Quadrilateral In Triangle

Let's define the following vectors:

$\vec{a} = \vec{BC}$

$\vec{b} = \vec{BA}$

$\vec{BG} = \vec{u_1} = \vec{a} + \dfrac{1}{5} (\vec{b} - \vec{a}) = \dfrac{4}{5} \vec{a} + \dfrac{1}{5} \vec{b}$

$\vec{BF} = \vec{u_2} = \vec{a} + \dfrac{4}{5} (\vec{b} - \vec{a}) = \dfrac{1}{5} \vec{a} + \dfrac{4}{5} \vec{b}$

and

$\vec{v_1} = \dfrac{1}{3} \vec{a} - \vec{b}$

$\vec{v_2} = \dfrac{2}{3} \vec{a} - \vec{b}$

Each of the points $P , Q, R$ and $S$ corresponds to certain multiples of the vectors that intersect at it.

1. Point P: Intersecting vectors $\vec{v_1}$ , and $\vec{u_2}$

We can write the following vector equation in $t$ and $s$

$t \vec{u_2} = \vec{b} + s \vec{v_1}$

This translates into

$t \dfrac{1}{5} \vec{a} + \dfrac{4}{5} \vec{b}) = \vec{b} + s (\dfrac{1}{3} \vec{a} - \vec{b})$

which implies

$t / 5 = s / 3$ and $4 t / 5 = 1 - s$ , for which the solution is

$s_1 = \dfrac{3}{7}$

$t_1 = \dfrac{5}{7}$

2: Point Q. Intersecting vectors: $\vec{u_1}$ and $\vec{v_1}$

$t \vec{u_1} = \vec{b} + s \vec{v_1}$

As before, this translates to:

$t( \dfrac{4}{5} \vec{a} + \dfrac{1}{5} \vec{b}) = \vec{b} + s ( \dfrac{1}{3} \vec{a} - \vec{b})$

which translates into,

$4 t/ 5 = s/3$

$t/5 = 1 - s$

from which,

$s_2 = \dfrac{12}{13}$

$t_2 = \dfrac{5}{4} (\dfrac{4}{13}) = \dfrac{5}{13}$

3: Point R. Intersecting vectors: $\vec{u_1}$ and $\vec{v_2}$

$t \vec{u_1} = \vec{b} + s \vec{v_2}$

The vector equation translates into,

$t ( \dfrac{4}{5} \vec{a} + \dfrac{1}{5} \vec{b}) = \vec{b} + s (\dfrac{2}{3} \vec{a} - \vec{b})$

which implies,

$4/5 t = 2/3 s$

$t/5 = 1 - s$

which has the solution:

$s_3 = \dfrac{6}{7}$

$t_3 = \dfrac{5}{7}$

1. Point S. Intersecting vectors: $\vec{u_2}$ and $\vec{v_2}$

Here, the vector equation is:

$t \vec{u_2} = \vec{b} + s \vec{v_2}$

This translates to:

$t (\dfrac{1}{5} \vec{a} + \dfrac{4}{5} \vec{b}) = \vec{b} + s (\dfrac{2}{3} \vec{a} - \vec{b})$

From which we generate two scalar equations:

$t/5 = 2/3 s$

$4/5 t = 1 - s$

The solution of which is:

$s_4 = \dfrac{3}{11}$

$t_4 = \dfrac{10}{11}$

Now comes the fun stuff !!

Note that the area of middle triangle = $[ ADE ] = \dfrac{1}{3} [ ABC ]$ (why ?)

In addition,

$[ APS ] = s_1 s_4 [ADE]$

$[ AQR ] = s_2 s_3 [ ADE ]$

and

[PQRS] = [AQR] - [APS]

This implies that,

$[ PQRS ] = \dfrac{1}{3} ( s_2 s_3 - s_1 s_4 ) [ ABC] = \dfrac{1}{3} \left( \left(\dfrac{12}{13} \right) \left( \dfrac{6}{7} \right) - \left( \dfrac{3}{7} \right) \left(\dfrac{3}{11} \right) \right) [ ABC ] = \dfrac{225}{1001} [ABC]$

Since $[ ABC] = 1001$ , then $[ PQRS] = \boxed{225}$ .

Note that we can verify this result, by using the $t$ values that we obtained. We have

$[ BFG] = \dfrac{3}{5} [ABC]$

In addition,

$[BPQ] = t_1 t_2 [BFG]$

$[BRS] = t_3 t_4 [BFG]$

and

$[PQRS] = [BRS] - [BPQ]$

Therefore,

$[PQRS] = \dfrac{3}{5} ( t_3 t_4 - t_1 t_2 ) [ABC] = \dfrac{3}{5} \left( \left(\dfrac{5}{7} \right) \left( \dfrac{10}{11} \right) - \left( \dfrac{5}{7} \right) \left(\dfrac{5}{13} \right) \right) [ ABC ] = \dfrac{225}{1001} [ABC]$

Which confirms our previous result.

Top left Fig proves the formula $\dfrac X {X+(1-X)a}$ how two intersecting lines from vertices to opposite sides are divided.
Bottom five sketches apply this to the given problem.
Top right Fig. leads us to the answer. Please correct the word box and fig for a mistake. Replace T with red B as shown in the bottom most Fig.