# BMO1 2006/07 Question 2 Geometry Problem

$2.$ In the convex quadrilateral $ABCD$, points $M,N$ lie on the side $AB$ such that $AM = MN = NB$, and points $P,Q$ lie on the side $CD$ such that $CP = PQ = QD$. Prove that Area of $AMCP=$ Area of $MNPQ = \frac{1} {3}$ Area of $ABCD$.

I know how to prove that $AMCP=\frac{1}{3}$ Area of $ABCD$, but not sure about the rest. Would it suffice to say that the other section is just a third of a square that has been stretched or does this need to be proven? Thanks in advance to anyone who can prove or come up with any hints.

It will suffice to prove that distance from $M$ to $CD$ $\rho(M,CD)=\frac{2}{3}\rho(A,CD)+\frac{1}{3}\rho(B,CD)$ and $\rho(N,CD)=\frac{1}{3}\rho(A,CD)+\frac{2}{3}\rho(B,CD)$ together with $\rho(P,AB)=\frac{2}{3}\rho(C,AB)+\frac{1}{3}\rho(D,AB)$ and $\rho(Q,AB)=\frac{1}{3}\rho(C,AB)+\frac{2}{3}\rho(D,AB)$, then make a triangulation.
I'll do about $NMPQ$: Area $S_{MNPQ}=S_{PQN}+S_{QMN}=\frac{1}{2}\left(PQ\cdot\rho(N,CD)+MN\cdot\rho(Q,AB)\right)= \frac{1}{2}\left(\frac{1}{3}CD\cdot\left(\frac{1}{3}\rho(A,CD)+\frac{2}{3}\rho(B,CD)\right)+\frac{1}{3}AB\cdot\left(\frac{1}{3}\rho(C,AB)+\frac{2}{3}\rho(D,AB)\right)\right)= \frac{1}{18}\left(CD\cdot \rho(A,CD)+2CD\cdot \rho(B,CD)+AB\cdot \rho(C,AB)+2AB\cdot \rho(D,AB) \right)= \frac{1}{9}\left(S_{ACD}+2S_{BCD}+S_{ABC}+2S_{ABD}\right)=\frac{1}{3}S_{ABCD}$
Now we want to prove about the distances. Wlog $D$ is closer to $AB$ than $C$ is (in the case $\rho(D,AB)=\rho(C,AB)$ $AB||CD$, and $\rho(X,AB)=const$ for every point $X$ on $CD$).
Let $DC_1||AB$ and the perpendiculars from $Q,P,C$ to $AB$ intersects $DC_1$ in the points $Q_1,P_1,C_1$ resp. $QQ_1||PP_1||CC_1\Rightarrow \Delta DQQ_1,\Delta DPP_1, \Delta DCC_1$ are similar $\Rightarrow QQ_1=\frac{1}{3}CC_1, PP_1=\frac{2}{3}CC_1$ hence we got the demanded equalities for distances.

• What exactly do you mean by the distance rho, is this the shortest distance between the two points? – MadChickenMan May 20 '15 at 21:39
• I mean the distance between two geometrical objects. In this case between a point and a line, i.e. length of the perpendicular, dropped from point to line. – Alexey Burdin May 20 '15 at 21:57
• Ah. I have actually just managed to come up with a different proof to the question, but I'm still interested in the first part of your solution. I'm not really sure how to prove the first statement. – MadChickenMan May 20 '15 at 22:03
• Edited, now you can see the full proof. :) – Alexey Burdin May 20 '15 at 22:39
• I see now. Thanks for all your help. – MadChickenMan May 21 '15 at 6:48

I would use vector notations (based on picture above). Use $\times$ as cross product. To make sure that all areas sum with correct sign - follow clockwise rotation. Then: $$S_{ADCB}*2 = |\vec{CA}\times\vec{CD} + \vec{AC}\times\vec{AB}| = |\vec{CA}\times\vec{CP*3} + \vec{AC}\times\vec{AM*3}|$$ and $$S_{ADCB} = 3 * S_{AMCP}$$ $$S_{MNPQ}*2 = |\vec{PM}\times\vec{PQ} + \vec{MP}\times\vec{MN}| = |(\vec{PC}+\vec{CA}+\vec{AM})\times\vec{CP} + (\vec{MA}+\vec{AC}+\vec{CP})\times\vec{AM}|$$ Now remember that cross product of parallel vectors is zero and cross product is anticommutative. $$S_{MNPQ}*2 = |(\vec{CA}\times\vec{CP}+\vec{AM}\times\vec{CP} + \vec{AC}\times\vec{AM}-\vec{AM}\times\vec{CP}|$$ $$S_{MNPQ} = S_{AMCP}$$

• Thanks for your answer. Afraid I'm not well versed in vector mathematics yet so I'm not able to full appreciate your proof though. – MadChickenMan May 21 '15 at 6:49