Area of triangle from a side length and bisector length

Question

Let there be a triangle $ABC$ such that the segment $BL$ bisects $\angle B$ and intersects $AC$ at $L$. Also, $BL = 3\sqrt{10}$, $AL = 2$ and $CL = 3$. Find the area of $ABC$.

My solving process:

1. By the Angle Bisector Theorem, $\frac{BA}{AL} = \frac{BC}{CL}$, meaning $2BC = 3BA$.
2. By the law of cosines on $\triangle{BLA}$: $$BA^2 = AL^2 + BL^2 - 2(AL)(BL)cos\angle{BLA} $$ $$ \frac{BA^2 - 94}{- 12\sqrt{10}} = cos\angle{BLA} $$
3. By the law of cosines on $\triangle{BLC}$: $$BC^2 = CL^2 + BL^2 - 2(CL)(BL)cos\angle{BLC} $$ $$ \frac{BC^2 - 99}{-18\sqrt{10}} = -cos\angle{BLA} $$
4. By the equations from steps 2 and 3 and using the relationship in step 1: $$\frac{99-BC^2}{-18\sqrt{10}} = \frac{BA^2 - 94}{- 12\sqrt{10}} $$ $$-18(BA^2 - 94) = -12(99-BC^2) $$ $$12BC^2 + 18BA^2 = (94)(18)+(12)(99) $$ $$20BC^2 = 6(480)$$ $$BC = 12, BA = 8$$
5. Lastly, by Heron's formula for a triangle with lengths 5, 8, 12: $$[ABC] = \sqrt{\frac{25}{2}*\frac{1}{2}*\frac{9}{2}*\frac{15}{2}} = \fbox{ $\frac{15\sqrt{15}}{4}$} $$

This is the correct answer, but as you can tell, it takes a bit of time to completely solve out, plus the time it takes to think about the question itself. In math contests, like one from where this question is from, it is obviously best to reduce the time spent solving per question. So, is there a quicker way to solve this? Thanks.

Answer 1

First Solution:

Draw $h=BH\perp AC$ then point $H$ is between the points $A$ and $L$.

$\frac{2}{3}=\frac{AL}{CL}=\frac{BA}{BC}=\frac{2y}{3y}$ (angle bisector theorem)

then

$(2-x)^2 + h^2 = 4y^2$ in $\triangle ABH$ (*)

$(3+x)^2 + h^2 = 9y^2$ in $\triangle CBH$ (**)

from (*) and (**) $y^2=2x+1$

$(2-x)^2 + h^2 = 4y^2 = 4\times (2x+1)$

$4-4x+x^2 + h^2 = 8x+4$

$h^2 = 12x - x^2$

and

$x^2+h^2=90$ in $\triangle BHL$

$x^2+12x - x^2=90$

$x=\frac{15}{2}, h=\frac{3\sqrt{15}}{2}$

$\text{Area} \triangle ABC= \frac{1}{2} \times 5 \times \frac{3\sqrt{15}}{2}=\frac{15\sqrt{15}}{4}$

Second Solution:

$\frac{2}{3}=\frac{AL}{CL}=\frac{BA}{BC}=\frac{2y}{3y}$ (angle bisector theorem)

Take point $D$ on $BC$ such that $BD=2y$ and $DC=y$

then

$\triangle ABD$ is an isosceles with $BA=BD$

$BL\perp AD$, let $E$ be intersection of $BL$ and $AL$

Draw $DF\parallel BL$ ($F$ is on $AC$)

$DF=2\times EL$ (midpoint theorem)

$\triangle CDF\sim \triangle CBL$ (AAA similarity theorem)

$\frac{CD}{CB}=\frac{1}{3}=\frac{2x}{3\sqrt{10}}=\frac{DF}{BL}$ (sides are in a propportion)

$DF=\sqrt{10}$ and $EL=\frac{\sqrt{10}}{2}$

$AE=\sqrt{2^2-(\frac{\sqrt{10}}{2})^2}=\frac{\sqrt6}{2}$ (paythagorean theorem in $\triangle AEL$)

$\text{Area} \triangle ABL = \frac{1}{2}\times 3\sqrt{10} \times \frac{\sqrt6}{2}=\frac{3\sqrt{15}}{2}$

$$\frac{\text{Area}\triangle ABL}{\text{Area}\triangle ABC}=\frac{\frac{3\sqrt{15}}{2}}{\text{Area}\triangle ABC}=\frac{2}{5}=\frac{AL}{AC} \text{(same height)}$$

$\text{Area}\triangle ABC=\frac{15\sqrt{15}}{4}$

Answer 2

Denote the sides of the triangle as $AB = c$, $BC = a$, $CA = b$. Then $b = AL+CL = 2+3 = 5$.

By the angle bisector theorem, $$\frac{c}{a} = \frac{AB}{CB} =\frac{AL}{CL} = \frac{2}{3}.$$

By the angle bisector formula, the length of the angle bisector $BL$ is $$BL = \frac{2}{a+c}\sqrt{acs(s-b)},$$ where $$s = \frac{a+b+c}{2} = \frac{a+5+\frac{2}{3}a}{2} = \frac{5}{6}a + \frac{5}{2}$$ is the semiperimeter. Consequently $$3\sqrt{10} = \frac{2}{a + \frac{2}{3}a}\sqrt{\tfrac{2}{3}a^2\left(\tfrac{5}{6}a+\tfrac{5}{2}\right)\left(\tfrac{5}{6}a - \tfrac{5}{2}\right)} = \sqrt{\tfrac{2}{3}}\sqrt{a^2-9}.$$

Solving for $a$ yields the unique positive solution $a = 12$, which in turn yields $c = 8$, and the remainder of the solution is the same as yours.