Ratio of Areas of Quadrilateral

Asked Sep 12 '15 at 08:42

Active Sep 12 '15 at 09:24

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Given a convex quadrilateral $ABCD$. We trisect $AB, BC, CD ,DA$ by the points $P_1, P_2, R_1, R_2, Q_1,Q_2,S_1,S_2$ respectively as shown below. Show that $$\frac{[KLMN]}{[ABCD]}=\frac{1}{9}.$$ Where $[ABCD]$ denotes the area of $ABCD$.

I think the question make use of the following fact: If the line $AB$ and $PQ$ intersect at $M$, then $[ABP]/[ABQ]=PM/QM$.

I tried to decompose the quadrilateral into triangle and make use of the theorem, but it seemed not very helpful, please helps.

asked Sep 12 '15 at 08:42

nayr ktn

I've just tested that with GeoGebra, and it's not true. – Intelligenti pauca Sep 12 '15 at 09:18
I agree with @Aretino, that cannot be true without further assumptions. It is clear that if $A$ and $D$ collapse, the area of $KLMN$ is way less than $\frac{1}{9}$ times the area of $ABCD$. – Jack D'Aurizio Sep 12 '15 at 09:32
Maybe the original problem was asking to prove the inequality $[KLMN]\color{red}{\leq} \frac{1}{9}[ABCD]$. – Jack D'Aurizio Sep 12 '15 at 09:35
http://db.math.ust.hk/notes_download/elementary/geometry/ge_G1.pdf
it's problem 4 in p.6 of the above pdf.
– nayr ktn Sep 12 '15 at 09:36
The first theorem there is true, but the second is not. The point is that $S_1R_1/KL\ne3$ in general. – Intelligenti pauca Sep 12 '15 at 09:48
It is not very clear how a "trisection point" is defined there, but maybe we are assuming that $AD,P_1 Q_1,P_2 Q_2,BC$ concur, and $AB,S_1 R_1,S_2 R_2,DC$ as well. – Jack D'Aurizio Sep 12 '15 at 09:49

Ratio of Areas of Quadrilateral

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