Triangle problem related to finding an area

Question

Given a triangle $\triangle ABC$ . Points $P, Q, R$ lie on sides $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$ respectively. $\overline{AP}$ bisects $\overline{BQ}$ at point $X$, $\overline{BQ}$ bisects $\overline{CR}$ at point $Y$, and $\overline{CR}$ bisects $\overline{AP}$ at point $Z$. Find the area of the triangle $\triangle XYZ$ in function area of $\triangle ABC$

Answer 1

I'll take "bisect" to mean "intersect". Also, for simplicity, I'll assume that each of $P$, $Q$, $R$ lies strictly between the endpoints of the corresponding edge. (Adapting the argument for the points being anywhere along the extended edges is straightforward.)

Define $p$, $q$, $r$ thusly:

$$p:=\frac{|BP|}{|PC|} \qquad q := \frac{|CQ|}{|QA|} \qquad r := \frac{|AR|}{|RB|}$$

Vectors $BP$ and $PC$ have the same direction (likewise $CQ$ and $QA$, as well as $AR$ and $RB$), so that we can write ...

$$P-B = p(C-P) \qquad Q-C = q(A-Q) \qquad R-A = r(B-R)$$

... and solve for $P$, $Q$, $R$:

$$P = \frac{B+p C}{1+p} \qquad Q = \frac{C+q A}{1+q} \qquad R = \frac{A+r B}{1+r}$$

For $X$, $Y$, $Z$, the algebra is somewhat more involved. We could break the vectors into coordinates, write equations for lines $AP$ and $BQ$, and then solve for intersection $X$, but here's a vector-only approach.

Note that $X = A + \frac{|AX|}{|AP|}(P-A)$. By the Law of Sines in $\triangle ABX$, we have

$$\begin{align}X = A + \frac{|AB| \sin B}{|AP|\sin X}(P-A) &= A + \frac{|AB|}{|AP|} \frac{|QB\times AB|}{|QB||AB|} \frac{|BQ||AP|}{|BQ \times AP|}(P-A) \\ &= A + \frac{|BQ\times AB|}{|BQ \times AP|}(P-A) \end{align}$$

Re-writing $P$, $Q$, $R$ in terms of $A$, $B$, $C$ gives

$$\begin{align} QB\times AB &= \frac{1}{(1+p)}(B\times A + C\times B + A\times C) \\ QB\times AP &= \frac{1+p+pq}{(1+p)(1+q)}(B\times A + C\times B + A\times C) \end{align}$$

so that

$$X = A + \frac{1+p}{1+p+pq}(P-A) = A + \frac{1+p}{1+p+pq}\left(\frac{B+pC}{1+p} -A\right) = \frac{B + C p+Apq}{1+p+p q}$$

Likewise, $$Y = \frac{C+Aq+Bqr}{1+q+qr} \qquad Z=\frac{A+Br+Crp}{1+r+rp}$$

... and with a bit more algebra, one gets this relation between the triangle areas ...

$$\frac{|\triangle XYZ|}{|\triangle ABC|} = \frac{\frac{1}{2}|YX\times ZX|}{\frac{1}{2}|BA\times CA|} = \frac{\left(1-pqr\right)^2}{\left(1+p+pq\right)\left(1+q+qr\right)\left(1+r+rp\right)}$$

Note that the area of $\triangle XYZ$ vanishes if (and only if) $pqr=1$. This re-captures Ceva's Theorem. ("Segments $AP$, $BQ$, $CR$ all meet at one point if and only if $pqr=1$.")

If I recall correctly from my college days, $\triangle XYZ$ is called a Korgen Triangle. Back then, I believe I knew of a reasonably slick derivation of the area formula; today, I just threw some expressions into Mathematica. If I can re-derive the slick version, I'll post it. (While I've edited to flesh-out my previous answer, the proof above isn't what I'm thinking of as "slick".)