Quick and cute summer Sangaku #8

Geometry Level 5

The diagram shows an equilateral triangle of side length 1. There is a point inside the triangle, this point $\boxed{P}$ is used to form 3 triangles with areas in arithmetic progression. Then we draw the incircle of each triangle.
We denote $\boxed{k}$ as the common difference of successive areas of the three triangles with $\boxed{0<k<β}$ , it means $\boxed{k}$ is inferior to this value called $\boxed{β}$ and you need to find it
HINT : the sum of the areas of the three triangles has to be EQUAL to the area of the whole equilateral triangle
We set $\boxed{k=\frac{β}{\sqrt{3}}}$
Question : The sum of the areas of the three circles is denoted as $A$ , calculate $\lfloor 10^7A \rfloor$

1 solution

Valentin Duringer
Jul 15, 2020

If the equilateral triangle has side length 1, then its area is equal to $\boxed{\frac{\sqrt{3}}{4}}$ . We denote $\boxed{x_1}$ as the area of the smallest triangle, we can then write :
$\boxed{x_1+(x_1+k)+(x_1+2k)=\frac{\sqrt{3}}{4}}$ $<=>$ $\boxed{x_1=\frac{1}{4\sqrt{3}} - k}$
$\boxed{x_1}$ can not be negative, then $\boxed{0<k<\frac{1}{4\sqrt{3}}}$
Then $\boxed{k=\frac{β}{\sqrt{3}}=\frac{1}{12}}$

We denote $\boxed{h_1}$ the height of the smallest triangle, $\boxed{h_2}$ the height of the intermediate triangle and $\boxed{x_2}$ its area
We can use the previous area equation knowing the value of $\boxed{k}$ .
$\boxed{x_1=\frac{\sqrt{3}-1}{12}}$ then $\boxed{h_1=\frac{\sqrt{3}-1}{6}}$
$\boxed{x_2=x_1+\frac{1}{12}}$ $<=>$ then $\boxed{x_2=\frac{\sqrt{3}-1}{12}+\frac{1}{12}=\frac{\sqrt{3}}{12}}$ $\boxed{h_2=\frac{\sqrt{3}}{6}}$
Now we can draw a coordinate system. If $\boxed{[AB]}$ is the base of the equilateral triangle, then we have $\boxed{A(0;0)}$ , $\boxed{B(1;0)}$ , $\boxed{C(\frac{1}{2};\frac{\sqrt{3}}{2})}$ and $\boxed{P(x_P;\frac{\sqrt{3}-1}{6})}$
To find $\boxed{x_P}$ we shall use the minimum distance formula between point $\boxed{P}$ and the line $\boxed{(BC)}$ , its equation being $\boxed{y=-\sqrt{3}x+\sqrt{3}}$ and we know this minimum distance is $\boxed{h_2}$ :
$\boxed{h_2=\frac{\left|\sqrt{3}x_P+y_P-\sqrt{3}\right|}{4}}$ $<=>$ $\boxed{x_P=\frac{9+\sqrt{3}}{18}}$
Now we can calculate the distances $\boxed{AP}$ , $\boxed{BP}$ and $\boxed{CP}$ using the distance formula. This will allow us to know every side of each triangle
After that we will be able to get the radius of each incircle using the formula $\boxed{r=\frac{2 \times Area}{perimeter}}$ .
Then we sum the areas and get $\boxed{1839073}$ as the final answer.