hairyberry

This looks intimidating, but it's completely fine!

$LO=LM=1$, and $\angle LOM={90}^{\circ}$, so because it is a right triangle is $1*1*\frac{1}{2}=\frac{1}{2}$.  The area is 1/2.

The side length of each of the equilateral triangles is 1. 

We see that 2 of such side lengths almost make the side length of the larger square, but there's still a little gap.

Each gap is half of a equilateral triangle, or a 30-60-90 triangle, and the length we want is $\frac{1}{2}$.

So the side length of the larger square is $2+\frac{1}{2}=\frac{5}{2}$.

The area is $({\frac{5}{2}})^{2}=\frac{25}{4}$. So the area of the larger square is 25/4.

$\overline{TR}=20*\frac{2}{5+2}=\frac{40}{7}$.

Set the value of $\overline{UV}=x$.

 $\because UVQP\sim TRVU$.

$\therefore \frac{TR}{UV}=\frac{UV}{PQ}$

$\therefore \frac{\frac{40}{7}}{x}=\frac{x}{20}$

${x}^{2}=\frac{800}{7}$

$x=\frac{\sqrt{800}}{\sqrt{7}}=\frac{20\sqrt{14}}{7}$

The interior angles of a regular decagon each measure $180-\frac{360}{10}={144}^{\circ}$

The angle in between two adjacent diagonals measure 36 degrees, for example $\angle P_{1}OP_{2}={36}^{\circ}$.

Set P1P2 as x, and since we don't see any good way of using 36 degrees, we turn to trigonometry. 

I chose to use the law of cosines.

Actually to be honest, I've been seeing more and more of 36 degrees with trigonometry in competition math, so it might be helpful memorizing some of those values . 

Remember that $\cos({36}^{\circ})=\frac{\sqrt{5}+1}{4}$

${x}^{2}=1+1-2*\frac{\sqrt{5}+1}{4}$.

${x}^{2}=2-\frac{\sqrt{5}+1}{2}$.

$x = \sqrt{2-\frac{\sqrt{5}+1}{2}}$

So the perimeter or $P_{1}+P_{2}\dots+P_{10}P_{1}$is $10\sqrt{2-\frac{\sqrt{5}+1}{2}}$.

The points that are 5 units away from the point (2, 7) form a circle, since the definition of a circle is the set of points so many units away from a point.

Remembering our circle formula, the equation of the circle is:

${(x-2)}^{2}+{(y-7)}^{2}=25$.

Any point that is on the circle and $y=5x-28$, must satisfy both equations, meaning it is the solution of the system:

$\begin{cases} {(x-2)}^{2}+{(y-7)}^{2}=25 \\ y=5x-28 \end{cases}$.

Substitute:

${(x-2)}^{2}+{(5x-35)}^{2}=25$

${(x-2)}^{2}+25{(x-7)}^{2}=25$

$26{x}^{2}-354x+1204=0$

$(13x-86)(x-7)=0$

$x_{1, 2}= 7, \frac{86}{13}$

Plugging back in,

$\begin{cases} x=7, y=7\\x=\frac{86}{13},y=\frac{66}{13}\end{cases}$

Our two points are $\bf{(7, 7), (\frac{86}{13}, \frac{66}{13})}$.

If we let 

$5x+5=3$, which solves to $x=-\frac{2}{5}$, then if we plug in -2/5, we get

$\frac{1}{5}*f(3)-5=\frac{4}{5}-5=-\frac{21}{5}$.

So our point is $(-\frac{2}{5}, -\frac{21}{5})$

By pythagorean theorem looking at a right triangle with the hypotenuse of the side length of the square,

${2}^{2}+{4}^{2}={c}^{2}$. c² is also the area of the square, so our area is 2²+4²=20.

$\begin{cases} a + b = 4\\ {a}^{2} b + {b}^{2} a = 6 \end{cases}$

$\begin{cases} a + b = 4\\ ab(a+b) = 6 \end{cases}$

$\begin{cases} a+b=4 \\ ab=\frac{3}{2}\end{cases}$

${a}^{3}+{b}^{3}=(a+b)({(a+b)}^{2}-3ab)$

${a}^{3}+{b}^{3}=4(16-3*\frac{3}{2})=46$