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Thanks Heureka.  :)

Guest if you join all of  your posts will be very easy to keep track of. 

So you should join.

Let F1 and F2  be the foci of the ellipse kx^2+y^2=1 where k is greatr than 1 is a constant.
Suppose that there is a circle which passes through F1  and F2  and
which lies tangent to the ellipse at two points on the x-axis.
Compute k.

We rearrange \(kx^2+y^2=1\):

\(\begin{array}{|rcll|} \hline \mathbf{kx^2+y^2} & \mathbf{=} & \mathbf{1} \\ \\ \dfrac{x^2}{\dfrac{1}{k}}+\dfrac{y^2}{1} &=& 1 \\\\ \dfrac{x^2}{ \left(\color{red}{\dfrac{1}{\sqrt{k}}} \right)^2 }+\dfrac{y^2}{{\color{red}1}^2} &=& 1 \quad | \quad \boxed{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1 } \\\\ \mathbf{a} &\mathbf{=} & \mathbf{\dfrac{1}{\sqrt{k}}} \\ \mathbf{b} &\mathbf{=} & \mathbf{1} \\ \hline \end{array} \)

The formula with the focus of this ellipse is \(c^2= b^2-a^2\)
where \(c\) is the distance from the focus to origin.

\(\begin{array}{|rcll|} \hline c^2 &=& b^2-a^2 \quad & | \quad b=1,\ a=\dfrac{1}{\sqrt{k}} \\\\ &=& 1-\dfrac{1}{k} \\\\ \mathbf{c} &\mathbf{=}& \mathbf{\sqrt{1-\dfrac{1}{k}} } \\ \hline \end{array} \)

Circle: \(r = c=a\)

\(\begin{array}{|rcll|} \hline \sqrt{1-\dfrac{1}{k}} &=& \dfrac{1}{\sqrt{k}} \\\\ 1-\dfrac{1}{k} &=& \dfrac{1}{k} \\\\ 1 &=& \dfrac{2}{k} \\\\ \mathbf{k} &\mathbf{=}& \mathbf{2} \\ \hline \end{array}\)

Let ABCD be a convex quadrilateral, and let M and N be the midpoints of sides AD and BC, respectively. Prove that MN<=(AB+CD)/2.

Let $ABCD$ be a convex quadrilateral, and let $M$ and $N$ be the midpoints of sides $\overline{AD}$ and $\overline{BC}$, respectively.
Prove that $MN \le (AB + CD)/2$.

When does equality occur?

Rotate the convex quadrilateral:

In the triangle  C'AB we must have \(\overline{AB}+\overline{C'D'} < 2\overline{NM}\)

Since \(\overline{C'D'} = \overline{CD}\)

\(\begin{array}{|rcll|} \hline \overline{AB}+\overline{CD} < 2\overline{NM} \\\\ \dfrac{\overline{AB}+\overline{CD}}{2} < \overline{NM} \\ \hline \end{array}\)

Equality occur if the convex quadrilateral is a trapezoid:

\(\begin{array}{|rcll|} \hline \overline{AB}+\overline{CD} &=& 2 \overline{NM} \\\\ \dfrac{\overline{AB}+\overline{CD}}{2}& =& \overline{NM} \\ \hline \end{array}\)

For all real numbers $r$ and $s$, define the mathematical operation $\#$ such that the following conditions apply:

$r\ \#\ 0 = r, r\ \#\ s = s\ \#\ r$, and $(r + 1)\ \#\ s = (r\ \#\ s) + s + 1$.

What is the value of $11\ \#\ 5$

\(\begin{array}{|rcll|} \hline \mathbf{(r + 1)\ \#\ s }& \mathbf{=} & \mathbf{(r\ \#\ s) + s + 1} \quad \text{ or } \quad \mathbf{r\ \#\ s = \Big((r-1)\ \#\ s\Big) + s + 1} \\ \hline \\ r\ \#\ s &=& \Big((r-1)\ \#\ s\Big) + s + 1 \quad | \quad (r-1)\ \#\ s = \Big((r-2)\ \#\ s\Big) + s + 1 \\ &=& \Big((r-2)\ \#\ s\Big) + s + 1 + s + 1 \\ &=& \Big((r-2)\ \#\ s\Big) + 2s + 2 \quad | \quad (r-2)\ \#\ s = \Big((r-3)\ \#\ s\Big) + s + 1 \\ &=& \Big((r-3)\ \#\ s\Big) + s + 1 + 2s + 2 \\ &=& \Big((r-3)\ \#\ s\Big) + 3s + 3 \quad | \quad (r-3)\ \#\ s = \Big((r-4)\ \#\ s\Big) + s + 1 \\ &=& \Big((r-4)\ \#\ s\Big) + s + 1 + 3s + 3 \\ &=& \Big((r-4)\ \#\ s\Big) + 4s + 4 \\ \ldots \\ \mathbf{r\ \#\ s} &\mathbf{=}& \mathbf{ \Big((r-n)\ \#\ s\Big) + ns + n} \qquad n\in \mathbb{Z} \\ \hline \end{array} \)

\(\mathbf{11\ \#\ 5 = \ ?}\)

\(\begin{array}{|rcll|} \hline \mathbf{r\ \#\ s} &\mathbf{=}& \mathbf{ \Big((r-n)\ \#\ s\Big) + ns + n} \quad &| \quad r=n=11,\ s=5 \\ 11\ \#\ 5 & = & \Big((11-11)\ \#\ 5\Big) + 11\cdot 5 + 11 \\ 11\ \#\ 5 & = & \Big(0\ \#\ 5\Big) + 11\cdot 5 + 1 \quad &| \quad 0\ \#\ 5 = 5\ \#\ 0 \\ 11\ \#\ 5 & = & \Big(5\ \#\ 0\Big) + 11\cdot 5 + 11 \quad &| \quad 5\ \#\ 0 = 5 \\ 11\ \#\ 5 & = & 5 + 11\cdot 5 + 11 \\ \mathbf{11\ \#\ 5} & \mathbf{=} & \mathbf{71} \\ \hline \end{array}\)

Tan 18 = 3/(5+x)

x = 4.23 ft

positive - length of a piece of metal and it's area (the metal will expand with temperature)

negative - the number of dates you get a year vs. the number of days you shower a year (shower less expect fewer dates)

you get the idea

The hypotenuse of a right triangle is the ramp's 7 ft length

sin 2  = opposite/hypotenuse

hypotenuse sin 2 = opposite

7 sin 2 = .24 ft  rise ~ .2 ft rise (nearest tenth)

\(|t+2i\sqrt{3}||6-4i|=26\\ |t+2i\sqrt{3}|^2|6-4i|^2=676\\ (t^2+12)(36+16)=676\\ t^2 + 12 = 13\\ t^2 = 1\\ t = \pm 1\\ \text{The positive value of }t \text{ satisfying above is }1\)

\(\text{Rational solutions means the discriminant is a positive perfect square}\\ D = 10^2 - 4(k)(k) = 10^2 - 4k^2 = 100-4k^2\\ \text{The perfect squares less than 100 are}\\ 81, 64, 49, 36, 25, 16, 9, 4, 1\\ 100-4k^2 \text{ is going to be even so we can ignore the odd squares}\\ 64 = 100- 36 = 100-4(3)^2,~k=3 \text{ works}\\ 36=100-64^2 = 100-4(4)^2,~k=4 \text{ works}\\ 16 = 100-84 = 100 - 4(21), \text{ nope}\\ 4 = 100-96 = 100-4(24),~\text{ nope}\\ \text{only }k=3,4\)

The derivation of the answer is based on a right triangle with opposite leg equal to x.

Triangles don't have negative length legs. 

But you are correct that it should be mentioned that 0<= x <=6

We all make mistakes. It is the askers responsibility to check the answer 

And the askers choice whether to accept or reject an answer.

So chill !  

I already said it all here

ARRRGH checking back I find a mistake I made. 

Second line should say 3.14 x 28 ft, not 24 

I swear, I did proof read it before I posted.

(sin(-4pi/3))(cos(11pi/6))-(tan(5pi/6))(cos(pi/3))

\(sin(\frac{-4\pi}{3})(cos(\frac{11\pi}{6})- tan(\frac{5\pi}{6})(cos\frac{\pi}{3})\\ =sin(2\pi+\frac{-4\pi}{3})(cos(2\pi-\frac{11\pi}{6}))- tan(\frac{5\pi}{6})(cos\frac{\pi}{3})\\ =sin(\frac{2\pi}{3})(cos(\frac{\pi}{6}))- tan(\frac{5\pi}{6})(cos\frac{\pi}{3})\\ =sin(\frac{\pi}{3})(cos(\frac{\pi}{6}))-- tan(\frac{\pi}{6})(cos\frac{\pi}{3})\\ =\frac{\sqrt3}{2}\times\frac{\sqrt3}{2}+ \frac{1}{\sqrt3}\times \frac{1}{2}\\ =\frac{3}{4}+ \frac{1}{2\sqrt3}\times \frac{\sqrt3}{\sqrt3}\\ =\frac{3}{4}+ \frac{\sqrt3}{6}\\ =\frac{9}{12}+ \frac{2\sqrt3}{12}\\ = \frac{9+2\sqrt3}{12}\\ \)

The answer is 33

Big Ben, the famous clock in London, has a minute hand that is 14 feet long. How far does the tip of the minute hand of Big Ben travel in 20 minutes? Round your answer to the nearest tenth.
 

The minute hand is the radius of the circle it traverses, so the diameter of that circle is 28 feet.

The circumference of a circle with a diameter of 28 feet = pi x 28 ft = 3.14 x 24 ft = 87.92 ft

20 minutes is a third of the way around a 60 minute clock, so the end of the minute hand travels 87.92/3 = 29.3 ft

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dgf.....

Is that how you thank someone when they do hours ( quite possibly)  of work for you at no cost to you ?  

I wonder how you would feel if it was the other way around?

I wonder if you even would ever give a stranger hours of your time for free? 

I notice that you gave yourself a point for your questions but never even gave Heureka one for his beautifully laid out answer.

No point, no thankyou.

It seems that my first impression of you was correct after all.

it says a convex quadrilateral

Is there a way to do this without calc?

Thanks you. This is a great solution! But is there a way to do it without calc, because in this competition, calculators are not allowed.