# Three cevians in a triangle create four sub-triangles of area $1$. Find the area of a non-triangular region.

I'm having trouble proving that all the white and green areas have the same area, from there on we can obtain the answer $$1+\sqrt5$$ by proving that the inner red triangle points are midpoints.

There is also the following way (Guy Rawe found).

Since $$QD||PC$$, $$FP||BR$$ and $$RE||AQ,$$ we obtain: $$\frac{QR}{RC}=\frac{QD}{PC}=\frac{BQ}{BP}=\frac{BQ}{BQ+PQ}=\frac{1}{1+\frac{PQ}{BQ}}=\frac{1}{1+\frac{FQ}{QR}}=\frac{QR}{QR+FQ}=\frac{QR}{FR},$$ which gives $$RC=FR.$$

Similarly, we obtain: $$AP=PD,$$ $$BQ=QE$$ and the rest is smooth.

Why does $$QD||PC?$$

Because $$S_{\Delta PQC}=S_{\Delta PQR}+S_{\Delta PRC}=1+S_{\Delta PRC}=S_{\Delta CDR}+S_{\Delta PRC}=S_{\Delta PDC}.$$

• how do we know that those lines are parallel? – SuperMage1 Jul 13 at 15:08
• @SuperMage1 I added something. See now. – Michael Rozenberg Jul 13 at 15:16

Let $$AD$$, $$BE$$ and $$CF$$ be our chevians of $$\Delta ABC$$.

Also, let $$AD\cap BE=\{P\}$$, $$AD\cap CF=\{R\}$$ and $$BE\cap CF=\{Q\}.$$

Thus, $$S_{\Delta APE}=S_{\Delta BQF}=S_{\Delta CRD}=S_{\Delta PQR}=1.$$ Let $$S_{AFQP}=x,$$ $$S_{BDRQ}=y$$ and $$S_{CEPR}=z$$.

Thus, $$S_{\Delta BDR}=\frac{S_{\Delta BDR}}{S_{\Delta CDR}}=\frac{BD}{DC}=\frac{x+y+2}{z+2},$$ which gives $$S_{\Delta BQR}=y-\frac{x+y+2}{z+2}=\frac{yz+y-x-2}{z+2}.$$ Now, $$S_{\Delta AFQ}=\frac{S_{\Delta AFQ}}{S_{\Delta BFQ}}=\frac{AF}{BF}=\frac{x+z+2}{y+2},$$ which gives $$S_{\Delta AQP}=x-\frac{x+z+2}{y+2}=\frac{xy+x-z-2}{y+2}.$$ Thus, $$\frac{\frac{x+z+2}{y+2}+1}{\frac{xy+x+z-2}{y+2}}=\frac{S_{\Delta AFQ}+S_{\Delta BQF}}{S_{\Delta AQP}}=\frac{S_{\Delta ABQ}}{S_{\Delta AQP}}=$$ $$=\frac{BQ}{PQ}=\frac{S_{\Delta BQR}}{S_{\Delta PQR}}=S_{\Delta BQR}=\frac{yz+y-x-2}{z+2}.$$ Id est, $$\frac{x+y+z+4}{xy+x-z-2}=\frac{yz+y-x-2}{z+2}$$ or $$(y+1)(z^2+x^2-xy+2x+4z-xyz+4)=0,$$ which gives $$z^2+x^2-xy+2x+4z=xyz-4.$$ By the same way we can obtain: $$x^2+y^2-yz+2y+4x=xyz-4$$ and $$y^2+z^2-zx+2z+4y=xyz-4.$$ Now, let $$x=\max\{x,y,z\}$$.

Thus, $$x^2+y^2-yz+2y+4x-(y^2+z^2-zx+2z+4y)=0$$ or $$(x-z)(x+z)+z(x-y)+4x-2y-2z=0$$ or $$(x-z)(x+z+2)+(x-y)(z+2)=0,$$ which gives $$x=y=z,$$ $$x^2+6x=x^3-4$$ or $$x^3-x^2-6x-4=0$$ or $$x^3+x^2-2x^2-2x-4x-4=0$$ or $$(x+1)(x^2-2x-4)=0$$ or $$x=1+\sqrt5$$ and we are done!