The trapped gemstone

Let the lengths of the sides of the golden triangle in increasing order be $c, d, 2d-c$ . Let the center of the circle on left be $O$ , that of the circle on the right be $P$ , and the vertices of the different colored triangles on the line $\overline {OP}$ from $O$ to $P$ be $A, B, C$ , and the common vertex of these triangles be $D$ .

Then $|\overline {AC}|=2c-d$

From the heights given, we can get the following relations

$d=\dfrac {146}{123}c$

$2d-c=\dfrac {169}{123}c$

Since lengths of all the sides of the golden triangle are integers, their minimum values are $123,146,169$

Then $|\overline {OB}|=\sqrt {123^2-(\frac{17520}{169})^2}$

$=\dfrac {11187}{169}$

$\implies |\overline {BC}|=\dfrac {9600}{169},|\overline {AC}|=100$ ,

$|\overline {AB}|=\dfrac {7300}{169},|\overline {OA}|=\dfrac {3887}{169}$ ,

$|\overline {CP}|=46$

Now, $\dfrac {R}{B}+\dfrac {G}{P}=\dfrac {|\overline {OA}|}{|\overline {CP}|}+\dfrac {|\overline {AB}|}{|\overline {BC}|}$

$=\dfrac 12+\dfrac {73}{96}=\dfrac {121}{96}$

So, $a=121,b=96$ and $\sqrt {a-b}=\boxed 5$

• We denote the sides of the golden triangle as : $x-k$ , $x$ , $x+k$ with k being the common difference between sides.
• We use this useful property : $\boxed{\frac{1}{r}=\frac{1}{h_1}+\frac{1}{h_2}+\frac{1}{h_3}}$ which allows us the solve for the inradius. : $\boxed{r=40}$
• Then we use Euler's theorem to calculate the circumradius, knowing the value of the inradius and the distance between them : $\boxed{(R-r)²=d²+r²}$ $<=>$ $\boxed{R=\frac{6929}{80}}$
• Now, we use two formulas : $\boxed{R=\frac{a\cdot b\cdot c}{4\cdot Area}}$ and $\boxed{r=\frac{2\cdot Area}{a+b+c}}$
• Then : $\boxed{R\cdot r=\frac{a\cdot b\cdot c}{2\cdot \left(a+b+c\right)}}$ $<=>$ $\boxed{\frac{\left(x-k\right)\cdot x\cdot \left(x+k\right)}{6\cdot x}=\frac{6929}{2}}$
• We obtain : $\boxed{x=\sqrt{k^2+20787}}$
• Since $\boxed{x}$ is an integer, then $\boxed{k^2+20787}$ has to be a perfect square and we find 5 values for $\boxed{x;k}$ and valid one is $\boxed{x=146}$ and $\boxed{k=23}$
• Then the sides of the triangles are $\boxed{123}$ , $\boxed{146}$ and $\boxed{169}$

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• The base a the red triangle has to be $\boxed{169-146=23}$
• The base a the blue triangle has to be $\boxed{169-123=46}$
• Using the altitude relative the base of the golden triangle and the pythagorean theorem we get :
• The base a the green triangle is $\boxed{\frac{7300}{169}}$
• The base a the green triangle is $\boxed{\frac{9600}{169}}$

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• Since the 4 small triangles have the same height, the ratio of their areas is the ratio of the bases.
• $\boxed{\frac{\color{#D61F06}R}{\color{#3D99F6}B}+\frac{\color{#20A900}G}{\color{#69047E}P}=\frac{23}{46}+\frac{\frac{7300}{169}}{\frac{9600}{169}}=\frac{121}{96}}$
• $\boxed{\sqrt{121-96}=5}$