heureka

Let
\(f(x)=(x^2+6x+9)^{50}-4x+3\), and let \(r_1,r_2,\ldots,r_{100}\) be the roots of \(f(x)\).

Compute \((r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}\).

\(\begin{array}{|rcll|} \hline \mathbf{f(x)} &=& \mathbf{(x^2+6x+9)^{50}-4x+3} \\ &=& \left((x+3)^2\right)^{50}-4x+3 \\ \mathbf{f(x)} &=& \mathbf{\left( x+3 \right)^{100}-4x+3} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & f(r_1) =0 &=& \left( r_1+3 \right)^{100}-4r_1+3 \\ & f(r_2) =0 &=& \left( r_2+3 \right)^{100}-4r_2+3 \\ & \ldots \\ & f(r_{100}) =0 &=& \left( r_{100}+3 \right)^{100}-4r_{100}+3 \\ & \hline \\ \text{sum} & 0 &=& (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} \\ & && -4(r_1+r_2+\ldots + r_{100}) + 3\cdot 100 \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ \hline \end{array}\)

\(\mathbf{\text{vieta:}}\)
For any polynomial equation
\(0=x^n+a_{n-1}·x^{n-1}+...+a_2·x^2+a_1·x^1+a_0\)
with the solutions \(r_1\dots r_n\), the relatively simple formulas for \(a_0\) and \(a_{n-1}\) are:
\(a_0=(-1)^n \prod\limits_{k=1}^{n} r_k \\ a_{n-1}= -\sum \limits_{k=1}^{n} r_k\)

\(\begin{array}{|rcll|} \hline && \left( x+3 \right)^{100} \\ \\ &=& x^{100} + \binom{100}{1}x^{99}\cdot 3 + \ldots \\ \\ &=& x^{100} + \underbrace{300}_{ \underbrace{=a_{n-1}}_{= -(r_1+r_2+\ldots + r_{100})}} x^{99} + \ldots \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(r_1+r_2+\ldots + r_{100}) - 300 \\ && \boxed{r_1+r_2+\ldots + r_{100} = -300} \\ (r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100} &=& 4(-300) - 300 \\ \mathbf{(r_1+3)^{100}+(r_2+3)^{100}+\cdots+(r_{100}+3)^{100}} &=& \mathbf{-1500} \\ \hline \end{array}\)

Four red candies and three green candies can be combined to make many different “flavors.”
Flavors are different if the percent red is different,
so “3 red / 0 green” is the same flavor as “2 red / 0 green,”
and likewise “4 red / 2 green” is the same flavor as “2 red / 1 green.”
If a flavor is to be made using some or all of the seven candies,
how many different flavors are possible?

\(\begin{array}{|c|c|c|c|c|} \hline & {\color{green}0} & {\color{green}1} &{\color{green}2} &{\color{green}3} \\ \hline {\color{red}0} & - & 0+1=1 & 0+2=2 & 0+3=3 \\ & & \frac{{\color{red}0}}{1}: \frac{{\color{green}1}}{1} &\frac{{\color{red}0}}{2}: \frac{{\color{green}2}}{2}&\frac{{\color{red}0}}{3}: \frac{{\color{green}3}}{3}\\ & & {\color{red}0}: {\color{green}1} & {\color{red}0}: {\color{green}1}&{\color{red}0}: {\color{green}1}\\ \hline {\color{red}1} & 1+0=1 & 1+1=2 & 1+2=3 & 1+3=4 \\ & \frac{{\color{red}1}}{1}: \frac{{\color{green}0}}{1} & \frac{{\color{red}1}}{2}: \frac{{\color{green}1}}{2} &\frac{{\color{red}1}}{3}: \frac{{\color{green}2}}{3}&\frac{{\color{red}1}}{4}: \frac{{\color{green}3}}{4}\\ & {\color{red}1}: {\color{green}0} & \\ \hline {\color{red}2} & 2+0=2 & 2+1=3 & 2+2=4 & 2+3=5 \\ & \frac{{\color{red}2}}{2}: \frac{{\color{green}0}}{2} & \frac{{\color{red}2}}{3}: \frac{{\color{green}1}}{3} &\frac{{\color{red}2}}{4}: \frac{{\color{green}2}}{4}&\frac{{\color{red}2}}{5}: \frac{{\color{green}3}}{5}\\ & {\color{red}1}: {\color{green}0} & &\frac{{\color{red}1}}{2}: \frac{{\color{green}1}}{2} \\ \hline {\color{red}3} & 3+0=3 & 3+1=4 & 3+2=5 & 3+3=6 \\ & \frac{{\color{red}3}}{3}: \frac{{\color{green}0}}{3} & \frac{{\color{red}3}}{4}: \frac{{\color{green}1}}{4} &\frac{{\color{red}3}}{5}: \frac{{\color{green}2}}{5}&\frac{{\color{red}3}}{6}: \frac{{\color{green}3}}{6}\\ & {\color{red}1}: {\color{green}0} & & &\frac{{\color{red}1}}{2}: \frac{{\color{green}1}}{2}\\ \hline {\color{red}4} & 4+0=4 & 4+1=5 & 4+2=6 & 4+3=7 \\ & \frac{{\color{red}4}}{4}: \frac{{\color{green}0}}{4} & \frac{{\color{red}4}}{5}: \frac{{\color{green}1}}{5} &\frac{{\color{red}4}}{6}: \frac{{\color{green}2}}{6}&\frac{{\color{red}4}}{7}: \frac{{\color{green}3}}{7}\\ & {\color{red}1}: {\color{green}0} & &\frac{{\color{red}2}}{3}: \frac{{\color{green}1}}{3} \\ \hline \end{array}\)

The percent red

\(\begin{array}{|c|c|c|c|c|} \hline & {\color{green}0} & {\color{green}1} &{\color{green}2} &{\color{green}3} \\ \hline {\color{red}0} & & {\color{red}0} & {\color{red}0}&{\color{red}0}\\ \hline {\color{red}1} & {\color{red}1} & \frac{{\color{red}1}}{2} &\frac{{\color{red}1}}{3}&\frac{{\color{red}1}}{4} \\ \hline {\color{red}2} & {\color{red}1} & \frac{{\color{red}2}}{3} &\frac{{\color{red}1}}{2}&\frac{{\color{red}2}}{5} \\ \hline {\color{red}3} & {\color{red}1} & \frac{{\color{red}3}}{4} &\frac{{\color{red}3}}{5} &\frac{{\color{red}1}}{2} \\ \hline {\color{red}4} & {\color{red}1} & \frac{{\color{red}4}}{5}&\frac{{\color{red}2}}{3}&\frac{{\color{red}4}}{7} \\ \hline \end{array}\)

The different flavors are \(\{ {\color{red}0},\ \frac{{\color{red}1}}{4},\ \frac{{\color{red}1}}{3},\ \frac{{\color{red}2}}{5},\ \frac{{\color{red}1}}{2},\ \frac{{\color{red}3}}{5},\ \frac{{\color{red}4}}{7},\ \frac{{\color{red}2}}{3},\ \frac{{\color{red}3}}{4},\ \frac{{\color{red}4}}{5},\ {\color{red}1} \} \)

The digits of a four-digit positive integer add up to 14.

The sum of the two middle digits is nine, and the thousands digit minus the units digit is one.

If the integer is divisible by 11,

what is the integer?

\(\begin{array}{|lrcll|} \hline \text{A four-digit positive integer}:& abcd \\ \hline \text{The digits add up to $14$}: & a+b+c+d=14 \\ \text{The sum of the two middle digits is nine}: & b+c=9 \\ \text{The thousands digit minus the units digit is one}: & a-d=1 \\ \text{Divisibility by Eleven: if the difference of the alternating sum is $0$}: & a+c-(b+d)=0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a+b+c+d &=& 14 \quad | \quad b+c=9 \\ a+9+d &=& 14 \quad | \quad a-d=1\quad \text{or} \quad a=1+d \\ 1+d+9+d &=& 14 \\ 2d+10&=& 14 \\ 2d &=& 4 \\ \mathbf{d} &=& \mathbf{2} \\ \hline a &=& 1+d \\ a &=& 1+2 \\ \mathbf{a} &=& \mathbf{3} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline a+c-(b+d) &=& 0 \\ a+c-b-d &=& 0 \\ a-d+c-b &=& 0 \quad | \quad a-d = 1 \\ 1+c-b &=& 0 \quad | \quad b+c=9\quad \text{or} \quad c=9-b \\ 1+9-b-b &=& 0 \\ 2b &=& 10 \\ \mathbf{b} &=& \mathbf{5} \\ \hline c &=& 9-b \\ c &=& 9-5 \\ \mathbf{c} &=& \mathbf{4} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \text{The four-digit positive integer $abcd$ is}:& \mathbf{3542} \\ & 3542 : 11 = 322 \\ \hline \end{array} \)

Please give an actual answer

In the diagram below, chords \(\overline{AB}\) and \(\overline{CD}\) are perpendicular, and meet at X.
If AX = 3, BX = 4, CX = 6, and DX = 2, then find the diameter of the circle.

\(\text{Let the center of the circle $C(x_c,\ y_c)$ }\)

\(\begin{array}{|rcll|} \hline 4+y_c &=& 3-y_c \\ 2y_c &=& -1 \\ \mathbf{y_c} &=& \mathbf{-\dfrac{1}{2}} \\ \hline \end{array} \begin{array}{|rcll|} \hline 6+x_c &=& 2-x_c \\ 2x_c &=& -4 \\ \mathbf{x_c} &=& \mathbf{-2} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline & (x-x_c)^2 + (y-y_c)^2 &=& r^2 \\ P(2,0): & (2-x_c)^2 + (0-y_c)^2 &=& r^2 \quad | \quad x_c = -2,\ y_c =-\dfrac{1}{2} \\ & (2+2)^2 + \left(0+\dfrac{1}{2} \right)^2 &=& r^2 \\ & 4^2 + \dfrac{1}{4} &=& r^2 \\ & 16 + \dfrac{1}{4} &=& r^2 \quad | \quad *4 \\ & 64+1 &=& 4r^2 \\ & 65 &=& 4r^2 \quad | \quad \text{sqrt both sides} \\ & \sqrt{65} &=& 2r \\ \hline \end{array}\)

The diameter of the circle is \(\mathbf{\sqrt{65}}\)

Find constants  and  such that
\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}\)

In an isosceles triangle, the sum of its base and its height is 10,
and the radius of its circumscribed circle is 25/8.
Find the sides of the triangle.

\(\begin{array}{|lrcll|} \hline & c+h &=& 10 \\ \text{or} & \mathbf{h} &=& \mathbf{10-c} \\ \hline \end{array} \begin{array}{|rcll|} \hline \left(\dfrac{c}{2}\right)^2+h^2 &=& a^2 \\ \left(\dfrac{c}{2}\right)^2+(10-c)^2 &=& a^2 \\ \dfrac{c^2}{4}+(10-c)^2 &=& a^2 \\ \mathbf{a^2} &=& \mathbf{\dfrac{c^2}{4}+(10-c)^2} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline A &=& \dfrac{ch}{2} \\\\ \mathbf{A} &=& \mathbf{\dfrac{c(10-c)}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{r} &=& \mathbf{\dfrac{abc}{4A}} \quad | \quad b=a \\\\ r &=& \dfrac{ca^2}{4A} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{4A} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{4*\dfrac{c(10-c)}{2}} \\\\ r &=& \dfrac{c* \left(\dfrac{c^2}{4}+(10-c)^2 \right)}{2c(10-c)} \\\\ r &=& \dfrac{ \dfrac{c^2}{4}+(10-c)^2 }{2(10-c)} \\\\ 2(10-c)r &=& \dfrac{c^2}{4}+(10-c)^2 \\\\ (10-c)^2 - 2r(10-c) + \dfrac{c^2}{4} &=& 0 \\\\ 100-20c+c^2 -20r + 2rc - \dfrac{c^2}{4} &=& 0 \\ \ldots \\ \mathbf{\dfrac{5}{4}c^2-c(20-2r) + 100-20r} &=& \mathbf{0} \\\\ c &=& \dfrac{20-2r \pm \sqrt{(20-2r)^2-4*\dfrac{5}{4}*(100-20r)} }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{20-2r \pm \sqrt{(20-2r)^2- 5(100-20r)} }{ 2*\dfrac{5}{4}} \\ && \ldots \\\\ c &=& \dfrac{20-2r \pm \sqrt{4(r^2+5r-25)} }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{20-2r \pm 2\sqrt{ r^2+5r-25 } }{ 2*\dfrac{5}{4}} \\\\ c &=& \dfrac{10-r \pm \sqrt{ r^2+5r-25 } }{ \dfrac{5}{4}}\quad | \quad r=\dfrac{25}{8} \\\\ c &=& \dfrac{6.875 \pm \sqrt{ 0.390625 } }{ 1.25} \\\\ c &=& \dfrac{6.875 \pm 0.625 }{ 1.25} \\\\ c_1 = \dfrac{7.5}{1.25} && c_2 = \dfrac{6.25}{1.25} \\\\ \mathbf{c_1=6} && \mathbf{c_2=5} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a^2} &=& \mathbf{\dfrac{c^2}{4}+(10-c)^2} \\ \hline \mathbf{a_1^2} &=& \mathbf{\dfrac{6^2}{4}+(10-6)^2} \quad | \quad c_1=6 \\ a_1^2 &=& \dfrac{36}{4}+4^2 \\ a_1^2 &=& 9+16 \\ a_1^2 &=& 25 \\ \mathbf{a_1} &=& \mathbf{5} \\ \hline \mathbf{a_2^2} &=& \mathbf{\dfrac{5^2}{4}+(10-5)^2} \quad | \quad c_2=5 \\ a_2^2 &=& \dfrac{25}{4}+5^2 \\ a_2^2 &=& \dfrac{25*5}{4} \\ \mathbf{a_2} &=& \mathbf{\dfrac{5}{2}\sqrt{5}} \\ \hline \end{array} \)

Let a b c d and e be positive integers.

The sum of the four numbers on each of the five segments connecting "points" of the star is 28.

What is the value of the sum \(a+b+c+d+e\)?

I assume:

\(\begin{array}{|lrcll|} \hline (1): & e+4+b &=& 28 \\ (2): & a+5+c &=& 28 \\ (3): & b+6+d &=& 28 \\ (4): & c+7+e &=& 28 \\ (5): & d+8+a &=& 28 \\ \hline \text{sum}: & 2(a+b+c+d+e) + 4+5+6+7+8 &=& 5*28 \\ & 2(a+b+c+d+e) + 30 &=& 140 \\ & 2(a+b+c+d+e) &=& 110 \\ & \mathbf{a+b+c+d+e} &=& \mathbf{55 } \\ \hline \end{array}\)

Solve

\(|x - 1| - 2|x| = 3|x + 1|\).

\(\begin{array}{|rcll|} \hline \text{zero in }~ |x - 1|: \\ \hline x - 1 &=& 0 \\ \hline \mathbf{x} &=&\mathbf{ 1 } \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{zero in }~ |x|: \\ \hline \mathbf{x} &=&\mathbf{0} \\ \hline \end{array} \begin{array}{|rcll|} \hline \text{zero in }~ |x + 1|: \\ \hline \hline x + 1 &=& 0 \\ \hline \mathbf{x} &=&\mathbf{ -1 } \\ \hline \end{array}\)

We create a sign table:

\(\begin{array}{|l|c|c\|\|c|c|c|c|c|} \hline \text{Interval or position} : & \left(-\infty,-1\right) & -1 & \left(-1,0\right) & 0 & \left(0,1\right) & 1 & \left(1,\infty\right) \\ \hline \text{sign of } (x+1): & - & 0 & + & + & + & + & + \\ \hline \text{sign of } (x): & - & - & - & 0 & +& +& + \\ \hline \text{sign of } (x-1): &- & - & - & - & - & 0 & + \\ \hline \end{array} \)

For the 4 cases  \((-\infty, -1] ,\ (-1 , 0] ,\ (0, 1] ,\ (1, \infty)\)
we can therefore always resolve all amounts and solve the corresponding equation without amounts,
taking into account the respective case condition.
The negative terms from the amounts are highlighted in green:

\(\begin{array}{|rcll|} \hline (-\infty, -1] \\ \hline {\color{green}-(x - 1)} - 2({\color{green}-x}) &=& 3({\color{green}-(x + 1)}) \\ -x+1 +2x &=& -3(x + 1) \\ -x+1 +2x &=& -3x -3 \\ 4x &=& -4 \\ \mathbf{x} &=& \mathbf{-1}\ \checkmark \qquad -\infty < x \le -1 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline (-1 , 0] \\ \hline {\color{green}-(x - 1)} - 2({\color{green}-x}) &=& 3(x + 1) \\ -x+1 +2x &=& 3x+3 \\ 2x &=& -2 \\ x &=& -1 \quad \mathbf{\text{no}} \qquad -1 < x \le 0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (0, 1] \\ \hline {\color{green}-(x - 1)} - 2(x) &=& 3(x + 1) \\ -x+1 -2x &=& 3x+3 \\ 6x &= & -2 \\ x &=& -\dfrac{1}{3} \quad \mathbf{\text{no}} \qquad 0 < x \le 1 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline (1, \infty) \\ \hline (x - 1) - 2(x) &=& 3(x + 1) \\ x-1 -2x &=& 3x+3 \\ 4x &= & -4 \\ x &=& -1 \quad \mathbf{\text{no}} \qquad 1 < x < \infty \\ \hline \end{array}\)

\(\mathbf{x = -1}\)

One of their common tangents is touching as shown in the figure. 

Find the radius of the circle with center R.

\(\text{Let $PR=p+r$} \\ \text{Let $PB=p-r$} \\ \text{Let $QR=q+r$} \\ \text{Let $QC=q-r$} \\ \text{Let $PQ=p+q$} \\ \text{Let $PA=p-q$} \\ \text{Let $DF=AQ$} \\ \text{Let $DF=DE+EF$}\)

\(\begin{array}{|rcll|} \hline \mathbf{DE^2 +PB^2} &=& \mathbf{PR^2} \\ DE^2 + (p-r)^2 &=& (p+r)^2 \\ DE^2 &=& (p+r)^2 - (p-r)^2 \\ DE^2 &=& p^2+2pr+r^2-p^2+2pr-r^2 \\ DE^2 &=& 2*2pr \\ \mathbf{DE} &=& \mathbf{2\sqrt{pr} } \\ \\ \hline \mathbf{EF^2 +QC^2} &=& \mathbf{QR^2} \\ EF^2 + (q-r)^2 &=& (q+r)^2 \\ EF^2 &=& (q+r)^2 - (q-r)^2 \\ EF^2 &=& q^2+2qr+r^2-q^2+2qr-r^2 \\ EF^2 &=& 2*2qr \\ \mathbf{EF} &=& \mathbf{2\sqrt{qr} } \\ \\ \hline \mathbf{DF^2 + PA^2} &=& \mathbf{PQ^2} \\ DF^2 + (p-q)^2 &=& (p+q)^2 \\ DF^2 &=& (p+q)^2 - (p-q)^2 \\ DF^2 &=& p^2+2pq+q^2-p^2+2pq-q^2 \\ DF^2 &=& 2*2pq \\ \mathbf{DF} &=& \mathbf{2\sqrt{pq} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{DF} &=& \mathbf{DE+EF} \\ 2\sqrt{pq} &=& 2\sqrt{pr}+2\sqrt{qr} \quad | \quad : 2 \\ \sqrt{pq} &=& \sqrt{pr}+\sqrt{qr} \\ \sqrt{p}\sqrt{q} &=& \sqrt{p}\sqrt{r}+\sqrt{q}\sqrt{r} \quad | \quad : \sqrt{r} \\\\ \dfrac{\sqrt{p}\sqrt{q}}{\sqrt{r}} &=& \sqrt{p} +\sqrt{q} \quad | \quad : \sqrt{p}\sqrt{q} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{\sqrt{p}}{\sqrt{p}\sqrt{q}} +\dfrac{\sqrt{q}}{\sqrt{p}\sqrt{q}} \\\\ \mathbf{\dfrac{1}{\sqrt{r}}} &=& \mathbf{\dfrac{1}{\sqrt{q}} +\dfrac{1}{\sqrt{p}}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{\sqrt{r}}} &=& \mathbf{\dfrac{1}{\sqrt{q}} +\dfrac{1}{\sqrt{p}}} \quad | \quad p=16,\ q=4 \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{1}{\sqrt{4}} +\dfrac{1}{\sqrt{16}} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{1}{2} +\dfrac{1}{4} \\\\ \dfrac{1}{\sqrt{r}} &=& \dfrac{3}{4} \\\\ \sqrt{r} &=& \dfrac{4}{3} \quad | \quad \text{square both sides}\\\\ r &=& \dfrac{16}{9} \\\\ r &=& 1.\bar{7}\ \text{cm} \\ \hline \end{array} \)

The radius of the circle with center R is \(\mathbf{1.\bar{7}\ \text{cm}}\)