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enter image description here

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 $S$

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  • $\begingroup$ Interesting ObTask: If $AD:DB=1:2$ show that $Q$ lies on the line $PR$. $\endgroup$ – Michael Hoppe Dec 8 '19 at 17:29
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$$S_{ABD}=S_{ACD}=\frac{1}{2} S $$

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, $$S_{PQR} = S_{QAP}+S_{QAR}-S_{APR} = S$$

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

enter image description here

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.

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A general solution:

Let$AP=k_1AB$, $AQ=k_2AD$ $AR=k_3AC$, and for $0<t<1$ 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 $t=\frac{q}{1+q}$.

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