Circles in a Triangle 01

Let the radius of the circles be $r$ .

Label the lower vertices of the triangle $A$ and $B$ going left to right.

Drop perpendiculars from the centers of the two lower circles to the base of the triangle. Label the two points of intersection $C$ and $D$ , again going left to right.

Then $CD = 2r$ and $AC = BD = \dfrac{4 - 2r}{2} = 2 - r$ .

Labeling the center of the lower left circle as $O$ , we then have that

$\tan(\frac{\pi}{6}) = \dfrac{OC}{AC} \Longrightarrow \dfrac{1}{\sqrt{3}} = \dfrac{r}{2 - r} \Longrightarrow$

$\sqrt{3}*r = 2 - r \Longrightarrow r = \dfrac{2}{1 + \sqrt{3}} = \boxed{0.732}$ to $3$ decimal places.

This splits the bottom side of the triangle into three parts: $2r$ and 2 parts of $r\sqrt{3}$ .

This means that $2r+2r\sqrt{3}=4$ ==> $r+r\sqrt{3}=2$

Divide by $1+\sqrt{3}$ and you get

$r=\frac{2}{1+\sqrt{3}}= \frac{2(1-\sqrt{3})} {(1+\sqrt{3})(1-\sqrt{3})}=\frac{2-2\sqrt{3}}{-2}=-1+\sqrt{3}$

$=\boxed{0.732}$

this is the exact half of the triangle . $$ we will be using the variable $\quad r$ for radius here, the $\quad r$ appear because they are from the center of the circle to a point on the circle, next,the $\quad 2r$ appear because it is the sum of the 2 radii. interestingly we have formed 3 parts that have a area that can be expressed i terms of $\quad r$ . let the remaining area be x, we know the height is $2\sqrt{3}$ by the Pythagorean theorem. $2\sqrt{3}=r^2+2r^2+\dfrac{r^2\sqrt{3}}{2}+x\longrightarrow 2\sqrt{3}= r^2(3+\dfrac{\sqrt{3}}{2})+x$ for x, we ca visualize, but here is a image we cut it off to form a different triangle. we get the values because the whole side is 4, but we did not take the 2r, so it is $\quad 4-2r\quad$ , a simlar case for $2-r$ . we get $(2-r)\sqrt{3}$ by the Pythagorean theorem. now we know its area is $\quad\dfrac{ (r^2 -4r+4)\sqrt{3}}{2}$ , we get that $2\sqrt{3}= r^2(3+\dfrac{\sqrt{3}}{2})+\dfrac{ (r^2 -4r+4)\sqrt{3}}{2} \rightarrow 4\sqrt{3}= 2r^2(3+\dfrac{\sqrt{3}}{2})+(r^2 -4r+4)\sqrt{3}$ $6r^2+2(r^2 -2r+2)\sqrt{3}=4\sqrt{3}\longrightarrow 3r^2+(r^2 -2r+2)\sqrt{3}=2\sqrt{3}$ $3r^2+r(r-2)\sqrt{3}=0\longrightarrow 3r+r\sqrt{3}=2\sqrt{3}$ $r=\dfrac{2\sqrt{3}}{3+\sqrt{3}}=\boxed{0.732}$

From center of lower left circle drop radius, r, to base of triangle. From center of lower left circle draw line to vertex of triangle. This is the hypotenuse of a small 30-60-90 triangle and must be equal to 2 r. The third side is (4 - 2r) / 2 or 2 - r. Using Pythagoreans Theorem: (2 r)^2 = r^2 + (2 - r)^2

4r^2 = r^2 + 4 - 4r + r^2

2r^2 + 4r - 4 = 0

r^2 + 2r - 2 = 0

r = ( - 2 + sqrt ( 4 - 4*(-2)))/2 = 0.732 Note: discarded second root because it is negative.

Length of altitude of given triangle =2* \sqrt{3}

let radius be r,

sin 30=r/x, where x= dist from point A to centre of 1st circle,

i.e, x=2r

now, inner circle also form a equilateral triangle of side 2r,

Length of altitude of inner equilateral triangle is \sqrt{3} r.

then total length of altitude of triangle ABC (in terms of r)= \sqrt{3} r +2r+r

i.e., \sqrt{3}r+2r+r=2 \sqrt{3}

on solving, r=0.732

I have a different solution.

Consider the triangle spanned by the centers of the circles. It has side length 2r and each side is at distance r from the side of the original triangle that is parallel to it.

Now consider in general a triangle with all sides at distance x from the original triangle's sides. The length of its sides depends linearly on the distance x. For x=0 you get the original triangle with side length 4, and for x=4sqrt(3)/6 you get a single point, side length 0. From that you get that the formula for the side length is 4 - 2*sqrt(3)*x.

Solving the equation 4 - 2*sqrt(3)*x=2*x gives the solution, 2/(1+sqrt(3)).