areacontest-mathgeometrytriangles
$ABCD$ is a parallelogram. Point $E$ divides $BC$ into two equal
lengths. If the area of $BEF$ is 126, what is the area of $ABCD$?
Source: Bangladesh Math Olympiad 2017 Junior Category.
I can not solve this problem. I am confused on the position of $'F'$. There is no information mentioned about it. Is there any lack of information in this question? Can anyone give me a hint?
Let $E$ be on $BC$ so $CE = CD$. Then $\angle DEC = 80^{\circ}$ so $CADE$ is cyclic, so we have by the power of the point with respect to $B$: $$a\cdot (a-d) = b\cdot (b-x)\;\;\;\;\;\;\;\;\;\;\;\;(1)$$
By angle bisector theorem we have $${b-x\over x} = {a\over b}\implies b(b-x)= ax\;\;\;\;\;\;\;\;(2)$$
By (1) and (2) we have $$a(a-d)= ax \implies a-d= x$$
Since we are searching for $d+x = a= 2017$ we are done.
Let $r$ denote the radius of the tiny circle (so $r=\frac mn$ in your notation, though it is not clear from the start that $r$ is rational).
Drop the perpendicular from the center of the big circle. We see quickly that it has length $12$.
Drop the perpendicular from the center of the tiny circle. We see that it has length $h$ where $$h^2+5^2=(5+r)^2$$
But in terms of $h,r$ the length of the first perpendicular we looked at is $8+r+h$, so $$8+r+h=12$$
Can you finish from here?
(note: I also get $\frac 89$)
Best Answer
Hint: