A Pre-RMO question! -24

This is a very good question!

Let $AP$ be $x$

$(8^2+x^2) + ( 8^2 + (20-x)^2 ) = 20^2$

$128 + 2x^2 - 40x + \cancel{400} = \cancel{400} \\ x^2 - 20x + 64 = 0 \implies (x-4)(x-16) = 0 \implies x = 4 or 16$

After finding the length of cutting at P, 4 and 16

• The length from B to the in-circle $= 8-r_1$
• The length from P to the in-circle $= 4-r_1$
• The length $BP = a = 8-r_1 + 4 - r_1 = 12 - 2r_1$

Similarly, we have the following equations:

• $12 - 2r_1 = a$
• $24 - 2r_2 = b$
• $a+b-2r_3= 20$

Sum up the 3 equations

$36 -2(r_1 + r_2 + r_3) + \cancel{a+b} = \cancel{a+b}+20 \\ 2(r_1 + r_2 + r_3) = 36 -20 = 16 \\ \therefore \boxed{r_1 + r_2 + r_3 = 8}$

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Finally, I found out that we can skip to find out the value of x

• The length from B to the in-circle $= 8-r_1$
• The length from P to the in-circle $= x-r_1$
• The length $BP = a = 8-r_1 + x - r_1 = 8 + x - 2r_1$

Similarly, we have the following equations:

• $8 + x - 2r_1 = a$
• $28 -x - 2r_2 = b$
• $a+b-2r_3= 20$

Sum up the 3 equations and solve for $\boxed{r_1 + r_2 + r_3 = 8}$

$\angle DPC = \angle PCB = 180^\circ - \angle PBC = \angle ABP$ , hence the right-angled triangles $\triangle PDC$ , $\triangle CPB$ , and $\triangle BAP$ are similar.

Thus, $\dfrac{{AP}}{{BP}} = \dfrac{{BP}}{{BC}} \Rightarrow \dfrac{x}{{\sqrt {{x^2} + 64} }} = \dfrac{{\sqrt {{x^2} + 64} }}{{20}} \Rightarrow 20x = {x^2} + 64 \Rightarrow x = 4{\text{ or }}x = 16$

Let $x = 4$ (if we take $x = 16$ we come to the same final result, since $r_1$ and $r_3$ just interchange their values).

In $\triangle APB$ we have:
$\left[ {APB} \right] = s \cdot {r_1} \Rightarrow {r_1} = \dfrac{{\left[ {APB} \right]}}{s} = \dfrac{{\frac{1}{2} \times 4 \times 8}}{{\frac{{8 + 4 + \sqrt {{4^2} + 64} }}{2}}} \Rightarrow {r_1} = \dfrac{8}{{3 + \sqrt 5 }}$

Due to similarity, $\dfrac{{{r_2}}}{{{r_1}}} = \dfrac{{BP}}{{AP}} = \dfrac{{4\sqrt 5 }}{4} \Rightarrow {r_2} = {r_1}\sqrt 5 \Rightarrow {r_2} = \dfrac{{8\sqrt 5 }}{{3 + \sqrt 5 }}$

and $\dfrac{{{r_3}}}{{{r_1}}} = \dfrac{{DC}}{{AP}} = \dfrac{8}{4} = 2 \Rightarrow {r_3} = 2{r_1} \Rightarrow {r_3} = \dfrac{{16}}{{3 + \sqrt 5 }}$

Hence, ${r_1} + {r_2} + {r_3} = \dfrac{8}{{3 + \sqrt 5 }} + \dfrac{{8\sqrt 5 }}{{3 + \sqrt 5 }} + \dfrac{{16}}{{3 + \sqrt 5 }} = \dfrac{{24 + 8\sqrt 5 }}{{3 + \sqrt 5 }} = \boxed{8}.$

Let $\angle {PBC}$ be $α$ . Then

$20\cos α=\dfrac {8}{\sin α}\implies \sin α=\dfrac {1}{\sqrt 5}$ ,

$\cos α=\dfrac {2}{\sqrt 5},\tan α=\dfrac {1}{2}$

$r_1=\dfrac {8\cos α}{1+\cos α+\sin α}=12-4\sqrt 5$

$r_2=\dfrac {16}{5\sin α\cos α+2\cos α+2\sin α}$

$=6\sqrt 5-10$

$r_3=\dfrac {8\tan α}{1+\sec α+\tan α}=6-2\sqrt 5$

Hence $r_1+r_2+r_3=12-4\sqrt 5+6\sqrt 5-10+6-2\sqrt 5=\boxed 8$ .