heureka

Let n and k be positive integers such that n<10^6  and
\(\dbinom{13}{13} + \dbinom{14}{13} + \dbinom{15}{13} + \dots + \dbinom{52}{13} + \dbinom{53}{13} + \dbinom{54}{13} = \dbinom{n}{k}\)

Let n and k be positive integers such that n<10^6  and
\(\dbinom{13}{13} + \dbinom{14}{13} + \dbinom{15}{13} + \dots + \dbinom{52}{13} + \dbinom{53}{13} + \dbinom{54}{13} = \dbinom{n}{k}\)

If x, y, and z are positive real numbers satisfying the system above, then find x, y, and z.
\(x + y + xy = 19, \quad y + z + yz = 29, \quad z + x + zx = 23\)

\(\begin{array}{|rcll|} \hline \mathbf{x + y + xy} &=& \mathbf{19} \\ x (1+y)+ y &=& 19 \\ x (1+y)+ y+1-1 &=& 19 \\ x (1+y)+ (1+y) &=& 20 \\ \mathbf{(1+y)(1+x)} &=& \mathbf{20}\quad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{y + z + yz } &=& \mathbf{29} \\ y (1+z)+ z &=& 29 \\ y (1+z)+ z+1-1 &=& 29 \\ y (1+z)+ (1+z) &=& 30 \\ \mathbf{(1+z)(1+y)} &=& \mathbf{30}\quad (2) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{z + x + zx } &=& \mathbf{23} \\ z (1+x)+ x &=& 23 \\ z (1+x)+ x+1-1 &=& 23 \\ z (1+x)+ (1+x) &=& 24 \\ \mathbf{(1+x)(1+z)} &=& \mathbf{24}\quad (3) \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(1)*(3)}{(2)}: & \dfrac{(1+y)(1+x)(1+x)(1+z)} {(1+z)(1+y)} &=& \dfrac{20*24}{30}\\ &&& 1+z \ne 0,\ \text{respectively}\ z\ne-1 \\\\ &&& 1+y \ne 0,\ \text{respectively}\ y\ne-1 \\\\ & (1+x)(1+x)^2 &=& \dfrac{20*24}{30} \\ & (1+x)^2 &=& 16 \\ & 1+x &=& 4 \quad | \quad x > 0 !\\ & \mathbf{x} &=& \mathbf{3} \\ \hline \end{array} \)

\(\begin{array}{|lrcll|} \hline \dfrac{(2)*(1)}{(3)}: & \dfrac{(1+z)(1+y)(1+y)(1+x) } {(1+x)(1+z) } &=& \dfrac{30*20}{24} \\ &&& 1+x \ne 0,\ \text{respectively}\ x\ne-1 \\\\ &&& 1+z \ne 0,\ \text{respectively}\ z\ne-1 \\\\ & (1+y)(1+y)^2 &=& \dfrac{30*20}{24} \\ & (1+y)^2 &=& 25 \\ & 1+y &=& 5 \quad | \quad y > 0 !\\ & \mathbf{y} &=& \mathbf{4} \\ \hline \end{array}\)

\(\begin{array}{|lrcll|} \hline \dfrac{(3)*(2)}{(1)}: & \dfrac{(1+x)(1+z)(1+z)(1+y)} {(1+y)(1+x)} &=& \dfrac{24*30}{20} \\ &&& 1+y \ne 0,\ \text{respectively}\ y\ne-1 \\\\ &&& 1+x \ne 0,\ \text{respectively}\ x\ne-1 \\\\ & (1+z)(1+z)^2 &=& \dfrac{24*30}{20} \\ & (1+z)^2 &=& 36 \\ & 1+z &=& 6 \quad | \quad z > 0 !\\ & \mathbf{z} &=& \mathbf{5} \\ \hline \end{array}\)

Fifteen unit circles are inscribed in an equilateral triangle in such a way that
each circle is externally tangent to its neighbors.
What is the area of the triangle?

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

\(\begin{array}{|rcll|} \hline A &=&\rho * s \quad | \quad s=\dfrac{a+a+a}{2},\ \rho = 1 \\ A &=&1 *\dfrac{3a}{2} \\ A &=& \dfrac{3a}{2} \\ && \boxed{ 2A = a^2*\sin{60^\circ},\ \sin{60^\circ} = \dfrac{\sqrt{3}}{2} \\ 2A = a^2*\dfrac{\sqrt{3}}{2} \\ A = a^2*\dfrac{\sqrt{3}}{4} }\\ a^2*\dfrac{\sqrt{3}}{4} &=& \dfrac{3a}{2} \\ a*\dfrac{\sqrt{3}}{4} &=& \dfrac{3}{2} \\ a &=& \dfrac{3}{2} *\dfrac{4}{\sqrt{3}} \\ a &=& \dfrac{6}{\sqrt{3}} * \dfrac{\sqrt{3}}{\sqrt{3}} \\ a &=& \dfrac{6\sqrt{3}}{3} \\ \mathbf{a} &=& \mathbf{2\sqrt{3}} \\ \hline \end{array}\)

The area of the triangle:

\(\begin{array}{|rcll|} \hline 2A &=& \left(8+2\sqrt{3} \right)^2 \sin{60^\circ} \quad | \quad \sin{60^\circ} = \dfrac{\sqrt{3}}{2} \\ 2A &=& \left(8+2\sqrt{3} \right)^2 \dfrac{\sqrt{3}}{2} \\ 4A &=& \sqrt{3}\left(8+2\sqrt{3} \right)^2 \\ 4A &=& \sqrt{3}(64+32\sqrt{3}+12) \\ 4A &=& \sqrt{3}(76+32\sqrt{3}) \\ 4A &=& 96+76\sqrt{3} \quad | \quad : 4 \\ \mathbf{A} &=& \mathbf{24+19\sqrt{3}} \\ \mathbf{A} &=& \mathbf{56.9089653438} \\ \hline \end{array}\)

The area of the triangle is \(\approx \mathbf{56.9}\)

In rectangle \(WXYZ\),
\(A\) is on side \(WX\) such that \(AX = 4\),
\(B\) is on side \(YZ\) such that \(BY = 18\), and
\(C\) is on side \(XY\) such that angle \(ACB= 90^\circ\) and
\(CY = 2CX\).
Find \(AB\).

\(\begin{array}{|lrcll|} \hline &AC^2 &=& 4^2+cx^2 \\ &CB^2 &=& (2cx)^2+18^2 \\ &x^2= AC^2+CB^2 &=& 4^2+cx^2+(2cx)^2+18^2 \\ &x^2 &=& 16+cx^2+ 4cx^2+ 324 \\ &x^2 &=& 5cx^2+ 340 \qquad (1) \\ \hline &x^2 &=& (18-4)^2 + (3cx)^2 \\ &x^2 &=& 14^2 + 9cx^2\\ &x^2 &=& 9cx^2 + 196 \qquad (2) \\ \hline (2)=(1): & 9cx^2 + 196 &=& 5cx^2+ 340 \\ & 9cx^2 - 5cx^2 &=& 340 - 196 \\ & 4cx^2 &=& 144 \\ & cx^2 &=& 36 \\ & \mathbf{cx} &=& \mathbf{6} \\ \hline (1): &x^2 &=& 5cx^2+ 340 \quad | \quad cx=6 \\ &x^2 &=& 5*6^2+ 340 \\ &x^2 &=& 520 \\ &x &=& 2\sqrt{130} \\ &\mathbf{x} &=& \mathbf{22.8035085020} \\ \hline \end{array}\)

AB is \(\approx \mathbf{22.8}\)

Compute

\(\dfrac 34+\dfrac3{28}+\dfrac3{70}+\dfrac3{130}+\cdots +\dfrac3{9700}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac 34+\dfrac3{28}+\dfrac3{70}+\dfrac3{130}+\cdots +\dfrac3{8554}+\dfrac3{9118}+\dfrac3{9700}} \\\\ &=& \sum \limits_{k=0}^{32} \dfrac{3}{(3k+1)(3k+4)} \\\\ &=& \sum \limits_{k=0}^{32}\left( \dfrac{1}{3k+1} -\dfrac{1}{3k+4} \right) \\\\ &=& \sum \limits_{k=0}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=0}^{32}\left(\dfrac{1}{3k+4}\right) \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=0}^{31}\left(\dfrac{1}{3k+4}\right) - \dfrac{1}{100} \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=0+1}^{31+1}\left(\dfrac{1}{3(k-1)+4}\right) - \dfrac{1}{100} \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=1}^{32}\left(\dfrac{1}{3k-3+4}\right) - \dfrac{1}{100} \\\\ &=& 1+\sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \sum \limits_{k=1}^{32}\left(\dfrac{1}{3k+1}\right) - \dfrac{1}{100} \\\\ &=& 1 - \dfrac{1}{100} \\\\ &=& \dfrac{100-1}{100} \\\\ &=& \mathbf{\dfrac{99}{100}} \\ \hline \end{array}\)

Find the number of real roots of  \(1 + x + x^{2} + x^{3} = x^{4} + x^{5}\)

\(\begin{array}{|lrcll|} \hline & 1 + x + x^2 + x^3 &=& x^4 + x^5 \\ & (1 + x)(x^2+1) &=& x^4(1+x) \\ & (1 + x)(x^2+1)-x^4(1+x) &=& 0 \\ & \mathbf{ (1 + x)(x^2+1-x^4) } &=& \mathbf{0} \\\\ 1) & x+1 &=& 0 \\ & \mathbf{x} &=& \mathbf{-1} \\\\ 2) & x^2+1-x^4 &=& 0 \\ & x^4-x^2-1 &=& 0 \\ & x^2 &=& \dfrac{1\pm \sqrt{1-4(-1)} }{2} \\ & \mathbf{x^2} &=& \mathbf{\dfrac{1\pm \sqrt{5} }{2}} \\\\ & x^2=\dfrac{1+\sqrt{5} }{2} &\text{or}& x^2=\dfrac{1-\sqrt{5} }{2} \quad < 0!\ \text{Complex solutions:} \\\\ & \mathbf{x=\sqrt{\dfrac{1+\sqrt{5}}{2}}} & \\\\ & \mathbf{x=-\sqrt{\dfrac{1+\sqrt{5} }{2}}} & \\\\ \hline \end{array}\)

Real solutions:
\(\mathbf{x=-1} \\ \mathbf{x=\sqrt{\dfrac{1+\sqrt{5}}{2}}} \\ \mathbf{x=-\sqrt{\dfrac{1+\sqrt{5} }{2}}}\)

A regular dodecahedron \(P_1 P_2 P_3 \dotsb P_{12}\) is inscribed in a circle with radius \(1\).
Compute  \((P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2\).
(The sum includes all terms of the form \((P_i P_j)^2\), where \(1 \le i < j \le 12\).

A regular dodecahedron \(P_1 P_2 P_3 \dotsb P_{12}\) is inscribed in a circle with radius \(1\).
Compute  \((P_1 P_2)^2 + (P_1 P_3)^2 + \dots + (P_{11} P_{12})^2\).
(The sum includes all terms of the form \((P_i P_j)^2\), where \(1 \le i < j \le 12\).

If

\(x + y + z = 26\) and \(\dfrac1x + \dfrac1y + \dfrac1z = 31\), then find \(\dfrac xy + \dfrac yx + \dfrac xz + \dfrac zx + \dfrac yz + \dfrac zy\).

\(\begin{array}{|rcll|} \hline && \dfrac xy + \dfrac yx + \dfrac xz + \dfrac zx + \dfrac yz + \dfrac zy \\\\ &=& \dfrac xy +\dfrac xz + \dfrac yx + \dfrac yz + \dfrac zx + \dfrac zy \\\\ &=& x\left(\dfrac 1y +\dfrac 1z \right) + y\left(\dfrac 1x + \dfrac 1z\right) + z\left(\dfrac 1x + \dfrac 1y\right) \\\\ && \boxed{\dfrac1x + \dfrac1y + \dfrac1z = 31\\ \dfrac1x + \dfrac1y = 31- \dfrac1z \\ \dfrac1x + \dfrac1z = 31- \dfrac1y \\ \dfrac1y + \dfrac1z = 31-\dfrac1x } \\\\ &=& x\left(31-\dfrac1x\right) + y\left(31- \dfrac1y\right) + z\left(31- \dfrac1z\right) \\\\ &=& 31x-1 +31y-1 + 31z-1 \\\\ &=& 31(x+y+z)-3\quad | \quad \mathbf{x + y + z = 26} \\ &=& \mathbf{803} \\ \hline \end{array}\)