MaxWong

Simplifying and completing the square, we have

\(t^2 -9t - 36 + 13t - 60\\ = t^2 + 4t - 96\\ = t^2 + 4t + 4 - 100\\ = (t + 2)^2 - 100\)

Note that \((t + 2)^2 \geq 0\) for any real t, and equality holds when \(t + 2 = 0\), i.e., \(t = -2\).

Hence, the smallest value is produced when t = -2.

I assume you meant largest.

Simplifying the expression and completing the square, we have

\(2s^2 - 8s + 19 - 7s^2 - 8s + 20\\ = -5s^2 - 16s + 39\\ = -5(s^2 + \frac{16}5s) + 39\\ = -5(s + \frac85)^2 + 5(\frac 85)^2 + 39\\ =-5(s + \frac85)^2 + \frac{259}5\)

Note that \(-5(s + \frac85)^2 \leq 0\) for any real s. Then the largest possible value is \(\dfrac{259}5\).

There is no smallest value. The expression can be arbitrarily small.

I assume you mean minimum.

Simplifying and completing the square,

\(-t^2 +8t-4+5t^2-4t+18\\ =4t^2 +4t + 14\\ =4t^2 +4t+1 +13\\ =(2t+1)^2 + 13\\ \)

Note that \((2t + 1)^2 \geq 0\) for all real t. Then the minimum is 13. 

The expression can be arbitrarily large so there is no maximum.

Sorry, there is a mistake in my argument: \(\frac n2\) can in fact be 729, so the lower bound is \(n \geq 1458\). The answer is then 365 instead of 364.

Note that since the rectangles are congruent, we have DB = BF. That would imply \(\angle BDF = \angle BFD\). Also, BE is perpendicular to EF, so \(\angle BED = \angle BEF = 90^\circ\). Together with the common side BE we have the pair of congruent triangles \(\triangle DEB \cong \triangle FEB\). Consequently we have the fact that E is the centre of the semi-circle. So we now have enough information to figure out that \(\triangle DAE\) is equilateral. The rest is done through some angle chasing, which is not hard.

I think you made a typo, the bottom-left corner is likely to be (-2, -5) instead of (-2, 5).

Key observation: If a line bisects the area of a rectangle, it must pass through the intersection of diagonals, i.e., the centre of the rectangle.

(It will divide the rectangle into two identical trapezoids that way.)

We first find the intersection of diagonals, i.e., the mid-point of a diagonal. That point is \(\left(\dfrac{-2 + 4}2, \dfrac{5 + 3}2\right) = (1, 4)\).

The line has slope 1/3 and passes through (1, 4). We can find the equation of the line through "point-slope form" of straight line equations.

The equation is 

\(\dfrac{y - 4}{x - 1} = \dfrac13\\ y - 4 = \dfrac13 x - \dfrac13\\ y = \dfrac13 x + \dfrac{11}3\)

Now that the equation is in slope-intercept form, we can directly read off the y-intercept from the equation. 

The y-intercept of the line is \(\dfrac{11}3\).

You are given the dimensions of the rectangle directly, the area is of course \(WX \cdot XY = 2 \cdot 3 = 6\).

Let \(CX = k\). Then, since AX is parallel to BY, we have

\(\tan^{-1} \dfrac{k}{15} + \tan^{-1} \dfrac{3k}{20} = 60^\circ\)

Taking tan on both sides and using compound angle formula,

\(\dfrac{\frac k{15} + \frac{3k}{20}}{1 - \frac{k}{15} \cdot \frac{3k}{20}} = \sqrt 3\\ \dfrac{\sqrt 3 k}{45} + \dfrac{\sqrt 3 k}{20} = 1- \dfrac{k^2}{100}\\ \dfrac{13\sqrt 3k}{180} = 1 - \dfrac{k^2}{100}\\ k^2 - 100 = -\dfrac{65 \sqrt 3 k}9\\ k^2 + \dfrac{65 \sqrt 3}9k - 100 = 0\\ \left(k + \dfrac{65 \sqrt 3}{18}\right)^2 = 100 + \left(\dfrac{65 \sqrt 3}{18}\right)^2 = \dfrac{15025}{108}\\ k = -\dfrac{65 \sqrt 3}{18} \pm \dfrac{5 \sqrt {601}}{6 \sqrt 3} \text{(reject negative)}\\ k = -\dfrac{65 \sqrt 3}{18} + \dfrac{5 \sqrt{1803}}{18}\)

Hence, by Pythagoras' theorem, 

\(AB = \sqrt{(20 - 15)^2 + k^2} = \dfrac{5}{18} \sqrt{2634 - 78 \sqrt{601}}\)

A natural question to ask is: Does the inverse exist?

This is the graph of the y = f(x). Obviously, the graph does not pass the horizontal line test, and hence is not 1-1. 

The inverse function does not exist, unless you restrict the domain even more.

Suppose X and Y are random variables which follow the continuous uniform distribution on [0, 6]. These random variables will represent the points where we break the stick at.

The probability in question is simply expressible as \(\mathcal P (\min(X, Y) \leq 5\text{ and }\max(X, Y) \geq 1\text{ and }|X - Y| \leq 5)\).

Let \(S = \{(x, y): 0\leq x \leq 6, 0 \leq y \leq 6\}\) and \(P = \{ (x, y) \in S\mid\min(x, y) \leq 5\text{ and }\max(X, Y) \geq 1\text{ and }|X - Y| \leq 5\}\). The required probability is \(\dfrac{\text{Area}(P)}{\text{Area}(S)}\). The following is a graph of P:

Now, the probability is \(\dfrac{\text{Area}(P)}{\text{Area}(S)} = \dfrac{(5 - 1)^2}{(6 - 0)^2} = \dfrac49\).