Find the ratio of areas in the $ABP$ and $PMCQ$ regions.

Question

For reference: In the figure, $(BM):(AQ)=(MC):(QC)$. calculate the ratio of areas in the $ABP$ and $PMCQ$ regions.

My progress

$\frac{S_{ABP}}{S_{PMCQ}}=?\\ \frac{x}{y}=\frac{m}{n}\\ \\ \frac{S_{ABM}}{S_{AMC}}=\frac{x}{y}\\ \frac{S_{ABQ}}{S_{BQC}}=\frac{m}{n}\\ \frac{S_{ABM}}{S_{AMC}}=\frac{S_{ABQ}}{S_{BQC}}$

I couldn't develop from here.

Answer 1

A couple of tricks which are useful for similar problems:

1. Affine maps preserve the ratios of areas and the ratios of lengths of segments on the same line, so you may assume without loss of generality that $\widehat{C}=90^\circ,AC=CB$ and $[ABC]=1$, for instance;
2. If we draw the diagonals of a trapezoid, there are important relations about the areas of the four resulting triangles.

The hyphotesis gives $MQ\parallel AB$, hence $CP$ meets $AB$ at its midpoint by (the converse of) Ceva's theorem. Let $\lambda=CQ/CA=CM/CB$. By Van Obel's theorem it follows that $BP/PQ=\frac{1-\lambda}{\lambda}+1=\frac{1}{\lambda}$. This leads to

$$ [AMB]=(1-\lambda)[ABC]=(1-\lambda),\qquad [APB]=\frac{1}{\lambda+1}[AMB]=\frac{1-\lambda}{1+\lambda}$$

$$[BMP]=[AQP]=\frac{\lambda}{1+\lambda}[AMB]=\frac{\lambda(1-\lambda)}{(1+\lambda)}$$

$$ [CQPM]=[ABC]-[AMB]-[AQP]= \lambda-\frac{\lambda(1-\lambda)}{1+\lambda}=\frac{2\lambda^2}{1+\lambda}$$

so

$$\frac{[APB]}{[CQPM]} = \frac{1-\lambda}{2\lambda^2}. $$

Answer 2

Given, $\frac{BM}{AQ}=\frac{MC}{QC}$ so
$$\frac{BM}{MC}=\frac{AQ}{QC}=\frac{λ}{1}$$ (say)
By Menelaus’ Theorem on triangle BQC,
$$\frac{AC}{AQ}\frac{QP}{PB}\frac{BM}{MC}=1$$ which implies that $\frac{QP}{PB}=\frac{1}{λ+1}$.Similarly by Menelaus’ Theorem on triangle AMC, $\frac{MP}{AP}=\frac{1}{λ+1}$.
By the property that ratio of areas of triangles with same height is equal to the ratio of their bases, we have $$\frac{ar(ABP)}{ar(APQ)}=\frac{λ+1}{1}$$,
$$\frac{ar(ABP)}{ar(MBP)}=\frac{λ+1}{1}$$
So ar(APQ)=ar(MBP). Let ar(APB)=(λ+1)k, then ar(APQ)=ar(MBP)=k. Now,
$$\frac{ar(ABQ)}{ar(BQC)}=\frac{λ}{1}$$
$$\frac{(λ+2)k}{k+ar(PMCQ)}=\frac{λ}{1}$$
$(λ+2)k= λk+λ.ar(PMCQ)$ so that ar(PMCQ)=$\frac{2k}{λ}$. This implies that
$$\frac{ar(ABP)}{ar(PMCQ)}=\frac{λ(λ+1)}{2}=\frac{AQ•AC}{2QC^2}=\frac{BM•BC}{2MC^2}$$ where $\frac{AQ}{QC}=\frac{λ}{1}$ .