# Triangle area problem from kangaroo KSA grade 9,10 Let ABC be a triangle with area $$S$$ such as in the image $$D$$ is the midpoint of $$BC$$ $$AP=2*AB,AQ=3*AD,AR=4*AC$$ what is the area of $$PQR$$ in terms of $$S$$

So what I tried is heron's formula, and a bit of angle chasing, but didn't get a very useful thing. I tried with geogebra and I think the awnser is simply just $$S$$

• Interesting ObTask: If $AD:DB=1:2$ show that $Q$ lies on the line $PR$. – Michael Hoppe Dec 8 '19 at 17:29

$$S_{ABD}=S_{ACD}=\frac{1}{2} S$$

Using the formula $$S_{XYZ} = \frac{1}{2} XY\cdot XZ \cdot sin(YXZ)$$, we get $$S_{QAP} = \frac{AP}{AB} \frac{AQ}{AD} S_{DAB} = 3 S$$ $$S_{QAR} = \frac{AQ}{AD} \frac{AR}{AC} S_{CAD} = 6 S$$ $$S_{APR} = \frac{AP}{AB} \frac{AR}{AC} S_{ABC} = 8 S$$ Now, $$S_{PQR} = S_{QAP}+S_{QAR}-S_{APR} = S$$

• That was my solution, too. – Michael Hoppe Dec 8 '19 at 11:51

Let X, Y, Z be the division points of RA. RQ is extended to meet YP at W. T, V are the division points of QA. ZV extended will meet the line through C parallel to BZ at U. After adding the above, we find (1) UVDC, VZBD and UZBC are //gms; and (2) W(Q)RT(P) and A(U)P(T)Y(Z) are triangles meeting the midpoint theorem requirements. [The proofs of the above will be skipped.]

[PQR] = [red] = 0.5 [pink] = 0.5 [grey] = 0.5 [green] = [orange] = [ABC] = S.

A general solution:

Let$$AP=k_1AB$$, $$AQ=k_2AD$$ $$AR=k_3AC$$, and for $$0 let $$BD=tBC$$. (In our special case we have $$t=1/2$$.)

From the well known formula for the area $$F$$ of a triangle, $$F=\frac12ab\sin(\gamma)$$, we derive that \begin{align} F(AQP)&=k_1k_2F(ABD)\\ F(AQR)&=k_2k_3F(ADC)\\ F(APR)&=k_1k_3F(ABC). \end{align} Furthermore, $$F(ABD)=tF(ABC)$$ and $$F(ADC)=(1-t)F(ABC)$$. Putting that together we have \begin{align} F(PQR)&=F(AQP)+F(AQR)-F(APR)\\ &=k_1k_2tF(ABC)+k_2k_3(1-t)F(ABC)-k_1k_3F(ABC), \end{align} that is

$$\frac{F(PQR)}{F(ABC)}=tk_1k_2+(1-t)k_2k_3-k_1k_3.$$

(That ratio may be negative, in which case the orientation of $$PQR$$ changes, but that doesn't really matter.)

Now one might investigate some special cases , for example $$t=1/2$$ and $$k_1$$, $$k_2$$ and $$k_3$$ are in arithmetic progression, that is $$k_2=k_1+d$$ and $$k_3=k_1+2d$$. A short computation shows that the ratio is $$d^2$$, independent of $$k_1$$. That solves the original problem since $$k_1=2$$ (which is irrelevant anyway) and $$d=1$$.

Or, given $$t=1/2$$ again, the ratio vanishes if $$k_2$$ is the harmonic mean of $$k_1$$ and $$k_2$$.

A last one. Given $$k_2=k_1q$$ and $$k_2=k_1q^2$$ for positive $$q$$ the area of $$PQR$$ vanishes if $$t=\frac{q}{1+q}$$.