Vtyjkyuk

In a square $ABCD$ , let $N$ and $M$ be the midpoints of sides $BC$ and $CD$ . If $AM$ meets $DN$ at $R$ and $BM$ meet $DN$ at $O$ , what is the ratio of area of $triangle RMO$ to the area of square $ABCD$?

Suppose $square ABCD$ is a unit square, there fore, $A_square=1$

Since $triangle ADM = triangle BCM$, therefore $angle AMD = angle BMC$.
Since $triangle ADM$ is a right triangle, therefore:
$$sinangle DAM = fracoverlineDMoverlineAM=frac0.5sqrt1.25$$
$$angle DAM = sin ^-1frac0.5sqrt1.25$$
Working your way around through the smaller triangles and you'll get: $$angle RDM = 90^circ - angle ADR = sin ^-1 frac0.5sqrt1.25$$
Since $angle AMD = angle BMC$, therefore $angle AMB = 2 sin ^-1frac0.5sqrt1.25$, and because $angle RDM = sin ^-1 frac0.5sqrt1.25 $, we know that $$overlineRM = overlineDMcdot sin Biggl(sin ^-1 frac0.5sqrt1.25Biggr)=frac0.5^2sqrt1.25$$
By the same token, we know that:
$$cos angle AMB = cos angle OMR = fracoverlineRMoverlineOM$$
$$overlineOM=fracoverlineRMcos angle OMR = fracfrac0.5^2sqrt1.25cos (2sin ^-1frac0.5sqrt1.25)=frac0.5^2 sqrt1.250.75$$
Now, $overlineOR$ must be equal to:
$$overlineOR=sqrtoverlineOM^2-overlineRM^2$$
$$overlineOR=sqrtBiggl(frac0.5^2 sqrt1.250.75Biggr)^2-Biggl(frac0.5^2sqrt1.25Biggr)^2=0.2981423997$$
Then, $A_triangle = frac12 b h = frac12 (overlineRM)(overlineOR)$, which is surprisingly:
$$A_triangle = frac130$$
$$therefore A_square : A_triangle rightarrow 1 : frac130$$

Though, I wonder, could there be a more intuitive way for solving for the ratio without any actual computation?

Lemma $1$: Let $ABCD$ be a parallelogram. Let $M$ be the midpoint of BC. Then $DM$ trisects $AC$.

Proof:

Let $N$ be the midpoint of $AD$. Let $AC$ intersect $NB$ at $P$ and $MD$ at $Q$. Then $Delta ANPsim Delta ADQ$, so $AP=PQ$. Similarly, $CQ=PQ$. So $AP=PQ=CQ$.

WLOG let the side of the square be $2$.

Notice that $C,O,A$ are collinear, so $CO$ is the diagonal. Be Lemma $1$, $CO=frac AC3=2sqrt2over3$. And $CN=1$. Then $[CMON]=2[CON]=2cdotfrac12cdot frac2sqrt23cdot1cdotsin 45^circ=frac23$.

Now, $Delta RDMsim Delta DAM$. So $[DRM]=frac15$.

But $[DNC]=[DRM]+[RMO]+[CMON]$. So, $1=frac15+frac23+[RMO]$. Therefore, $[RMO]=frac215$. Since $[ABCD]=4$, we get $$[RMO]over[ABCD]=fracfrac2154=frac130$$

$boxedtiny Z$

Let angle$BMC$ = angle$AMD$ = $α$.

Let angle$AMB$ = $β$.

Let the lenght of the side of the square be $a$.

Construct $MX$ perpendicular to $AB$.

So, $XB= fraca2$ and $BM=fracasqrt52$.

$sinfracβ2=frac1sqrt5$.

So, $ β= 2arcsinfrac1sqrt5$.

Also,
$tanα=2$.

So, $α=arctan2$.

In right $ΔDMR$, $fracRMfraca2=cosα$.

This leads to $RM=fracacdotcos[arctan(2)]2=fraca2sqrt5$.

In right $ΔRMO$,

$tan[2arcsinfrac1sqrt(5)]=fracROfraca2sqrt5$.

This leads to $RO= frac2a3sqrt5$
using tan double angle formula.

Thus,
area of $ΔRMO=fracRO×RM2= fraca^230$.

And finally,

The ratio is of the area of the triangle $Delta RMO$ to the area of the square is $1:30$.

This is method without any actual computation..