heureka

A regular heptagon and a regular 18-gon share a common edge AB, as shown.

Find angle BCD, in degrees.

\(\begin{array}{|rcll|} \hline \mathbf{\angle 1} &=& \dfrac{(7-2)180^\circ}{7}\mathbf{=\dfrac{5*180^\circ}{7}} \\ \hline 2*\angle 2 + \angle 1 &=& 180^\circ \\ 2*\angle 2 &=& 180^\circ - \angle 1 \\ 2*\angle 2 &=& 180^\circ - \dfrac{5*180^\circ}{7} \\ 2*\angle 2 &=& 180^\circ \left(1 - \dfrac{5}{7} \right) \\ 2*\angle 2 &=& 180^\circ \left( \dfrac{2}{7} \right) \\ \mathbf{\angle 2} &=& \mathbf{\dfrac{180^\circ}{7}} \\ \hline \mathbf{\angle 3} &=& \dfrac{(18-2)180^\circ}{18} =\dfrac{16*180^\circ}{18} \mathbf{=160^\circ } \\ \hline \angle 4 &=& 180^\circ - \angle 3 \\ \angle 4 &=& 180^\circ - 160^\circ \\ \mathbf{\angle 4} &=& \mathbf{20^\circ} \\ \hline \mathbf{\angle 5} &=& 180^\circ - \angle 1 \mathbf{= 180^\circ \left( \dfrac{2}{7} \right)} \\ \hline 2*\angle 6 + \angle 4 + \angle 5 &=& 180^\circ \\ 2*\angle 6 + 20^\circ + 180^\circ \left( \dfrac{2}{7} \right) &=& 180^\circ \\ 2*\angle 6 + 180^\circ \left( \dfrac{2}{7} \right) &=& 160^\circ \\ 2*\angle 6 &=& 160^\circ - 180^\circ \left( \dfrac{2}{7} \right) \\ \mathbf{\angle 6} &=& \mathbf{80^\circ - \dfrac{180^\circ}{7}} \\ \hline \angle BCD &=& \angle 2 + \angle 6 \\ \angle BCD &=& \dfrac{180^\circ}{7}+ 80^\circ - \dfrac{180^\circ}{7} \\ \mathbf{\angle BCD} &=& \mathbf{80^\circ} \\ \hline \end{array}\)

At what time between 2 and 3' O clock are the hand of clock together exactly ?

\(\begin{array}{|rcll|} \hline t_h &=& \dfrac{12}{11} \times 2 \\ t_h &=& 2.\overline{18}\ \text{hours} \\ &=& 2^h 10^m 54.\overline{54}^s \\ \hline \end{array} \)

Quadrilateral PQRS is a square with a side length of \(13\).

Points \(X\) and \(Y\) lie outside the square so that \(PX=RY=12\) and \(QX=SY=5\).

Find \(\overline{XY}^2\).

\(\begin{array}{|rcll|} \hline \mathbf{\overline{XY}^2} &=& \mathbf{(13-2z)^2+(13+2h)^2} \quad &| \quad z=\sqrt{5^2-h^2} \\\\ &=& \left(13-2\sqrt{5^2-h^2}\right)^2+(13+2h)^2 \quad &| \quad h= \dfrac{5*12}{13} \\ &=& \left(13-2\sqrt{5^2-\dfrac{5^2*12^2}{13^2} }\right)^2 +\left(13+ \dfrac{10*12}{13}\right)^2 \\ &=& \left(13-2*5\sqrt{\dfrac{13^2-12^2}{13^2} }\right)^2 +\left(13+ \dfrac{120}{13}\right)^2 \\ &=& \left(13-2*5\sqrt{\dfrac{5^2}{13^2} }\right)^2 +\left(\dfrac{13^2+120}{13}\right)^2 \\ &=& \left(13-\dfrac{50}{13}\right)^2+\left(\dfrac{13^2+120}{13}\right)^2 \\ &=& \left( \dfrac{13^2-50}{13}\right)^2+\left(\dfrac{13^2+120}{13}\right)^2 \\ &=& \dfrac{ \left(13^2-50\right)^2+ \left(13^2+120\right)^2}{13^2} \\ &=& \dfrac{ 13^4-2*13^2*50+50^2+13^4+2*13^2*120+120^2}{13^2} \\ &=& \dfrac{ 13^2 \left( 2*13^2-100+240\right) } {13^2} +\dfrac{50^2+120^2}{13^2} \\ &=& 2*13^2-100+240 + 100 \\ &=& 2*13^2 +240 \\ &=& \mathbf{578} \\ \hline \end{array}\)

Find angle PRQ,

\(\text{Let $\angle PRQ = x $}\)

\(\begin{array}{|rcll|} \hline y+2*90^\circ + \theta &=& 360^\circ \\ y + \theta &=& 180^\circ \\ \mathbf{y} &=& \mathbf{180^\circ-\theta} \\\\ 120^\circ+130^\circ+y+x &=& 360^\circ \\ 250^\circ+y+x &=& 360^\circ \\ y+x &=& 110^\circ \quad | \quad \mathbf{y=180^\circ-\theta} \\ 180^\circ-\theta+x &=& 110^\circ \\ -\theta+x &=& -70^\circ \\ \mathbf{\theta} &=& \mathbf{70^\circ +x} \\\\ x &=& \dfrac{\theta}{2} \quad | \quad \mathbf{\theta=70^\circ +x} \\\\ x &=& \dfrac{70^\circ +x}{2} \\\\ 2x &=& 70^\circ + x \\ \mathbf{ x} &=& \mathbf{70^\circ} \\ \hline \end{array}\)

***delete***

Find the area of a cyclic quadrilateral with sides 2, 2, 3, 1.

Formula: \(\boxed{ A = \sqrt{(s-a)(s-b)(s-c)(s-d)}\\ \text{where }s \text{ is the semiperimeter }\\s=\dfrac{a+b+c+d}{2} } \)

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

\(\begin{array}{|rcll|} \hline A &=& \sqrt{(4-2)(4-2)(4-3)(4-1)} \\ A &=& \sqrt{2*2*1*3} \\ \mathbf{A} &=& \mathbf{2\sqrt{3}} \\ \hline \end{array}\)

Let \(\omega\) be a complex number such that \(\omega^5 = 1\) and \(\omega \neq 1\).

Compute \(\dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^2}{1 - \omega^4} + \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^4}{1 - \omega^3}\)

\(\begin{array}{|rcll|} \hline && \mathbf{ \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^2}{1 - \omega^4} + \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^4}{1 - \omega^3} } \quad &| \quad \omega^4 = \dfrac{\omega^5}{\omega}=\dfrac{1}{\omega} \\\\ &=& \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^2}{1 - \dfrac{1}{\omega}} + \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^4}{1 - \omega^3} \\\\ &=& \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^2}{\dfrac{\omega-1}{\omega}} + \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^4}{1 - \omega^3} \\\\ &=& \dfrac{\omega}{1 - \omega^2} - \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^3}{1 - \omega} + \dfrac{\omega^4}{1 - \omega^3} \\\\ &=& \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^4}{1 - \omega^3} \quad &| \quad \omega^3 = \dfrac{\omega^5}{\omega^2}=\dfrac{1}{\omega^2} \\\\ &=& \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^4}{1 - \dfrac{1}{\omega^2}} \\\\ &=& \dfrac{\omega}{1 - \omega^2} + \dfrac{\omega^4}{\dfrac{\omega^2-1}{\omega^2}} \\\\ &=& \dfrac{\omega}{1 - \omega^2} - \dfrac{\omega^6}{1 - \omega^2} \quad | \quad \omega^6 = \omega^5 \omega=\omega \\\\ &=& \dfrac{\omega}{1 - \omega^2} - \dfrac{\omega}{1 - \omega^2} \\\\ &=& \mathbf{0} \\ \hline \end{array}\)

How many positive two-digit integers are there in which each of the two digits is prime?

\(\begin{array}{|c|c|c|c|c|} \hline & \mathbf{2} & \mathbf{3} & \mathbf{5} & \mathbf{7} \\ \hline \mathbf{2} & 22 & 23 & 25 & 27 \\ \hline \mathbf{3} & 32 & 33 & 35 & 37 \\ \hline \mathbf{5} & 52 & 53 & 55 & 57 \\ \hline \mathbf{7} & 72 & 73 & 75 & 77 \\ \hline \end{array}\)

If ab = 15 and a^2 + b^2 = 40, then find (a + b)^4 - (a - b)^4.

\(\begin{array}{|rcll|} \hline (a+b)^2 &=& a^2+b^2+2ab \\ &=& 40+2*15 \\ &=& 70 \\ \mathbf{(a+b)^4} &=& \mathbf{70^2} \\\\ (a-b)^2 &=& a^2+b^2-2ab \\ &=& 40-2*15 \\ &=& 10 \\ \mathbf{(a-b)^4} &=& \mathbf{10^2} \\ \\ (a+b)^4 - (a-b)^4 &=& 70^2-10^2 \\ \mathbf{ (a+b)^4 - (a-b)^4 } &=& \mathbf{4800} \\ \hline \end{array}\)

If the numbers on the third diagonal of Pascal's Triangle are triangular numbers,

what is the value of the 50th triangular number?

(The nth triangular number is \(1+2+3+\ldots+n\).)

The 50th triangular number is \(\dbinom{51}{2} = 1275\)