Let ABCDE be a convex pentagon. If P, Q, R and S are the respective centroids of the triangles ABE, BCE, CDE and DAE, show that PQRS is a parallelogram and its area is 2/9 of that of ABCD.
Problem
Source: Central American Olympiad 2000, problem 3
Tags: geometry, parallelogram, trigonometry, analytic geometry, graphing lines, slope, complex numbers
08.08.2007 09:09
Let P′, Q′, R′, S′ be the midpoints of AB, BC, CD, DA, respectively. To prove that P′Q′R′S′ is a parallelogram, draw AC and BD and note that P′S′∥BD∥Q′R′ and P′Q′∥AC∥S′R′. EPEP′=EQEQ′=ERER′=ESES′=23⟹ PQ∥P′Q′,QR∥Q′R′,RS∥R′S′,SP∥S′P′⟹ PQRS is a parallelogram We know [P′Q′R′S′]=12[ABCD], and since PQP′Q′=QRQ′R′=RSR′S′=SPS′P′=23, [PQRS][P′Q′R′S]=49⟹[PQRS]=29[ABCD].
08.08.2007 09:39
Jutaro wrote: Let ABCDE be a convex pentagon. Denote the centroids P, Q, R, S of the triangles ABE, BCE, CDE, DAE respectively. Show that the quadrilateral PQRS is a parallelogram and 9⋅[PQRS]=2⋅[ABCD]. PROOF. Denote the midpoints X,Y,Z,U,V of the sides [AB], [BC], [CD], [DE], [EA] respectively. Prove easily that {PQ∥XY∥AC , PQ=23⋅XY=13⋅ACQR∥YZ∥BD , QR=23⋅YZ=13⋅BDRS∥AC , RS=13⋅ACSP∥BD , SP=13⋅BD‖ \implies \left\{\begin{array}{c}PQ\parallel RS\parallel AC\ ,\ PQ=RS=\frac{1}{3}\cdot AC\\\\ QR\parallel SP\parallel BD\ ,\ QR=SP=\frac{1}{3}\cdot BD\end{array}\right\| \implies the quadrilateral PQRS is parallelogram and m\left(\widehat{AC,BD}\right)=m(\widehat{PQR})=\phi \implies [PQRS]=QP\cdot QR\cdot\sin \phi= \left(\frac{1}{3}\cdot AC\right)\cdot\left(\frac{1}{3}\cdot BD\right)\cdot\sin\phi = \frac{2}{9}\cdot\left(\frac{1}{2}\cdot AC\cdot BD\cdot\sin\phi\right)= \frac{2}{9}\cdot [ABCD] \implies 9\cdot [PQRS]=2\cdot [ABCD]. Remark I. I"ll use the complex numbers. Denote X(x)- the point X with the affix x\in \mathcal C Therefore, \left\{\begin{array}{c}\left\{\begin{array}{c}3p=a+b+e\\\ 3r=c+d+e\end{array}\right\|\implies 3(p+r)=a+b+c+d+2e\\\\ \left\{\begin{array}{c}3q=b+c+e\\\ 3s=a+d+e\end{array}\right\|\implies 3(q+s)=a+b+c+d+2e\end{array}\right\| \implies p+r=q+s \implies PQRS is parallelogram. Remark II. Consider the hexagon ABCDEE, i.e. suppose that we have identical six balls such that only one ball in each from A,B,C,D and two balls in E. P is centroid of \triangle ABE and R is centroid of \triangle CDE \implies the centroid of the hexagon ABCDEE is the midpoint of [PR]. Q is centroid of \triangle BCE and S is centroid of \triangle ADE \implies the centroid of the hexagon ABCDEE is the midpoint of [QS]. Thus, the centroid of the hexagon ABCDEE is both the midpoint of the segment [PR] and the midpoint of the segment [QS] ! Therefore, the quadrilateral PQRS is parallelogram.
08.08.2007 10:02
Interesting. It is a nice problem, but I don't think that this problem is appropriately placed on an olympiad...It takes a total of 'no thinking' to coordinate bruteforce with the shoelace formula. The parallelogram thing can be shown by vectors.
09.08.2007 00:37
Altheman wrote: Interesting. It is a nice problem, but I don't think that this problem is appropriately placed on an olympiad...It takes a total of 'no thinking' to coordinate bruteforce with the shoelace formula. The parallelogram thing can be shown by vectors. Sure, Altheman! But you should take into account that the problem belongs to one of the first Olympiads ever organized in our region (by the way, my country hosted this one, in 2000 ). Also, the competition is aimed to high school students under 16, so the problems are not supposed to be really difficult.
28.06.2012 06:35
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