Tangent bisectors

Although the point is not mentioned in the problem, the circle of radius $\frac{5}{3}$ also intersects side $BC$ at a point. We shall call that point $M$ .

Now let $AK = x$ and $KC = 3\sqrt{3} - x$ and we shall make the additional restriction that $AK \leq KC$ . By the given ratio, $BL = \frac{5}{6}x$ . Through power of point, $AK^2 = (AL)(AB)$ . Solving this relationship yields $AL = \frac{2}{3}x$ and $AB = \frac{3}{2}x$ .

Next, we proceed to angle chase. We see that $\angle BAC = \angle BKM$ and $\angle BCA = \angle BKL$ . However, since quadrilateral $BMKL$ is cyclic, we have that $\angle BKM = \angle BLM$ and $\angle BKL = \angle BML$ so $LM$ is parallel to $AC$ !

We can now use the fact that Triangle $BLM$ is similar to Triangle $BAC$ . Thus we have that $\frac{\frac{5}{6}x}{\frac{3}{2}x} = \frac{LM}{3\sqrt{3}}$ , so $LM = \frac {5\sqrt{3}}{3}$ . We are given that the radius of the circle is $\frac{5}{3}$ so if we construct Triangle $OLM$ , then we see that $\angle LOM = 120 ^ \circ$ . Thus, we get that $\angle ABC = 60 ^ \circ$ .

Since we have the value of $\angle ABC$ , we might expect to find the area of Triangle $ABC$ by using $\frac{(AB)(BC)\sin\angle ABC)}{2}$ . Now in order to do so, we must find the value of $x$ .

Using the Angle Bisector Theorem we have that $\frac{\frac{3}{2}x}{x} = \frac{BC}{3\sqrt{3} - x}$ . Evaluating yields $BC = \frac{9\sqrt{3}}{2} - \frac{3}{2}x$ . With the same similar triangle relations as before, we get that $BM = \frac{5\sqrt{3}}{2} - \frac{5}{6}x$ . Using Ptolemy's Theorem, we can find that $BK = \frac{5}{2}$ . Now we use Law of Cosines on Triangle $ABK$ to get that $x = \frac{3\sqrt{3} - \sqrt{7}}{2}$ since we wanted $AK \leq KC$ . Now we can use the area formula that was mentioned earlier. Plugging in the value for $x$ , we get that the area is $\frac{9\sqrt{3} - 3\sqrt{7}}{4} \frac{9\sqrt{3} + 3\sqrt{7}}{4} \frac{\sqrt{3}}{2}\frac{1}{2}$ which simplifies to $\frac{45\sqrt{3}}{16}$ so our answer is $45+3+16$ or $64$ .

First, assume that, the circle intersect the line $BC$ at $M$ , and then define $O$ is the center of circle. I will divide into many steps.

$1$ . Proof that $LM \parallel AC$ . $$ Proof : Since the line $AC$ tangent the circle at $K$ . So, power point in $A$ and $C$ . We obtain that :

$AK^2 = AL(AB) \ldots \ldots (1)$ $CK^2 = CM(CB) \ldots \ldots (2)$

In the other hand, since $BK$ is angle bisector. We obtain that, by angle bisector theorem, that is :

$AK. BC = CK. AB \implies \frac{AK}{CK} = \frac{AB}{CB} \ldots \ldots (3)$

Then, divide equation $(1)$ into $(2)$ , we obtain : $\frac{AK^2}{BK^2} = \frac{AL(AB)}{CM(BC)}$ , combine with equation $(3)$ , obatain that

$\frac{AB^2}{BC^2} = \frac{AL(AB)}{CM(BC)} \implies \frac{AB}{BC} = \frac{AL}{CM} \implies \frac{AB}{AL} = \frac{BC}{CM}$ $\implies \frac{AL + BL}{AL} = \frac{BM + CM}{CM} \implies \frac{BL}{AL} = \frac{BM}{CM}$ . Implies that, since $\angle ABC = \angle LBM$ , that is $\triangle ABC \sim \triangle LBM$ . Implies that $LM \parallel AC$ .

$2$ . Find the length of $LM$ . $$ From the problem, we can assume that $AK = a$ , then, $BL = \frac{5}{6}. AK = \frac{5a}{6}$ . From, the equation $(1)$ , and solve the equation by putting the values of $AK$ and $BL$ , we obatin that $AL = \frac{2a}{3}$ . Then, by similarity, obtain that :

$\frac{LM}{AC} = \frac{BL}{AB} \implies \frac{LM}{3 \sqrt{3}} = \frac{\frac{5a}{6}}{\frac{3a}{2}} \implies LM = \frac{5 \sqrt{3}}{3}.$

Assume that $BK$ intersect $LM$ at $N$ . Then, assume that $\angle ABC = \alpha$ . We obtain that,

$\angle MOK = \angle LOK = 2 \angle ABK = \angle ABC = \alpha$ .

It's easy to see that $LN = NM$ , since $LOM$ is isosceles triangle. And, since $OK \perp AC$ , then $OK \perp LM$ , since $AC \parallel LM$ . Then,

$3$ . Find the length of $NK$ . $$ So, See the right triangle $OMN$ , then, we obtain

$\sin \alpha = \sin (\angle NOM) = \frac{NM}{OM} = \frac{\frac{5}{6} \sqrt{3}}{\frac{5}{3}} = \frac{1}{2} \sqrt{3}$ , then, $\frac{1}{2} = \cos \alpha = \frac{ON}{OM} = \frac{ON}{\frac{5}{3}}$ , then $ON = \frac{5}{6}.$ So,

$NK = OK - ON = \frac{5}{3} - \frac{5}{6} = \frac{5}{6}$

$4$ . Find the height of triangle $ABC$ . Assume that $BP$ is height of triangle $ABC$ . Then, assume that $BP$ intersect at $LM$ at $Q$ . Then, it's easy to see that $PQ \parallel NK$ , implies $BQ = NK = \frac{5}{6}$ . Since, $BK \perp LM$ and $BQ \perp LM$ . So, by similarity we obtain :

$\frac{BQ}{QP} = \frac{BQ}{QP} \implies \frac{\frac{5}{6}}{\frac{2}{3}} = \frac{h - \frac{5}{6}}{\frac{5}{6}} \implies h = \frac{15}{8}$

$5$ . Find the area of triangle $ABC$ $$ $[ABC] = \frac{1}{2}. AC. BP = \frac{1}{2}. 3 \sqrt{3}. \frac{15}{8} = \frac{45 \sqrt{3}}{16}.$

So, $a + b + c = 45 + 3 + 16 = \boxed{64}$

Let $\angle ABK = \alpha$ and $\angle LBO = \beta$ (where O is the center of the circle passing through $B$ , $K$ and $L$ ), then we can calculate the following angles in terms of $\alpha$ and $\beta$ : $\angle AKL = \alpha, \angle BAC = 90-2\alpha+\beta, \angle AKB=90+\alpha-\beta$ , $\angle ACB = 90-\beta$ .

Notice that the two triangles $\triangle ABK$ and $\triangle AKL$ are similar, hence: $AK^2=AB \times AL=AB(AB-BL)=AB^2-\frac56AB \times AK$ , which can be solved to obtain $AK=\frac23AB$ .

Apply the law of sines on $\triangle ABK$ and recall that $\angle ABK=\alpha$ and $\angle AKB=90+\alpha-\beta$ : $AB \times \sin(\alpha)=AK \times \sin(90+\alpha-\beta)=\frac23AB \times \cos(\beta-\alpha)$ , thus $\frac32\sin(\alpha)=\cos(\beta-\alpha)$ .

Next, apply the law of sines on $\triangle BLO$ and recall that $\angle LBO = \beta$ : $BL \times \sin(\beta)=OL \times \sin(180-2\beta)=OL\times \sin(2\beta)$ , thus $BL= \frac{10}3\cos(\beta)$ , and hence $AK=\frac65BL=4\cos(\beta)$ .

Lastly, apply the law of sines on the two triangles $\triangle ABK$ and $\triangle BCK$ and recall that $\angle ABK = \angle CBK = \alpha, \angle BAC = 90-2\alpha+\beta$ , $\angle ACB = 90-\beta$ : $\sin(90-2\alpha+\beta) \times AK = \sin(\alpha) \times BK = \sin(90-\beta) \times CK$ , thus $cos(2\alpha-\beta) \times AK = \cos(\beta) \times (3\sqrt{3}-AK)$ .

Now recall that $\frac32\sin(\alpha)=\cos(\beta-\alpha)$ and $AK=4\cos(\beta)$ , we obtain: $3\sqrt{3}=4\cos(2\alpha-\beta)+4\cos(\beta)=8cos(\alpha)\cos(\alpha-\beta)$ $=8\cos(\alpha)\frac32\sin(\alpha)=6\sin(2\alpha)$ , which yields $2\alpha=60^\circ, \alpha=30^\circ$ .

We calculate the area of the triangle $\triangle ABC$ : $Area(\triangle ABC)=\frac12AB \times AC \times \sin(\angle BAC)$ $=\frac12 \times (\frac32 \times 4\cos(\beta)) \times 3\sqrt{3} \times \cos(60^\circ-\beta)$ $=\frac{9\sqrt{3}}2(\cos(60^\circ)+\cos(60^\circ-2\beta))$ $=\frac{9\sqrt{3}}2(\frac12+2\cos^2(30^\circ-\beta))-1)$

Suppose the circle, described in the problem, intersects $BC$ at a point $M$ . Suppose $\measuredangle ABK =x.$ Because $BK$ is an angle bisector, $\measuredangle CBK =x.$ From the inscribed angle theorem, $\measuredangle MLK=\measuredangle MBK=x.$ Because the circle is tangent to $AC$ at $K$ , $\measuredangle AKL=\measuredangle LBK=x.$ Similarly, $\measuredangle LMK=x$ and $\measuredangle MKC =x.$

The above implies that the line $LM$ is parallel to $AC$ . Note also that the triangles $AKL$ and $ABK$ are similar, so $\frac{|AL|}{|AK|}=\frac{|AK|}{|AB|}$ . Because $|AK|:|BL|=6:5,$ we can say that $|AK|=6a,\ |BL|=5a$ for some positive length $a$ . Then $\frac{|AL|}{|AK|}=\frac{|AK|}{|AB|}$ can be rewritten as $|AL|\cdot |AB|=|AK|^2,$ or $|AL|\cdot (|AL|+5a)=36a^2$ . Solving this quadratic equation for $|AL|,$ we get $|AL|=4a$ or $|AL|=-9a.|$ Because $|AL|>0,$ $|AL|=4a.$

$$

The triangles $ABC$ and $LBM$ are similar, so $|LM|=|AC|\cdot \frac{|LB|}{|AB|}=3\sqrt{3}\frac{5}{9}=\frac{5\sqrt{3}}{3}.$ From the Sine Law, considering that the diameter of a circumscribed circle of $LBM$ is $\frac{10}{3},$ $\sin \angle LBM = \frac{5\sqrt{3}}{3} / \frac{10}{3}=\frac{\sqrt{3}}{2}.$ So $\measuredangle LBM=60^{\circ}$ and $x=30^{\circ}.$

$$

Because $|LM|=\frac{5\sqrt{3}}{3}$ and $\measuredangle KLM = \measuredangle KML =30^{\circ},$ the altitude from $K$ has length $\frac{1}{2}|LM|\cdot \tan 30^{\circ}= \frac{5}{6}.$ Because $LM$ is parallel to $AC$ and $|AB|:|AL|=9:4,$ the altitude of the triangle $ABC$ from $B$ has length $\frac{9}{4}\cdot \frac{5}{6}=\frac{15}{8}.$ From this, the area of the triangle $ABC$ is $\frac{1}{2}\cdot 3\sqrt{3}\frac{15}{8}= \frac{45\sqrt{3}}{16}.$ Therefore, the answer is $45+3+16=64.$