4 Easy Circles

We note that $\triangle CDE$ and $\triangle CDF$ are congruent. This means that $CD$ bisects $\angle C$ and by angle bisector theorem ,

$\frac {DB}{AB} = \frac {BC}{CA} \implies \frac {DB}{14-DB} = \frac {13}{15} \implies DB = \frac {13 \cdot 14}{15+13} = 6.5$

We also note that a $\text{13-14-15}$ triangle is made up of a $\text{3-4-5}$ right triangle and a $\text{5-12-13}$ right triangle sharing an altitude of $12$ ; and that $\triangle BDF$ is a $\text{5-12-13}$ right triangle. Since $DB = 6.5$ , $BF = \frac 5{13} \cdot 6.5 = 2.5$ and $DF = \frac {12}{13} \cdot 6.5 = 6$ . And the radius of the incircle of $\triangle BDF$ ,

$r = \frac {\frac 12 BF \cdot DF}{\frac 12(BF+DF+DB)} = \frac {2.5 \cdot 6}{2.5+6+6.5} = \boxed 1$

By Angle bisector theorem ,

$\dfrac{DB}{DA}=\dfrac{a}{b}\Rightarrow \dfrac{DB}{DA+DB}=\dfrac{a}{a+b}\Rightarrow \dfrac{DB}{AB}=\dfrac{a}{a+b}\Rightarrow DB=\dfrac{ac}{a+b}\Rightarrow DB=\dfrac{13\times 14}{13+15}\Rightarrow DB=6.5$ The height from $C$ of the $13$ - $14$ - $15$ $\triangle ABC$ is ${{h}_{c}}=12$ .

Using two different expressions for the area of $\triangle BCD$ we have

$\cancel{\dfrac{1}{2}}BC\cdot DF=\cancel{\dfrac{1}{2}}DB\cdot {{h}_{c}}\Rightarrow 13\times DF=6.5\times 12\Rightarrow DF=6$ By Pythagorean theorem on right $\triangle FDB$ ,

$FB=\sqrt{D{{B}^{2}}-D{{B}^{2}}}=\sqrt{{{6.5}^{2}}-{{6}^{2}}}\Rightarrow FB=2.5$ Finally, for the radius $r$ of the incircle of $\triangle FDB$ ,

$r=\dfrac{\left[ FDB \right]}{\dfrac{1}{2}\left( DF+FB+BD \right)}=\dfrac{\dfrac{1}{2}DF\cdot FB}{\dfrac{1}{2}\left( DF+FB+BD \right)}=\dfrac{\dfrac{1}{2}6\times 2.5}{\dfrac{1}{2}\left( 6+2.5+6.5 \right)}=1$ Working similarly we find that the incircle of $\triangle EAD$ equals $1.5$ , thus the smaller circle is the incircle of $\triangle FDB$ , hence, the answer is $\boxed{1}$ .

The congruency of the two largest circles gives ED = DF. The two smaller right triangles would still be similar to 9-12-15 (left RT) and 5-12-13 (right RT). Considering that both their longer legs are still the same now, and that their hypothenuses have to share the base of length 14 between them, we got the scale of 1/2 (from 14/(13+15) = 1/2) so now what we're looking for is just the incircle of 2.5-6-6.5. Sketching 2+3+10 for the original, the answer is just 2 (original's incircle) times 1/2 (the scale) = 1.

Draw altitude $CG$ . Since the $13$ - $14$ - $15$ triangle is a $9$ - $12$ - $15$ right triangle combined with a $5$ - $12$ - $13$ right triangle at side $12$ , $CG = 12$ .

Now $\triangle AGC \sim \triangle AED$ and $\triangle BGC \sim \triangle BFD$ by AA similarity, so let $AE = 3m$ , $ED = 4m$ , and $AD = 5m$ (because it is similar to a $9$ - $12$ - $15$ or $3$ - $4$ - $5$ triangle) and let $BF = 5n$ , $DF = 12n$ , and $DB = 13n$ (because it is similar to a $5$ - $12$ - $13$ triangle).

From segment addition, $AB = AD + DB = 5m + 13n = 14$ , and by symmetry of the two large circles, $ED = DF = 4m = 12n$ .

Solving $5m + 13n = 14$ and $4m = 12n$ gives $m = \frac{3}{2}$ and $n = \frac{1}{2}$ .

Therefore, the radius of the smallest circle is $r = \cfrac{2A_{\triangle BFD}}{P_{\triangle BFD}} = \cfrac{2 \cdot \frac{1}{2} \cdot BF \cdot DF}{BF + DF + DB} = \cfrac{5n \cdot 12n}{5n + 12n + 13n} = \cfrac{60n^2}{30n} = 2n = 2 \cdot \frac{1}{2} = \boxed{1}$ .