Ratio of the area of a triangle and the area of a square containing it

Question

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 $ABCD$?

Answer 1

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: $$\sin\angle DAM = \frac{\overline{DM}}{\overline{AM}}=\frac{0.5}{\sqrt{1.25}}$$ $$\angle DAM = \sin ^{-1}\frac{0.5}{\sqrt{1.25}}$$ Working your way around through the smaller triangles and you'll get: $$\angle RDM = 90^\circ - \angle ADR = \sin ^{-1} \frac{0.5}{\sqrt{1.25}}$$ Since $\angle AMD = \angle BMC$, therefore $\angle AMB = 2 \sin ^{-1}\frac{0.5}{\sqrt{1.25}}$, and because $\angle RDM = \sin ^{-1} \frac{0.5}{\sqrt{1.25}} $, we know that $$\overline{RM} = \overline{DM}\cdot \sin \Biggl(\sin ^{-1} \frac{0.5}{\sqrt{1.25}}\Biggr)=\frac{0.5^2}{\sqrt{1.25}}$$ By the same token, we know that: $$\cos \angle AMB = \cos \angle OMR = \frac{\overline{RM}}{\overline{OM}}$$ $$\overline{OM}=\frac{\overline{RM}}{\cos \angle OMR} = \frac{\frac{0.5^2}{\sqrt{1.25}}}{\cos (2\sin ^{-1}\frac{0.5}{\sqrt{1.25}})}=\frac{0.5^2 \sqrt{1.25}}{0.75}$$ Now, $\overline{OR}$ must be equal to: $$\overline{OR}=\sqrt{\overline{OM}^2-\overline{RM}^2}$$ $$\overline{OR}=\sqrt{\Biggl(\frac{0.5^2 \sqrt{1.25}}{0.75}\Biggr)^2-\Biggl(\frac{0.5^2}{\sqrt{1.25}}\Biggr)^2}=0.2981423997$$ Then, $A_\triangle = \frac12 b h = \frac12 (\overline{RM})(\overline{OR})$, which is surprisingly: $$A_\triangle = \frac1{30}$$ $$\therefore A_\square : A_\triangle \rightarrow 1 : \frac1{30}$$

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

Answer 2

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, $XB= \frac{a}{2}$ and $BM=\frac{a\sqrt5}{2}$.

$\sin{\frac{β}{2}}=\frac{1}{\sqrt5}$.

So, $ β= 2\arcsin{\frac{1}{\sqrt5}}$.

Also, $\tan{α}=2$.

So, $α=\arctan{2}$.

In right $ΔDMR$, $\frac{RM}{\frac{a}{2}}=\cos{α}$.

This leads to $RM=\frac{a\cdot\cos[\arctan(2)]}{2}=\frac{a}{2\sqrt5}$.

In right $ΔRMO$,

$\tan[2\arcsin{\frac{1}{\sqrt(5)}}]=\frac{RO}{\frac{a}{2\sqrt5}}$.

This leads to $RO= \frac{2a}{3\sqrt5}$ using tan double angle formula.

Thus, area of $ΔRMO=\frac{RO×RM}{2}= \frac{a^2}{30}$.

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..

Answer 3

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 $\Delta ANP\sim \Delta ADQ$, so $AP=PQ$. Similarly, $CQ=PQ$. So $AP=PQ=CQ$.

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

Notice that $C,O,A$ are collinear, so $CO$ is the diagonal. Be Lemma $1$, $CO=\frac {AC}3={2\sqrt2\over3}$. And $CN=1$. Then $[CMON]=2[CON]=2\cdot\frac12\cdot \frac{2\sqrt2}3\cdot1\cdot\sin 45^\circ=\frac23$.

Now, $\Delta RDM\sim \Delta DAM$. So $[DRM]=\frac15$.

But $[DNC]=[DRM]+[RMO]+[CMON]$. So, $1=\frac15+\frac23+[RMO]$. Therefore, $[RMO]=\frac2{15}$. Since $[ABCD]=4$, we get $${[RMO]\over[ABCD]}=\frac{\frac2{15}}4=\frac1{30}$$

$\boxed{\tiny Z}$

Answer 4

Let use the plough horse for this kind of problems, i.e., barycentric coordinates with respect to triangle $ACD$ which contains triangle $MRO$.

(we will use the fact that $O$ belongs to line segment $AC$, due to symmetry wrt to this axis of $BM$ and $DN$).

The barycentrical coordinates of points $M,R,O$ are readily computed as:

$$M=\begin{pmatrix}0\\ \frac12 \\ \frac12 \end{pmatrix}, \ \ \ R=\begin{pmatrix}\frac15\\ \frac25 \\ \frac25 \end{pmatrix} \ \ \ O=\begin{pmatrix}\frac13\\ \frac23 \\ 0\end{pmatrix}$$

Their determinant is $\frac{1}{15}$, accounting for the areas' ratio $\frac{[MRO]}{[ACD]}$.

Therefore, the final areas' ratio is:

$$\dfrac{[MRO]}{2 \times [ACD]}=\dfrac{[MRO]}{[ABCD]}=\dfrac{1}{30}$$