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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 $\triangle ABC$

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too general-- the solution goes by finding the intersections for the proportion of the triangle it is cutting at each step and is too specific. easier to work the numbers along the way, not worth a general formula. –  ashley Dec 19 '12 at 2:35

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up vote 5 down vote accepted

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".)

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How did you arrive at $P$, $Q$, $R$? and to arrive at $X$, $Y$, $Z$ –  John Chang Dec 19 '12 at 9:16
    
@JohnChang to get from $P-B = p(C-P)$ to $P = \frac{B+p C}{1+p}$, just expand the parenthesis, collect terms in $P$, and divide. As for $X = A + \frac{|AX|}{|AP|}(P-A)$, that's just saying to get to $X$, start at $A$, head in the direction of $P$, cover the correct proportion of the distance to $P$. –  AakashM Dec 20 '12 at 12:53
    
@AakashM: John's comment-question referred to a previous version of my answer, where the vectors appeared without explanation. I hope my edit has clarified things for him. –  Blue Dec 20 '12 at 15:54

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