---

title: Mathex Special 2024 01
tags: Mathex Specials
description: Mathex Specials No. 1
type: pdf
slideOptions:
theme: white

---

###### tags: `OTM`

## Q20_M2020


This question needs some extra knowledge about the area of a triangle under fixed perimeters.

The question is asking for the difference between the area of the maximum and the area of the mininum. Let's use an expression to clarify it:

$$A_{max} - A_{min}$$

For $A_{max}$, you can now just remember this simple rule:

- **For triangles with fixed perimeters, the equilateral triangle has the largest erea**.

So in our case, each side using 5 units of sticks, so the Max erea is $A_{max} = \frac{1}{2}\times 5\times \sqrt{5^2 - 2.5^2} = 10.825\ unit^2$

For $A_{min}$, we can use the Heron's formula:

$$A = \sqrt{S(S-a)(S-b)(S-c)}$$, where $a, b, c$ are lengths of three sides, $S=\frac{a+b+c}{2}$

The fomula above is a general formula for triangle given lengths of three sides. To work out the $A_{min}$, we need some arrangement and trial.

Firstly, we can rearrange the S to become $c = 2S - a -b$, then we plug-in the c into the Heron's Formula:

$$A = \sqrt{S(S-a)(S-b)[S-(2S - a -b)]}$$
$$=\sqrt{S(S-a)(S-b)(a+b-S)}$$
$$=\sqrt{7.5(7.5-a)(7.5-b)(a+b-7.5)}$$
$$=\sqrt{7.5(a-7.5)(b-7.5)(a+b-7.5)}$$

Here $S = 7.5$ as it is the half of the perimeter.

We want the area as small as posible, we need:

- a + b as small as possible
- product of three bracket must be positive

Now let's try some combination of a and b:

- if a = 1, b = 1, we have a+b = 2, and the other two bracket must be negative, so three negatives, give us negative.

- So we have to make $(a+b)>7.5$, so the first choice is $(a+b) = 8$. Then we can try different combination of a and b, for example, a = 1 and b = 7. Then the equation of area become

$$\sqrt{7.5(1-7.5)(7-7.5)(8-7.5)}=3.491\ unit^2$$

- Actually this is the $A_{min}$, if you don't believe me, you can try other combinations, such as 2 and 6, 3 and 5, 4 and 4.

So the final answer would be $10.825-3.491=7.334\ unit^2$

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## Q18_M2020


This is also a hard question. I think it is too hard for Mathex level.

- For a chessboard, the only possible dimension is 8*8,
- So we have 9 horizental lines and 9 vertical lines,
- We firstly pick up 2 lines from 9 horizentoal, we have 36 choices;
- Then we pick up 2 lines from 9 verticals, we have 36 choices
- then any 2 vercicals and 2 horizentals form a square or rectangles, so we have 1296 squares or rectangles, minus 204 squares, we have 1092 rectangles.