Internal Triangle Area

Extend $BA$ to $F$ such that $BF=BC$ and connect $CF$ . Triangle $BCF$ is an equilateral triangle of side length 1. Let $G$ be a point on $AF$ such that angle $ACG$ is $20^\circ$ . We see triangle $CDE$ is similar to triangle $CGA$ by AAA. Since $CD=\frac {1}{2} CA)$ , thus $2\times [CDE]=\frac {1}{2} [CGA]$ . Also, $ABC$ and $GFC$ are congruent triangles by AAA and $BC=CF$ .

As such, $[ ABC] + 2 \times [CDE] = \frac {1}{2} [CFB]$ . $Given CB=1$ , $[CFB]= \frac {\sqrt{3}}{8}$ , so $a=3$ and $b=8$ and the required sum is 11.

[Latex edits - Calvin]

Upon drawing a simple diagram of the Triangle, we see immediately that, as Angle A = 100 and Angle B = 60, Angle C must equal 20 degrees.

The Point E is then chosen such that the angle EDC is 80. But, we just established that Angle C is 20, therefore, Angle DEC must be 80 degrees, making Triangle EDC an Isosceles triangle, with EC = DC.

Now, using the sine rule, we see \frac {BC} {sin 100} = \frac {AC} {sin 60}.

As BC = 1, this leads us to AC = \frac {sin 60} {sin 100} .

Now, please bear with the following method. It may seem long-winded for calculating the area of triangle ABC, but it helps simplify things later;

First extend the line AB and draw the perpendicular from the AB to C. Let the foot of this perpendicular on AB be F.

We now have 2 right-angle triangles AFC and BFC, with angle BCF = 30.

Now, BF= 1 sin 30, and FC = 1 cos 30, while, AF = AC sin 10 and FC = AC cos 10.

Now, [ABC] = [BFC] - [AFC].

[BFC] = \frac {1} {2} * BF * FC = 0.5 sin 30 cos 30 = 0.25 sin 60, using the sine double-angle formula. (sin 2x = 2 sin x cos x).

Similarly,

[AFC] = 0.5 AC^2 sin 10 cos 10 = 0.25 AC^2 sin 20.

Now, finally, therefore,

[ABC] = \frac {1}{4} (sin 60 - AC^2 sin 20) .

Now for Triangle CDE, which, much more simply,

[CDE] = 0.5 CD CE sin 20, where CD = CE = 0.5 AC.

So, [CDE] = 0.5 (0.25 AC^2) sin 20.

Now, to find [ABC]+2 [CDE], simply add our 2 expressions together;

[ABC]+2[CDE] = \frac {1} {4} sin 60 - \frac {1} {4} AC^2 sin 20 + 2*\frac {1} {8} AC^2 sin 20

\Rightarrow [ABC]+2[CDE] = \frac {1} {4} sin 60 - \frac {1} {4} AC^2 sin 20 + \frac {1} {4} AC^2 sin 20

\Rightarrow [ABC]+2[CDE] = \frac {1} {4} sin 60 ,

But sin 60 = \frac {\sqrt{3}} {2},

So, [ABC]+2[CDE] = \frac {\sqrt{3}} {8},

where a = 3 and b = 8.

Thus, a+b = 11.

Note- The long, terrible way of finding the Area of ABC above is simply so that the AC^2 terms cancel out so neatly when computing [ABC]+2[CDE], to simply hand us our final answer. Otherwise, we might have to wrestle with some pretty annoying trigonometric terms.

Thank You.

Draw a line parallel to $ED$ passing through $A$ , to intersect the line $BC$ at $F$ . Then the triangles $CAF$ and $CDE$ are similar, with $[CAF] = 4[CDE]$ , and moreover the triangles $ABC$ and $FBA$ are similar, with $[FBA] = \big(\tfrac{AB}{BC}\big)^2[ABC] = AB^2 [ABC]$ .

Using the Sine Rule, $AB \; =\; \frac{AB}{BC} \; = \; \frac{\sin20^\circ}{\sin100^\circ} \; = \; \frac{\sin20^\circ}{\sin80^\circ} \; = \; \frac{\sin20^\circ}{\cos10^\circ} \; = \; 2\sin10^\circ$ and we recall the identity $\sin3\theta = 3\sin\theta - 4\sin^3\theta$ . Thus $\begin{array}{rcl} [ABC] + 2[CDE] & = & [ABC] + \tfrac12[CAF] \; = \; [ABC] + \tfrac12\big([ABC] - [FBA]\big) \\ & = & \tfrac32[ABC] - \tfrac12[FBA] \; = \; \tfrac12\big(3 - AB^2\big)[ABC] \\ & = & \tfrac12(3 - 4\sin^210^\circ)[ABC] \\ & = & \tfrac12(3 - 4\sin^210^\circ)\times\tfrac12 \times AB \times BC \times \sin60^\circ \\ & = & \tfrac12(3 - 4\sin^210^\circ) \times \tfrac12 \times 2\sin10^\circ \times 1 \times \tfrac12\sqrt{3} \\ & = & \tfrac14\sqrt{3} \times \sin30^\circ \; = \; \tfrac18\sqrt{3} \end{array}$ and hence we deduce that $a=3$ and $b=8$ , so that $a+b=11$ .

I have simplier solution:

alt

$\bigtriangleup FEC = \bigtriangleup FCD = \bigtriangleup DCG = \bigtriangleup DGH = \bigtriangleup HDI = \bigtriangleup DIA$

Now, fun part is that area what is looking for $[ABC] + 2[CDE]$ is exactly $[HBC]$ .

Why is that? Because,

$[DGC] + [DGH] = [DEC]$ and

$[HDI] + [DIA] = [DEC]$ so

$2[DEC] = [DGC]+[DGH]+[HDI]+[DIA]$

Now, area $[HDI]$ is half of area of equilateral triangle where side is 1 (because angles are 60,30,90) So:

$[HDI]=\frac{\frac{1\sqrt{3}}{4}}{2}=\frac{\sqrt{3}}{8}$ .

Solution is $3+8=11$ .