A Quadrilateral and A Triangle in a Trapizium

Question

In the above diagram, $ABCD$ is a Trapizium with $AD || BC$ and $BC \perp AB$
$AB = 20, \; AD = 6,\; BC = 30$
$M$ is a point on $DC$ such that $[ADMB] = [BMC]$, where $[x]$ denotes the area of $x$.
The length of $MC$ can be expressed as $\dfrac{a}{b} \sqrt{c}$.
Find the length of $MC$.
I can't find a way to start. Any hint will be helpful.

Answer 1

Hint: You can set up coordinates with the origin at A, D at (6,0), B at (0,20), C at (30,20). Then you can let t be the ratio DM/DC so that M becomes D+t(C-D). Note if you find t you can finish.

The other thing is to express each area in terms of t and equate the two expressions. There is a formula for area of a convex polygon once one lists the vertices in counterclockwise order. It is 1/2 times the "downs" minus the "ups". Here each down or up is a product of some coordinates. For example for vertices (1,3), (2,4), (0,2) the downs are 1*4,2*2,0*3 while the ups are 1*2,2*3,0*4. This works for more points, there are then just more "downs" and "ups"

Added note: The equation of equal areas will be linear in t with integer coefficients, so t will end up rational. Also side DC has length the squareroot of an integer. So getting an answer $(a/b)\sqrt{c}$ will follow.

Some more details:

The formula for area mentioned above, if $(a_1,b_1),(a_2,b_2),...,(a_k,b_k)$ is a list of the vertices of a convex polygon going around the boundary in the counterclockwise direction, is $A=(1/2)[D-U]$ where $$D=a_1b_2+a_2b_3+...+a_{n-1}b_n+a_nb_1,$$ and $$U=a_1b_n+a_2b_1+...+a_n b_{n-1}.$$

Now consider a general version of your diagram, where $A=(0,0),\ B=(0,b),\ C=(c,b),\ D=(a,0).$ Then the point $M=D+t(C-D)$ has coordinates $(a+t(c-a),tb).$ When these coordinates are put in to find the areas of the quadrilateral part and the triangle, we find that twice their areas come out $b(ct+a)$ for the quadrilateral and $b(c-ct)$ for the triangle. Equating these (doubled) areas then gives $t=(c-a)/2c.$ [Interesting here is no $b$ occurs in this $t.$]

For the actual length of $MC$ we need to use $(1-t)$ since I set up the ratio as $DM/DC$ rather than what I should have done, namely $MC/DC.$ Then since $1-t=(c+a)/2c,$ it is this fraction which is to be multiplied by the length of $DC$ and that length is $\sqrt{(c-a)^2+b^2}.$

Note the answer expected may assume that one removes the square part of the integer inside the radical, and combining it with the fraction for $1-t.$

Answer 2

Note:- In the following, quantities a ~ g can easily be found.

1. [⊿ABD] = a

2. [Trap ABCD] = b

3. [⊿BDC] = c

4. CD = g

Since BM divides the whole figure into two equal halves (of area = d each), [⊿BDM] = e (= d - a).

If this is so, we can let [⊿BMC] = f. M should be at a place that divides [⊿BDC] in the ratio $e : f$. $MC = \dfrac {f}{e + f} g$.