Yellow Area

[Answer to Bonus]

From the preface of this daily challenge , we see that if the base of the triangle is $\frac{1}{n} \times s$ , then it's height will always be $\frac{1}{n+1} \times s$ . Together the area is:

$\begin{aligned} \color{#E81990}{\text{Pink Triangle}} &= \dfrac{1}{2}\bigg(\dfrac{1}{n} \cdot s \cdot \dfrac{1}{n+1} \cdot s\bigg) \\ \\ &= \dfrac{s^2}{2(n)(n+1)} \end{aligned}$

$\begin{aligned} \color{#CEBB00}{\text{Yellow Quadrilateral}} &= \dfrac{s^2}{2} - \color{#E81990}{\text{Pink Triangle}} \\ \\ &= \dfrac{s^2}{2} - \dfrac{s^2}{2(n)(n+1)} \end{aligned}$

Label the figure as above. Since the side length of the square is $s$ , $DG = \dfrac sn$ . Let $EF$ be perpendicular to $CD$ . Then $\triangle DEF$ and $\triangle DBC$ are similar. If $DF=a$ , $EF=DF=a$ . Also $\triangle EFG$ and $\triangle ADG$ are similar. Then $\dfrac {FG}{EF} = \dfrac {DG}{AD} = \dfrac 1n$ $\implies FG = \dfrac {EF}n = \dfrac an$ . Now $DG=DF+FG = a + \dfrac an = \dfrac sn$ , $\implies a = \dfrac s{n+1}$ .

Then the area of the pink $\triangle DEG$ , $[DEG]= \dfrac as \left(a+\dfrac an\right) = \dfrac {(n+1)a^2}{2n} = \dfrac {s^2}{2n(n+1)} = \dfrac {A_\square}{2n(n+1)}$ , where $A_\square=s^2$ is the area of the square.

The area of the yellow quadrilateral $BCGE$ , $A_{\color{#CEBB00}\text{yellow}} = [BCD]-[DEG] = \dfrac {A_\square}2 - \dfrac {A_\square}{2n(n+1)} = \dfrac {n^2+n-1}{2n(n+1)}A_\square$ .

For $n=2$ and $A_\square$ , $A_{\color{#CEBB00}\text{yellow}} = \dfrac {4+2-1}{2(2)(2+1)} \times 12 = \boxed 5$ .

I've come up with an interesting solution:

And then the answer is

$\int_0^{\sqrt{12}}\min(\frac{x}{2}+\sqrt{3},-x+\sqrt{12})\rm dx$

Calculate it, and you get

$\boxed5$

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Another way:

Suppose the blue triangle on the right is $A$ , and the blue triangle on the left is $B$ .

Because the distance from the point on the bisector of the angle to both sides of the angle is equal, the length of the orange line is equal to the length of the black line.

So the area of $B$ is $2$ times the area of the pink triangle (the height is equal, and the base is twice).

And because $A$ is similar to the pink triangle, the ratio of the sides is $2:1$ , so the area ratio is $4:1$ , that is, the area of $A$ is the pink triangle $4$ times.

So the area of the entire big blue triangle (that is, $A+B$ ) is six times that of the pink triangle, and because its area is equal to half the area of the square, which is $6$ . So the area of the pink triangle is $1$ .

So the area of the yellow triangle is half the area of the square minus the area of the pink triangle:

$6-1=\boxed{5}$

For any pink triangle with the base 1/n of the square's side, the yellow quadrilateral is always n²+n-1 times larger.

This can also be solved using coordinate geometry.

Let the bottom left corner of the square be the origin. Then the top left corner of the square has coordinates $(0, s)$ and the bottom right corner of the pink triangle has coordinates $(\frac{s}{n}, 0)$ , and the line through these two points is $y = -nx + s$ . Since the diagonal line has an equation of $y = x$ , these two lines intersect at $(\frac{s}{n + 1}, \frac{s}{n + 1})$ .

Since the pink triangle has a base of $\frac{s}{n}$ and a height of $\frac{s}{n + 1}$ , its area is $A_{\text{pink}} = \frac{s^2}{2n(n + 1)}$ .

Since the yellow quadrilateral is half the square minus the pink triangle, its area is $A_{\text{yellow}} = \frac{s^2}{2} - \frac{s^2}{2n(n + 1)}$ .

In this problem, $s^2 = 12$ and $n = 2$ , so $A_{\text{yellow}} = \frac{12}{2} - \frac{12}{2 \cdot 2(2 + 1)} = \boxed{5}$ .

As $\overline{BD}$ is the diagonal $\Rightarrow$ area of $\blue{blue}$ region $=\frac{12}{2}=\boxed{6}$

Area of square = $\overline{AB}^2=12\Rightarrow AB=2\sqrt{3}=\overline{BC}$

As $\angle DFE$ and $\angle BFA$ are opposite to each other $\therefore \boxed{\blue{\angle DFE=\angle BFA}}$

As $ABCD$ is a square $\therefore AB || CD \Rightarrow \boxed{\blue{\angle FDE=\angle FBA}}$

From the above two statements we deduce that $\triangle FDE$ ~ $\triangle FBA$ $.......[1]$

Let the ratio of the altitude of $\triangle FAB$ to $\triangle FDE$ be $k$

From $[1]$ : $\frac{\overline{AB}}{\overline{DE}}=k \Rightarrow 2=k$

Let the altitude of be 0 $x$

Therefore altitude of $\triangle FAB=2x$

From the diagram: $2x+x=\overline{AB}\Rightarrow 3x=2\sqrt{3}\Rightarrow \boxed{\blue{x=\frac{2}{3}\sqrt{3}}}$

Area of $\pink{pink}$ region $=\frac{1}{\cancel{2}}\sqrt{3}\times \frac{\cancel{2}}{3}\sqrt{3}=\frac{3}{3}=\boxed{\blue{1}}$

Area of ${yellow}$ region $=12-1-6=12-7=\boxed{5}$