m\right)^m}{\left(\sqrt2\right)^{14}}\) .

\(\begin{array}{|rcll|} \hline \left(2^m\cdot3^5\right)^m\cdot 9^7 &=& \dfrac{\left(256\cdot3^m\right)^m}{\left(\sqrt2\right)^{14}} \quad &| \quad \left(\sqrt2\right)^{14}=2^{\frac{14}{2}} = 2^7 \\ \left(2^m\cdot3^5\right)^m\cdot 9^7 &=& \dfrac{\left(256\cdot3^m\right)^m}{2^7} \quad &| \quad 256 = 2^8 \\ \left(2^m\cdot3^5\right)^m\cdot 9^7 &=& \dfrac{\left(2^8\cdot3^m\right)^m}{2^7} \quad &| \quad 9^7=(3^2)^7=3^{2\cdot 7}=3^{14} \\ \left(2^m\cdot3^5\right)^m\cdot 3^{14} &=& \dfrac{\left(2^8\cdot3^m\right)^m}{2^7} \quad &| \quad \cdot 2^7 \\ \left(2^m\cdot3^5\right)^m\cdot 3^{14}\cdot2^7 &=& \left(2^8\cdot3^m\right)^m \\ 2^{(m^2)}\cdot 3^{5m}\cdot 3^{14}\cdot2^7 &=& 2^{8m}\cdot3^{(m^2)} \\ \mathbf{2^{\overbrace{(m^2-8m+7)}^{=0}}} &=& \mathbf{3^{\overbrace{(m^2-5m-14)}^{=0}}} \quad |&\quad \text{The only solution is } 2^0 = 3^0 = 1 \\ 2^0 &=& 3^0 \\ \hline \end{array} \)

\(\text{Is there an equal $m$ that sets $m^2-8m +7$ and $m^2-5m-14$ to zero at the same time?} \)

\(\begin{array}{|rcll|} \hline m^2-8m+7 &=& 0 \\\\ m &=& \dfrac{ 8\pm \sqrt{64-4\cdot 7} }{2} \\ m &=& \dfrac{ 8\pm 6 }{2} \\\\ m_1 &=& \dfrac{ 8+ 6 }{2} \\ \mathbf{m_1} &=& \mathbf{7} \\\\ m_2 &=& \dfrac{ 8- 6 }{2} \\ \mathbf{m_2} &=& \mathbf{1} \\ \hline \end{array} \begin{array}{|rcll|} \hline m^2-5m-14 &=& 0 \\\\ m &=& \dfrac{ 5\pm \sqrt{25-4\cdot(-14)} }{2} \\ m &=& \dfrac{ 5\pm 9 }{2} \\\\ m_1 &=& \dfrac{ 5+ 9 }{2} \\ \mathbf{m_1} &=& \mathbf{7} \\\\ m_2 &=& \dfrac{ 5- 9 }{2} \\ \mathbf{m_2} &=& \mathbf{-2} \\ \hline \end{array} \)

\(\text{So $m$ is $7$ and the sum of all the integer values of $m$ is also $\mathbf{7}$} .\)

Find the sum of all integral values of \(c\) with \(c\le 25\) for which the equation \(y=x^2-7x-c\) has two rational roots.

\(\begin{array}{|lrcll|} \hline & x^2-7x-c &=& 0 \\ & x &=& \dfrac{7\pm \sqrt{49-4(-c)}}{2} \\\\ & x &=& \dfrac{7\pm \sqrt{49+4c}}{2} \\\\ \text{Two rational roots} & 49+4c &>& 0 \\ & 4c &>& -49 \\ & c &>& -\frac{49}{4} \\ & \mathbf{ c } &>& \mathbf{-12.25} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \text{Integral values} & c &=& \{ -12,\ -11,\ -10,\ \ldots \ ,\ -1,\ 0,\ 1,\ \ldots \ ,\ 25\} \\\\ \text{The sum of all integral values of $c$ with $c\le 25$} & &=& \left(\dfrac{-12+25}{2}\right)\cdot (25+13) \\ & &=& \dfrac{13\cdot 38}{2} \\ & &=& \mathbf{247} \\ \hline \end{array} \)

Suppose \(f(x)=x+1\).
\(\text{For what value of $x$ is }\overbrace{f(f(\cdots(f}^\text{2015 total $f$s}(x))\cdots)=0 ?\)

\(\begin{array}{|lrcll|} \hline f \\ \hline 1 & f(x) &=& x+1 \\ 2 & f\Big(f(x)\Big) = f(x+1) = (x+1)+1 &=& x+2 \\ 3 & f\Big(f\Big(f(x)\Big)\Big) = f(x+2) = (x+2)+1 &=& x+3 \\ \ldots \\ 2015 & \overbrace{f(f(\cdots(f}^\text{2015 total $f$s}(x))\cdots) &=& x + 2015 \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline x+2015 &=& 0 \\ \mathbf{x} &=& \mathbf{-2015} \\ \hline \end{array}\)

Thank you, CPhill !