heureka

A square is inscribed in the ellipse \(x^2 + 4y^2 = 4\), with two side parallel to the x-axis.  
What is the area of the square.

\(\text{Let the side of the square $s$}\)

\(\begin{array}{|rcll|} \hline x^2 + 4y^2 &=& 4 \quad | \quad x=y=\dfrac{s}{2} \\ \left(\dfrac{s}{2}\right)^2 + 4*\left(\dfrac{s}{2}\right)^2 &=& 4 \\ 5*\left(\dfrac{s}{2}\right)^2 &=& 4 \\ \left(\dfrac{s}{2}\right)^2 &=& \dfrac{4}{5} \\ \dfrac{s^2}{4} &=& \dfrac{4}{5} \\ \mathbf{s^2} &=& \mathbf{\dfrac{16}{5}} \\ \mathbf{s^2} &=& \mathbf{3.2} \\ \hline \end{array}\)

The area of the square is \(\mathbf{3.2}\)

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 \(1 \le i < j \le 12\).)

Factor \(ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a.\)

\(\begin{array}{|rcll|} \hline && \mathbf{ab^3 - a^3 b + bc^3 - b^3 c + ca^3 - c^3 a} \\ &=& -a^3 b+ ab^3 + ca^3 - b^3 c - c^3 a + bc^3 \\ &=& -ab(a^2-b^2) + c(a^3 - b^3) - c^3( a -b) \\ &=& -ab(a-b)(a+b) + c(a-b)(a^2+ab+b^2) - c^3( a -b) \\ &=& (a-b)\Big( -ab(a+b) + c(a^2+ab+b^2) - c^3 \Big) \\ &=& (a-b)( -a^2b-ab^2 + ca^2+cab+cb^2 - c^3) \\ &=& (a-b)( ca^2 -a^2b +cab -ab^2 - c^3 +cb^2 ) \\ &=& (a-b)\Big( a^2(c-b) + ab(c-b)-c(c^2-b^2) \Big) \\ &=& (a-b)\Big( a^2(c-b) + ab(c-b)-c(c-b)(c+b) \Big) \\ &=& (a-b)(c-b)\Big( a^2 + ab-c(c+b) \Big) \\ &=& (a-b)(c-b)(a^2 + ab-c^2-cb) \\ &=& (a-b)(c-b)(a^2-c^2 + ab-cb) \\ &=& (a-b)(c-b)\Big((a-c)(a+c) + b(a-c) \Big) \\ &=& (a-b)(c-b)(a-c)(a+c+b) \\ &=& \mathbf{(a-b)(c-b)(a-c)(a+b+c)} \\ \hline \end{array}\)

Factor \(9u^2 + 16v^2 - 4w^2 - 24uv\).

\(\begin{array}{|rcll|} \hline && \mathbf{9u^2 + 16v^2 - 4w^2 - 24uv} \\ &=& 9u^2 - 24uv + 16v^2 - 4w^2 \\ &=& (9u^2 - 24uv + 16v^2) - 4w^2 \\ &=& (3u-4v)^2 - 4w^2 \\ &=& \Big( (3u-4v) - 2w \Big)\Big( (3u-4v) + 2w \Big) \\ &=& \mathbf{(3u-4v- 2w)(3u-4v + 2w)} \\ \hline \end{array}\)

In rectangle EFGH, EH = 3 and EF = 4. Let M be the midpoint of \(\overline{EF}\),
and let X be a point such that MH = MX and \angle \(MHX = 72^\circ\), as shown below.
Find \(\angle XGH\), in degrees.

\(\begin{array}{|rcll|} \hline 2x &=& 180^\circ-2*72^\circ \quad | \quad : 2\\ x &=& 90^\circ-72^\circ \\ \mathbf{x} &=& \mathbf{18^\circ} \\ \hline \end{array}\)

There are four unequal, positive integersa, b, candNsuch thatN= 5a+ 3b+ 5c.
It is also true that N= 4a+ 5b+ 4c and N is between 131 and 150.
What is the value of a+b+c?

\(\begin{array}{|rcll|} \hline N &=& 5a+ 3b+ 5c \\ -N &=& 4a+ 5b+ 4c \\ \hline 0 &=& a-2b+c \qquad \text{or} \qquad \mathbf{a+c= 2b} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline N &=& 5a+ 3b+ 5c \\ N &=& 5(a+c) + 3b \quad | \quad a+c= 2b \\ N &=& 5*2b + 3b \\ \mathbf{N} &=& \mathbf{13b} \qquad \text{, where $b$ is an integer} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \text{Multiples of $13$ between $131$ and $150$:} \\ \hline 13\times10 &=& 130 \\ 13\times11 &=& 143\ \leftarrow \ \text{ the only one} \\ 13\times12 &=& 156 \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline \mathbf{N} &=& \mathbf{13b} \quad | \quad \mathbf{N=143} \\ 143 &=& 13b \quad | \quad : 13 \\ 11 &=& b \\ \mathbf{b} &=& \mathbf{11} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{a+c} &=& \mathbf{2b} \quad | \quad \mathbf{b=11} \\ a+c &=& 2*11 \\ \mathbf{a+c} &=& \mathbf{22} \\ \hline a+b+c &=& (a+c) +b \\ a+b+c &=& 22 +11 \\ \mathbf{a+b+c} &=& \mathbf{33} \\ \hline \end{array}\)

On \(\triangle ABC\), BC is extended to E such that \(CE=\dfrac{BC}{2}\).

IF D is the mid point of AC and ED cuts AB at F,

Find the value of \(\dfrac{AB}{AF}\).

Ceva's theorem

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{BC}{CE}*\dfrac{EG}{GA}*\dfrac{AF}{BF}} &=& \mathbf{1} \\\\ \dfrac{2}{1}*\dfrac{EG}{GA}*\dfrac{AF}{BF} &=& 1 \\\\ \dfrac{EG}{GA} &=& \dfrac{1}{2}* \dfrac{BF}{AF} \\\\ \mathbf{\dfrac{GA}{EG}} &=& \mathbf{2* \dfrac{AF}{BF}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{AD}{DC}} &=& \mathbf{\dfrac{GA}{EG}+\dfrac{AF}{BF}} \\\\ \dfrac{1}{1} &=& 2* \dfrac{AF}{BF}+\dfrac{AF}{BF} \\\\ 1 &=& 3* \dfrac{AF}{BF} \\\\ \dfrac{AF}{BF} &=& \dfrac{1}{3} \\\\ \mathbf{\dfrac{BF}{AF}} &=& \mathbf{\dfrac{3}{1}} \\ \hline BF &=& \dfrac{3}{4}AB \\ AF &=& \dfrac{1}{4}AB \\\\ \dfrac{AB}{AF} &=& \dfrac{AB}{\dfrac{1}{4}AB } \\\\ \mathbf{\dfrac{AB}{AF}} &=& \mathbf{4} \\ \hline \end{array}\)

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

Find the value of n and k.

Solve system  \(\dfrac{1}{x} - \dfrac{1}{y} = 1,\ 2x + y = 7xy \quad| \quad x \ne 0,\ y\ne 0\).

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{x} - \dfrac{1}{y}} &=& \mathbf{1} \quad| \quad *xy \\\\ \dfrac{xy}{x} - \dfrac{xy}{y} &=& xy \\\\ y - x &=& xy \\ y &=& xy +x \\ y &=& x(y +1) \\\\ \mathbf{x} &=& \mathbf{\dfrac{y}{y+1}} \qquad (1) \\ \hline \end{array} \begin{array}{|rcll|} \hline \mathbf{2x+y} &=& \mathbf{7xy} \quad | \quad \mathbf{x=\dfrac{y}{y+1}} \\\\ 2*\dfrac{y}{(y+1)}+y &=& 7*\dfrac{y}{(y+1)}*y \\\\ \dfrac{2y}{(y+1)}+y &=& \dfrac{7y^2}{(y+1)}\quad| \quad *\dfrac{(y+1)}{y} \\\\ 2+y+1 &=& 7y \\ 3 &=& 6y \\\\ y &=& \dfrac{3}{6} \\\\ \mathbf{y} &=& \mathbf{\dfrac{1}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x} &=& \mathbf{\dfrac{y}{y+1}} \quad | \quad y=\dfrac{1}{2} \\\\ x &=& \frac{1}{2}\above 1pt \frac{1}{2}+1 \\\\ x &=& \frac{1}{2}\above 1pt \frac{3}{2} \\\\ x &=& \dfrac{1}{2}*\dfrac{2}{3} \\\\ \mathbf{x} &=& \mathbf{\dfrac{1}{3}} \\ \hline \end{array}\)