heureka

Find the ordered pair (a, b) of real numbers for which
\((ax + b)(x^5 + 1) - (5x + 1)\)
is divisible by \(x^{2}+1\).

Thank you, CPhill !

Let \(n\) and \(k\) be positive integers such that  \(\dbinom{n}{k}:\dbinom{n}{k + 1} = 4:11\).
Find the smallest possible value of \(n\)

\(\begin{array}{|rcll|} \hline \mathbf{\dbinom{n}{k + 1}} &=& \dfrac{n!}{(k+1)!(n-k-1)!} \quad &| \quad k!(k+1)=(k+1)! \\\\ &=& \dfrac{1}{k+1} *\dfrac{n!}{k!(n-k-1)!} \quad &| \quad (n-k-1)!(n-k)=(n-k)! \\\\ &=& \dfrac{n-k}{k+1} * \dfrac{n!}{k!(n-k)!} \quad &| \quad \dfrac{n!}{k!(n-k)!} = \dbinom{n}{k} \\\\ &=& \mathbf{ \dfrac{n-k}{k+1} \dbinom{n}{k} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \dbinom{n}{k}:\dbinom{n}{k + 1} &=& 4:11 \\\\ \dfrac{ \dbinom{n}{k}} { \dbinom{n}{k + 1}} &=& \dfrac{4}{11} \quad | \quad \boxed{ \mathbf{\dbinom{n}{k + 1} = \dfrac{n-k}{k+1} \dbinom{n}{k} } } \\\\ \dfrac{ \dbinom{n}{k}} { \dfrac{n-k}{k+1} \dbinom{n}{k} } &=& \dfrac{4}{11} \\\\ \dfrac{ 1} { \dfrac{n-k}{k+1} } &=& \dfrac{4}{11} \\\\ \dfrac{k+1} {n-k} &=& \dfrac{4}{11} \\\\ 11(k+1) &=& 4(n-k) \\ 11k+11 &=& 4n-4k \\ \cdots \\ \mathbf{4n} &=& \mathbf{11+15k} \quad | \quad \text{Diophantine equation} \\ \hline \end{array}\)

Euler's method for linear diophantine equations  \(\mathbf{4n =11+15k}\)

\(\begin{array}{|rclrcl|} \hline \mathbf{4n} &=& \mathbf{11+15k} \\ n &=& \dfrac{11+15k}{4} \\ n &=& \dfrac{12-1+16k-k}{4} \\ n &=& \dfrac{12+16k-1-k}{4} \\ n &=& 3+4k-\underbrace{\dfrac{1+k}{4}}_{=a} \\ \mathbf{n} &=& \mathbf{3+4k-a} & a&=& \dfrac{1+k}{4} \\ & & & 4a&=& 1+k \\ & & & \mathbf{k} &=& \mathbf{4a-1} \\ \mathbf{n} &=& \mathbf{3+4(4a-1)-a} \\ n &=& 3+16a-4-a \\ \mathbf{n} &=& \mathbf{-1+15a} \qquad a\in \mathbf{Z}\\ && n_{min},~ \text{if}~ a=1 \\ n_{min} &=& -1+15 \\ \mathbf{n_{min}} &=& \mathbf{14} \\\\ \mathbf{k} &=& \mathbf{4a-1} \quad | \quad a=1 \\ k &=& 4 -1 \\ \mathbf{k} &=& \mathbf{3} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \dbinom{14}{3}: \dbinom{14}{4} &=& 4:11 \\ \hline \end{array}\)

Two unit circles are externally tangent at B. Let AB and BC be diameters of the two circles.

A tangent is drawn from A to the circle with diameter BC, and a tangent is drawn from C to the circle with diameter AB,

so that the two tangent lines are parallel.

Find the distance between the two lines of tangency.

I assume:

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{x-1}{1}} &=& \mathbf{\dfrac{1}{3}} \\\\ x-1 &=& \dfrac{1}{3} \\ x &=& 1+ \dfrac{1}{3} \\ \mathbf{x} &=& \mathbf{\dfrac{4}{3}} \\ \hline \end{array}\)

The polynomial x^{101} + Ax + B  is divisible by x^2 + x + 1 for some real numbers A and B.
Find A+B

The polynomial \(x^{101} + Ax + B\)  is divisible by \(x^2 + x + 1\) for some real numbers \(A\) and \(B\).
Find \(A+B\)

\(\begin{array}{|rcll|} \hline \mathbf{x^2+x+1} &=& \mathbf{0} \\ x &=& \dfrac{-1\pm \sqrt{1-4*1} }{2} \\ x &=& \dfrac{-1\pm i\sqrt{3} }{2} \\ \hline \mathbf{x_1} = \mathbf{-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}} && \mathbf{x_2} = \mathbf{-\left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline & x^{101} + Ax + B &=& q(x)(x^2 + x + 1) \\ \hline x_1 = -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}: & \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& q(x_1)\times 0 \\ &\mathbf{ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B } &=& \mathbf{ 0 } \qquad (1) \\ \hline x_2 = -\left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right): & \left[-\left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)\right]^{101} - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& q(x_2)\times 0 \\ &\mathbf{ -\left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B }&=& \mathbf{ 0 } \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{1} &=& -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2} \\ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{2} &=& -\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2} \\ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{3} &=& 1 \\ \hline \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{4}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{1} \\ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{5}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{2} \\ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{6}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{3} \\ \ldots \\ \hline \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} &=& \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{33*3+2} \\ &=& \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{2}\\ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} &=& -\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2} \\ \mathbf{ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} } &=& \mathbf{ -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) } \\ \hline \end{array} \begin{array}{|rcll|} \hline \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{1} &=& \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2} \\ \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{2} &=& -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2} \\ \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{3} &=& -1 \\ \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{4} &=& -\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2} \\ \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{5} &=& \dfrac{1}{2}- \dfrac{i\sqrt{3}}{2} \\ \left(+\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{6} &=& 1 \\ \hline \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{7}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{1} \\ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{8}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{2} \\ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{9}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{3} \\ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{10}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{4} \\ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{11}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{5} \\ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{12}&=&\left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{6} \\ \ldots \\ \hline \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} &=& \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{16*6+5} \\ &=& \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{5}\right)^{5}\\ \mathbf{ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} } &=& \mathbf{ \dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (1): & \mathbf{ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B } &=& \mathbf{ 0 } \quad | \quad \mathbf{ \left(-\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} = -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) } \\ & -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& 0 \qquad (3) \\ \hline (2): & \mathbf{ -\left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B }&=& \mathbf{ 0 } \quad | \quad \mathbf{ \left(\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right)^{101} = \dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}} \\ & -\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& 0 \qquad (4) \\ \hline (3)-(4): & -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B \\ & -\Bigg(-\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B\Bigg) &=& 0 \\\\ & -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B \\ & +\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) + A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) - B &=& 0 \\\\ & -\left(\dfrac{1}{2}+\dfrac{i\sqrt{3}}{2}\right) +A\left( -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) \\ & +\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) + A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) &=& 0 \\\\ & -\dfrac{1}{2}-\dfrac{i\sqrt{3}}{2} -\dfrac{A}{2}+ \dfrac{Ai\sqrt{3}}{2} \\ & +\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}+\dfrac{A}{2}+ \dfrac{Ai\sqrt{3}}{2} &=& 0 \\\\ & -\dfrac{i\sqrt{3}}{2}+ \dfrac{Ai\sqrt{3}}{2} - \dfrac{i\sqrt{3}}{2}+ \dfrac{Ai\sqrt{3}}{2} &=& 0 \\\\ & -i\sqrt{3} + Ai\sqrt{3} &=& 0 \\\\ & Ai\sqrt{3} &=& i\sqrt{3} \\\\ & \mathbf{ A } &=& \mathbf{1} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (4): & -\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) - A\left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& 0 \quad | \quad \mathbf{A=1} \\\\ & -\left(\dfrac{1}{2}- \dfrac{i\sqrt{3}}{2}\right) - \left( \dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2}\right) + B &=& 0 \\\\ & -\dfrac{1}{2}+ \dfrac{i\sqrt{3}}{2} - \dfrac{1}{2}- \dfrac{i\sqrt{3}}{2} + B &=& 0 \\\\ & -\dfrac{1}{2} - \dfrac{1}{2} + B &=& 0 \\\\ & -1 + B &=& 0 \\\\ & \mathbf{ B } &=& \mathbf{1} \\ \hline \end{array}\)

\(A+B = 1+1\\ \mathbf{A+B = 2} \)

Let

\(A = (4,-1),\ B = (6,2)\), and \(C = (-1,2)\).
There exists a point \(X\) and a constant \(k\) such that for any point \(P\):
\(PA^2 + PB^2 + PC^2 = 3PX^2 + k\).
Find the constant \(k\)

My attempt:

\(\small{ \begin{array}{|rcll|} \hline \mathbf{PA^2 + PB^2 + PC^2} &=& \mathbf{3PX^2 + k} \\ \boxed{PA = P-A\\ PB=P-B\\PC=P-C\\PX=P-X } \\ (P-A)^2 + (P-B)^2 + (P-C)^2 &=& 3(P-X)^2 + k \\ P^2-2PA+A^2 + P^2-2PB+B^2 + P^2-2PC+C^2 &=& 3\left(P^2-2PX+X^2 \right)+ k \\ 3P^2+A^2+B^2+C^2-2P(A+B+C) &=& 3P^2-6PX+3X^2 + k \\ A^2+B^2+C^2-2P(A+B+C) &=& k+3X^2-6PX \\ \boxed{A=\dbinom{4}{-1},\ B=\dbinom{6}{2},\ C=\dbinom{-1}{2} } \\ \dbinom{4}{-1}\dbinom{4}{-1}+\dbinom{6}{2}\dbinom{6}{2}+\dbinom{-1}{2}\dbinom{-1}{2}-2P\Bigg(\dbinom{4}{-1}+\dbinom{6}{2}+\dbinom{-1}{2}\Bigg) &=& k+3X^2-6PX \\ (16+1)+(36+4)+(1+4)-2P\Bigg(\dbinom{4+6-1}{-1+2+2}\Bigg) &=& k+3X^2-6PX \\ 62-2P \dbinom{9}{3} &=& k+3X^2-6PX \\ 62-2P \dbinom{3*3}{3*1} &=& k+3X^2-6PX \\ 62-2*3P \dbinom{3}{1} &=& k+3X^2-6PX \\ 62-6P \dbinom{3}{1} &=& k+3X^2-6PX \\ \text{compare} && \text{compare} \\ {\color{red}62}-6P {\color{blue}\dbinom{3}{1}} &=& {\color{red}k+3X^2}-6P{\color{blue}X} \\ \hline \mathbf{{\color{blue}X}} &=& \mathbf{{\color{blue}\dbinom{3}{1}}} \\\\ {\color{red}k+3X^2} &=& {\color{red}62} \\ k+3\dbinom{3}{1}\dbinom{3}{1} &=& 62 \\ k+3(3*3+1*1) &=& 62 \\ k+3*10 &=& 62 \\ k+30 &=& 62 \\ k &=& 62-30 \\ \mathbf{k} &=& \mathbf{32} \\ \hline \end{array} }\)

The constant k is 32

Help please

Thank you, CPhill !

Let O denote the centroid of triangle ABC.

Let M and N be points on sides AB and AC, respectively, so that M, O, and N are collinear, and AM/MB = 5/2.

Find AN/NC.

\(\begin{array}{|rcll|} \hline \text{Let } A=(x_A,\ y_A)&=&(0,\ 0) \\ \text{Let } B=(x_B,\ y_B)&=&(7,\ 0) \\ \text{Let } C=(x_C,\ y_C) \\ \text{Let } M=(x_M,\ y_M)&=&(5,\ 0) \\ \text{Let } N=(x_N,\ y_N) \\ \text{Let } O=(x_O,\ y_O)&=&\left(\dfrac{x_A+x_B+x_C}{3},\ \dfrac{y_A+y_B+y_C}{3} \right) \\ &=&\left(\dfrac{0+7+x_C}{3},\ \dfrac{0+0+y_C}{3} \right) \\ &=&\left(\dfrac{7+x_C}{3},\ \dfrac{y_C}{3} \right) \\ \hline \end{array}\)

M, O, and N are collinear

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{y_O-y_M}{x_O-x_M}} &=& \mathbf{\dfrac{y_N-y_O}{x_N-x_O}} \\\\ (x_N-x_O)(y_O-y_M) &=& (y_N-y_O)(x_O-x_M) \quad | \quad y_M=0,\ x_M=5 \\ (x_N-x_O)(y_O-0) &=& (y_N-y_O)(x_O-5) \\ (x_N-x_O)y_O &=& (y_N-y_O)(x_O-5) \\ x_Ny_O-x_Oy_O &=& y_Nx_O-5y_N-y_Ox_O+5y_O \\ x_Ny_O &=& y_Nx_O-5y_N+5y_O \quad | \quad y_O=\dfrac{y_C}{3},\ x_O=\dfrac{7+x_C}{3} \\ \dfrac{x_Ny_C}{3} &=& \dfrac{y_N(7+x_C)}{3}-5y_N+\dfrac{5y_C}{3} \\ \dfrac{x_Ny_C}{3} &=& \dfrac{7y_N}{3}+\dfrac{y_Nx_C}{3}-5y_N+\dfrac{5y_C}{3} \\ && \boxed{ \dfrac{y_N}{x_N}= \dfrac{y_C}{x_C}\quad \text{ or } \quad y_Nx_C=x_Ny_C } \\ 0 &=& \dfrac{7y_N}{3}-5y_N+\dfrac{5y_C}{3} \quad | \quad * 3 \\ 0 &=& 7y_N -15y_N+ 5y_C \\ 8y_N &=& 5y_C \\\\ \mathbf{\dfrac{y_N}{y_C}} &=& \mathbf{\dfrac{5}{8}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{AN}{y_N}} &=& \mathbf{\dfrac{AN+NC}{y_C}} \\\\ \dfrac{AN}{AN+NC} &=& \dfrac{y_N}{y_C} \quad | \quad \mathbf{\dfrac{y_N}{y_C}=\dfrac{5}{8}} \\\\ \dfrac{AN}{AN+NC} &=& \dfrac{5}{8} \\\\ 8AN &=& 5(AN+NC) \\ 8AN &=& 5AN+5NC \\ 3AN &=& 5NC \\\\ \mathbf{\dfrac{AN}{NC}} &=& \mathbf{\dfrac{5}{3}} \\ \hline \end{array}\)