heureka

Find 1 + (1 + 2) + (1 + 2 + 3) + (1 + 2 + 3 + 4) + ... + (1 + 2 + 3 + ... + n).

\(\mathbf{s_n=\ ?}\)

\(\begin{array}{|lcccccccccc|} \hline \text{First row}: & \color{red}d_0=1 & &3 & &6 & &10 & & 15 & \ldots \\ \text{Second Row}: & &\color{red}d_1=2 & &3 & &4 & &5 & \ldots \\ \text{Third row}: & & &\color{red}d_2=1 & &1 & & 1& \ldots \\ \hline \end{array} \)

\(\begin{array}{rcl} s_n &=& \dbinom{n}{1}\cdot {\color{red}d_0 } + \dbinom{n}{2}\cdot {\color{red}d_1 } + \dbinom{n}{3}\cdot {\color{red}d_2 }\\ s_n &=& \dbinom{n}{1}\cdot {\color{red}1 } + \dbinom{n}{2}\cdot {\color{red}2} + \dbinom{n}{3}\cdot {\color{red}1}\\ \\ \hline \binom{n}{1} &=& n \\ \binom{n}{2} &=& ( \dfrac{n}{2} ) \cdot ( \dfrac{n-1}{1} ) \\ \binom{n}{3} &=& ( \dfrac{n}{3} ) \cdot ( \dfrac{n-1}{2} )\cdot ( \frac{n-2}{1} ) \\ \hline \\ s_n &=& (n)\cdot {\color{red}1} + ( \dfrac{n}{2} ) \cdot ( \dfrac{n-1}{1} )\cdot {\color{red}2} + ( \dfrac{n}{3} ) \cdot ( \dfrac{n-1}{2} )\cdot ( \dfrac{n-2}{1} )\cdot {\color{red}1} \quad | \quad \cdot 6\\ \\ 6\cdot s_n &=& n\cdot 6 + n \cdot ( n-1 )\cdot 6 + n \cdot ( n-1 )\cdot ( n-2 ) \\ 6\cdot s_n &=& n \left[~ 6 + ( n-1 )\cdot 6 + ( n-1 )\cdot ( n-2 ) ~\right] \\ 6\cdot s_n &=& (n) \left(~ 6 + 6n-6 + n^2 - 3n + 2 ~\right) \\ 6\cdot s_n &=& (n) \left(~ n^2 + 3n + 2 ~\right) \\ 6\cdot s_n &=& (n) \cdot (n+1) \cdot ( n+2 ) \\\\ \mathbf{s_n} &=& \mathbf{ \dfrac{ n \cdot (n+1) \cdot ( n+2 ) }{6} } \qquad \text{or} \qquad \mathbf{s_n} = \dbinom{n+2}{3} \\\\ s_1 &=& 1 = \dfrac{ 1 \cdot 2 \cdot 3}{6} = 1\\ s_2 &=& 1+3 = \dfrac{ 2 \cdot 3 \cdot 4 }{6} = 4 \\ s_3 &=& 1+3+6 = \dfrac{ 3 \cdot 4 \cdot 5 }{6} = 10 \\ s_4 &=& 1+3+6+10 = \dfrac{ 4 \cdot 5 \cdot 6 }{6} = 20\\ s_5 &=& 1+3+6+10+15 = \dfrac{ 5 \cdot 6 \cdot 7 }{6} = 35\\ \cdots \end{array}\)

For a certain value of r, the system

     x + y + 3z = 10, 

-4x + 2y + 5z = 7, 
            rx + z = 3
has no solutions. What is this value of r?

There is no solution, if the determinant \(\begin{vmatrix} 1&1&3\\-4&2&5\\r&0&1\end{vmatrix} = 0 \)

\(\begin{array}{|rcll|} \hline \begin{vmatrix} 1&1&3\\-4&2&5\\r&0&1\end{vmatrix} &=& 0 \\ 1\cdot2\cdot1+r\cdot1\cdot5+(-4)\cdot0\cdot3-r\cdot2\cdot3-(-4)\cdot1\cdot1-1\cdot5\cdot0 &=& 0 \\ 2+5r+0-6r+4-0&=& 0 \\ 6-r &=& 0 \\ \mathbf{r} &=& \mathbf{6} \\ \hline \end{array} \)

Camy made a list of every possible distinct five-digit positive integer that can be formed using each of the digits 1, 3, 4, 5 and 9 exactly once in each integer. What is the sum of the integers on Camy's list?

I assume, we count the numbers of all permutations of 13459.

The sum is:

\(\begin{array}{|rcll|} \hline && \dfrac{5!}{5} \times (1+3+4+5+9)\times (10^0+10^1+10^2+10^3+10^4) \quad | \quad 5!=120 \\ &=& \dfrac{120}{5} \times 22\times 11111 \\ &=& 24 \times 22\times 11111 \\ &=&\mathbf{ 5 866 608 } \\ \hline \end{array} \)

The sum of the integers on Camy's list is 5866608

In the diagram below, points A, B, C, and P are situated so that PA=2, PB=3, PC=4, and BC=5.
What is the maximum possible area of \(\triangle ABC\)?

\(\text{Let $\triangle BPC = 90^\circ$ } \\ \text{Let $\triangle BPA = \alpha$ } \\ \text{Let $\triangle CPA = 270^\circ-\alpha$ } \\ \text{Let $ A_1 = \text{area }\triangle BPA$ } \\ \text{Let $ A_2 = \text{area }\triangle CPA$ } \\ \text{Let $ A_3 = \text{area }\triangle BPC=\dfrac{3\cdot 4}{2}= 6 $ } \\ \)

\(\begin{array}{|lrcll|} \hline & A_1 &=& \dfrac{2\cdot 3 \cdot \sin(\alpha)}{2} \\ (1) & A_1 &=& 3 \cdot \sin(\alpha) \\\\ & A_2 &=& \dfrac{2\cdot 4 \cdot \sin(270^\circ-\alpha)}{2} \\ & A_2 &=& 4 \cdot \sin(270^\circ-\alpha) \quad | \quad \sin(270^\circ-\alpha) = -\cos(\alpha) \\ (2) & A_2 &=& -4 \cdot \cos(\alpha) \\ \hline & f(\alpha) &=& A_1+A_2 \\ & &=& 3 \sin(\alpha)-4 \cos(\alpha) \\ & f'(\alpha) &=& 3 \cos(\alpha)+4 \sin(\alpha) = 0 \\ & 3 \cos(\alpha)+4 \sin(\alpha)& =& 0 \\ & 4 \sin(\alpha)& =& -3 \cos(\alpha) \\ & \tan(\alpha)& =& -\dfrac{3}{4} \\ & \alpha &=& \arctan \left(-\dfrac{3}{4} \right) +180^\circ\\ & \alpha &=& -\arctan (\dfrac{3}{4}) +180^\circ\\ & \alpha &=& -36.8698976458 +180^\circ\\ &\mathbf{ \alpha } &=& \mathbf{143.130102354^\circ \quad | \quad \text{area }(A_1+A_2)_{\text{max.}} }\\ \hline \end{array}\)

Area of  \(\triangle ABC_{\text{max.}}\)

\(\begin{array}{|rcll|} \hline A_{\text{max.}} &=& A_1+A_2+A_3 \\ &=& 3 \sin(\alpha)-4 \cos(\alpha) + \dfrac{3\cdot 4}{2} \\ &=& 3 \sin(143.130102354^\circ)-4 \cos(143.130102354^\circ) + \dfrac{3\cdot 4}{2} \\ &=& 3\cdot (0.6) -4\cdot (-0.8) + 6 \\ &=& 1.8 + 3.2 + 6 \\ &=& 5 + 6 \\ \mathbf{A_{\text{max.}}} &=& \mathbf{ 11 } \\ \hline \end{array}\)

The maximum possible area of \(\triangle ABC =11\)

***delete***

There exist positive integers n and k such that

\(32 \dbinom{6}{6} + 31 \dbinom{7}{6} + 30 \dbinom{8}{6} + \dots + 3 \dbinom{35}{6} + 2 \dbinom{36}{6} + \dbinom{37}{6} = \dbinom{n}{k}\)
Enter the ordered pair \((n, k)\)

Using the hockey stick identity: \(\sum \limits_{i=r}^{n} \dbinom{i}{r} = \dbinom{n+1}{r+1} \qquad \text{for } n,r\in \mathbb{N},\quad n \geq r\)

What is the greatest positive integer n such that 3 to the power of n is a factor of 200!

\(\begin{array}{|rcll|} \hline n &=& \lfloor \dfrac{200}{3}\rfloor + \lfloor \dfrac{200}{3^2}\rfloor+\lfloor \dfrac{200}{3^3}\rfloor+\lfloor \dfrac{200}{3^4}\rfloor \\\\ n &=& 66+ 22+7+2 \\ \mathbf{n} &=& \mathbf{97} \\ \hline \end{array}\)

The graph of \(y=ax^2 + bx + c\) is given below, where \(a\), \(b\), and \(c\) are integers. Find \(a\).

You get three Points from the picture above:

• Point 1 ( Vertex ): \(P_1 (-1,-2)\)
• Point 2 : \(P_2(0,-1)\)
• Point 3 ( symmetric to Point 2) : \(P_3(-2,-1)\)

\(\begin{array}{|lrcll|} \hline & y &=& ax^2 + bx + c \\ \hline P_2(0,-1): & -1 &=& a*0+b*0 + c \\ & \mathbf{c} &=& \mathbf{-1} \\ \hline P_1(-1,-2): & -2 &=& a*(-1)^2+b*(-1) -1\\ & -2 &=& a -b -1 \\ & b &=& a + 1 \\ & \mathbf{b} &=& \mathbf{a+1} \\ \hline P_3(-2,-1): & -1 &=& a*(-2)^2+b*(-2) -1 \\ & -1 &=& 4a -2b -1 \\ & 4a-2b &=& 0 \quad | \quad :2 \\ & 2a-b &=& 0 \quad | \quad b=a+1 \\ & 2a-(a+1) &=& 0 \\ & 2a-a-1 &=& 0 \\ & a-1 &=& 0 \\ & \mathbf{a} &=& \mathbf{1} \quad | \quad b= a+1=1+1=2 \\ \hline \end{array} \)

\(y=x^2+2x-1\)

Thank you, Melody !

An aeroplane is observed at the same instant from three stations A,B, C in a horizontal straight line

but not in a vertical plane through the aeroplane.
If \(AB=BC =c\) and the angles of elevation from A,B C are respectively  \(\alpha\), \(\beta\) and \(\gamma\),

prove that the height of the aeroplane is

\(\dfrac{c\sqrt{2}}{\sqrt{\cot^2(\alpha)+\cot^2(\gamma)-2\cot^2(\beta) }}\)

\(\text{The height of the aeroplane $=h $} \\ \text{$F$ is the nadir point of the aeroplane } \\ \text{$r = \overline{AF}$ } \\ \text{$s = \overline{BF}$ } \\ \text{$t = \overline{CF}$ } \\ \text{$\angle CBF = \epsilon$ } \\ \text{$\angle ABF = 180^\circ-\epsilon$ } \)

\(\begin{array}{|rclcrcl|} \hline \cot(\alpha) &=& \dfrac{r}{h} &\text{or}& r &=& h\cot(\alpha) \qquad (1) \\ \cot(\beta) &=& \dfrac{s}{h} &\text{or}& s &=& h\cot(\beta) \qquad (2) \\ \cot(\gamma) &=& \dfrac{t}{h} &\text{or}& t &=& h\cot(\gamma) \qquad (3) \\ \hline \end{array} \)

cos-rule:

\(\begin{array}{|lrcll|} \hline (1) & t^2 &=& s^2+c^2-2sc\cos(\epsilon) \\\\ & r^2 &=& s^2+c^2-2sc\cos(180^\circ-\epsilon) \\ &&& \boxed{ \cos(180^\circ-\epsilon)=- \cos(\epsilon)} \\ (2) & r^2 &=& s^2+c^2+2sc\cos(\epsilon) \\ \hline (2)+(1): & r^2+t^2 &=& s^2+c^2+2sc\cos(\epsilon)+s^2+c^2-2sc\cos(\epsilon)\\ & r^2+t^2 &=& 2s^2+2c^2 \\ & r^2+t^2-2s^2 &=& 2c^2\\ &&& \boxed{ r^2=h^2\cot^2(\alpha)\\ s^2=h^2\cot^2(\beta)\\ t^2=h^2\cot^2(\gamma) }\\ & h^2\cot^2(\alpha)+h^2\cot^2(\gamma)-2h^2\cot^2(\beta) &=& 2c^2 \\ & h^2\left(\cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta)\right) &=& 2c^2 \\ & h^2 &=& \dfrac{2c^2 }{ \cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta) } \\\\ &\mathbf{ h } &=& \mathbf{ \dfrac{c\sqrt{2} }{ \sqrt{\cot^2(\alpha)+ \cot^2(\gamma)-2 \cot^2(\beta)} } } \\ \hline \end{array}\)