A geometry problem by A Former Brilliant Member

Recall that the areas of triangles with equal altitudes are proportional to the bases of the triangles.

$\dfrac{35+84+K}{M+40+30}=\dfrac{84}{M}$

$\dfrac{119+K}{M+70}=\dfrac{84}{M}$

$119M+MK=84(M+70)$

$35M+MK=5880$ $\color{#D61F06}(1)$

$\dfrac{84+40+M}{30+35+K}=\dfrac{40}{30}$

$(124+M)(30)=40(65+K)$

$3720+30M=2600+40K$

$112+3M=4K$

$K=\dfrac{112+3M}{4}=28+\dfrac{3}{4}M$ $\color{#D61F06}(2)$

Substitute $\color{#D61F06}(2)$ in $\color{#D61F06}(1)$ .

$35M+M\left(28+\dfrac{3}{4}M\right)=5880$

$35M+28M+\dfrac{3}{4}M^2=5880$

$\dfrac{3}{4}M^2+63M-5880=0$

$M=56$

It follows that

$B=28+\dfrac{3}{4}(56)=70$

The desired answer is $\dfrac{56}{70}=$ $\boxed{0.8}$

Let $M$ be the area of the yellow region and $K$ be the area of the green region. Note that triangle $XPB$ and triangle $ZPB$ share the same altitude from $P$ , so the ratio of their areas is the same as the ratio of their bases. Similarly, triangle $XYB$ and triangle $ZYB$ share the same altitude from $Y$ , so the ratio of their areas is the same as the ratio of their bases. Moreover, the two pairs of bases are actually the same, and thus in the same ratio. As a result, we have:

$\dfrac{40}{30}=\dfrac{124+M}{65+K}$ $\implies$ $4K=3M+112$

Applying identical reasoning to the triangles with bases $YC$ and $ZC$ , we get

$\dfrac{K}{35}=\dfrac{M+K+84}{105}$ $\implies$ $2K=M+84$

Substituting from this equation into the previous one gives $M=56$ , from which we get $K=70$ .

Finally,

$\dfrac{M}{K}=\dfrac{56}{70}=\boxed{0.8}$ .