MaxWong

We first find the radius,

Let r be the radius.

\(\dfrac{40^\circ}{360^\circ} \cdot \pi r^2 = 4\pi\\ r = 6\)

Then we can use the formula \(P = \dfrac{\theta}{360^\circ} \cdot \left(2\pi r\right) + 2r\) to calculate the perimeter of the sector.

I believe you can finish it on your own.

PROBABILITY has 11 letters.

The most straightforward thought would be the factorial of 11. But there are repeating letters, so we would have double-counted if we say the number of ways is 11!.

In fact, we need to divide by the factorial of the number of each type of repeating letters.

The number of ways would be \(\dfrac{11!}{\color{green}2!\color{blue}2!\color{black}} = 9979200\)

The green one is for repeating 'B's, and the blue one is for repeating 'I's.

We can parameterize the coordinates of P.

\(P = (t, 5t + 3)\) for \(t\in\mathbb R\)

We use the formula to calculate the midpoint of PQ.

\(M = \left(\dfrac{t + 3}{2}, \dfrac{5t + 3 - 2}{2}\right) = \left(\dfrac{t + 3}{2}, \dfrac{5t +1}{2}\right)\)

Then, we let M_x be the x-coordinate of M and M_y be the y-coordinate of M. 

We then find the relationship between M_x and M_y.

Consider 10M_x.

\(10M_x = 5(t + 3) = 5t + 15 = (5t + 1) + 14 = 2M_y + 14\)

\(10M_x - 2M_y - 14 = 0\\ 5M_x - M_y - 7 = 0\)

Therefore the point M must lie on the line \(5x - y - 7 = 0\)

Have you heard of the "inverse Pythagorean theorem"?

Let a and b be the legs of a right triangle, and h be the altitude to the hypotenuse.

The inverse Pythagorean theorem states that \(h^{-2} = a^{-2} + b^{-2}\).

We can use this equation to solve this problem.

Let h be the altitude to the hypotenuse.

\(\dfrac1{h^2} = \dfrac1{8^2} + \dfrac1{15^2}\\ h^2 = \dfrac{120^2}{8^2 + 15^2} = \left(\dfrac{120}{17}\right)^2\\ h = \dfrac{120}{17}\)

Assuming \(x_y\) means "x in base y".

The equation given implies 

\(7\Diamond + 4 = 8\Diamond + 1\)

The above result is obtained by changing both sides to base 10.

Solving, we have \(\Diamond = 3\).

Let \(H\in AB\) such that \(PH \perp AB\) and let \(T\in CD\) such that \(PT \perp CD\)

\(\dfrac{AB}2 = \dfrac{AB\cdot PH}{2}\\ PH = 1\)

Let x be the side length of the square and considering similar triangles,

\(x = \dfrac1{x - 1}\)

I will leave the solving process for you as an exercise.

At the end, we get a beautiful result, which is \(x = \varphi\), where \(\varphi\) is the golden ratio.

Let BC = a, AC = b.

Now, \(\overline{AM} = \sqrt{b^2 + \left(\dfrac a2\right)^2} = \dfrac12 \sqrt{4b^2 + a^2}\)

and \(\overline{BN} = \sqrt{a^2+ \left(\dfrac{b}{2}\right)^2} = \dfrac12 \sqrt{4a^2 + b^2}\)

According to the question, AM = 19 and BN = 13.

\(4b^2 + a^2 = 38^2 \\4a^2 + b^2 = 26^2\)

Adding gives

\(a^2 + b^2 = \dfrac{38^2 + 26^2}{5}\)

The length of the hypotenuse is \(\sqrt{a^2+ b^2} = 2\sqrt{106}\)

Let AZ = a, ZW = b.

We know that \(b^2 - a^2 = 273\)

The yellow area, using the formula for area of trapezoid, is \(\dfrac{(a + b)(b - a)}2\).

You can continue from here. This is just one step away from the answer.

‖a⃗‖ = \(\sqrt{(4 - (-2))^2 + (-2 - (-5))^2} = \sqrt{6^2 + 3^2} = 3\sqrt 5\)