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Relevant wiki: Length and Area - Composite Figures - Intermediate

Solution 1

Area of blue triangle at the bottom = quarter of square * length ratio squared * 2 $=\frac{1\times 1}{2}\times\left(1-\sin\frac{\pi}{6}\right)^2\times 2=\frac{1}{4}$

Area of triangle $=ab\sin C\\=\frac{1}{2}a\left(\frac{a\sin B}{\sin A}\right)\sin C\\=\frac{a^2\sin B\sin C}{2\sin A}\\=\frac{a^2\sin B\sin C}{2\sin\left(\pi-B-C\right)}\\=\frac{a^2\sin B\sin C}{2\sin\left(B+C\right)}\\=\frac{a^2\sin B\sin C}{2\left(\sin B\cos C+\cos B\sin C\right)}\\=\mathbf{\frac{a^2}{2\left(\cot B+\cot C\right)}}$

Area of each triangle at the top $=\frac{\left(\sqrt{2}\right)^2}{2\left(\cot\frac{\pi}{12}+\cot\frac{\pi}{2}\right)}\\=2-\sqrt{3}$

Overlapping area $=2-\frac{1}{4}-\left(2-\sqrt{3}\right)\times 2\\\mathbf{=2\sqrt{3}-\frac{9}{4}}$

Answer $=2\times 3 + \left(-9\right)\times 4\\\mathbf{=-30}$

Solution 2

Similarly,

Overlapping area = Area of equilateral triangle - Area of red acute triangle * 2

Area of equilateral triangle $=\left(\frac{1}{2}\times 1\times 1\times\sin\frac{2\pi}{3}\times 3\right)=\frac{3\sqrt{3}}{4}$

Base of each red acute triangle (horizontal side) $=\frac{1\times\sin\frac{\pi}{3}\times 2-1}{2}=\frac{\sqrt{3}-1}{2}$

Area of each red acute triangle $=\frac{\left(\frac{\sqrt{3}-1}{2}\right)^2}{2\left(\cot\frac{\pi}{3}+\cot\frac{\pi}{4}\right)}=\frac{9}{8}-\frac{5\sqrt{3}}{8}$

Overlapping area $=\frac{3\sqrt{3}}{4}-\left(\frac{9}{8}-\frac{5\sqrt{3}}{8}\right)\times 2\\\mathbf{=2\sqrt{3}-\frac{9}{4}}$

Answer $=2\times 3 + \left(-9\right)\times 4\\\mathbf{=-30}$

Area of square = As = 2

Area of lateral triangle = At = $\frac{1}{2} \sqrt2 \times \sqrt2 \tan 15 = 2 - \sqrt3$

Area of bottom triangle = Ab = $\frac{1}{2} \times \frac{1}{2} \times 1 = \frac{1}{4}$

Overlapping area = As - Ab - 2At = $2 - \frac{1}{4} - 2 (2 - \sqrt3) = 2 \sqrt3 - \frac{9}{4}$

So ab + cd = (2)(3) + (-4)(9) = -30

The same solution has been posted above by my friend vishwash but the solution is not easy to read as it has small letters so i have posted the same solution here with large letters .

I try to solve this using Analytic Geometry, also known as Coordinate Geometry.

Let the center point of the circle is $(0,0)$ and satisfy the equation $x ^ 2 + y ^ 2 = 1$ . To find the area of polygon $ACEGI$ we need to find the coordinate of point $A$ , $C$ , $E$ , $G$ and $I$ .

From the picture above we see that the coordinate of point $A$ is $(0,1)$ .

First, find the equation of line $\overline{DH}, \overline{AD}, \overline{AH}, \overline{BF} and \overline{FJ}.$ .

Since $\overline{OD} = 1$ , $\angle ODK = 30 ^ \circ$ , and $\sin 30^\circ = 0.5$ , we know that $\overline{OK} = 0.5$ .

So, the equation of line $\overline{DH}$ is $y = -0.5$ .

Take $y = -0.5$ and substitute to $x ^ 2 + y ^ 2 = 1$ , we found that point $D (-0.5\sqrt{3}, -0.5)$ and $H (0.5\sqrt{3}, -0.5)$ .

From $A (0,1)$ and $D (-0.5\sqrt{3}, -0.5)$ , equation of line $\overline{AD}$ is $y = \sqrt{3} x + 1$ .

From $A (0,1)$ and $H (0.5\sqrt{3}, -0.5)$ , equation of line $\overline{AH}$ is $y = - \sqrt{3} x + 1$ .

From $B (-1,0)$ and $F (0,-1)$ , equation of line $\overline{BF}$ is $y = - x - 1$ .

From $J (1,0)$ and $F (0,-1)$ , equation of line $\overline{FJ}$ is $y = x - 1$ .

The intersection of line $\overline{AD}$ and $\overline{BF}$ produce point $C (- \sqrt{3} + 1 , \sqrt{3} - 2)$ .

The intersection of line $\overline{BF}$ and $\overline{DH}$ produce point $E (-5, -5)$ .

The intersection of line $\overline{DH}$ and $\overline{FJ}$ produce point $G (5, -5)$ .

The intersection of line $\overline{AH}$ and $\overline{FJ}$ produce point $I ( \sqrt{3} - 1 , \sqrt{3} - 2)$ .

Using the method Area of Irregular Polygon - Coordinate Geometry by using coordinate of point $A$ , $C$ , $E$ , $G$ , and $I$ .

So, we found that the area of polygon $ACEGI$ which is the overlapping area of square and triangle is

Overlapping Area $= \begin{vmatrix} 0 & 1 & 1 \\ \sqrt{3}-1 & \sqrt{3}-2 & 1 \\ 0.5 & -0.5 & 1 \\ -0.5 & -0.5 & 1 \\ -\sqrt{3}+1 & \sqrt{3}-2 & 1 \\ 0 & 1 & 1 \end{vmatrix}$

Overlapping Area $= 2 \sqrt{3} - \frac {9}{4}$ .

Answer: $ab + cd = 2 \cdot 3 + (-9) \cdot 4 = \textbf{-30}$ .

There are two main approaches to this problem.
Consider the Square and take away from the areas of the three triangles that do not overlap. Ujjwal Rane and Lolly Lau has done it below.
Second approach to take away from the areas of the equilateral triangle, the areas of the two identical triangles not overlapping. There are three diffrent solutions for this given below as of to day. I add the fourth below.

This is a more detailed version of @Lolly Lau 's Solution 2

First of all, we need to find the lengths of the square and the triangle. Refer to the diagram below (ignore the A at the bottom left)

Remember, a unit circle has a radius of $1$ .

First, notice that $AO$ and $DO$ are the radius of the circle and the diagonals of the square $ADEF$ . These two lines form a right angle at $O$ , and we can use Pythagoras' Theorem to find the side of the square:

$AD = \sqrt{DO^2+AO^2} = \sqrt{1^2+1^2} = \sqrt{2}$

Next, notice that $AO$ and $CO$ are the angle bisectors of $\angle BAC$ and $\angle BCA$ respectively. This implies that $\angle OAC = \angle OCA = 30^{\circ}$ . Therefore, $\angle AOC = 120^{\circ}$ , and we can use sine rule to find the side of the triangle:

$\dfrac{AC}{\sin \angle AOC}=\dfrac{OC}{\sin \angle OAC} \implies AC = \dfrac{1 \times \sin 120^{\circ}}{\sin30^{\circ}}= \dfrac{\sqrt{3}}{2} \div \dfrac{1}{2} = \sqrt{3}$

Now, to find the overlapping area, we can calculate the area of the equilateral triangle, and subtract the two small red triangles (one of them is labelled as $\triangle BGH$ in the diagram)

Overlapping area $=$ Area of $\triangle ABC - 2 \times$ Area of $\triangle BGH$

Finding the area of $\triangle ABC$ is easy:

Area of $\triangle ABC = \dfrac{1}{2}\left(\sqrt{3}\right)^2 \sin60^{\circ} = \dfrac{3}{2} \left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{3\sqrt{3}}{4}$

Finding the area of $\triangle BGH$ is NOT easy, but you can do it step by step. First, we will need to find the length of $AG$ :

$\angle DAG = \dfrac{\angle DAF - \angle BAC}{2} = \dfrac{90^{\circ}-60^{\circ}}{2} = 15^{\circ}$

Therefore, $\angle DGA = 75^{\circ}$

Using sine rule and the formula for addition of angles , you can find the length of $AG$ in terms of square roots:

$\sin 75^{\circ}\\ = \left(\sin(45^{\circ}+30^{\circ})\right)\\ =\left(\sin45^{\circ}\cos30^{\circ}+\sin30^{\circ}\cos45^{\circ}\right) \\ =\left(\dfrac{1}{\sqrt{2}} \times \dfrac{\sqrt{3}}{2} + \dfrac{1}{2} \times \dfrac{1}{\sqrt{2}}\right)\\ =\dfrac{\sqrt{3}+1}{2\sqrt{2}}$

$\dfrac{AG}{\sin \angle ADG} = \dfrac{AD}{\sin \angle DGA}\\ AG = \dfrac{\sqrt{2} \sin 90^{\circ}}{\sin 75^{\circ}}\\ =\dfrac{\sqrt{2}(1)}{\frac{\sqrt{3}+1}{2\sqrt{2}}}\\ =\dfrac{4}{\sqrt{3}+1}\\ =\dfrac{4(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}\\ =\dfrac{4(\sqrt{3}-1)}{2}\\ =2(\sqrt{3}-1)\\ =2\sqrt{3}-2$

With this, we can calculate the lengths of sides of $\triangle BGH$ :

$BG = AB - AG = \sqrt{3} - (2\sqrt{3}-2) = 2 -\sqrt{3}$

$\angle BGH$ and $\angle DGA$ are opposite angles, therefore $\angle BGH = 75^{\circ}$ , and $\angle BHG = 45^{\circ}$

$\dfrac{BH}{\sin \angle BGH} = \dfrac{BG}{\sin \angle BHG}\\ BH=\dfrac{(2-\sqrt{3})\sin 75^{\circ}}{\sin 45^{\circ}}\\ =(2-\sqrt{3})\left(\dfrac{\sqrt{3}+1}{2\sqrt{2}}\right) \div \dfrac{1}{\sqrt{2}}\\ =\dfrac{(2-\sqrt{3})(\sqrt{3}+1)(\sqrt{2})}{2\sqrt{2}}\\ =\dfrac{2\sqrt{3}+2-\sqrt{3}-3}{2}\\ =\dfrac{\sqrt{3}-1}{2}$

Therefore, area of $\triangle BGH = \dfrac{1}{2} \left(\dfrac{\sqrt{3}-1}{2}\right)(2-\sqrt{3})\sin60^{\circ} = \dfrac{2\sqrt{3}-2-3+\sqrt{3}}{4} \left(\dfrac{\sqrt{3}}{2}\right)=\dfrac{\sqrt{3}(3\sqrt{3}-5)}{8}=\dfrac{9-5\sqrt{3}}{8}$

Overlapping area $=$ Area of $\triangle ABC - 2 \times$ Area of $\triangle BGH$

$=\dfrac{3\sqrt{3}}{4}-2\left(\dfrac{9-5\sqrt{3}}{8}\right)\\ =\dfrac{3\sqrt{3}-(9-5\sqrt{3})}{4}\\ =\dfrac{8\sqrt{3}-9}{4}\\ =2\sqrt{3}-\dfrac{9}{4}$

$\implies a=2,\;b=3,\;c=-9,\;d=4,\;ab+cd = 2(3)+(-9)(4)=6-36=\boxed{-30}$